CBSE Class 10 All Subjects Revision Worksheet Set 06

ENGLISH

Question 1. From the front page of the newspaper choose any one news item and underline the verbs. Also write down their tense. Paste the newspaper cuttings in your grammar note book along with the newspaper date. (Do this for 30 days).
Answer: To successfully complete this task, choose one prominent news article from the front page of a daily newspaper each day. Carefully identify and underline all action words (verbs) within the text. Then, create a table in your grammar notebook indicating each verb and its corresponding tense. Neatly cut out the article and paste it next to your analysis, making sure to write the publication date clearly at the top. Below is a sample layout for one day:
- News Clip Date: June 1, 2026
- Sentence: "The government has announced a new environmental policy today."
- Verb: "has announced" - Present Perfect Tense.
In simple words: Pick a news story every day for a month, paste it in your notebook, find the verbs, and write down what tense they are in.

Exam Tip: Consistent daily work is better than trying to paste all thirty clippings at the last minute, as regular practice helps you identify complex tenses easily.

Question 2. Read Unit 1, 2,3 & 4 from your Main Course Book and complete the following worksheets from BBC
(i) Worksheet 44 on Pg 107, Grammar: Worksheet 100 - Pg 243
(ii) Worksheet 45 on Pg 109, Worksheet 99 - Pg 241
(iii) Worksheet 47 on Pg 113, Worksheet 93 - Pg 229, 228
(iv) Worksheet 49 on Pg 117, Worksheet 106 - Pg 255, 256
(v) Worksheet 66 on Pg 163
(vi) Worksheet 67 on Pg 165
(vii) bbc class test on Pg 179
Answer: Carefully read through the designated Units (1, 2, 3, and 4) in your CBSE Class X Main Course Book (MCB). Once you understand the themes and comprehension patterns, open your BBC Compacta practice book and systematically solve the specified worksheets on reading comprehension and grammar. Ensure all answers are written in a neat, legible hand directly in the workbook spaces provided.
In simple words: Read the first four units of your English book, then complete the listed grammar and reading sheets in your BBC practice book.

Exam Tip: Read the passage questions first before scanning the text in the worksheets; this helps you locate the correct answers much faster.

Question 3. Conduct a survey of your society and find out about child labour. Write an article about any one such child you have met.
Answer: Conduct informal interactions in your local neighborhood (such as at local tea stalls, construction sites, or small auto repair shops) to observe and gather information regarding child labor. Based on your findings, write an article structured with a catchy heading, an introduction detailing the issue of child labor, a case study of a child you observed, and a concluding call to action. Below is a sample article:

The Stolen Childhood: A Close Look at Child Labor
Child labor remains a silent plague in our neighborhoods, stripping children of their fundamental right to education. During a recent local survey, I encountered twelve-year-old Amit, who works at a local roadside tea stall. Instead of carrying a school bag, Amit carries heavy trays of tea cups from dawn until dusk. His family's dire financial situation forced him to drop out of school to earn a meager daily wage of fifty rupees. Amit's eyes still light up when he sees children his age in school uniforms, highlighting his unfulfilled desire to learn. It is the collective responsibility of citizens and local authorities to report such instances and support organizations that help rehabilitate these children back into schools.
In simple words: Find out about child labor in your area and write an article about a working child's daily struggle and dreams.

Exam Tip: To score maximum marks in article writing, strictly follow the format: Title, By-line (your name), Introduction, Body Paragraphs, and Conclusion.

Question 4. Make a power point presentation taking ideas from the last unit of MCB. Add new concepts and your own views.
Answer: Create an engaging digital presentation containing 8 to 10 slides based on the theme of the final unit of your Main Course Book (typically focused on National Integration or Travel and Tourism). Use the following slide-by-slide layout:
- Slide 1: Title slide with your name, class, and topic.
- Slide 2: Introduction to the core theme.
- Slide 3: Key issues highlighted in the textbook unit.
- Slide 4-5: New contemporary concepts related to the topic (e.g., modern digital connectivity aiding national unity, or sustainable eco-tourism).
- Slide 6-7: Your personal observations and analytical views on the subject.
- Slide 8: Summary and concluding recommendations.
- Slide 9: Bibliography/Sources.
In simple words: Design a slide show based on the last unit of your English book, adding fresh ideas and your own thoughts.

Exam Tip: Avoid overcrowding slides with text; use brief bullet points and high-quality images to keep your presentation visually appealing.

Question 5. Read any collection of short stories by RUSKIN BOND and write a review on that.
Answer: Read a short story collection by Ruskin Bond, such as "The Blue Umbrella and Other Stories" or "The Night Train at Deoli", and write a book review. Below is a sample format:

Book Review: "The Blue Umbrella and Other Stories" by Ruskin Bond
- Introduction: This collection brings together some of the most heartwarming tales set in the scenic Himalayan foothills of Garhwal. Bond’s simple yet evocative writing makes these stories highly relatable.
- Plot Summary: The title story revolves around a young village girl named Binya, who trades her lucky leopard-claw pendant for a beautiful blue silk umbrella. Her prized possession soon becomes the envy of the entire village, especially the local shopkeeper, Ram Bharosa. The narratives explore human greed, innocence, kindness, and the quiet beauty of mountain life.
- Themes: Nature conservation, human relationships, and finding joy in simple things.
- Personal Opinion: The stories are delightful and full of warmth. Bond's gentle narrative style is perfect for readers of all ages. I highly recommend this collection to anyone looking for a comforting read.
In simple words: Read a book of short stories by Ruskin Bond and write a review explaining what the book is about and why you liked it.

Exam Tip: A stellar book review must always mention the book's title, the author's name, the primary theme, key characters, and your critical recommendation.

Question 6. Read stories from READERS DIGEST.
Answer: Spend time reading various real-life inspirational narratives, humor columns, and educational articles from Reader's Digest magazines. To document this activity, maintain a small logbook containing the titles of 5 to 6 articles you read, a brief 2-sentence summary of each, and a list of new vocabulary words you discovered along with their dictionary definitions.
In simple words: Read articles from Reader's Digest to improve your reading habits and note down any new words you learn.

Exam Tip: Pay attention to the concise, lively writing style used in Reader's Digest articles to help improve your own essay-writing skills.

SECOND LANGUAGE - HINDI

सभी कार्य 'मी एन माइन' अभ्यास पुस्तिका में किया जाना चाहिए ।

Question 1. पत्र लेखन व अनुच्छेद लेखन (अभ्यास कार्य २७ से ३५ तक) (पृष्ठ ५७ से ७५)
Answer: अपनी 'मी एन माइन' अभ्यास पुस्तिका के पृष्ठ संख्या ५७ से ७५ पर दिए गए सभी पत्र लेखन और अनुच्छेद लेखन संबंधी अभ्यासों को पूरा करें।
- औपचारिक पत्र लिखते समय कार्यालयी शिष्टाचार, स्पष्ट विषय और सटीक भाषा का प्रयोग करें।
- अनौपचारिक पत्र में आत्मीयता और उचित संबोधन का ध्यान रखें।
- अनुच्छेद लिखते समय दी गई शब्द-सीमा (८० से १०० शब्द) के भीतर विचारों को क्रमबद्ध रूप में प्रस्तुत करें।
In simple words: अपनी वर्कबुक में दिए गए सभी पत्र और अनुच्छेद लिखने के अभ्यासों को सुंदर लिखावट में पूरा करें।

Exam Tip: पत्र लेखन में प्रारूप (Format) के लिए विशेष अंक निर्धारित होते हैं, इसलिए प्रेषक का पता, तिथि और संबोधन का सही स्थान अवश्य याद रखें।

Question 2. अपठित गद्यांश तथा पद्यांश (अभ्यास कार्य १ से ५ तक) (पृष्ठ १ से १०)
Answer: पृष्ठ १ से १० तक के अपठित गद्यांशों और पद्यांशों को ध्यानपूर्वक पढ़ें। प्रत्येक गद्यांश/पद्यांश के नीचे दिए गए प्रश्नों को समझें और उनके उत्तर संक्षिप्त व अपनी भाषा में लिखें। पद्यांशों के उत्तर लिखते समय कविता की पंक्तियों को सीधे उतारने के बजाय उनका भाव स्पष्ट करते हुए उत्तर दें।
In simple words: पहले दस पन्नों के बिना पढ़े हुए गद्यांश और पद्यांश को पढ़कर उनके नीचे लिखे सवालों के जवाब दें।

Exam Tip: प्रश्नों के उत्तर हमेशा गद्यांश की सीमा के भीतर से ही दें, अपने मन से कोई अतिरिक्त जानकारी न जोड़ें।

Question 3. शब्द, पद और पदबंध (अभ्यास कार्य ४४ से ४६ तक) (पृष्ठ ६३ से १००)
Answer: व्याकरण के इस भाग के अंतर्गत शब्द और पद के बीच का अंतर स्पष्ट रूप से समझें। जब कोई सार्थक ध्वनि समूह स्वतंत्र होता है तो वह शब्द कहलाता है, और जब वही शब्द वाक्य में प्रयुक्त होता है तो पद बन जाता है। पदबंध के पाँचों भेदों (संज्ञा, सर्वनाम, विशेषण, क्रिया, क्रियाविशेषण पदबंध) का अभ्यास कार्य ४४ से ४६ में दिए गए वाक्यों के माध्यम से अच्छे से अभ्यास करें।
In simple words: शब्द, पद और पदबंध के व्याकरण नियमों को समझकर वर्कबुक के दिए गए पेजों को हल करें।

Exam Tip: पदबंध की पहचान करते समय वाक्य के अंतिम रेखांकित पद को ध्यान से देखें कि वह संज्ञा है, विशेषण है या क्रिया, उसी के आधार पर भेद तय होता है।

Question 4. वाक्य विचार, संधि (अभ्यास कार्य ५२ से ५३ तक) (पृष्ठ १०६ से ११२)
Answer: रचना के आधार पर वाक्य के तीन भेदों (सरल, संयुक्त, मिश्र) का अध्ययन करें और उन्हें आपस में बदलने (वाक्य रूपांतरण) का अभ्यास करें। संधि के नियमों (विशेषकर स्वर, व्यंजन और विसर्ग संधि) को समझें और शब्दों के संधि व संधि-विच्छेद संबंधी प्रश्नों को अभ्यास कार्य ५२ व ५३ में हल करें।
In simple words: वाक्य बदलने के नियम और शब्दों को जोड़ने-तोड़ने (संधि) के अभ्यास पूरे करें।

Exam Tip: मिश्र वाक्य में मुख्य उपवाक्य और आश्रित उपवाक्य की पहचान करने के लिए 'कि', 'जो', 'क्योंकि', 'जैसे ही' जैसे योजकों पर ध्यान दें।

Question 5. साहित्य (स्पर्श और संचयन में से) (अभ्यास कार्य ६६ से ८७ तक) (पृष्ठ १३६ से १८२)
Answer: अपनी मुख्य पाठ्यपुस्तकों 'स्पर्श' और 'संचयन' के पाठों को गहनता से पढ़ें। इसके पश्चात अभ्यास पुस्तिका के पृष्ठ १३६ से १८२ तक दिए गए सभी प्रश्नों के उत्तर लिखें। चरित्र-चित्रण और मूल्य-परक प्रश्नों के उत्तर देते समय पाठ के मुख्य संदर्भों और नैतिक मूल्यों का उल्लेख अवश्य करें।
In simple words: अपनी हिंदी किताबों की कहानियों और कविताओं को पढ़कर वर्कबुक के प्रश्न-उत्तर हल करें।

Exam Tip: साहित्य के उत्तरों में मुख्य शब्दों और लेखक/कवि के नाम को रेखांकित करने से परीक्षक पर बहुत अच्छा प्रभाव पड़ता है।

Question 6. मॉडल टेस्ट पेपर (कुल दो टेस्ट पेपर) (पृष्ठ २८१ से ३१२)
Answer: संपूर्ण पाठ्यक्रम का पुनरावलोकन करने के बाद अभ्यास पुस्तिका के अंत में दिए गए दोनों मॉडल टेस्ट पेपरों को एक शांत वातावरण में, तीन घंटे की समय-सीमा निर्धारित करके हल करें। इससे आपको परीक्षा में समय प्रबंधन करने में मदद मिलेगी।
In simple words: परीक्षा की तैयारी के लिए पुस्तक के अंत में दिए गए दोनों परीक्षा प्रश्नपत्रों को स्वयं हल करें।

Exam Tip: मॉडल पेपर हल करते समय मुख्य परीक्षा की तरह साफ लिखावट और उचित मार्जिन का प्रयोग करें।

GERMAN

Question 1. Make a PowerPoint presentation on any one of the topic Modalverben, Adjectiv endungen, Nebensatz (weil, obwohl), Themen. 1st slide should be of Introduction (Vorstellung). 2nd slide should be of small information about Germany.
Answer: Choose one grammar topic, for example, "Modalverben" (Modal Verbs: müssen, können, wollen, dürfen, sollen, mögen). Design a presentation containing 6 to 8 slides in German:
- Slide 1: Vorstellung (Introduction) - Write: "Guten Tag! Mein Name ist..., ich bin in der Klasse X und heute präsentiere ich das Thema 'Modalverben'."
- Slide 2: Information about Germany - Write quick facts: Capital (Berlin), Population (ca. 84 Millionen), Currency (Euro), and National Flag (Schwarz-Rot-Gold).
- Slide 3: Introduction to Modalverben - Define what modal verbs are and how they affect word order (the main verb goes to the end of the sentence in infinitive form).
- Slide 4: Conjugation of 'können' and 'müssen' with clear examples.
- Slide 5: Conjugation of 'dürfen' and 'sollen' with sentences showing permission and obligation.
- Slide 6: Interactive practice sentences for class review.
- Slide 7: Vielen Dank! (Thank you slide).
In simple words: Create a German slide show about modal verbs. Start with your introduction, add some fun facts about Germany, and then explain the grammar rules.

Exam Tip: Make sure you double-check verb conjugations in German, as irregular forms like "ich kann" or "er darf" are very common evaluation targets.

Question 2. Solve CBSE SAMPLE PAPER (2005, 2006, 2007)
Answer: Obtain copies of the Class X German CBSE board/sample papers for the years 2005, 2006, and 2007. Solve them on loose sheets or in your German practice notebook. Focus especially on reading comprehension (Leseverstehen), letter writing (Brief schreiben), and grammar sections containing tenses, prepositions, and clause connectors.
In simple words: Practice solving the German CBSE board exam papers from 2005, 2006, and 2007 in your notebook.

Exam Tip: Time yourself while writing the letter-writing section to ensure you do not spend more than twenty minutes on it.

FRENCH

Question 1. Complete Chapter 1-6 in Entre Jeunes (Yellow Book) and from Get Ready All sections and revise them.
Answer: Thoroughly read Chapters 1 through 6 in your French textbook "Entre Jeunes." Solve all the exercises at the end of each chapter, and complete the corresponding sections in the supplementary practice book "Get Ready." Pay close attention to vocabulary, dialogue completion, and comprehension passages.
In simple words: Read and complete all activities for Chapters 1 to 6 in your French books.

Exam Tip: Create flashcards for new French verbs and their irregular conjugations in the present and past tenses (passé composé).

Question 2. Prepare interactive charts, placards, Brochures, PPT on vocabulary and tenses in cartoon form.
Answer: Create a fun visual resource, such as a poster or a digital slideshow, where French grammar and vocabulary are explained using cartoon characters and speech bubbles. For example, draw a cartoon character explaining "Le Futur Proche" using the formula: Aller (present tense) + Infinitive Verb, accompanied by a character saying, "Je vais manger une pizza !"
In simple words: Make a poster or a presentation explaining French grammar words and tenses using funny cartoon drawings.

Exam Tip: Use vibrant colors and clear, bold fonts for your cartoon text to ensure your charts are easy to read from a distance.

Question 3. Charts on format of Dialogue, Letters, C.V. and Advertisement give one example of each.
Answer: Prepare four separate charts outlining the standard layouts used in French writing assessments:
- Dialogue Chart: Show conversational turn-taking using dashes (-) or character names, focusing on greeting and closing phrases.
- Letter Chart: Label the correct placement of city/date (top right), greeting (left), body paragraphs, and formal sign-offs (e.g., "Amicalement" or "Cordialement").
- C.V. Chart: Divide into clear sections: État civil (Personal info), Formation (Education), Expérience (Work history), and Langues (Languages).
- Advertisement Chart: Create a sample promotional flyer featuring a catchy headline, descriptions, prices, and contact details.
In simple words: Make posters showing the correct layout for writing conversations, letters, resumes, and advertisements in French.

Exam Tip: Memorize standard opening and closing salutations for letters, as using them correctly automatically secures format marks.

Question 4. Revise Covered Syllabus
Answer: Systematically review all grammar topics, vocabulary lists, and writing formats taught during school terms. Practice writing short paragraphs on common topics such as "My daily routine," "My family," or "My summer vacation" to build writing fluency.
In simple words: Study all the French lessons you have learned so far to keep your skills sharp.

Exam Tip: Spend fifteen minutes daily listening to simple French audio clips or reading short stories to improve your comprehension and pronunciation skills.

SANSKRIT

Question 1. सी बी एस ई वार्षिक परीक्षा प्रश्न पत्र वर्ष २००६, २००७, २००८ हल कीजिए ।
Answer: वर्ष २००६, २००७ और २००८ के सीबीएसई संस्कृत के बोर्ड प्रश्न पत्रों को प्राप्त करें और उन्हें अपनी अभ्यास पुस्तिका में हल करें। विशेष रूप से अपठित-अवबोधनम्, रचनात्मक-कार्यम् (पत्र और चित्र वर्णन), तथा अनुप्रयुक्त-व्याकरणम् (संधि, शब्दरूप, धातुरूप) खंडों को ध्यानपूर्वक हल करें।
In simple words: पुराने सालों (२००६ से २००८) के संस्कृत बोर्ड पेपर अपनी कॉपी में हल करें।

Exam Tip: शब्दरूप (जैसे बालक, लता, फल) और धातुरूप (लट्, लृट्, लङ् लकार) का नियमित अभ्यास करें, क्योंकि व्याकरण खंड में इनसे सीधे प्रश्न पूछे जाते हैं।

Question 2. पुस्तक 'मी एन माइन' (Me n Mine) से १० अपठित गद्यांश और १० चित्र वर्णन कीजिए ।
Answer: अपनी 'मी एन माइन' संस्कृत अभ्यास पुस्तिका से किन्हीं १० अपठित गद्यांशों को पढ़कर उनके नीचे दिए गए एकपदेन और पूर्णवाक्येन प्रश्नों के उत्तर लिखें। इसके अतिरिक्त, १० चित्र वर्णनों को पूरा करें जिसमें मंजूषा में दिए गए शब्दों की सहायता से चित्र पर आधारित ५-५ सरल संस्कृत वाक्यों का निर्माण करें।
In simple words: वर्कबुक से १० बिना पढ़े पैराग्राफ और १० फोटो देखकर उनपर वाक्य लिखने का अभ्यास पूरा करें।

Exam Tip: चित्र वर्णन करते समय 'अस्मिन् चित्रे...', 'अस्ति/सन्ति' जैसे सरल वाक्य ढाँचों का प्रयोग करें जिससे व्याकरण संबंधी अशुद्धियाँ न हों।

Question 3. संस्कृत अभ्यास पुस्तिका से 'प्रत्यय' अभ्यास कार्य पूर्ण करें ।
Answer: अपने संस्कृत पाठ्यक्रम में शामिल मुख्य प्रत्ययों जैसे क्त्वा, तुमुन्, ल्यप्, क्त, क्तवतु, मतुप्, ठक् और टाप् आदि के नियमों को समझें। इसके बाद अभ्यास पुस्तिका में प्रत्ययों को जोड़ने (संयोग) और अलग करने (विभाग/विच्छेद) से संबंधित सभी अभ्यास कार्य पूरे करें।
In simple words: संस्कृत व्याकरण में प्रत्यय वाले पाठ के सारे अभ्यास अपनी वर्कबुक में पूरे करें।

Exam Tip: प्रत्यय लगाते समय शब्दों में होने वाले मूल परिवर्तनों (जैसे ल्यप् प्रत्यय में 'य' बचना और शुरू में उपसर्ग होना) को ध्यान में रखें।

MATHEMATICS

Activity 1. To verify experimentally whether a given sequence is an arithmetic progression or not.
Answer: Write this activity in your Maths Practical File using the following structure:
- Objective: To verify experimentally whether a given sequence of numbers is an arithmetic progression (AP) or not.
- Materials Required: Grid paper, colored paper strips, scissors, glue, and a geometry box.
- Procedure:
1. Let the given sequence be \( 2, 5, 8, 11, \dots \)
2. Cut rectangular strips of colored paper of uniform width, with lengths proportional to the terms of the sequence: 2 cm, 5 cm, 8 cm, 11 cm, etc.
3. Paste these strips vertically adjacent to one another on a grid paper, starting from a common base line.
4. Draw a line connecting the top-right corners of these strips.
- Observation: The line joining the top corners of the adjacent strips is a straight slanted line. The difference in height between any two consecutive strips is constant at 3 cm.
- Conclusion: Since the consecutive heights have a common difference, the sequence is verified to be an Arithmetic Progression.
In simple words: Cut strips of paper according to the numbers in your sequence and paste them side-by-side. If their tops form a perfect straight slope, the sequence is an AP.

Exam Tip: Always make sure the width of all the strips is exactly identical so that the visual slope represents the common difference accurately.

Activity 2. To solve a system of linear equation graphically and investigate the conditions for a unique solutions.
Answer: Write this activity in your Maths Practical File:
- Objective: To solve a system of linear equations graphically and verify the conditions for a unique solution.
- Materials Required: Graph paper, ruler, pencils, and eraser.
- Procedure:
1. Let the given system of equations be:
\( 2x - y = 1 \)
\( x + y = 5 \)
2. Find at least three solution points \( (x, y) \) for each equation.
- For equation 1: Points are \( (0, -1), (1, 1), (2, 3) \).
- For equation 2: Points are \( (0, 5), (2, 3), (5, 0) \).
3. Plot these coordinates on the graph paper and draw a straight line through each set.
- Observation: The two lines intersect at exactly one point, which is \( (2, 3) \). This intersection represents the unique solution of the system.
- Mathematical Check: For a system \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), unique solution exists if \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). Here, \( \frac{2}{1} \neq \frac{-1}{1} \), which satisfies the condition.
- Conclusion: The system of linear equations has a unique solution at \( (2, 3) \).
In simple words: Draw lines for both equations on a graph paper. The point where they cross each other is the unique answer to the system.

Exam Tip: Clearly label the axes, write down the equations along their respective lines, and highlight the intersection point with its coordinates.

ASSIGNMENTS - QUADRATIC EQUATION

Question 1. Given that one root of quadratic equation \( ax^2 + bx + c = 0 \) is three times of other. Show that \( 3b^2 = 16ac \).
Answer: Let the roots of the given quadratic equation be \( \alpha \) and \( 3\alpha \).
From the relationship between roots and coefficients:
Sum of roots:
\( \alpha + 3\alpha = -\frac{b}{a} \)
\( \implies 4\alpha = -\frac{b}{a} \)
\( \implies \alpha = -\frac{b}{4a} \) (Equation 1)
Product of roots:
\( \alpha \cdot (3\alpha) = \frac{c}{a} \)
\( \implies 3\alpha^2 = \frac{c}{a} \) (Equation 2)
Substitute the value of \( \alpha \) from Equation 1 into Equation 2:
\( 3 \left( -\frac{b}{4a} \right)^2 = \frac{c}{a} \)
\( \implies 3 \left( \frac{b^2}{16a^2} \right) = \frac{c}{a} \)
\( \implies \frac{3b^2}{16a^2} = \frac{c}{a} \)
Multiply both sides by \( 16a^2 \):
\( \implies 3b^2 = \frac{16a^2 c}{a} \)
\( \implies 3b^2 = 16ac \)
Hence Proved.
In simple words: Name the roots x and 3x. Use the sum and product formulas to eliminate x and arrive at the final required equation.

Exam Tip: Writing down standard relations like sum \( = -b/a \) and product \( = c/a \) clearly before substituting values earns step-marking credit.

Question 2. Solve for x: \( (x - 3) (x + 9) (x - 7) (x + 5) = 1680 \)
Answer: Group the factors strategically to obtain a common quadratic expression:
\( [ (x - 3)(x + 5) ] [ (x + 9)(x - 7) ] = 1680 \)
Expand both groups:
\( (x^2 + 2x - 15)(x^2 + 2x - 63) = 1680 \)
Let \( y = x^2 + 2x \). The equation simplifies to:
\( (y - 15)(y - 63) = 1680 \)
\( \implies y^2 - 78y + 945 = 1680 \)
\( \implies y^2 - 78y - 735 = 0 \)
Solve for \( y \) using the quadratic formula:
\( y = \frac{-(-78) \pm \sqrt{(-78)^2 - 4(1)(-735)}}{2(1)} \)
\( \implies y = \frac{78 \pm \sqrt{6084 + 2940}}{2} \)
\( \implies y = \frac{78 \pm \sqrt{9024}}{2} \)
Since \( \sqrt{9024} = 8\sqrt{141} \):
\( \implies y = \frac{78 \pm 8\sqrt{141}}{2} = 39 \pm 4\sqrt{141} \)
Now substitute \( y = x^2 + 2x \):
\( x^2 + 2x = 39 \pm 4\sqrt{141} \)
\( \implies x^2 + 2x - (39 \pm 4\sqrt{141}) = 0 \)
Using the quadratic formula to solve for \( x \):
\( x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-(39 \pm 4\sqrt{141}))}}{2} \)
\( \implies x = \frac{-2 \pm \sqrt{4 + 156 \pm 16\sqrt{141}}}{2} \)
\( \implies x = -1 \pm \sqrt{40 \pm 4\sqrt{141}} \)
In simple words: Pair up the bracket terms to make a common expression. Replace that expression with a single letter, solve the quadratic equation, and then find the final values for x.

Exam Tip: When grouping bracket terms, pair them such that the sum of the constant numbers is equal, like \( -3 + 5 = 2 \) and \( 9 - 7 = 2 \), to ensure the middle term remains common.

Question 3. Find the value of ‘k’ so that the roots of the quadratic equation \( (k+1)x^2 + 2kx + 4 = 0 \) is equal to the product of the roots.
Answer: Let the roots of the given quadratic equation be \( \alpha \) and \( \beta \). The question implies that the sum of the roots is equal to the product of the roots, meaning \( \alpha + \beta = \alpha\beta \).
From the coefficients:
Sum of roots: \( \alpha + \beta = -\frac{2k}{k+1} \)
Product of roots: \( \alpha\beta = \frac{4}{k+1} \)
Setting them equal:
\( -\frac{2k}{k+1} = \frac{4}{k+1} \)
Assuming \( k \neq -1 \), we can multiply both sides by \( (k+1) \):
\( \implies -2k = 4 \)
\( \implies k = -2 \)
In simple words: Set the sum of the roots formula equal to the product of the roots formula, cancel out the bottom parts, and solve for k.

Exam Tip: Remember to state the restriction \( k \neq -1 \), because if \( k = -1 \), the equation ceases to be a quadratic equation as the \( x^2 \) term becomes zero.

Question 4. Find the real values of x and y which make: \( (2x - 3y - 13)^2 + (3x + 5y + 9)^2 = 0 \)
Answer: Since the squares of real numbers are always non-negative, the sum of two squares can only equal zero if both individual terms are simultaneously equal to zero.
Therefore:
\( 2x - 3y - 13 = 0 \implies 2x - 3y = 13 \) (Equation 1)
\( 3x + 5y + 9 = 0 \implies 3x + 5y = -9 \) (Equation 2)
Solve this system of linear equations using elimination:
Multiply Equation 1 by 5 and Equation 2 by 3:
\( 10x - 15y = 65 \) (Equation 3)
\( 9x + 15y = -27 \) (Equation 4)
Add Equation 3 and Equation 4:
\( (10x + 9x) + (-15y + 15y) = 65 - 27 \)
\( \implies 19x = 38 \)
\( \implies x = 2 \)
Substitute \( x = 2 \) into Equation 1:
\( 2(2) - 3y = 13 \)
\( \implies 4 - 3y = 13 \)
\( \implies -3y = 9 \)
\( \implies y = -3 \)
The real values are \( x = 2 \) and \( y = -3 \).
In simple words: Since squared numbers cannot be negative, both parts inside the brackets must be exactly zero. Set up two simple equations and solve them.

Exam Tip: Be careful with sign changes when transferring numbers across the equals sign during the substitution steps.

Question 5. For what value of ‘k’ the given equation \( kx^2 – 4x + 2 = 0 \) has real and equal roots?
Answer: For a quadratic equation to have real and equal roots, its discriminant \( D \) must equal zero.
Here, \( a = k \), \( b = -4 \), and \( c = 2 \).
\( D = b^2 - 4ac = 0 \)
\( \implies (-4)^2 - 4(k)(2) = 0 \)
\( \implies 16 - 8k = 0 \)
\( \implies 8k = 16 \)
\( \implies k = 2 \)
In simple words: Find the value of k that makes the discriminant formula \( b^2 - 4ac \) equal to zero.

Exam Tip: In similar problems, if the resulting value of k makes the lead coefficient zero, that value must be rejected. Here, \( k = 2 \) is a valid solution.

Question 6. A plane left 30 minutes later than its scheduled time. In order to reach its destination 1500 km away on time, it has to increase its speed by 250 km/hr than its usual speed. Find its usual speed.
Answer: Let the usual speed of the plane be \( v \) km/hr.
The distance is 1500 km.
Time taken at the usual speed: \( t_1 = \frac{1500}{v} \) hours.
Time taken at the increased speed: \( t_2 = \frac{1500}{v + 250} \) hours.
According to the problem, the difference in time is 30 minutes, which is \( \frac{1}{2} \) hour:
\( \frac{1500}{v} - \frac{1500}{v + 250} = \frac{1}{2} \)
\( \implies 1500 \left( \frac{(v + 250) - v}{v(v + 250)} \right) = \frac{1}{2} \)
\( \implies 1500 \left( \frac{250}{v^2 + 250v} \right) = \frac{1}{2} \)
\( \implies \frac{375000}{v^2 + 250v} = \frac{1}{2} \)
Cross-multiply:
\( \implies v^2 + 250v = 750000 \)
\( \implies v^2 + 250v - 750000 = 0 \)
Solve by splitting the middle term:
\( \implies v^2 + 1000v - 750v - 750000 = 0 \)
\( \implies v(v + 1000) - 750(v + 1000) = 0 \)
\( \implies (v - 750)(v + 1000) = 0 \)
So \( v = 750 \) or \( v = -1000 \).
Since speed cannot be negative, we reject \( v = -1000 \).
Thus, the usual speed of the plane is 750 km/hr.
In simple words: Write an equation comparing the travel time at normal speed and fast speed. Solve the quadratic equation to find the plane's speed, ignoring any negative answers.

Exam Tip: Always convert minutes into hours (e.g., 30 minutes \( = 0.5 \) hours) to keep your units consistent with km/hr speed measurements.

Question 7-8. Factorise \( x^2(y – z) + y^2(z – x) + z^2(x – y) \)
Answer: Expand the expression and group terms according to the powers of \( x \):
\( x^2(y - z) + y^2 z - y^2 x + z^2 x - z^2 y \)
Rearrange the terms:
\( x^2(y - z) - x(y^2 - z^2) + yz(y - z) \)
Factor the middle term using \( y^2 - z^2 = (y - z)(y + z) \):
\( = x^2(y - z) - x(y - z)(y + z) + yz(y - z) \)
Take out the common factor \( (y - z) \):
\( = (y - z) [ x^2 - x(y + z) + yz ] \)
\( = (y - z) [ x^2 - xy - xz + yz ] \)
Group terms inside the brackets:
\( = (y - z) [ x(x - y) - z(x - y) ] \)
\( = (y - z)(x - y)(x - z) \)
Rearranging into cyclic order:
\( = -(x - y)(y - z)(z - x) \)
In simple words: Open the brackets and group the parts with similar letters. Take out common factors step-by-step to get the final cyclic multiplication form.

Exam Tip: Cyclic expressions of this type always result in symmetrical factor groups like \( (x-y)(y-z)(z-x) \) up to a sign change.

Question 9. Find p for real roots \( 3x^2 + px – 2 = 0 \)
Answer: For a quadratic equation to have real roots, the discriminant \( D \) must be greater than or equal to zero.
Here, \( a = 3 \), \( b = p \), and \( c = -2 \).
\( D = b^2 - 4ac \ge 0 \)
\( \implies p^2 - 4(3)(-2) \ge 0 \)
\( \implies p^2 + 24 \ge 0 \)
Since the square of any real number \( p^2 \) is always non-negative (\( p^2 \ge 0 \)), the term \( p^2 + 24 \) is always greater than or equal to 24, which is strictly greater than 0 for all real values of \( p \).
Therefore, the given equation has real roots for all real values of \( p \) (i.e., \( p \in \mathbb{R} \)).
In simple words: Calculate the discriminant. Because it is a squared variable added to a positive number, the result is always positive, meaning the equation always has real roots.

Exam Tip: Be sure to write a brief sentence explaining why a squared variable like \( p^2 \) makes the expression positive for all real numbers to secure full marks.

Question 10. For what value of K, \( (4 – k) x^2 + (2k + 4) x + ( 8k+1) = 0 \) is a perfect square. (0, 3)
Answer: A quadratic equation is a perfect square when its discriminant \( D \) is equal to zero.
Here, \( a = 4 - k \), \( b = 2(k + 2) \), and \( c = 8k + 1 \).
\( D = b^2 - 4ac = 0 \)
\( \implies [2(k + 2)]^2 - 4(4 - k)(8k + 1) = 0 \)
\( \implies 4(k^2 + 4k + 4) - 4(32k + 4 - 8k^2 - k) = 0 \)
Divide the entire equation by 4:
\( \implies (k^2 + 4k + 4) - (-8k^2 + 31k + 4) = 0 \)
\( \implies k^2 + 4k + 4 + 8k^2 - 31k - 4 = 0 \)
Combine like terms:
\( \implies 9k^2 - 27k = 0 \)
\( \implies 9k(k - 3) = 0 \)
This gives:
\( 9k = 0 \implies k = 0 \)
or
\( k - 3 = 0 \implies k = 3 \)
Thus, the values of \( K \) are 0 and 3.
In simple words: Set the discriminant formula to zero, expand the brackets, simplify the equation, and solve for k to get the two values.

Exam Tip: Make sure both values of k satisfy the condition that the leading term \( (4-k) \) does not become zero. Since neither 0 nor 3 equals 4, both are valid solutions.

Question 11. solve for x : \( 36x^2 – 12ax + ( a^2 – b^2 ) = 0 \)
Answer: Solve the quadratic equation using the quadratic formula:
Here, \( A = 36 \), \( B = -12a \), and \( C = a^2 - b^2 \).
\( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
\( \implies x = \frac{-(-12a) \pm \sqrt{(-12a)^2 - 4(36)(a^2 - b^2)}}{2(36)} \)
\( \implies x = \frac{12a \pm \sqrt{144a^2 - 144a^2 + 144b^2}}{72} \)
\( \implies x = \frac{12a \pm \sqrt{144b^2}}{72} \)
\( \implies x = \frac{12a \pm 12b}{72} \)
Simplify by dividing the numerator and denominator by 12:
\( \implies x = \frac{12(a \pm b)}{72} = \frac{a \pm b}{6} \)
Thus, the solutions are \( x = \frac{a + b}{6} \) and \( x = \frac{a - b}{6} \).
In simple words: Put the variables into the quadratic formula. The complex parts under the square root cancel out, leaving a simple fraction answer.

Exam Tip: Alternatively, you can solve this by completing the square, noting that \( 36x^2 - 12ax + a^2 = (6x - a)^2 \), which simplifies the math significantly.

Question 12. In the following equation, determine the set of values of p for which the quadratic equation has real roots: \( px^2 + 4x + 1 = 0 \)
Answer: For the equation to have real roots, the discriminant \( D \) must satisfy \( D \ge 0 \).
Here, \( a = p \), \( b = 4 \), and \( c = 1 \).
\( D = b^2 - 4ac \ge 0 \)
\( \implies (4)^2 - 4(p)(1) \ge 0 \)
\( \implies 16 - 4p \ge 0 \)
\( \implies 16 \ge 4p \)
\( \implies p \le 4 \)
Additionally, for this equation to remain a quadratic equation, the coefficient of \( x^2 \) must not be zero:
\( \implies p \neq 0 \)
So, the set of values of \( p \) is \( p \le 4 \) and \( p \neq 0 \).
In simple words: Find the range of p that keeps the discriminant positive or zero. Make sure p is not equal to zero so the equation stays quadratic.

Exam Tip: Forgetting the condition that the leading coefficient \( a \neq 0 \) is a very common trap in board exams; always explicitly state that \( p \neq 0 \).

Question 13. Solve for x: \( \frac{x}{x+1} + \frac{x+1}{x} = \frac{34}{15} \) (where \( x \neq 0, -1 \))
Answer: To simplify the equation, let \( y = \frac{x}{x+1} \). Then, its reciprocal is \( \frac{x+1}{x} = \frac{1}{y} \).
The equation becomes:
\( y + \frac{1}{y} = \frac{34}{15} \)
\( \implies \frac{y^2 + 1}{y} = \frac{34}{15} \)
Cross-multiply to form a quadratic equation:
\( \implies 15(y^2 + 1) = 34y \)
\( \implies 15y^2 - 34y + 15 = 0 \)
Solve by splitting the middle term:
\( \implies 15y^2 - 25y - 9y + 15 = 0 \)
\( \implies 5y(3y - 5) - 3(3y - 5) = 0 \)
\( \implies (5y - 3)(3y - 5) = 0 \)
This gives:
\( y = \frac{3}{5} \) or \( y = \frac{5}{3} \)
Now substitute \( y = \frac{x}{x+1} \) back to solve for \( x \):
Case 1: If \( y = \frac{3}{5} \)
\( \frac{x}{x+1} = \frac{3}{5} \)
\( \implies 5x = 3(x + 1) \)
\( \implies 5x = 3x + 3 \)
\( \implies 2x = 3 \implies x = \frac{3}{2} \)
Case 2: If \( y = \frac{5}{3} \)
\( \frac{x}{x+1} = \frac{5}{3} \)
\( \implies 3x = 5(x + 1) \)
\( \implies 3x = 5x + 5 \)
\( \implies -2x = 5 \implies x = -\frac{5}{2} \)
Thus, the solutions are \( x = \frac{3}{2} \) and \( x = -\frac{5}{2} \).
In simple words: Substitute a simpler letter for the fractional parts, solve the resulting equation, and then substitute back to get the final values for x.

Exam Tip: Substitution is a powerful tool to prevent tedious algebraic expansions that often lead to calculation mistakes.

Question 14. Solve for a and b: \( 2^a + 3^b = 17 \) and \( 2^{a+2} - 3^{b+1} = 5 \)
Answer: Let \( x = 2^a \) and \( y = 3^b \).
Rewrite the given equations in terms of \( x \) and \( y \):
Equation 1: \( x + y = 17 \)
Equation 2: \( 2^{a+2} - 3^{b+1} = 5 \implies (2^2 \cdot 2^a) - (3^1 \cdot 3^b) = 5 \)
\( \implies 4x - 3y = 5 \)
Now, solve this linear system using elimination:
Multiply Equation 1 by 3:
\( 3x + 3y = 51 \) (Equation 3)
Add Equation 2 and Equation 3:
\( (4x - 3y) + (3x + 3y) = 5 + 51 \)
\( \implies 7x = 56 \)
\( \implies x = 8 \)
Substitute \( x = 8 \) into Equation 1:
\( 8 + y = 17 \implies y = 9 \)
Now substitute back to find \( a \) and \( b \):
\( 2^a = x \implies 2^a = 8 \implies 2^a = 2^3 \implies a = 3 \)
\( 3^b = y \implies 3^b = 9 \implies 3^b = 3^2 \implies b = 2 \)
The solutions are \( a = 3 \) and \( b = 2 \).
In simple words: Change the exponential terms into single variables, solve the easy system of equations, and then find the values of exponents.

Exam Tip: Always separate composite exponents like \( 2^{a+2} \) into \( 4 \cdot 2^a \) before attempting substitution.

Question 15-16. Solve for x and y: \( (a-b)x + (a+b)y = a^2 - 2ab - b^2 \) and \( (a+b)(x+y) = a^2 + b^2 \)
Answer: Expand the second equation:
\( (a+b)x + (a+b)y = a^2 + b^2 \) (Equation 2)
The first equation is:
\( (a-b)x + (a+b)y = a^2 - 2ab - b^2 \) (Equation 1)
Subtract Equation 1 from Equation 2:
\( [ (a+b)x + (a+b)y ] - [ (a-b)x + (a+b)y ] = (a^2 + b^2) - (a^2 - 2ab - b^2) \)
\( \implies [ (a+b) - (a-b) ] x = a^2 + b^2 - a^2 + 2ab + b^2 \)
\( \implies 2bx = 2ab + 2b^2 \)
Factor out \( 2b \) from both sides:
\( \implies 2bx = 2b(a+b) \)
\( \implies x = a + b \)
Now substitute \( x = a + b \) into Equation 2:
\( (a+b)(a+b) + (a+b)y = a^2 + b^2 \)
\( \implies (a+b)^2 + (a+b)y = a^2 + b^2 \)
\( \implies (a^2 + 2ab + b^2) + (a+b)y = a^2 + b^2 \)
Subtract \( a^2 + b^2 \) from both sides:
\( \implies 2ab + (a+b)y = 0 \)
\( \implies (a+b)y = -2ab \)
\( \implies y = -\frac{2ab}{a+b} \)
Thus, the solutions are \( x = a + b \) and \( y = -\frac{2ab}{a+b} \).
In simple words: Subtract the two equations to eliminate the y-term, find x, and then substitute the value of x back to find y.

Exam Tip: Do not expand the bracket coefficient terms too early; keeping them grouped as \( (a+b) \) makes it easy to subtract and eliminate terms.

Question 17. The age of father is 3 years more than 3 times the son’s age. 3 years hence the age of the father will be 10 years more than twice the age of the son. Find their present age.
Answer: Let the present age of the son be \( s \) years, and the present age of the father be \( f \) years.
From the first condition:
\( f = 3s + 3 \) (Equation 1)
From the second condition (3 years in the future):
Son's age will be \( s + 3 \) and Father's age will be \( f + 3 \).
\( f + 3 = 2(s + 3) + 10 \)
\( \implies f + 3 = 2s + 6 + 10 \)
\( \implies f = 2s + 13 \) (Equation 2)
Equate Equation 1 and Equation 2:
\( 3s + 3 = 2s + 13 \)
\( \implies s = 10 \)
Substitute \( s = 10 \) back into Equation 1:
\( f = 3(10) + 3 = 33 \)
The present age of the father is 33 years and the son's present age is 10 years.
In simple words: Set up two algebraic age equations. Match them up to find the son's age first, then use it to calculate the father's age.

Exam Tip: Remember to add years to both the father and son when constructing future age equations, as a common error is only increasing one person's age.

Question 18. Two places A & B are 120 km apart from each other on a highway. A car starts from A and another from B at the same time. If they move in the same direction, they meet in 6 hours, and if they move in opposite directions, they meet in 1 hr 12 min. find the speed of each car.
Answer: Let the speed of the car starting from A be \( u \) km/hr, and the speed of the car starting from B be \( v \) km/hr (where \( u > v \)).
Convert 1 hour 12 minutes to hours:
\( 1 \text{ hr } 12 \text{ min } = 1 + \frac{12}{60} = 1 + \frac{1}{5} = \frac{6}{5} \text{ hours} = 1.2 \text{ hours} \)
Case 1: Moving in the same direction (Relative speed \( = u - v \)):
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
\( \implies u - v = \frac{120}{6} = 20 \) (Equation 1)
Case 2: Moving in opposite directions (Relative speed \( = u + v \)):
\( \implies u + v = \frac{120}{1.2} = 100 \) (Equation 2)
Add Equation 1 and Equation 2:
\( 2u = 120 \implies u = 60 \text{ km/hr} \)
Substitute \( u = 60 \) into Equation 2:
\( 60 + v = 100 \implies v = 40 \text{ km/hr} \)
The speed of the first car is 60 km/hr and the speed of the second car is 40 km/hr.
In simple words: Write equations using relative speeds. When moving together, subtract speeds; when moving towards each other, add speeds. Solve to find the speed of both cars.

Exam Tip: Clearly state which car is faster at the beginning of your solution to ensure the direction difference equation is correct.

Question 19. Find the value of a and b such that the polynomial \( P(x )= (x^2 + 3x + 2) (x^2 + 2x + a) \) and \( Q(x)= (x^2 + 7x + 12) (x^2 + 7x + b) \) have \( (x +1) (x + 3) \) as their HCF.
Answer: Factor the known parts of both polynomials first:
\( x^2 + 3x + 2 = (x + 1)(x + 2) \)
\( x^2 + 7x + 12 = (x + 3)(x + 4) \)
So, we have:
\( P(x) = (x + 1)(x + 2)(x^2 + 2x + a) \)
\( Q(x) = (x + 3)(x + 4)(x^2 + 7x + b) \)
Since the Highest Common Factor (HCF) is \( (x+1)(x+3) \), both factors must divide both polynomials.
1. \( (x+3) \) must be a factor of \( P(x) \). Since it does not divide \( (x+1)(x+2) \), it must divide \( x^2 + 2x + a \).
Using the Factor Theorem, substitute \( x = -3 \) into that expression:
\( (-3)^2 + 2(-3) + a = 0 \)
\( \implies 9 - 6 + a = 0 \)
\( \implies a = -3 \)
2. \( (x+1) \) must be a factor of \( Q(x) \). Since it does not divide \( (x+3)(x+4) \), it must divide \( x^2 + 7x + b \).
Substitute \( x = -1 \) into that expression:
\( (-1)^2 + 7(-1) + b = 0 \)
\( \implies 1 - 7 + b = 0 \)
\( \implies b = 6 \)
Thus, the values are \( a = -3 \) and \( b = 6 \).
In simple words: Factor the easy parts first. Since the shared factors must divide both polynomials, substitute the root values into the other parts of the equations to find a and b.

Exam Tip: Polynomial HCF questions are solved easily using the Factor Theorem instead of long division method.

Question 20. A passenger train takes 3 hours less for a journey of 360 km if its speed is increased by 10 kmph. What is the usual speed?
Answer: Let the usual speed of the passenger train be \( x \) kmph.
The total distance is 360 km.
Time taken at usual speed: \( \frac{360}{x} \) hours.
Time taken at increased speed: \( \frac{360}{x + 10} \) hours.
According to the problem:
\( \frac{360}{x} - \frac{360}{x+10} = 3 \)
Divide the entire equation by 3:
\( \implies \frac{120}{x} - \frac{120}{x+10} = 1 \)
\( \implies 120 \left( \frac{(x+10) - x}{x(x+10)} \right) = 1 \)
\( \implies \frac{1200}{x^2 + 10x} = 1 \)
\( \implies x^2 + 10x - 1200 = 0 \)
Factorize by splitting the middle term:
\( \implies x^2 + 40x - 30x - 1200 = 0 \)
\( \implies x(x + 40) - 30(x + 40) = 0 \)
\( \implies (x - 30)(x + 40) = 0 \)
Since speed cannot be negative, we reject \( x = -40 \).
Thus, the usual speed of the passenger train is 30 kmph.
In simple words: Write an equation for the travel times. Solve the quadratic equation to get the speed of the train, choosing the positive value.

Exam Tip: Clearly specify why you are discarding any negative values when dealing with real-world quantities like speed, time, or distance.

ARITHMETIC PROGRESSION

Question 1. If the 8th term of an AP is 31 and 15th term is 16 more than the 11th term, find the AP.
Answer: Let the first term of the AP be \( a \) and the common difference be \( d \).
The formula for the \( n \)-th term is \( a_n = a + (n - 1)d \).
Given:
1. \( a_8 = 31 \implies a + 7d = 31 \) (Equation 1)
2. \( a_{15} = a_{11} + 16 \)
\( \implies a + 14d = (a + 10d) + 16 \)
\( \implies 4d = 16 \implies d = 4 \)
Substitute \( d = 4 \) into Equation 1:
\( a + 7(4) = 31 \)
\( \implies a + 28 = 31 \implies a = 3 \)
The AP is given by \( a, a+d, a+2d, a+3d, \dots \)
Thus, the AP is: 3, 7, 11, 15, 19, ...
In simple words: Use the formulas for different terms of the AP to set up simple equations. Solve them to find the starting number and the gap value, then write out the sequence.

Exam Tip: Show the step-by-step subtraction of terms on both sides of the equation to avoid minor mathematical slip-ups.

Question 2. The nth term of an AP is 7 – 4n. Find its common difference.
Answer: Given the general term \( a_n = 7 - 4n \).
Find the first two terms of the sequence by substituting \( n = 1 \) and \( n = 2 \):
For \( n = 1 \): \( a_1 = 7 - 4(1) = 3 \)
For \( n = 2 \): \( a_2 = 7 - 4(2) = -1 \)
The common difference \( d \) is:
\( d = a_2 - a_1 = -1 - 3 = -4 \)
In simple words: Find the first and second terms of the sequence by plugging in 1 and 2, then subtract the first term from the second.

Exam Tip: You can quickly find the common difference of any linear nth-term expression because it is always equal to the coefficient of n (which is \( -4 \) here).

Question 3. Write the next two term of the AP: \( \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)
Answer: Simplify the radical terms of the given AP:
\( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \)
\( \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \)
\( \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \)
This shows the AP is \( 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots \), where:
First term \( a = 2\sqrt{2} \)
Common difference \( d = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2} \)
The next two terms are the 4th and 5th terms:
4th term \( = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50} \)
5th term \( = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{36 \times 2} = \sqrt{72} \)
The next two terms are \( \sqrt{50} \) and \( \sqrt{72} \).
In simple words: Simplify the square roots to see the pattern. Add the common difference to find the next two terms, then write them back as simple square roots.

Exam Tip: Always write your final answers in the same format as the question (using unsimplified radicals like \( \sqrt{50} \)) to secure full presentation marks.

Question 4. The first term of an AP is p and its common difference is q. Find its 10th term.
Answer: The general formula for the \( n \)-th term of an AP is \( a_n = a + (n - 1)d \).
Here, the first term \( a = p \) and the common difference \( d = q \).
To find the 10th term (\( a_{10} \)), substitute \( n = 10 \):
\( a_{10} = p + (10 - 1)q \)
\( \implies a_{10} = p + 9q \)
In simple words: Put the given letters into the standard term formula to get the final algebraic expression.

Exam Tip: Do not worry if your answer is an algebraic expression rather than a number; literal values are common in board exams.

Question 5. The sum of the 5th and 7th terms of AP is 52 and the 10th term is 46. Find the AP.
Answer: Using the formula \( a_n = a + (n-1)d \):
Given:
1. \( a_5 + a_7 = 52 \)
\( \implies (a + 4d) + (a + 6d) = 52 \)
\( \implies 2a + 10d = 52 \)
Divide by 2:
\( \implies a + 5d = 26 \) (Equation 1)
2. \( a_{10} = 46 \)
\( \implies a + 9d = 46 \) (Equation 2)
Subtract Equation 1 from Equation 2:
\( (a + 9d) - (a + 5d) = 46 - 26 \)
\( \implies 4d = 20 \implies d = 5 \)
Substitute \( d = 5 \) into Equation 1:
\( a + 5(5) = 26 \)
\( \implies a + 25 = 26 \implies a = 1 \)
The AP is: 1, 6, 11, 16, 21, ...
In simple words: Form equations using the terms of the AP. Solve for the first term and the gap to write out the sequence.

Exam Tip: Check your final AP against the original question parameters to verify your math is correct.

Question 6-10. (These questions are duplicates of Questions 1 to 5 as printed in the homework sheet. Please refer to the solutions of Questions 1 to 5 above.)

Question 11. Solve the following sub-parts:
(a) How many two-digit numbers are divisible by 3?
(b) How many three-digit numbers are divisible by 7?
(c) How many multiples of 4 lie between 10 and 250?
(d) How many natural numbers between 100 and 200, which are divisible by 4.
(e) How many natural numbers less than 200 which are divisible by 6.
(f) How many natural numbers between 100 and 500, which are divisible by 8
Answer:
(a) The two-digit numbers divisible by 3 form an AP: 12, 15, 18, ..., 99.
Here, \( a = 12, d = 3, a_n = 99 \).
\( 99 = 12 + (n - 1)3 \implies 87 = (n - 1)3 \implies n - 1 = 29 \implies n = 30 \)
There are 30 such numbers.

(b) The three-digit numbers divisible by 7 form an AP: 105, 112, 119, ..., 994.
Here, \( a = 105, d = 7, a_n = 994 \).
\( 994 = 105 + (n - 1)7 \implies 889 = (n - 1)7 \implies n - 1 = 127 \implies n = 128 \)
There are 128 such numbers.

(c) Multiples of 4 between 10 and 250 form an AP: 12, 16, 20, ..., 248.
Here, \( a = 12, d = 4, a_n = 248 \).
\( 248 = 12 + (n - 1)4 \implies 236 = (n - 1)4 \implies n - 1 = 59 \implies n = 60 \)
There are 60 such numbers.

(d) Natural numbers strictly between 100 and 200 divisible by 4 form an AP: 104, 108, ..., 196.
Here, \( a = 104, d = 4, a_n = 196 \).
\( 196 = 104 + (n - 1)4 \implies 92 = (n - 1)4 \implies n - 1 = 23 \implies n = 24 \)
There are 24 such numbers.

(e) Natural numbers less than 200 divisible by 6 form an AP: 6, 12, 18, ..., 198.
Here, \( a = 6, d = 6, a_n = 198 \).
\( 198 = 6 + (n - 1)6 \implies 192 = (n - 1)6 \implies n - 1 = 32 \implies n = 33 \)
There are 33 such numbers.

(f) Natural numbers strictly between 100 and 500 divisible by 8 form an AP: 104, 112, ..., 496.
Here, \( a = 104, d = 8, a_n = 496 \).
\( 496 = 104 + (n - 1)8 \implies 392 = (n - 1)8 \implies n - 1 = 49 \implies n = 50 \)
There are 50 such numbers.
In simple words: Find the first and last numbers for each series. Use the general term formula of an AP to solve for the total count of numbers.

Exam Tip: Pay close attention to wordings like "between" (do not include the boundary values) vs. "from... to..." (include boundary values if divisible).

Question 12. For what value of n, are the nth terms of two APs: 63, 65, 67,………… and 3, 10, 17,……………….. equal ?
Answer: For the first AP: 63, 65, 67, ...
First term \( a_1 = 63 \), Common difference \( d_1 = 2 \)
\( a_n = 63 + (n - 1)2 = 2n + 61 \)
For the second AP: 3, 10, 17, ...
First term \( a_2 = 3 \), Common difference \( d_2 = 7 \)
\( a_n' = 3 + (n - 1)7 = 7n - 4 \)
Set their \( n \)-th terms equal:
\( 2n + 61 = 7n - 4 \)
\( \implies 65 = 5n \)
\( \implies n = 13 \)
Thus, the 13th terms of both APs are equal.
In simple words: Write down the nth term formula for both sequences, make them equal to each other, and solve to find the value of n.

Exam Tip: Since n represents a position in a sequence, your final value for n must always be a positive integer.

Question 13. Subba Rao started work in 2001 at annual salary of Rs. 10,000 and received an increment of Rs. 500 each year. In which year did his income reach Rs. 15,000?
Answer: This situation forms an AP representing annual salary:
First term \( a = 10000 \), Common difference \( d = 500 \), Target term \( a_n = 15000 \)
Using \( a_n = a + (n - 1)d \):
\( 15000 = 10000 + (n - 1)500 \)
\( \implies 5000 = (n - 1)500 \)
\( \implies n - 1 = 10 \implies n = 11 \)
Since \( n = 11 \), the target salary is reached in the 11th year.
The starting year is 2001 (which is the 1st year).
The 11th year is:
\( 2001 + (11 - 1) = 2011 \).
His income reached Rs. 15,000 in the year 2011.
In simple words: Use the AP formula to find how many years it takes to reach the target salary, then add that count to the starting year.

Exam Tip: Be careful when calculating the final calendar year; the formula \( \text{Year} = \text{Start Year} + n - 1 \) prevents off-by-one errors.

Question 14. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n,
Answer: The weekly savings form an AP:
First term \( a = 5 \), Common difference \( d = 1.75 \), Target term \( a_n = 20.75 \)
Using the formula \( a_n = a + (n - 1)d \):
\( 20.75 = 5 + (n - 1)1.75 \)
\( \implies 15.75 = (n - 1)1.75 \)
\( \implies n - 1 = \frac{15.75}{1.75} \)
\( \implies n - 1 = 9 \)
\( \implies n = 10 \)
In simple words: Write down the savings numbers as an AP. Solve for the week count using the standard term formula.

Exam Tip: Remove the decimal points by multiplying the terms of the fraction by 100 to make division easier: \( \frac{1575}{175} = 9 \).

Question 15. In a flower bed, there are 23 rose plants in the first row, 21 in the second row, 19 in the third row, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
Answer: The number of rose plants in each row forms a decreasing AP:
23, 21, 19, ..., 5
Here, \( a = 23 \), \( d = 21 - 23 = -2 \), and last term \( a_n = 5 \).
Using \( a_n = a + (n - 1)d \):
\( 5 = 23 + (n - 1)(-2) \)
\( \implies 5 - 23 = -2(n - 1) \)
\( \implies -18 = -2(n - 1) \)
\( \implies n - 1 = 9 \)
\( \implies n = 10 \)
There are 10 rows in the flower bed.
In simple words: Write the sequence of rose plants, calculate the common difference (which is negative), and find the row count using the AP term formula.

Exam Tip: Since the sequence is decreasing, ensure your common difference \( d \) is negative to avoid wrong row counts.

COORDINATE GEOMETRY

Question 1. Find the mid – point of the line segment joining the points (4, 3) and (2, 1).
Answer: Let the points be \( A(4, 3) \) and \( B(2, 1) \).
The mid-point formula is:
\( M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
Substitute the values:
\( M(x, y) = \left( \frac{4 + 2}{2}, \frac{3 + 1}{2} \right) \)
\( \implies M(x, y) = \left( \frac{6}{2}, \frac{4}{2} \right) = (3, 2) \)
The coordinates of the mid-point are (3, 2).
In simple words: Average the x-coordinates and the y-coordinates of the given points to find the middle point.

Exam Tip: Always state the general formula before substituting numbers to show your methodology clearly.

Question 2. Find the coordinates of the point which divides the line segment joining the points (1, 3) and (2, 7) in the ratio 3 : 4.
Answer: Let the points be \( A(1, 3) \), \( B(2, 7) \), and the ratio be \( m : n = 3 : 4 \).
Using the Section Formula:
\( P(x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \)
Substitute the values:
\( P(x, y) = \left( \frac{3(2) + 4(1)}{3+4}, \frac{3(7) + 4(3)}{3+4} \right) \)
\( \implies P(x, y) = \left( \frac{6 + 4}{7}, \frac{21 + 12}{7} \right) \)
\( \implies P(x, y) = \left( \frac{10}{7}, \frac{33}{7} \right) \)
The coordinates of the division point are \( \left( \frac{10}{7}, \frac{33}{7} \right) \).
In simple words: Put the given points and the ratio values into the section formula to find the exact coordinates.

Exam Tip: Avoid converting fractional coordinates into decimals unless specifically asked in the question.

Question 3. Show that the points (1, 1); (3, – 2) and (– 1, 4) are collinear.
Answer: Let the points be \( A(1, 1) \), \( B(3, -2) \), and \( C(-1, 4) \).
The points are collinear if the area of the triangle formed by them is equal to zero.
\( \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
Substitute the coordinates:
\( \text{Area} = \frac{1}{2} |1(-2 - 4) + 3(4 - 1) + (-1)(1 - (-2))| \)
\( \implies \text{Area} = \frac{1}{2} |1(-6) + 3(3) - 1(3)| \)
\( \implies \text{Area} = \frac{1}{2} |-6 + 9 - 3| \)
\( \implies \text{Area} = \frac{1}{2} |0| = 0 \)
Since the area of the triangle is 0, the given points are collinear.
In simple words: Calculate the area of the triangle using the three points. Since the result is zero, it proves they lie on the same straight line.

Exam Tip: You can also prove collinearity by showing that the sum of the distances between two pairs of points equals the distance between the outermost pair: \( AB + BC = AC \).

Question 4. Find the centroid of the triangle whose vertices are (3, – 5); (– 7, 4) and (10, – 2).
Answer: Let the vertices be \( A(3, -5) \), \( B(-7, 4) \), and \( C(10, -2) \).
The formula for the coordinates of the centroid \( G(x, y) \) is:
\( G(x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
Substitute the values:
\( G(x, y) = \left( \frac{3 + (-7) + 10}{3}, \frac{-5 + 4 + (-2)}{3} \right) \)
\( \implies G(x, y) = \left( \frac{6}{3}, \frac{-3}{3} \right) = (2, -1) \)
The coordinates of the centroid are (2, -1).
In simple words: Add all the x-values and divide by 3, then add all the y-values and divide by 3 to find the centroid.

Exam Tip: Be careful with positive and negative integers when performing quick additions in the numerator.

Question 5. Find the area of a triangle whose vertices are A (1, 2); B (3, 5) and C (– 4, – 7)
Answer: Given vertices: \( A(1, 2) \), \( B(3, 5) \), and \( C(-4, -7) \).
Use the triangle area formula:
\( \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
Substitute the coordinates:
\( \text{Area} = \frac{1}{2} |1(5 - (-7)) + 3(-7 - 2) + (-4)(2 - 5)| \)
\( \implies \text{Area} = \frac{1}{2} |1(12) + 3(-9) - 4(-3)| \)
\( \implies \text{Area} = \frac{1}{2} |12 - 27 + 12| \)
\( \implies \text{Area} = \frac{1}{2} |-3| \)
\( \implies \text{Area} = 1.5 \text{ sq. units} \)
In simple words: Put the three corner points into the area formula and compute the absolute value to get the triangle's area.

Exam Tip: The absolute value bars \( | \dots | \) mean that if the calculation yields a negative number, you must make it positive because area can never be negative.

Question 6. If the distance of the point P(x, y) from the point A (5, 1) and (– 1, 5) are equal, show that 3x = 2y.
Answer: Let \( A(5, 1) \) and \( B(-1, 5) \).
Given that the distance of \( P(x, y) \) from \( A \) and \( B \) is equal:
\( PA = PB \implies PA^2 = PB^2 \)
Using the distance formula:
\( (x - 5)^2 + (y - 1)^2 = (x - (-1))^2 + (y - 5)^2 \)
\( \implies (x - 5)^2 + (y - 1)^2 = (x + 1)^2 + (y - 5)^2 \)
Expand the brackets:
\( \implies x^2 - 10x + 25 + y^2 - 2y + 1 = x^2 + 2x + 1 + y^2 - 10y + 25 \)
Cancel \( x^2, y^2, 25 \), and 1 from both sides:
\( \implies -10x - 2y = 2x - 10y \)
Rearrange the variables:
\( \implies 10y - 2y = 2x + 10x \)
\( \implies 8y = 12x \)
Divide both sides by 4:
\( \implies 2y = 3x \)
\( \implies 3x = 2y \)
Hence Proved.
In simple words: Make the distance squared from P to A equal to the distance squared from P to B. Expand the terms, cancel the duplicates, and simplify to get the relationship.

Exam Tip: Squaring both sides of the distance equation immediately gets rid of the radical signs, making calculations much cleaner.

Question 7. In what ratio does the point (– 4, 6) divide the line segment joining the points A (– 6, 10) and B (3, – 8).
Answer: Let the point \( P(-4, 6) \) divide the line segment joining \( A(-6, 10) \) and \( B(3, -8) \) in the ratio \( k : 1 \).
Using the Section Formula for the x-coordinate:
\( x = \frac{kx_2 + 1x_1}{k + 1} \)
Substitute the values:
\( -4 = \frac{k(3) + 1(-6)}{k + 1} \)
\( \implies -4(k + 1) = 3k - 6 \)
\( \implies -4k - 4 = 3k - 6 \)
\( \implies 6 - 4 = 3k + 4k \)
\( \implies 2 = 7k \)
\( \implies k = \frac{2}{7} \)
Thus, the ratio is \( \frac{2}{7} : 1 \), which is equivalent to 2 : 7.
In simple words: Assume the ratio is k to 1. Solve for k using only the x-coordinates to get the final division ratio.

Exam Tip: You can verify your calculated ratio by checking if it yields the correct y-coordinate using the section formula: \( \frac{2(-8) + 7(10)}{2+7} = \frac{54}{9} = 6 \), which matches perfectly.

Question 9. For what value of m, the points (4, 3); (m, 1) and (1, 9) are collinear.
Answer: Let the points be \( A(4, 3) \), \( B(m, 1) \), and \( C(1, 9) \).
For the points to be collinear, the area of the triangle formed by them must be zero.
\( \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0 \)
\( \implies |4(1 - 9) + m(9 - 3) + 1(3 - 1)| = 0 \)
\( \implies |4(-8) + m(6) + 1(2)| = 0 \)
\( \implies |-32 + 6m + 2| = 0 \)
\( \implies |6m - 30| = 0 \)
\( \implies 6m - 30 = 0 \)
\( \implies 6m = 30 \)
\( \implies m = 5 \)
Thus, the value of \( m \) is 5.
In simple words: Set the area equation to zero and solve to find the value of m.

Exam Tip: Always make sure to write the complete collinearity condition statement before initiating calculations.

Question 10. Prove that the coordinates of the centroid of a triangle ABC with vertices A(x1, y1); B(x2, y2); C(x3, y3) are given by (x1 + x2 + x3)/3, (y1 + y2 + y3)/3.
Answer: Let \( AD \) be the median of the triangle \( ABC \) drawn from vertex \( A(x_1, y_1) \) to the side \( BC \) with endpoints \( B(x_2, y_2) \) and \( C(x_3, y_3) \).
The coordinates of the midpoint \( D \) of \( BC \) are:
\( D = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) \)
The centroid \( G(x, y) \) is a point on the median \( AD \) that divides it internally in the ratio 2 : 1.
Using the Section Formula with points \( A(x_1, y_1) \), \( D\left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) \), and ratio \( m : n = 2 : 1 \):
\( x = \frac{2 \left( \frac{x_2 + x_3}{2} \right) + 1(x_1)}{2 + 1} = \frac{x_2 + x_3 + x_1}{3} = \frac{x_1 + x_2 + x_3}{3} \)
\( y = \frac{2 \left( \frac{y_2 + y_3}{2} \right) + 1(y_1)}{2 + 1} = \frac{y_2 + y_3 + y_1}{3} = \frac{y_1 + y_2 + y_3}{3} \)
Thus, the coordinates of the centroid are:
\( G(x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
Hence Proved.
In simple words: Find the midpoint of the base of the triangle. Use the section formula in the ratio 2 to 1 on the line connecting the top corner to this midpoint to get the centroid formula.

Exam Tip: Clearly draw a simple reference triangle labeling the median and division ratio to make your proof easy to understand.

Question 11. Prove that the diagonals of a rectangle bisect each other and are of equal length. Find the coordinates of the points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
Answer: Part 1: Let the vertices of a rectangle be situated at \( A(0,0) \), \( B(a,0) \), \( C(a,b) \), and \( D(0,b) \).
- Length of diagonals:
\( AC = \sqrt{(a - 0)^2 + (b - 0)^2} = \sqrt{a^2 + b^2} \)
\( BD = \sqrt{(0 - a)^2 + (b - 0)^2} = \sqrt{a^2 + b^2} \)
Since \( AC = BD \), the diagonals are of equal length.
- Bisection:
Mid-point of \( AC = \left( \frac{0 + a}{2}, \frac{0 + b}{2} \right) = \left( \frac{a}{2}, \frac{b}{2} \right) \)
Mid-point of \( BD = \left( \frac{a + 0}{2}, \frac{0 + b}{2} \right) = \left( \frac{a}{2}, \frac{b}{2} \right) \)
Since both mid-points are identical, the diagonals bisect each other.

Part 2: In any triangle \( ABC \), the points \( Q \) and \( R \) on medians \( BE \) and \( CF \) dividing them in the ratio 2 : 1 are, by definition, the centroid of the triangle.
Since a triangle has only one unique centroid, both points \( Q \) and \( R \) coincide at the centroid.
Thus, the coordinates for both \( Q \) and \( R \) are:
\( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \).
In simple words: Assign simple coordinates to the rectangle's corners to verify that the diagonals are equal and cross in the middle. The points on the medians are the centroid of the triangle.

Exam Tip: Placing one vertex of the rectangle at the origin \( (0,0) \) simplifies distance calculations significantly.

Question 12. In what ratio does the line 4x + y = 11 divide the line segment joining the points (1, 3) and (2, 7).
Answer: Let the line \( 4x + y = 11 \) divide the segment joining \( A(1, 3) \) and \( B(2, 7) \) at point \( P \) in the ratio \( k : 1 \).
The coordinates of \( P \) are:
\( P = \left( \frac{2k + 1}{k + 1}, \frac{7k + 3}{k + 1} \right) \)
Since point \( P \) lies on the line \( 4x + y = 11 \), its coordinates must satisfy the line's equation:
\( 4 \left( \frac{2k + 1}{k + 1} \right) + \left( \frac{7k + 3}{k + 1} \right) = 11 \)
Multiply the entire equation by \( (k + 1) \):
\( \implies 4(2k + 1) + (7k + 3) = 11(k + 1) \)
\( \implies 8k + 4 + 7k + 3 = 11k + 11 \)
\( \implies 15k + 7 = 11k + 11 \)
\( \implies 4k = 4 \)
\( \implies k = 1 \)
The ratio is \( 1 : 1 \), which means the line bisects the segment.
In simple words: Express the division point coordinates with k, put them into the line's equation, and solve for k to get the ratio.

Exam Tip: A ratio of \( 1 : 1 \) means the intersection point is exactly the mid-point of the segment, which you can verify by checking if \( (1.5, 5) \) lies on the line.

Question 13. PQRS is a square of side ‘b’ units. If P lies at the origin, sides PQ and PS lie along x – axis and y – axis respectively, find the coordinates of the vertices of the square PQRS.
Answer: - Since \( P \) lies at the origin: \( P = (0, 0) \)
- Since side \( PQ \) of length \( b \) lies along the positive x-axis: \( Q = (b, 0) \)
- Since side \( PS \) of length \( b \) lies along the positive y-axis: \( S = (0, b) \)
- The fourth vertex \( R \) is located \( b \) units along the x-axis and \( b \) units along the y-axis: \( R = (b, b) \)
Thus, the coordinates of the vertices are \( P(0,0) \), \( Q(b,0) \), \( R(b,b) \), and \( S(0,b) \).
In simple words: Use the side length of the square to determine coordinates starting from the origin along the coordinate axes.

Exam Tip: Sketching a coordinate grid and plotting the square visually takes a few seconds and prevents coordinate swaps.

Question 14. If the points (5, 4) and (x, y) are equidistant from the point (4, 5); then show that x2 + y2 – 8x – 10y + 39 = 0
Answer: Let \( A(5, 4) \), \( B(x, y) \), and the center point be \( C(4, 5) \).
Given that \( A \) and \( B \) are at equal distances from \( C \):
\( BC = AC \implies BC^2 = AC^2 \)
Using the distance formula:
\( AC^2 = (5 - 4)^2 + (4 - 5)^2 = 1^2 + (-1)^2 = 1 + 1 = 2 \)
\( BC^2 = (x - 4)^2 + (y - 5)^2 = x^2 - 8x + 16 + y^2 - 10y + 25 \)
Equate the two:
\( x^2 + y^2 - 8x - 10y + 41 = 2 \)
Subtract 2 from both sides:
\( x^2 + y^2 - 8x - 10y + 39 = 0 \)
Hence Proved.
In simple words: Find the distance squared from the center to the first point, make it equal to the distance squared to the second point, and simplify the equation.

Exam Tip: Expanding terms like \( (x-4)^2 \) using \( (a-b)^2 = a^2 - 2ab + b^2 \) carefully ensures no signs are incorrect.

Question 15. The line segment joining the points (3, – 4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, – 2) and (5/3, q) respectively, find the value of p and q.
Answer: Trisection implies the segment \( AB \) with points \( A(3, -4) \) and \( B(1, 2) \) is divided into three equal parts by points \( P \) and \( Q \).
- Point \( P \) divides \( AB \) in the ratio 1 : 2.
Using the Section Formula for the x-coordinate of \( P \):
\( p = \frac{1(1) + 2(3)}{1 + 2} = \frac{1 + 6}{3} = \frac{7}{3} \)
- Point \( Q \) divides \( AB \) in the ratio 2 : 1.
Using the Section Formula for the y-coordinate of \( Q \):
\( q = \frac{2(2) + 1(-4)}{2 + 1} = \frac{4 - 4}{3} = 0 \)
Thus, the values are \( p = \frac{7}{3} \) and \( q = 0 \).
In simple words: Use the ratios 1 to 2 for the first dividing point and 2 to 1 for the second dividing point to solve for the missing coordinates.

Exam Tip: Remember that "trisection" always corresponds to dividing ratios of \( 1:2 \) and \( 2:1 \).

TRIANGLE

Question 1. State and prove Pythagoras theorem.
Answer: - Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
- Given: A right-angled triangle \( ABC \), right-angled at \( B \).
- To Prove: \( AC^2 = AB^2 + BC^2 \)
- Construction: Draw \( BD \perp AC \).
- Proof:
In \( \triangle ADB \) and \( \triangle ABC \):
\( \angle ADB = \angle ABC = 90^\circ \)
\( \angle A = \angle A \) (Common)
By AA similarity:
\( \triangle ADB \sim \triangle ABC \implies \frac{AD}{AB} = \frac{AB}{AC} \)
\( \implies AB^2 = AD \times AC \) (Equation 1)
Similarly, in \( \triangle BDC \) and \( \triangle ABC \):
\( \triangle BDC \sim \triangle ABC \implies \frac{CD}{BC} = \frac{BC}{AC} \)
\( \implies BC^2 = CD \times AC \) (Equation 2)
Add Equation 1 and Equation 2:
\( AB^2 + BC^2 = (AD \times AC) + (CD \times AC) \)
\( \implies AB^2 + BC^2 = AC(AD + CD) \)
Since \( AD + CD = AC \):
\( \implies AB^2 + BC^2 = AC(AC) \)
\( \implies AC^2 = AB^2 + BC^2 \)
Hence Proved.
In simple words: Draw a line straight down from the right angle to the opposite side. Use similar triangles to relate the sides and add the resulting equations together.

Exam Tip: Draw the perpendicular line clearly in your diagram, as it is the key construction needed to prove this theorem.

Question 2. State and prove Thale’s theorem.
Answer: - Statement (Basic Proportionality Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
- Given: A triangle \( ABC \) where a line \( DE \) is parallel to \( BC \), intersecting \( AB \) at \( D \) and \( AC \) at \( E \).
- To Prove: \( \frac{AD}{DB} = \frac{AE}{EC} \)
- Construction: Join \( BE \), \( CD \), and draw \( DM \perp AC \), \( EN \perp AB \).
- Proof:
The area of a triangle is given by \( \frac{1}{2} \times \text{base} \times \text{height} \).
\( \text{Area}(\triangle ADE) = \frac{1}{2} \times AD \times EN \)
\( \text{Area}(\triangle BDE) = \frac{1}{2} \times DB \times EN \)
Divide these areas:
\( \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \) (Equation 1)
Similarly:
\( \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \) (Equation 2)
Note that \( \triangle BDE \) and \( \triangle CDE \) are on the same base \( DE \) and between the same parallel lines \( DE \) and \( BC \).
Therefore, their areas are equal:
\( \text{Area}(\triangle BDE) = \text{Area}(\triangle CDE) \)
This makes the left side of Equation 1 and Equation 2 equal.
Thus, we get:
\( \frac{AD}{DB} = \frac{AE}{EC} \)
Hence Proved.
In simple words: Draw lines connecting the bottom corners to the opposite sides. Calculate the ratio of the triangular areas on each side to prove the ratio of the side segments is identical.

Exam Tip: Clearly write down the geometric rule that triangles on the same base and between same parallel lines have equal areas to secure full marks.

Question 3. P and Q are the points on the sides AB and AC respectively of triangle ABC. If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.
Answer: Let us compute the ratios of the segments on sides \( AB \) and \( AC \):
\( \frac{AP}{PB} = \frac{3}{6} = \frac{1}{2} \)
\( \frac{AQ}{QC} = \frac{5}{10} = \frac{1}{2} \)
Since \( \frac{AP}{PB} = \frac{AQ}{QC} \), by the Converse of the Basic Proportionality Theorem, \( PQ \parallel BC \).
Now, compare \( \triangle APQ \) and \( \triangle ABC \):
\( \angle A = \angle A \) (Common)
\( \angle APQ = \angle ABC \) (Corresponding angles)
By AA similarity, \( \triangle APQ \sim \triangle ABC \).
Thus, the ratio of their corresponding sides must be equal:
\( \frac{PQ}{BC} = \frac{AP}{AB} \)
Since \( AB = AP + PB = 3 + 6 = 9 \) cm:
\( \frac{PQ}{BC} = \frac{3}{9} = \frac{1}{3} \)
Cross-multiply:
\( \implies BC = 3 PQ \)
Hence Proved.
In simple words: Show that the line cuts the sides in equal proportions, making the small triangle similar to the big triangle. Use side ratios to prove the base relationship.

Exam Tip: Remember to calculate the full side length \( AB \) by adding the individual segments together before computing the scale factor.

Question 4. Find the length of the tangent drawn from a point P whose distance from the centre of the circle is 25 cm. It is given that the radius of the circle is 7 cm.
Answer: Let the circle have its center at \( O \). Let \( T \) be the point of contact of the tangent drawn from external point \( P \).
The radius \( OT \) is perpendicular to the tangent \( PT \) at the point of contact, making \( \angle OTP = 90^\circ \).
Thus, \( \triangle OTP \) is a right-angled triangle.
Using the Pythagoras Theorem:
\( OP^2 = OT^2 + PT^2 \)
Substitute the values:
\( 25^2 = 7^2 + PT^2 \)
\( \implies 625 = 49 + PT^2 \)
\( \implies PT^2 = 625 - 49 \)
\( \implies PT^2 = 576 \)
\( \implies PT = \sqrt{576} = 24 \text{ cm} \)
The length of the tangent is 24 cm.
In simple words: The radius and tangent line meet at a right angle, forming a right triangle. Use the Pythagoras theorem to solve for the missing tangent length.

Exam Tip: Tangent problems are almost always solved by using the right-angle property between radius and tangent lines.

Question 5. Prove that the lengths of the two tangents drawn from an external point to a circle are equal.
Answer: - Given: A circle with center \( O \), an external point \( P \), and two tangents \( PA \) and \( PB \) with contact points \( A \) and \( B \).
- To Prove: \( PA = PB \)
- Construction: Join \( OA \), \( OB \), and \( OP \).
- Proof:
In \( \triangle OAP \) and \( \triangle OBP \):
\( \angle OAP = \angle OBP = 90^\circ \) (Radius is perpendicular to tangent at point of contact)
\( OA = OB \) (Radii of the same circle)
\( OP = OP \) (Common side)
By RHS (Right angle - Hypotenuse - Side) congruence:
\( \triangle OAP \cong \triangle OBP \)
Therefore, by CPCT (Corresponding Parts of Congruent Triangles):
\( PA = PB \)
Hence Proved.
In simple words: Draw lines from the center to the contact points and the external point. This forms two right-angled triangles that are identical, proving the tangents are equal.

Exam Tip: Remember to cite "RHS congruence" and "CPCT" as justifications to receive full marks for your geometric proofs.

Question 6. Draw a circle of diameter 4 cm. Take a point P, 5 cm away from the centre of the circle. From P, draw a pair of tangents to the circle.
Answer: Follow these steps of construction to draw the figure using geometry instruments:
1. Mark a point \( O \) as the center. Using a compass set to a radius of 2 cm (since diameter is 4 cm), draw a circle.
2. Draw a straight line from center \( O \) to a point \( P \) such that the length \( OP = 5 \) cm.
3. Draw the perpendicular bisector of the line segment \( OP \). Let \( M \) be the midpoint of \( OP \) where the bisector intersects it.
4. Set your compass radius to \( OM \) (or \( MP \)), place the metal tip on midpoint \( M \), and draw a secondary circle.
5. Let the secondary circle intersect the primary circle at two points, say \( A \) and \( B \).
6. Draw straight lines connecting \( P \) to \( A \), and \( P \) to \( B \).
The lines \( PA \) and \( PB \) are the required pair of tangents.
In simple words: Draw your circle, mark point P at 5 cm from the center. Find the middle of line OP, draw a new circle from this midpoint, and connect the crossing points to P.

Exam Tip: Always keep your construction lines thin and do not erase the arcs used for finding the perpendicular bisector, as examiners grade them.

Question 7. Prove converse of Pythagoras theorem.
Answer: - Statement: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.
- Given: A triangle \( ABC \) in which \( AC^2 = AB^2 + BC^2 \).
- To Prove: \( \angle B = 90^\circ \)
- Construction: Construct another triangle \( PQR \), right-angled at \( Q \), such that \( PQ = AB \) and \( QR = BC \).
- Proof:
In right-angled triangle \( PQR \), using the Pythagoras Theorem:
\( PR^2 = PQ^2 + QR^2 \)
Substitute the constructed values \( PQ = AB \) and \( QR = BC \):
\( PR^2 = AB^2 + BC^2 \) (Equation 1)
But we are given:
\( AC^2 = AB^2 + BC^2 \) (Equation 2)
Compare Equation 1 and Equation 2:
\( PR^2 = AC^2 \implies PR = AC \)
Now, compare \( \triangle ABC \) and \( \triangle PQR \):
\( AB = PQ \) (By construction)
\( BC = QR \) (By construction)
\( AC = PR \) (Proved above)
By SSS (Side - Side - Side) congruence:
\( \triangle ABC \cong \triangle PQR \)
Therefore, by CPCT:
\( \angle B = \angle Q \)
Since \( \angle Q = 90^\circ \) by construction:
\( \angle B = 90^\circ \)
Hence Proved.
In simple words: Construct a secondary right triangle identical to the first two sides. Prove the third sides must also be equal, making the triangles congruent and showing the angle is indeed 90 degrees.

Exam Tip: Be sure to write down the construction steps clearly, as the entire proof relies on the comparison with the constructed right-angled triangle.

Question 8. Prove that the ratio of the corresponding altitudes of two similar triangles is equal to the ratio of their corresponding sides.
Answer: - Given: Two similar triangles \( ABC \) and \( DEF \) such that \( \triangle ABC \sim \triangle DEF \). Let \( AP \) and \( DQ \) be the altitudes drawn from vertices \( A \) and \( D \) to the opposite sides \( BC \) and \( EF \), respectively.
- To Prove: \( \frac{AP}{DQ} = \frac{AB}{DE} \)
- Proof:
Since \( \triangle ABC \sim \triangle DEF \), their corresponding angles are equal:
\( \angle B = \angle E \) (Equation 1)
Now, compare the two smaller triangles \( \triangle ABP \) and \( \triangle DEQ \):
\( \angle B = \angle E \) (From Equation 1)
\( \angle APB = \angle DQE = 90^\circ \) (Since \( AP \) and \( DQ \) are altitudes)
By AA similarity:
\( \triangle ABP \sim \triangle DEQ \)
Since the triangles are similar, their corresponding sides are in proportion:
\( \frac{AP}{DQ} = \frac{AB}{DE} \)
Hence Proved.
In simple words: Use the angle properties of similar triangles to prove the small triangles containing the altitudes are also similar. Write down their side proportions.

Exam Tip: The AA similarity criterion is the easiest way to relate sub-elements of similar triangles, such as medians, altitudes, or angle bisectors.

Question 9. In an equilateral triangle, prove that the three times the square on one side is equal to four times the square of its altitude.
Answer: Let \( ABC \) be an equilateral triangle with side length \( a \). Let \( AD \) be the altitude from \( A \) to \( BC \).
Since \( AD \perp BC \) and the triangle is equilateral, \( AD \) bisects the base \( BC \).
Therefore, \( BD = \frac{a}{2} \).
In right-angled triangle \( ABD \), using the Pythagoras Theorem:
\( AB^2 = AD^2 + BD^2 \)
Substitute the side lengths:
\( a^2 = AD^2 + \left( \frac{a}{2} \right)^2 \)
\( \implies a^2 = AD^2 + \frac{a^2}{4} \)
\( \implies a^2 - \frac{a^2}{4} = AD^2 \)
\( \implies \frac{3a^2}{4} = AD^2 \)
Cross-multiply by 4:
\( \implies 3a^2 = 4AD^2 \)
Since \( a \) is the length of any side (e.g., \( AB \)):
\( \implies 3(AB)^2 = 4(AD)^2 \)
Hence Proved.
In simple words: The altitude splits the base of an equilateral triangle in half. Apply the Pythagoras theorem to one half to get the required side-to-height ratio.

Exam Tip: Remember to specify that the altitude of an equilateral triangle bisects the base; this fact is the core of your proof.

Question 10. Prove that the area of the ∆ BCE described on one side BC of a square ABCD as base is one half the area of the similar ∆ ACF described on the diagonal AC as base ?
Answer: Let the side of the square \( ABCD \) be \( x \) units.
The diagonal of a square of side \( x \) is given by:
\( AC = x\sqrt{2} \) units.
We are given two similar triangles: \( \triangle BCE \) (with base \( BC = x \)) and \( \triangle ACF \) (with base \( AC = x\sqrt{2} \)).
According to the theorem on areas of similar triangles, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
\( \frac{\text{Area}(\triangle BCE)}{\text{Area}(\triangle ACF)} = \left( \frac{BC}{AC} \right)^2 \)
Substitute the base values:
\( \frac{\text{Area}(\triangle BCE)}{\text{Area}(\triangle ACF)} = \left( \frac{x}{x\sqrt{2}} \right)^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} \)
Cross-multiply:
\( \implies \text{Area}(\triangle BCE) = \frac{1}{2} \text{ Area}(\triangle ACF) \)
Hence Proved.
In simple words: The diagonal of a square is square-root of two times longer than its side. Since similar triangles scale with the square of their sides, the area ratio becomes one half.

Exam Tip: Be sure to write the complete area-ratio theorem statement clearly before substituting the side variables.

SOCIAL SCIENCE - ECONOMICS

Very Short Answer Type Questions

Question 1. What do you mean by economic development?
Answer: Economic development is a sustained, long-term process that leads to a real increase in per capita income along with improvements in the overall standard of living, reduction in poverty, and better access to health and education.
In simple words: It means a country is getting richer while also improving the daily lives and well-being of all its citizens.

Exam Tip: Mention both income increase and qualitative aspects like health/education to get full marks.

Question 2. Mention one important characteristic of development.
Answer: One essential feature of development is that different people can have varying, and sometimes even conflicting, goals regarding what progress means to them.
In simple words: What looks like progress to one person might not be seen as progress by someone else.

Exam Tip: Use the classic textbook example of dams providing electricity to industrialists while displacing local tribal communities.

Question 3. Why has Kerala a low infant mortality rate?
Answer: Kerala maintains a remarkably low infant mortality rate because of its strong public focus on basic healthcare services, high female literacy, and effective food distribution channels.
In simple words: It has excellent public health facilities and high levels of education among mothers.

Exam Tip: Keep the terms "basic health" and "educational facilities" as core keywords in your answer.

Question 4. What are renewable and non-renewable resources?
Answer: Renewable resources are naturally replenished over time (such as solar or wind energy), whereas non-renewable resources are finite and exhaustible (such as coal or oil) because they take millions of years to form.
In simple words: Renewable resources never run out, but non-renewable resources are gone forever once we use them up.

Exam Tip: Always provide clear, distinct examples for both types of resources to strengthen your answer.

Question 5. How does the agricultural sector dominate the other two sectors in spite of its lowest contribution to the GDP of the nation?
Answer: Agriculture remains dominant because it continues to employ nearly half of the country's entire workforce, making it vital for public livelihood security despite its low share in national economic output.
In simple words: Even though farming brings in less money, it is still crucial because it provides jobs to more than half of the population.

Exam Tip: Highlight the contrast between the share of employment (high) and the share of GDP (low) to show a clear understanding.

Question 6. Give two demerits of the unorganized sector.
Answer: Two major drawbacks are:
1. Low and irregular wages with no provisions for paid leave or holidays.
2. A complete lack of job security, meaning employers can dismiss workers without any prior notice.
In simple words: Workers get paid very little with no job security or paid sick leave.

Exam Tip: Present your demerits as clear, numbered points for better presentation.

Question 7. What are the bases on which we divide the Indian economy into different sectors?
Answer: The economy is categorized based on:
1. The nature of production activities (Primary, Secondary, Tertiary).
2. The conditions of employment (Organized and Unorganized).
3. The ownership of assets (Public and Private).
In simple words: We sort the economy by the type of work done, how secure the jobs are, and who owns the businesses.

Exam Tip: Mentioning all three criteria shows a comprehensive grasp of the chapter.

Question 8. Give one of the drawbacks of the organized sector.
Answer: A primary limitation of the organized sector is that its job creation rate is quite slow, leaving many job seekers with no choice but to work in the unorganized sector.
In simple words: There are very few new jobs created in secure, formal offices and factories.

Exam Tip: Emphasize how this slow growth leads to underemployment in informal sectors.

Question 9. Identify the most important drawback of the barter system.
Answer: The biggest obstacle is the necessity for a "double coincidence of wants," meaning both trading parties must want exactly what the other person is offering.
In simple words: You can only trade if you find someone who has what you need and wants exactly what you have.

Exam Tip: "Double coincidence of wants" is the single most important term for this answer; underline it.

Question 10. What is the chief function of money?
Answer: The primary function of money is to act as a universally accepted medium of exchange, which removes the difficulties of trading under the barter system.
In simple words: Money's main job is to make buying and selling simple for everyone.

Exam Tip: Keep the answer brief and focus on the term "medium of exchange."

Question 11. Name two facilities offered by the demand deposits.
Answer: Two key features are:
1. The facility to withdraw money from your account whenever required.
2. The ability to make direct payments using checks instead of physical cash.
In simple words: You can take your money out whenever you want, and you can pay people using bank checks.

Exam Tip: Mentioning "cheque payments" and "easy withdrawals" covers the essential features examiners look for.

Question 12. Define cheque and credit.
Answer: - Cheque: A paper document that instructs a bank to pay a specific sum from a person's account to the person named on it.
- Credit: An agreement where a lender provides money, goods, or services to a borrower in return for a promise of future repayment with interest.
In simple words: A cheque is a written note telling your bank to pay someone, and credit means borrowing money now to pay it back later.

Exam Tip: Define both terms separately in short paragraphs to keep your presentation clean.

Question 13. What do you mean by terms of credit?
Answer: Terms of credit comprise the complete set of conditions for a loan, which includes the interest rate, collateral requirements, supporting documentation, and the chosen method of repayment.
In simple words: These are the rules and requirements you must agree to when borrowing money.

Exam Tip: List the four components (interest, collateral, documents, repayment mode) as a brief list for clarity.

Question 14. Mention one disadvantage of a high cost of borrowing.
Answer: A high cost of borrowing leaves the borrower with very little income after paying off the high interest rates, which often traps them in a cycle of debt.
In simple words: High interest rates eat up your profits and can trap you in a dangerous cycle of unpaid loans.

Exam Tip: Use the term "debt-trap" as it is a crucial term in development economics assessments.

Question 15. What is a SHG?
Answer: A Self-Help Group (SHG) is a small, informal association of 15 to 20 rural residents (mostly women) who meet regularly to pool their savings and provide low-interest loans to its members.
In simple words: It is a small group of village neighbors who save money together to help each other with cheap loans.

Exam Tip: Mentioning "rural women" and the typical group size of "15-20 members" shows precise textbook knowledge.

Short Answer Type Questions

Question 1. Distinguish between modern and traditional types of money.
Answer: - Modern Money: Consists of paper notes, coins, and digital transactions. It is authorized by the government and has no intrinsic value of its own (it is not made of precious metals).
- Traditional Money: Consisted of items that had everyday value, such as grains, cattle, or coins minted from precious metals like gold, silver, and copper.
In simple words: Modern money is just paper and digital digits backed by the government, whereas traditional money was made of useful things like grain or gold coins.

Exam Tip: Creating a simple comparative table is the best way to present distinctions clearly.

Question 2. Why is the Rupee accepted as a medium of exchange in the entire India?
Answer: The Indian Rupee is universally accepted across the country because it is officially authorized by the central government of India. Furthermore, by national law, no individual or organization in the country can legally refuse payments made in Rupees for settling transactions.
In simple words: The government backs the Rupee, and the law says no one in India can reject it as payment.

Exam Tip: Mention the role of the Reserve Bank of India (RBI) in issuing currency on behalf of the central government to make your answer more authoritative.

Question 3. What is collateral?
Answer: Collateral is an asset owned by the borrower (such as land, buildings, vehicles, livestock, or bank deposits) that is pledged to the lender as a security guarantee until the loan is fully repaid. If the borrower fails to repay the loan, the lender has the legal right to sell this asset to recover the money.
In simple words: It is an asset you own that the bank can take and sell if you are unable to pay back your loan.

Exam Tip: Give common examples like "land papers" or "gold jewelry" to show practical application.

Question 4. Distinguish between formal and informal sources of credit.
Answer: - Formal Sources: Include banks and cooperative societies. They charge reasonable interest rates, require collateral, and are strictly supervised by the Reserve Bank of India (RBI).
- Informal Sources: Include money lenders, friends, relatives, and traders. They often charge extremely high interest rates, require minimal paperwork, and operate with no external regulation.
In simple words: Formal credit comes from official banks with low interest rates, while informal credit comes from local lenders who charge very high rates.

Exam Tip: Contrast the role of the RBI (which supervises formal sources but has no control over informal ones) as a key point of difference.

Question 5. Why is a loan sanctioned to the SHGs? Give one example.
Answer: Banks grant loans to Self-Help Groups because their regular savings habit proves their financial discipline, and the group takes collective responsibility for repayment, eliminating the need for individual collateral.
- Example: A bank may grant a group loan to an SHG so its members can buy sewing machines to start a small tailoring business.
In simple words: Banks trust these groups because they save money together regularly and guarantee to pay back the loan as a team.

Exam Tip: Emphasize that "collective responsibility" is what overcomes the "lack of collateral" obstacle for poor borrowers.

Question 6. Why are the poor households depend upon the informal sources of credit?
Answer: Poor households often rely on informal lenders because:
1. They lack the assets required as collateral for formal bank loans.
2. Banks are not easily accessible in many remote rural regions.
3. The application process for bank loans involves complex documentation that is difficult for illiterate borrowers to complete.
In simple words: They use local lenders because they do not have land or gold to show as security, and banks require too much complicated paperwork.

Exam Tip: Clearly list these as three separate points to make your answer highly readable.

Question 7. What do you understand by disguised unemployment? Explain with an example.
Answer: Disguised unemployment is a situation where more people are working on a task than are actually required. If a few workers are removed, the total output remains completely unchanged, indicating that their actual contribution is zero.
- Example: If a small family farm requires only three people to cultivate it, but all five family members work there, the extra two members are under disguised unemployment.
In simple words: It is when people look busy working, but their work does not actually add anything to the final output because there are too many workers.

Exam Tip: Define the term clearly first, then provide a simple agricultural example to illustrate.

Question 8. How can underemployment occur in the service sector? Give example.
Answer: Underemployment occurs in the service sector when people are forced to do jobs that do not utilize their skills fully or do not offer full-time work, leading to low earnings.
- Example: Thousands of casual painters, plumbers, and daily wage cart pullers search for work all day but might only find a few hours of work, earning very little.
In simple words: It happens when skilled workers can only find casual, low-paying daily jobs instead of proper full-time positions.

Exam Tip: Do not limit underemployment to agriculture; showing its presence in urban services earns extra credit.

Question 9. With economic growth, the occupational structure of the country has changed. Explain this in the context of India.
Answer: While India’s GDP has seen a massive shift away from agriculture toward the service (tertiary) sector, the share of employment has not changed in equal proportion. Agriculture still employs nearly 45% of the workforce, despite contributing less than 20% to the national income, showing that the shift in jobs has been much slower than the shift in production.
In simple words: Even though services and factories produce most of our wealth now, a massive portion of our population is still stuck working in farming.

Exam Tip: Highlight this discrepancy between "production share" and "employment share" to write a sophisticated answer.

Question 10. How can job opportunities can be created in the service sector?
Answer: Livelihoods in the service sector can be expanded by:
1. Providing low-interest loans to help people start small services like transport, repair shops, or training centers.
2. Improving tourism infrastructure to generate jobs for guides, hotels, and transport operators.
3. Investing in public education and healthcare, which naturally creates new jobs for teachers and health workers.
In simple words: We can create jobs by funding schools and hospitals, promoting tourism, and giving cheap loans to small business owners.

Exam Tip: Mentioning government planning and policy initiatives (like promotion of regional tourism) adds value to your answer.

Question 11. How are the farmers affected in the rural areas where the money lenders are unorganized?
Answer: In areas dominated by unorganized money lenders, farmers suffer because they are charged usurious interest rates. If crops fail due to bad weather, farmers are unable to repay the debt, which forces them to sell their lands, dragging them deep into a spiral of unpaid loans.
In simple words: Without proper banks, farmers borrow from local lenders at high rates and risk losing their land if their crops fail.

Exam Tip: Explain how high interest rates directly lead to landlessness among small farmers.

Question 12. State few measures to protect the laboureres in the unorganized sector?
Answer: Key steps to protect casual workers include:
1. Setting and strictly enforcing minimum wage laws to prevent exploitation.
2. Providing cheap loans to help workers escape exploitative employers.
3. Setting up safety nets like health insurance and pension schemes for informal workers.
In simple words: The government must ensure fair minimum wages, provide simple credit options, and offer basic healthcare support.

Exam Tip: Focus on the twin goals of legal protection (wages) and economic support (loans/insurance).

Question 13. Human development is the essence of social development. Explain.
Answer: Human development shifts the focus from simple economic growth to the quality of human life. Social progress remains incomplete if citizens are rich but lack clean air, safety, and basic literacy. Thus, true development is about expanding human capabilities, health, and freedom of choice.
In simple words: Real progress is not just about having more money; it is about ensuring everyone is healthy, educated, and free to live a good life.

Exam Tip: Use the concept of "capabilities expansion" as it is central to human development theory.

Question 14. Distinguish between human and economic development.
Answer: - Economic Development: Focuses primarily on quantitative indicators such as GDP, national wealth, and per capita income.
- Human Development: A much broader concept that focuses on qualitative indicators like life expectancy, literacy levels, and access to clean water and healthcare.
In simple words: Economic development is about a nation's wealth, while human development is about the quality of life of its citizens.

Exam Tip: Point out that economic development is a means to achieve the ultimate end, which is human development.

Question 15. What are the factors that contribute to human development?
Answer: The primary pillars of human development are:
1. Quality education and high literacy rates.
2. Access to affordable healthcare facilities, which leads to higher life expectancy.
3. Adequate per capita income to afford a decent standard of living.
4. Social equality, personal freedom, and clean environmental conditions.
In simple words: Good schools, reliable hospitals, fair incomes, and a clean environment are the key ingredients for human development.

Exam Tip: Connect these points to the three dimensions of the Human Development Index (HDI): Knowledge, Long Life, and Decent Income.

Long Answer Questions

Question 1. Enumerate the economic and non-economic factors contributing to development.
Answer: Development depends on a mix of economic and non-economic factors:
Economic Factors:
1. Natural Resources: Availability of fertile land, water, and minerals forms the base of production.
2. Capital Formation: Regular investments in machinery, factories, and technology boost productivity.
3. Infrastructure: Roads, electricity grids, and communication networks facilitate trade.
Non-Economic Factors:
1. Education and Health: A skilled and healthy workforce is more productive.
2. Social Structure: Caste-free and equal societies grow faster as they prevent wastage of human talent.
3. Political Stability: Transparent governance and stable laws encourage long-term business investments.
In simple words: Economic growth requires resources, machinery, and roads, but it also depends on good schools, reliable hospitals, and stable governments.

Exam Tip: Structure long answers under bold subheadings to make them easy for the examiner to read.

Question 2. What do you mean by national development? What are the aspects covered in it?
Answer: National development is a comprehensive term that includes improvements in the living standards of a country's citizens, a rise in per capita income, and equitable distribution of resources. It covers several aspects:
1. Economic Growth: Increasing national output through industrialization and modern agriculture.
2. Social Justice: Ensuring the benefits of development reach poor and marginalized groups, reducing inequality.
3. Institutional Progress: Strengthening legal and administrative systems to provide quick justice and transparent services.
4. Sustainable Environment: Managing natural resources carefully so that the development of today does not compromise the needs of future generations.
In simple words: National development means the whole country is moving forward, providing fair opportunities, better services, and a clean environment to all its citizens.

Exam Tip: Emphasize that "equitable distribution" is just as important as "growth" in national development assessments.

Question 3. Why is the public sector enterprises so important to us?
Answer: Public sector enterprises (owned by the government) are crucial for several reasons:
1. Infrastructure Development: They undertake massive capital projects like railways, dams, and power plants that private companies avoid due to low profits.
2. Affordable Services: They provide essential services like health, education, and transport at cheap rates to help poor families.
3. Job Creation: They act as major employers, providing secure jobs with fair wages and social security benefits.
4. Balanced Regional Growth: The government sets up factories in backward and rural regions to promote local development and reduce regional inequalities.
In simple words: Government companies are important because they build big national projects, keep essential services cheap, create secure jobs, and develop poor areas.

Exam Tip: Contrast the "profit motive" of the private sector with the "welfare motive" of the public sector to highlight their different roles.

Question 4. Why are the secondary and the tertiary sectors could not create employment opportunities?
Answer: Despite their rapid economic growth, the secondary and tertiary sectors have failed to generate enough jobs because:
1. Capital-Intensive Technology: Factories are increasingly using automated machines and robotics instead of human labor, reducing the need for new workers.
2. Skill Mismatch: Modern services like IT, finance, and telecommunications require highly specialized technical skills, which most rural migrants do not possess.
3. Informal Growth: A large portion of service growth has been in informal services like street vending and casual labor, which do not offer secure, full-time jobs.
In simple words: Factories are using machines instead of people, and modern offices need high-tech skills that many job seekers do not have.

Exam Tip: Use terms like "capital-intensive production" and "skill gap" to describe this economic problem accurately.

Question 5. Distinguish between the primary and the secondary sectors?
Answer:

In simple words: The primary sector extracts raw materials straight from nature, while the secondary sector processes them into useful manufactured products in factories.

Exam Tip: Presenting comparisons in a table with clear "basis of distinction" column ensures maximum marks.

Question 6. How can unemployment be removed by creating more jobs?
Answer: Livelihood opportunities can be expanded through targeted investments:
1. Rural Infrastructure: Constructing canals, roads, and storage cold chains creates immediate construction jobs and helps farmers sell crops easily.
2. Promoting Small Industries: Setting up food-processing units and cottage industries in rural areas creates non-farm jobs close to home.
3. Expanding Tourism and Local Crafts: Developing tourist spots and helping regional artisans can create over 35 lakh jobs according to planning estimates.
4. Vocational Training: Establishing training centers to teach modern technical skills helps youth find jobs easily.
In simple words: We can tackle unemployment by building rural roads, setting up local food factories, promoting tourism, and teaching useful job skills to youth.

Exam Tip: Mentioning government schemes like MGNREGA (100 days of guaranteed work) provides a great real-world example for this answer.

Question 7. Define commercial banks. State their functions.
Answer: Commercial banks are financial institutions that accept deposits from the public and grant loans for consumption and investment purposes to earn a profit.
Their primary functions include:
1. Accepting Deposits: They keep public savings secure in savings, current, or fixed accounts.
2. Providing Loans: They grant credit to farmers, business owners, and consumers at reasonable interest rates.
3. Facilitating Payments: They offer payment services like check clearance, online transfers, and demand drafts.
4. Credit Creation: They expand the money supply in the economy through the process of lending out deposits.
In simple words: Commercial banks are financial businesses that keep your money safe, lend money to borrowers, and make payments simple through checks and cards.

Exam Tip: Clearly distinguish between "accepting deposits" and "advancing loans" as the two core pillars of commercial banking.

COMPUTER

Question 1. Solve Question No 2 from CBSE Question Bank (2002, 2003, 2004, 2005 and 2006) from the last page the Computer Text Book.
Answer: To complete this task, locate Question Number 2 in the designated CBSE Question Bank section at the end of your Class X computer textbook. Solve all the sub-parts, focusing on core programming questions (such as string manipulation, loops, or basic database queries depending on the syllabus) in your computer science assignment notebook. Write the code blocks or database statements clearly, showing both the inputs and the expected terminal outputs.
In simple words: Open the back of your computer book, find Question 2 in the CBSE section, and write down the answers in your notebook.

Exam Tip: Dry-run your programs on paper with simple test inputs before writing the final code blocks to catch logical errors early.

CHEMISTRY

Question 1. Balance the following equations.
(a) \( \text{Fe} + \text{Cl}_2 \rightarrow \text{FeCl}_3 \)
(b) \( \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{HCl} \)
(c) \( \text{Ca(OH)}_2 + \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} \)
(d) \( \text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \)
(e) \( \text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \rightarrow \text{AlCl}_3 + \text{BaSO}_4 \)
Answer:
(a) \( 2\text{Fe} + 3\text{Cl}_2 \rightarrow 2\text{FeCl}_3 \)
(b) \( 2\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HCl} \)
(c) \( \text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} \)
(d) \( \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \)
(e) \( 3\text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \rightarrow 2\text{AlCl}_3 + 3\text{BaSO}_4 \)
In simple words: Add coefficients in front of the molecules so that the count of each atom is identical on both the left and right sides.

Exam Tip: Always write a quick final tally table showing that the atom count is equal on both sides to verify your balanced equation.

Question 2. A copper coin fell in silver nitrate solution. After few days when it was taken out, it was distorted. Explain why?
Answer: Copper is more reactive than silver on the reactivity series. Therefore, when a copper coin is placed in a silver nitrate solution, a displacement reaction occurs. Copper displaces silver from its solution, forming copper nitrate while metallic silver deposits on the coin. This chemical erosion distorts the coin.
The chemical equation is:
\( \text{Cu(s)} + 2\text{AgNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + 2\text{Ag(s)} \)
The solution also changes from colorless to light blue due to the formation of \( \text{Cu}^{2+} \) ions.
In simple words: Copper is more active than silver, so it pushes silver out of the solution. This eats away the copper coin, changing its shape and turning the liquid blue.

Exam Tip: Mentioning both the coin distortion and the blue color change of the solution shows a complete practical understanding.

Question 3. Consider the reaction:
\( \text{CuSO}_4(\text{aq.}) + \text{Fe(s)} \rightarrow \text{FeSO}_4(\text{aq.}) + \text{Cu(s)} \)
\( \text{FeSO}_4(\text{aq.}) + \text{Zn(s)} \rightarrow \text{ZnSO}_4(\text{aq.}) + \text{Fe(s)} \)
Indicate which is most reactive and which is least reactive metal out of Zn, Cu and Fe.
Answer: - In the first reaction, Iron (\( \text{Fe} \)) displaces Copper (\( \text{Cu} \)) from its solution, showing that \( \text{Fe} \) is more reactive than \( \text{Cu} \) (\( \text{Fe} > \text{Cu} \)).
- In the second reaction, Zinc (\( \text{Zn} \)) displaces Iron (\( \text{Fe} \)) from its solution, showing that \( \text{Zn} \) is more reactive than \( \text{Fe} \) (\( \text{Zn} > \text{Fe} \)).
By combining these observations, the reactivity order is:
\( \text{Zn} > \text{Fe} > \text{Cu} \)
Thus, Zinc is the most reactive metal and Copper is the least reactive metal.
In simple words: Zinc pushes iron out, and iron pushes copper out. This shows Zinc is the strongest (most reactive) and Copper is the weakest (least reactive).

Exam Tip: Use the ">" symbols to write out the reactivity order clearly, as this format is highly favored by board examiners.

Question 4. What is the difference between displacement and Redox reactions? Give an example of each type.
Answer: - Displacement Reaction: A reaction where a more reactive element displaces a less reactive element from its salt solution.
- Example: \( \text{Fe(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{FeSO}_4\text{(aq)} + \text{Cu(s)} \)
- Redox Reaction: A reaction where both oxidation (loss of electrons or gain of oxygen) and reduction (gain of electrons or loss of oxygen) occur at the same time.
- Example: \( \text{CuO(s)} + \text{H}_2\text{(g)} \rightarrow \text{Cu(s)} + \text{H}_2\text{O(l)} \) (where \( \text{CuO} \) is reduced and \( \text{H}_2 \) is oxidized).
In simple words: Displacement is when a strong metal replaces a weaker one in a solution, while Redox is when one substance loses oxygen and another gains it at the same time.

Exam Tip: Write down both definitions along with balanced chemical equations, clearly labeling which parts are being oxidized and reduced.

Question 5. Identify the substance oxidized, the substance reduced, oxidizing agent and reducing agent in each of the following reactions.
(a) \( \text{Zn (s)} + 2\text{ Ag NO}_3\text{ (aq)} \rightarrow \text{Zn (NO}_3)_2 + 2\text{Ag (s)} \)
(b) \( \text{Fe} + \text{S} \rightarrow \text{FeS} \)
Answer:
(a) For reaction \( \text{Zn} + 2\text{AgNO}_3 \rightarrow \text{Zn(NO}_3)_2 + 2\text{Ag} \):
- Substance Oxidized: \( \text{Zn} \) (loses electrons to form \( \text{Zn}^{2+} \))
- Substance Reduced: \( \text{AgNO}_3 \) (specifically \( \text{Ag}^+ \) gains electrons)
- Oxidizing Agent: \( \text{AgNO}_3 \) (since it helps \( \text{Zn} \) oxidize)
- Reducing Agent: \( \text{Zn} \) (since it reduces \( \text{Ag}^+ \))

(b) For reaction \( \text{Fe} + \text{S} \rightarrow \text{FeS} \):
- Substance Oxidized: \( \text{Fe} \) (loses electrons to form \( \text{Fe}^{2+} \))
- Substance Reduced: \( \text{S} \) (gains electrons to form \( \text{S}^{2-} \))
- Oxidizing Agent: \( \text{S} \)
- Reducing Agent: \( \text{Fe} \)
In simple words: The element that loses electrons is oxidized, and the one that gains them is reduced. The oxidized substance acts as the reducing agent, and the reduced one is the oxidizing agent.

Exam Tip: Remember that both the substance oxidized and the reducing agent refer to the exact same chemical reactant on the left side of the equation.

Question 6. Is the reaction \( 2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl} \) an oxidation reduction reaction. Why?
Answer: Yes, it is a redox reaction. According to electronic concept, Sodium (\( \text{Na} \)) loses electrons to form \( \text{Na}^+ \) ions, meaning it undergoes oxidation.
At the same time, Chlorine gas (\( \text{Cl}_2 \)) gains those electrons to form \( \text{Cl}^- \) ions, meaning it undergoes reduction.
Oxidation half-reaction: \( 2\text{Na} \rightarrow 2\text{Na}^+ + 2\text{e}^- \)
Reduction half-reaction: \( \text{Cl}_2 + 2\text{e}^- \rightarrow 2\text{Cl}^- \)
Since both processes occur together, it is a Redox reaction.
In simple words: Yes, because sodium gives away electrons (oxidation) and chlorine takes them in (reduction) to form salt.

Exam Tip: Explain redox reactions using the electron-transfer definition (loss and gain of electrons) when oxygen is not present in the reaction.

Question 7. The molecular formula of chloride of a cation X is \( \text{XCl}_2 \). Write the formula of sulphate of X.
Answer: In the chloride compound \( \text{XCl}_2 \), since the valency of Chlorine (\( \text{Cl} \)) is \( -1 \), the cation \( \text{X} \) must have a valency of \( +2 \) (written as \( \text{X}^{2+} \)).
The sulphate polyatomic ion has a valency of \( -2 \) (written as \( \text{SO}_4^{2-} \)).
Using the criss-cross method to write the chemical formula:
\( \text{X}^{2+} \) and \( \text{SO}_4^{2-} \) combine in a \( 1 : 1 \) ratio.
Thus, the formula of its sulphate is \( \text{XSO}_4 \).
In simple words: The compound formula tells us the cation has a charge of \( +2 \). Since sulphate has a charge of \( -2 \), they combine in equal parts to form \( \text{XSO}_4 \).

Exam Tip: Always reduce the subscripts to their simplest whole-number ratio when writing final chemical formulas (e.g., write \( \text{XSO}_4 \) instead of \( \text{X}_2(\text{SO}_4)_2 \)).

Question 8. What do you mean by indicators give examples?
Answer: Indicators are weak organic acids or bases that change color depending on whether they are in an acidic or basic medium, helping us identify solutions.
- Examples:
1. Natural Indicators: Litmus paper (turns red in acid, blue in base), turmeric paste.
2. Synthetic Indicators: Phenolphthalein (colorless in acid, pink in base), Methyl orange.
In simple words: Indicators are substances that change their color to show whether a liquid is an acid or a base.

Exam Tip: Mentioning the distinct colors changes of at least one indicator in both acid and base media makes your answer complete.

Question 9. What do you mean by Basicity of an Acid?
Answer: The basicity of an acid refers to the total number of hydronium ions (\( \text{H}_3\text{O}^+ \) or \( \text{H}^+ \)) that a single molecule of the acid can release upon complete ionization in water.
- Examples:
- \( \text{HCl} \) is monobasic (releases one \( \text{H}^+ \) ion).
- \( \text{H}_2\text{SO}_4 \) is dibasic (releases two \( \text{H}^+ \) ions).
In simple words: It is the count of hydrogen ions that one molecule of an acid can release when dissolved in water.

Exam Tip: Do not confuse "basicity of an acid" with pH; basicity is strictly about the count of replaceable hydrogen atoms.

Question 10. What do you mean by strong and weak acid? Give example.
Answer: - Strong Acid: An acid that undergoes complete ionization in aqueous solution, producing a high concentration of hydronium ions (\( \text{H}^+ \)).
- Example: Hydrochloric acid (\( \text{HCl} \)), Sulphuric acid (\( \text{H}_2\text{SO}_4 \)).
- Weak Acid: An acid that only undergoes partial ionization in water, resulting in a low concentration of hydronium ions.
- Example: Acetic acid (\( \text{CH}_3\text{COOH} \)), Carbonic acid (\( \text{H}_2\text{CO}_3 \)).
In simple words: Strong acids split up completely in water to release lots of hydrogen ions, whereas weak acids only split up partially.

Exam Tip: Use the term "degree of ionization" to explain the fundamental difference between strong and weak acids.

Question 11. What do you mean by weak and strong base? Give suitable example?
Answer: - Strong Base: A base that completely dissociates in water to release a high concentration of hydroxide ions (\( \text{OH}^- \)).
- Example: Sodium hydroxide (\( \text{NaOH} \)), Potassium hydroxide (\( \text{KOH} \)).
- Weak Base: A base that only partially dissociates in water, producing a low concentration of hydroxide ions.
- Example: Ammonium hydroxide (\( \text{NH}_4\text{OH} \)), Calcium hydroxide (\( \text{Ca(OH)}_2 \)).
In simple words: Strong bases dissolve completely to release lots of hydroxide ions, while weak bases release very few.

Exam Tip: Clearly write the chemical dissociation formulas (e.g., \( \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \)) to illustrate complete dissociation.

Question 12. What is neutralization reaction?
Answer: A neutralization reaction is a chemical reaction where an acid reacts with a base to form salt and water, accompanied by the release of heat.
In terms of ionic theory:
\( \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)} \)
- Example:
\( \text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)} \)
In simple words: It is the chemical reaction that occurs when you mix an acid with a base, which cancels out their properties to produce salt and water.

Exam Tip: Neutralization reactions are always exothermic; mentioning this heat release shows excellent chemistry skills.

PHYSICS - CHAPTER: LIGHT

Question 1. Why are the danger signals of red colour?
Answer: Red light has the longest wavelength among all the visible colors. According to Rayleigh's scattering law, the intensity of scattered light is inversely proportional to the fourth power of its wavelength. Due to its long wavelength, red light is scattered the least by air molecules, dust, or smoke, enabling it to travel long distances and remain clearly visible even in foggy or dusty conditions.
In simple words: Red has a long wavelength and does not get scattered easily by dust or fog, so it remains visible from very far away.

Exam Tip: Citing "Rayleigh’s scattering law" and the term "least scattered" secures maximum marks.

Question 2. Magnification produced by a convex mirror for an object of size 5 cm is 1/2. What is the size of the image?
Answer: The formula for magnification \( m \) is:
\( m = \frac{h_i}{h_o} \)
Here, \( h_o = 5 \) cm and \( m = \frac{1}{2} \).
\( \frac{1}{2} = \frac{h_i}{5} \)
\( \implies h_i = \frac{5}{2} = 2.5 \text{ cm} \)
The size of the image is 2.5 cm.
In simple words: Since the magnification is one half, the image size is simply half of the object's original height.

Exam Tip: Since convex mirrors always produce upright virtual images, the magnification value is positive, which results in a positive image height.

Question 3. Refractive index of water is 1.33. Calculate the speed of light in water.
Answer: The formula for refractive index \( n \) is:
\( n = \frac{c}{v} \)
Where:
- \( c = 3 \times 10^8 \) m/s (speed of light in vacuum)
- \( n = 1.33 \approx \frac{4}{3} \) (refractive index of water)
- \( v \) is the speed of light in water.
Rearranging the formula to solve for \( v \):
\( v = \frac{c}{n} = \frac{3 \times 10^8}{\frac{4}{3}} \)
\( \implies v = \frac{9 \times 10^8}{4} = 2.25 \times 10^8 \text{ m/s} \)
The speed of light in water is \( 2.25 \times 10^8 \) m/s.
In simple words: Divide the speed of light in a vacuum by the water's refractive index to find the speed of light inside water.

Exam Tip: Substituting \( 1.33 \) as the fraction \( \frac{4}{3} \) makes calculations much simpler and faster.

Question 4. The radius of curvature of a concave mirror is 50 cm. Where an object should be held from the mirror so as to form its image at infinity?
Answer: To form the image of an object at infinity using a concave mirror, the object must be placed exactly at its principal focus (\( F \)).
The focal length \( f \) is related to the radius of curvature \( R \) by the formula:
\( f = \frac{R}{2} \)
Given, \( R = 50 \) cm:
\( f = \frac{50}{2} = 25 \text{ cm} \)
Therefore, the object should be placed at a distance of 25 cm in front of the concave mirror.
In simple words: An image is formed at infinity when the object is at the focus point. The focus point is always half of the radius of curvature, which is 25 cm.

Exam Tip: Always specify that the object should be placed "in front of the mirror" to describe its position completely.

Question 5. A 3 cm high object is placed at a distance of 20 cm from a concave mirror. A real image is formed at 40 cm from the mirror. Calculate the focal length of concave mirror and size of the image formed.
Answer: Given data:
- Height of object \( h_o = 3 \) cm
- Object distance \( u = -20 \) cm (using Cartesian sign convention)
- Image distance \( v = -40 \) cm (negative because a real image is formed in front of the mirror)
Using the Mirror Formula:
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
\( \implies \frac{1}{f} = \frac{1}{-40} + \frac{1}{-20} \)
\( \implies \frac{1}{f} = \frac{-1 - 2}{40} = -\frac{3}{40} \)
\( \implies f = -\frac{40}{3} \approx -13.33 \text{ cm} \)
The focal length of the concave mirror is \( -13.33 \) cm.
Now, using the Magnification Formula:
\( m = -\frac{v}{u} = \frac{h_i}{h_o} \)
\( \implies -\left( \frac{-40}{-20} \right) = \frac{h_i}{3} \)
\( \implies -2 = \frac{h_i}{3} \)
\( \implies h_i = -6 \text{ cm} \)
The size of the image formed is 6 cm, and the negative sign indicates that it is real and inverted.
In simple words: Plug the distances into the mirror formula to calculate the focal length, then use the magnification ratio to find the height of the inverted image.

Exam Tip: Always apply the correct sign conventions at the start of your calculation; real images in mirrors always have a negative \( v \) value.

Question 6. An object 45 cm from the lens gives a virtual image at a distance 15 cm in front of the lens. What is the focal length of the lens? Which type of lens is it?
Answer: Given data:
- Object distance \( u = -45 \) cm
- Image distance \( v = -15 \) cm (negative because the virtual image is formed on the same side as the object, in front of the lens)
Using the Lens Formula:
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
\( \implies \frac{1}{f} = \frac{1}{-15} - \frac{1}{-45} \)
\( \implies \frac{1}{f} = -\frac{1}{15} + \frac{1}{45} \)
\( \implies \frac{1}{f} = \frac{-3 + 1}{45} = -\frac{2}{45} \)
\( \implies f = -22.5 \text{ cm} \)
Since the focal length is negative, it is a diverging (concave) lens.
In simple words: Use the lens formula with the given distances to find the focal length. A negative focal length shows that it is a concave lens.

Exam Tip: A virtual image formed by a lens is always on the same side as the object, so its image distance \( v \) must be negative.

Question 7. An 8 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 25 cm. The distance of the object from the lens is 30 cm. Find the (a) position, (b) nature and (c) size of the image formed.
Answer: Given data:
- Height of object \( h_o = 8 \) cm
- Focal length of convex lens \( f = +25 \) cm
- Object distance \( u = -30 \) cm
Using the Lens Formula:
\( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
\( \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \)
\( \implies \frac{1}{v} = \frac{1}{25} + \frac{1}{-30} \)
\( \implies \frac{1}{v} = \frac{6 - 5}{150} = \frac{1}{150} \)
\( \implies v = +150 \text{ cm} \)

(a) Position of Image: The image is formed at a distance of 150 cm on the other side of the lens.
(b) Nature of Image: Since \( v \) is positive, the image is real and inverted.
(c) Size of Image: Using the magnification formula:
\( m = \frac{v}{u} = \frac{h_i}{h_o} \)
\( \implies \frac{150}{-30} = \frac{h_i}{8} \)
\( \implies -5 = \frac{h_i}{8} \implies h_i = -40 \text{ cm} \)
The size of the image is 40 cm (inverted).
In simple words: Calculate the image distance using the lens equation. Use the magnification ratio to find the final height of the real, upside-down image.

Exam Tip: The positive value of \( v \) confirms that a real image is formed on the opposite side of the lens from the object.

Question 8. A ray of light enters from air to kerosene. Calculate the speed of light in kerosene, if refractive index of kerosene with respect to air is 1.41.
Answer: Given data:
- Refractive index of kerosene \( n = 1.41 = \sqrt{2} \)
- Speed of light in air/vacuum \( c = 3 \times 10^8 \) m/s
Using the Refractive Index Formula:
\( n = \frac{c}{v} \)
\( \implies v = \frac{c}{n} = \frac{3 \times 10^8}{1.41} \approx 2.12 \times 10^8 \text{ m/s} \)
The speed of light in kerosene is \( 2.12 \times 10^8 \) m/s.
In simple words: Divide the speed of light in air by the kerosene's index value to calculate the light speed inside kerosene.

Exam Tip: Always include the proper unit of speed (m/s) in your final numerical answer to avoid losing marks.

Question 9. We wish to obtain a real, inverted image of the same size as that of object by a thin convex lens of focal length 20 cm. Where the object should be placed? Draw the ray diagram to show the image formation.
Answer: To obtain a real, inverted image of the same size as the object using a convex lens, the object must be placed at a distance of \( 2F_1 \) (twice the focal length) in front of the lens.
Given, focal length \( f = 20 \) cm:
\( \text{Object distance } u = -2f = -2(20) = -40 \text{ cm} \)
The object should be placed 40 cm in front of the lens.
- Ray Diagram Description:
Place the object \( AB \) at \( 2F_1 \). Draw a ray parallel to the principal axis that passes through the focus \( F_2 \) on the other side after refraction. Draw a second ray through the optical center \( O \) that goes straight. The two rays intersect exactly at \( 2F_2 \) on the other side, forming a real, inverted image \( A'B' \) of the same size as the object.
In simple words: To get an image of the same size, place the object at twice the focal length distance, which is 40 cm. The light rays will cross at the same distance on the other side.

Exam Tip: When drawing ray diagrams, use a sharp pencil and a ruler to ensure your rays are perfectly straight and intersect at the exact \( 2F \) point.

Question 10. A converging mirror forms a real image of height 4 cm of an object of height 1 cm placed 20 cm away from the mirror. Calculate the image distance. What is the focal length of the mirror?
Answer: Given data:
- Height of object \( h_o = 1 \) cm
- Height of real image \( h_i = -4 \) cm (negative because it is real and inverted)
- Object distance \( u = -20 \) cm
Using the Magnification Formula for mirrors:
\( m = -\frac{v}{u} = \frac{h_i}{h_o} \)
\( \implies -\frac{v}{-20} = \frac{-4}{1} \)
\( \implies \frac{v}{20} = -4 \implies v = -80 \text{ cm} \)
The image distance is 80 cm in front of the mirror.
Now, using the Mirror Formula:
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
\( \implies \frac{1}{f} = \frac{1}{-80} + \frac{1}{-20} = \frac{-1 - 4}{80} = -\frac{5}{80} \)
\( \implies f = -16 \text{ cm} \)
The focal length of the mirror is \( -16 \) cm.
In simple words: Use the magnification ratio to find the image distance, then use the mirror formula to calculate the focal length.

Exam Tip: A "converging mirror" refers to a concave mirror, which means your final focal length must be negative.

Question 11. Speed of light in three transparent media is as given:- a) 2 × 10^8 m/sec b) 2.25 × 10^8 m/sec c) 1.20 × 10^8 m/sec. Which of the medium has the highest refractive index?
Answer: The refractive index \( n \) of a medium is inversely proportional to the speed of light \( v \) in that medium (\( n = \frac{c}{v} \)).
Therefore, the medium with the lowest speed of light will have the highest refractive index.
Comparing the given speeds:
- Medium a: \( 2 \times 10^8 \) m/s
- Medium b: \( 2.25 \times 10^8 \) m/s
- Medium c: \( 1.20 \times 10^8 \) m/s
Since Medium c has the lowest speed of light (\( 1.20 \times 10^8 \) m/s), it has the highest refractive index.
In simple words: Light travels slowest in the material with the highest refractive index. Since speed is lowest in C, C has the highest index.

Exam Tip: Always write down the inverse relationship formula (\( n \propto 1/v \)) to justify your theoretical choice.

Question 12. Complete the path of the ray of light after reflecting at the mirror in the given diagram.
Answer: In the given diagram of a concave mirror, the incident ray is parallel to the principal axis.
According to the rules of reflection for spherical mirrors, any light ray that travels parallel to the principal axis of a concave mirror will pass through its principal focus (\( F \)) after reflecting from the mirror.
To complete the diagram:
Draw a straight line starting from the point of impact on the mirror surface, passing directly through the focus point \( F \), and add an arrow pointing away from the mirror to indicate the reflected ray's direction.
In simple words: Draw the reflected ray passing straight through the focus point F of the concave mirror.

Exam Tip: Never forget to draw arrows on your light rays to indicate their direction of travel, as diagrams without arrows lose marks.

Question 13. A ray strikes the mirror at 20°. What is the angle of reflection?
Answer: There are two possible interpretations depending on how the angle is measured:
- Case 1: If the ray makes an angle of 20° with the normal (i.e., angle of incidence \( i = 20^\circ \)):
According to the law of reflection, the angle of reflection \( r \) is equal to the angle of incidence:
\( \angle r = \angle i = 20^\circ \)
- Case 2: If the ray makes a glancing angle of 20° with the mirror surface:
The angle of incidence is:
\( \angle i = 90^\circ - 20^\circ = 70^\circ \)
Using the law of reflection:
\( \angle r = \angle i = 70^\circ \)
In simple words: If the angle is measured from the perpendicular normal line, the reflection angle is 20 degrees. If it is from the mirror surface, the reflection angle is 70 degrees.

Exam Tip: State both cases in your answer if the question does not specify whether the angle is with the normal or the mirror surface.

Question 14. Radius of curvature of a mirror is 24 cm. What type of mirror is it and what type of image will be formed by it?
Answer: The type of mirror cannot be determined solely by the magnitude of its radius of curvature.
- If the mirror is convex (positive sign):
It will always form a virtual, erect, and diminished image regardless of the object's position.
- If the mirror is concave (negative sign):
The image type depends on the object's position relative to its focal length (\( f = 12 \) cm):
1. If the object is placed beyond 12 cm, the image will be real and inverted.
2. If the object is placed within 12 cm, the image will be virtual, erect, and magnified.
In simple words: The number alone does not tell us the mirror type. If it is a convex mirror, it always forms small, upright images. If it is concave, the image can be real or virtual depending on where the object is placed.

Exam Tip: Explicitly analyze both convex and concave mirror scenarios to demonstrate a thorough understanding of the topic.

Question 15. When we hold a concave mirror towards the sun and focus it to a sheet of paper, the paper catches fire. Give reason.
Answer: The sun is at an infinite distance, so the light rays coming from it are parallel to the principal axis. When these parallel rays strike a concave mirror, the mirror converges them all onto its principal focus. This concentrates all the solar energy and heat onto a single tiny spot on the paper. The intense heat quickly raises the temperature of the paper to its ignition point, causing it to catch fire.
In simple words: The concave mirror converges all the sun's parallel rays onto a single tiny spot, creating intense heat that burns the paper.

Exam Tip: Use the term "converging action" of the concave mirror to explain the concentration of light energy. Balanced diagrams of parallel rays meeting at the focus are also helpful.

Question 16. Draw a ray diagram to show the complete path of the light ray that travels through a rectangular glass slab.
Answer: To draw this ray diagram, follow these guidelines:
1. Draw a rectangular box representing the glass slab.
2. Draw an oblique incident ray striking the top surface. Draw a normal (perpendicular) line at the point of incidence.
3. Show the ray bending towards the normal as it enters the optically denser glass medium (angle of refraction \( r < \) angle of incidence \( i \)).
4. At the bottom surface, draw another normal line. Show the ray bending away from the normal as it exits back into the rarer air medium.
5. The emerging ray should be drawn perfectly parallel to the original incident ray path (drawn as a dashed projection line), showing a clear "lateral displacement."
In simple words: Draw a ray hitting the glass, bending inside towards the normal, and then bending away as it exits, running parallel to its original path.

Exam Tip: Clearly label the "lateral displacement" distance between the original path projection and the emergent ray in your diagram.

Question 17. The absolute refractive index of any medium is more than one. Why?
Answer: The absolute refractive index \( n \) of a medium is defined as the ratio of the speed of light in a vacuum \( (c) \) to the speed of light in that specific medium \( (v) \):
\( n = \frac{c}{v} \)
According to the laws of physics, the speed of light is at its absolute maximum in a vacuum \( (c \approx 3 \times 10^8\text{ m/s}) \). In any other material medium, light travels slower because of interactions with the matter. Since the value of \( v \) is always less than \( c \), the ratio \( \frac{c}{v} \) must always be greater than one.
In simple words: Light travels fastest in a vacuum. Since it slows down in any other material, dividing the speed in a vacuum by the slower speed always gives a number bigger than one.

Exam Tip: Mention that the refractive index is a ratio of identical quantities, which means it has no units.

Question 18. The following table gives the values of refractive indices of a few media: Medium: Water, Crown Glass, Rock Salt, Ruby. Refractive Index: 1.33, 1.52, 1.54, 1.71. Use this table to give an example of (a) a media pair so that light speeds up when it goes from one of these media to another, (b) a media pair so that light slows down when it goes from one medium to another.
Answer:
(a) Light speeds up when it travels from an optically denser medium (having a higher refractive index) to an optically rarer medium (having a lower refractive index).
- *Example pair:* From Ruby (1.71) to Water (1.33).
(b) Light slows down when it moves from an optically rarer medium (having a lower refractive index) to an optically denser medium (having a higher refractive index).
- *Example pair:* From Water (1.33) to Crown Glass (1.52).
In simple words: Light goes faster when moving from a high-index material to a low-index one, like Ruby to Water. It goes slower when moving from a low-index material to a high-index one, like Water to Crown Glass.

Exam Tip: Remember that optical density is directly related to the refractive index - a higher index means higher optical density and slower speed of light.

Question 19. A 10 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 30 cm. The distance of the object from the lens is 20 cm. Find the (i) position, (ii) nature and (iii) size of the image formed.
Answer: Given parameters:
Height of the object, \( h_o = 10\text{ cm} \)
Focal length of the convex lens, \( f = +30\text{ cm} \)
Object distance, \( u = -20\text{ cm} \)
Using the lens formula:
\( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
Substitute the values:
\( \frac{1}{v} - \frac{1}{-20} = \frac{1}{30} \)
\( \implies \frac{1}{v} + \frac{1}{20} = \frac{1}{30} \)
\( \implies \frac{1}{v} = \frac{1}{30} - \frac{1}{20} \)
\( \implies \frac{1}{v} = \frac{2 - 3}{60} \)
\( \implies \frac{1}{v} = -\frac{1}{60} \)
\( \implies v = -60\text{ cm} \)
Now, let's determine each required parameter:
(i) **Position:** The image is formed at a distance of 60 cm in front of the lens, on the same side as the object.
(ii) **Nature:** Since the image distance \( v \) is negative, the image is virtual and erect.
(iii) **Size:** Using the magnification formula for lenses:
\( m = \frac{v}{u} = \frac{h_i}{h_o} \)
\( \implies \frac{-60}{-20} = \frac{h_i}{10} \)
\( \implies 3 = \frac{h_i}{10} \)
\( \implies h_i = 30\text{ cm} \)
The height of the image is 30 cm.
In simple words: Solve using the lens formula to find the image position at 60 cm on the object side. Because it is virtual, it is upright and three times larger, reaching 30 cm in height.

Exam Tip: A negative image distance in lenses always signifies a virtual and erect image.

Question 20. A convex mirror used as a rear view mirror in a truck had a radius of curvature of 2.5 m. If a car is at a distance of 5 m from the mirror, calculate the distance at which the image is formed and what the size of the image is, if the height of the car is 2 m.
Answer: Given parameters:
Radius of curvature, \( R = +2.5\text{ m} \)
Focal length of the convex mirror, \( f = \frac{R}{2} = \frac{2.5}{2} = +1.25\text{ m} \)
Object distance, \( u = -5\text{ m} \)
Object height, \( h_o = 2\text{ m} \)
Using the mirror formula:
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
Substitute the values:
\( \frac{1}{1.25} = \frac{1}{v} + \frac{1}{-5} \)
\( \implies \frac{4}{5} = \frac{1}{v} - \frac{1}{5} \)
\( \implies \frac{1}{v} = \frac{4}{5} + \frac{1}{5} \)
\( \implies \frac{1}{v} = \frac{5}{5} = 1 \)
\( \implies v = +1\text{ m} \)
The image is formed 1 m behind the mirror.
Now, using the magnification formula:
\( m = -\frac{v}{u} = \frac{h_i}{h_o} \)
\( \implies -\frac{1}{-5} = \frac{h_i}{2} \)
\( \implies \frac{1}{5} = \frac{h_i}{2} \)
\( \implies h_i = \frac{2}{5} = 0.4\text{ m} \)
The size (height) of the image is 0.4 m.
In simple words: Using the mirror equations, we find the image forms 1 meter behind the mirror. The height of the car's image is scaled down to 0.4 meters.

Exam Tip: Rearview mirrors are always convex because they provide a wider field of view and always form upright, smaller images of the vehicles behind.

Assignment - 5

Chapter - Human Eye and the Colourful World

Question 1. Why are the danger signals of red colour?
Answer: Red light has the longest wavelength in the visible light spectrum. According to Rayleigh's law of scattering, the intensity of scattered light is inversely proportional to the fourth power of its wavelength. Because of its long wavelength, red light suffers the least amount of scattering by air molecules, dust, and smoke particles. This allows the red light to travel long distances without losing much intensity, making it clearly visible even in fog or rain.
In simple words: Red light has a long wavelength and does not get scattered easily by fog or dust. This keeps it bright and visible from far away.

Exam Tip: Use the term "Rayleigh's scattering" to explain the scientific reason for least scattering of red light.

Question 2. What is the far point and near point of the human eye with normal vision?
Answer: For a person with healthy, normal vision:
- **Near Point:** This is the closest distance at which an object can be seen clearly without straining the eye. For a normal eye, the near point is \( 25\text{ cm} \).
- **Far Point:** This is the maximum distance up to which the eye can see objects with clear focus. For a normal eye, the far point is at infinity \( (\infty) \).
In simple words: A person with normal vision can see things clearly from as close as 25 cm up to infinitely far away.

Exam Tip: The near point is also termed the "least distance of distinct vision" (LDDV), which is commonly tested in short questions.

Question 3. State two main causes of near sightedness.
Answer: Near-sightedness (also known as myopia) is caused by:
1. **Excessive curvature of the eye lens:** The cornea or eye lens becomes too curved, which makes the focal length of the eye too short.
2. **Elongation of the eyeball:** The distance between the eye lens and the retina becomes too long, causing light rays to focus in front of the retina instead of directly on it.
In simple words: Myopia happens when the lens is too curved or the eyeball is too long, making light focus before it reaches the back of the eye.

Exam Tip: Always list both causes clearly: excessive lens curvature and elongation of the eyeball.

Question 4. Why does it take sometime to see objects in a cinema hall, when we just enter the hall?
Answer: In a brightly lit area, the pupil of the eye contracts to limit the amount of light entering, protecting the retina. When we step into a dimly lit cinema hall, very little light is available. It takes a few moments for the iris to expand the pupil to let more light in. During this adjustment period, we are temporarily unable to see clearly in the dark.
In simple words: In bright light, our pupils shrink. When we enter a dark theater, it takes a few moments for our pupils to open wider to let in enough light to see.

Exam Tip: Mention the roles of both the "iris" (controlling muscle) and the "pupil" (the opening) to explain this adaptation process.

Question 5. State one way how an eye differs from a camera?
Answer: The eye adjusts its focus by changing the shape and thickness of its flexible crystalline lens using ciliary muscles (adjusting the focal length). In contrast, a camera has a rigid glass lens and adjusts its focus by physically moving the lens closer to or further from the film or sensor (changing the image distance).
In simple words: The eye changes the thickness of its lens to focus, whereas a camera moves its hard lens back and forth.

Exam Tip: Use the term "power of accommodation" when describing how the eye lens changes its focal length.

Question 6. What is presbyopia? What are the causes and how is it corrected?
Answer: Presbyopia is an age-related vision defect where a person loses the ability to focus clearly on nearby objects due to the gradual aging of the eye.
- **Causes:** It is caused by the gradual weakening of the ciliary muscles and the loss of flexibility in the crystalline lens over time.
- **Correction:** This condition is typically corrected using bifocal lenses. The upper part of the lens is concave for distant vision, and the lower part is convex for reading near objects.
In simple words: Presbyopia is old-age near-sightedness caused by stiff eye muscles. It is corrected using double-focus glasses that help with both reading and looking far away.

Exam Tip: Clarify that presbyopia is similar to hypermetropia in symptoms, but its cause is muscle aging rather than eyeball shape.

Question 7. Why stars appear slightly higher in position than their actual position when viewed near the horizon. Explain with the help of a figure.
Answer: This phenomenon occurs due to continuous atmospheric refraction. The earth's atmosphere is made of layers of air with different densities and temperatures. As we go closer to the earth's surface, the air becomes denser and its refractive index increases.
When starlight enters the atmosphere obliquely, it travels from rarer to denser layers, bending continuously toward the normal. Because of this bending, the light reaches our eyes along a curved path. When our brain projects this path straight back, the star appears to be at a higher position (its apparent position) than where it actually is.

In simple words: The earth's air bends starlight downward as it travels to the ground. Our eyes project this light straight back, which makes the stars look higher than they really are.

Exam Tip: Draw a simple sketch showing the earth, the bending light ray, the actual star position, and the apparent star position higher up.

Question 8. What do you understand by the power of accommodation of the eye? Draw a diagram to show how hypermetropia is corrected?
Answer: The power of accommodation is the ability of the eye lens to adjust its focal length by contracting or relaxing the ciliary muscles, allowing it to focus clearly on both near and distant objects on the retina.
Hypermetropia (far-sightedness) is corrected using a convex lens of appropriate power. The convex lens converges the incoming light rays slightly before they enter the eye, helping the eye lens focus the rays precisely on the retina.

In simple words: The power of accommodation is how the eye adjusts its focus for close and far objects. Far-sightedness is corrected using a magnifying convex lens to bring the focus forward onto the retina.

Exam Tip: In your correction diagram, ensure that the light rays after passing through the correction lens and eye lens meet exactly on the retina.

Question 9. A myopic person has been using spectacles of power -1.00D for clear vision. During old age he also needs to use separate reading glasses of power +2.00D. Explain what may have happened to his eye sight?
Answer: The person originally suffers from myopia (near-sightedness), which is corrected using a concave lens with a power of \( -1.00\text{ D} \) to focus distant objects clearly on the retina.
As the person ages, they also develop presbyopia. The ciliary muscles weaken, and the crystalline lens loses its flexibility, making it difficult to focus on close objects (loss of accommodation). To read comfortably, they need a convex lens with a power of \( +2.00\text{ D} \) to assist in focusing near objects.
Consequently, the person now suffers from both myopia and presbyopia, requiring bifocal lenses for complete vision correction.
In simple words: The person was initially near-sighted and needed minus power glasses. As they aged, their eye muscles grew weak, making it hard to read up close, so they now need plus power reading glasses as well.

Exam Tip: Mention the term "bifocal lens" as the standard solution for people suffering from both near and distant vision issues.

Question 10. What is long / far sightedness? What causes it? How is it corrected?
Answer: Long-sightedness (hypermetropia) is a vision defect where a person can see distant objects clearly but cannot focus on nearby objects comfortably.
- **Causes:** It is caused by:
1. **Shortening of the eyeball:** The eyeball becomes too short along the optical axis, causing the retina to be too close to the lens.
2. **Large focal length of the eye lens:** The eye lens is too thin or unable to curve sufficiently, making its focal power too weak.
- **Correction:** It is corrected by using spectacles with a convex (converging) lens of suitable power, which pre-converges the rays so they focus directly on the retina.
In simple words: Far-sightedness means you can see far away but not up close. It happens when the eyeball is too short or the lens is too weak, and is corrected using a convex lens.

Exam Tip: Write down both causes clearly, as each cause holds separate weight in mark distribution.

Question 11. What is meant by ‘total internal reflection’? State an application of total internal reflection with the help of a ray diagram.
Answer: Total Internal Reflection (TIR) is an optical phenomenon that occurs when a ray of light traveling in an optically denser medium strikes the boundary of an optically rarer medium at an angle of incidence greater than the critical angle for that pair of media. Instead of refracting, the entire light ray is reflected back into the denser medium.
- **Conditions for TIR:**
1. Light must travel from a denser medium to a rarer medium.
2. The angle of incidence must be greater than the critical angle \( (i > \theta_c) \).
- **Application:** Optical fibers are a key application of TIR. Light signals introduced into the glass core of the fiber at a high angle are continuously reflected internally along its length with minimal loss of signal.

In simple words: Total internal reflection is when light inside a dense material hits the boundary at a flat angle and bounces back inside completely, acting like a perfect mirror. It is used to carry internet signals inside optical fibers.

Exam Tip: Draw a simple three-step diagram: refraction away from the normal, refraction at 90 degrees (critical angle), and total internal reflection.

Question 12. A 14 year old student is not able to clearly see the questions written on the black board placed at a distance of 5m from her. a) Name the defect of vision she is suffering from and the type of lens used to correct this defect. b) Write two main causes of this defect. c) With the help of a labeled diagram show how this defect can be rectified.
Answer:
(a) The student is unable to see distant objects clearly (the blackboard is at 5m, which is beyond the normal near point), but can likely see near objects clearly. Thus, she is suffering from **Myopia (near-sightedness)**. This defect is corrected by using a **concave (diverging) lens** of appropriate focal length.
(b) The two primary causes of myopia are:
1. **Excessive curvature of the eye lens:** The lens is too thick, making its focal length too short.
2. **Elongation of the eyeball:** The distance from the lens to the retina is longer than normal, causing the image of distant objects to form in front of the retina.
(c) **Correction Diagram:**
A concave lens is placed in front of the eye. Parallel rays coming from a distant object are diverged slightly by the concave lens. When these diverged rays pass through the eye's natural lens, they are focused precisely on the retina, allowing the student to see the blackboard clearly.

In simple words: The student has near-sightedness (myopia) because her eyes focus distant light in front of the retina. She needs a concave lens to diverge the light slightly so it focuses correctly on the back of the eye.

Exam Tip: Clearly label the "concave lens," "parallel rays," and "retina" in your correction diagram to ensure you score full marks on 5-point questions.

BIOLOGY

Question 1. Complete the assignments and the sample papers given to you in your bio practice note book
Answer: To complete this assignment, systematically review and solve all the practice questions, worksheets, and board sample papers provided by your teacher. Write all responses neatly in your Biology practice notebook. Ensure you include labeled diagrams for topics like life processes (nutrition, respiration, transportation, and excretion) to make your answers complete.
In simple words: Complete all the biology worksheets and practice exam papers neatly in your notebook.

Exam Tip: Focus on practicing neat pencil diagrams with clear labels, as biology answers accompanied by diagrams always score higher.

Question 2. Complete all the NCERT questions
Answer: Solve all the in-text questions and end-of-chapter exercises from your NCERT Biology textbook for the covered chapters. Write down the complete answers in your homework notebook, explaining physiological processes, differences between systems, and experimental setups in detail.
In simple words: Write down and learn the answers to all the questions in your science textbook for the biology chapters.

Exam Tip: Board exam questions are often taken directly from NCERT exercises, so mastering these questions is crucial for scoring well.