Maharashtra Board Class 7 Maths Miscellaneous Problems Set 1 Solutions

Question 1. Solve the following:
(i) (-16) × (-5)
(ii) (72) ÷ (-12)
(iii) (-24) × (2)
(iv) 125 ÷ 5
(v) (-104) ÷ (-13)
(vi) 25 × (-4)
Answer:
Solution:
(i) (-16) × (-5) = 80
(ii) \(72 \div (-12) = \frac{72}{-12}\)
\( = \frac{1}{(-1)} \times \frac{72}{12}\)
\( = (-1) \times 12\)
\( = -6\)
(iii) (-24) × 2 = -48
(iv) \(125 \div 5 = \frac{125}{5}\)
\( = 25\)
(v) \( (-104) \div (-13) = \frac{-104}{-13}\)
\( = \frac{104}{13}\)
\( = 8\)
(vi) 25 × (-4) = -100
In simple words: This question involves basic arithmetic operations like multiplication and division with positive and negative integers. Remember that multiplying or dividing two negative numbers results in a positive number, while one positive and one negative results in a negative number.

🎯 Exam Tip: Pay close attention to the signs of the numbers. A common mistake is to get the sign incorrect in multiplication and division, which can lead to incorrect answers. Practice integer operations thoroughly.

Question 2. Find the prime factors of the following numbers and find their LCM and HCF:
(i) 75,135
(ii) 114,76
(iii) 153,187
(iv) 32,24,48
Answer:
Solution:
(i) 75 = 3 x 25
= 3 × 5 × 5
135 = 3 x 45
= 3 x 3 x 15
= 3 × 3 × 3 × 5
\( \therefore \) HCF of 75 and 135 = 3 × 5
= 15
LCM of 75 and 135 = 3 × 5 × 5 × 3 × 3
= 675
(ii) 114 = 2 × 57
= 2 x 3 x 19
76 = 2 × 38
= 2 x 2 x 19
\( \therefore \) HCF of 114 and 76 = 2 × 19
= 38
LCM of 114 and 76 = 2 × 19 × 3 × 2
= 228
(iii) 153 = 3 × 51
= 3 x 3 x 17
187 = 11 x 17
\( \therefore \) HCF of 153 and 187 = 17
LCM of 153 and 187 = 17 × 3 × 3 × 11
= 1683
(iv) 32 = 2 × 16
= 2 × 2 × 8
= 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2
24 = 2 x 12
= 2 × 2 × 6
= 2 × 2 × 2 × 3
48 = 2 × 24
= 2 x 2 x 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
\( \therefore \) HCF of 32, 24 and 48 = 2 × 2 × 2
= 8
LCM of 32,24 and 48 = 2 × 2 × 2 × 2 × 2 × 3
= 96
In simple words: This question requires finding the prime factorization of numbers and then using those factors to calculate their Highest Common Factor (HCF) and Least Common Multiple (LCM). HCF is found by multiplying common prime factors, while LCM involves multiplying all prime factors, taking the highest power for each.

🎯 Exam Tip: Always list all prime factors for each number clearly. For HCF, identify only the common factors with their lowest power. For LCM, identify all distinct factors with their highest power. Double-check your calculations, especially when dealing with multiple numbers.

Question 3. Simplify:
(i) \(\frac{322}{391}\)
(ii) \(\frac{247}{209}\)
(iii) \(\frac{117}{156}\)
Answer:
Solution:
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक लंबी विभाजन प्रक्रिया को दर्शाता है जिसका उपयोग दो संख्याओं, 391 और 322 का महत्तम समापवर्तक (HCF) ज्ञात करने के लिए किया जाता है। विभाजन के चरणों को स्पष्ट रूप से दिखाया गया है, जहाँ प्रत्येक शेषफल अगले चरण में भाजक बन जाता है, जब तक शेषफल शून्य न हो जाए।
\( \therefore \) HCF of 322 and 391 = 23
\( \therefore \frac{322}{391} = \frac{322 \div 23}{391 \div 23} = \frac{14}{17}\)
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक लंबी विभाजन प्रक्रिया को दर्शाता है जिसका उपयोग दो संख्याओं, 247 और 209 का महत्तम समापवर्तक (HCF) ज्ञात करने के लिए किया जाता है। चरणों को स्पष्ट रूप से दिखाया गया है, जहाँ शेषफल शून्य होने तक भाजक और शेषफल का उपयोग किया जाता है।
\( \therefore \) HCF of 247 and 209 = 19
\( \therefore \frac{247}{209} = \frac{247 \div 19}{209 \div 19} = \frac{13}{11}\)
(iii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक लंबी विभाजन प्रक्रिया को दर्शाता है जिसका उपयोग दो संख्याओं, 156 और 117 का महत्तम समापवर्तक (HCF) ज्ञात करने के लिए किया जाता है। विभाजन के चरणों को स्पष्ट रूप से दिखाया गया है, जहाँ प्रत्येक शेषफल अगले चरण में भाजक बन जाता है, जब तक शेषफल शून्य न हो जाए।
\( \therefore \) HCF of 117 and 156 = 39
\( \therefore \frac{117}{156} = \frac{117 \div 39}{156 \div 39} = \frac{3}{4}\)
In simple words: This question involves simplifying fractions by dividing both the numerator and the denominator by their Highest Common Factor (HCF). The HCF is found using the long division method, which systematically identifies the largest number that divides both parts of the fraction.

🎯 Exam Tip: To simplify fractions efficiently, always find the HCF of the numerator and denominator first. The long division method is reliable for finding HCF, especially for larger numbers. Make sure your division steps are accurate.

Question 4.
(i) 784
(ii) 225
(iii) 1296
(iv) 2025
(v) 256
Answer:
Solution:
(i) 784

\( \therefore 784 = 2 \times 2 \times 2 \times 2 \times 7 \times 7\)
\( \therefore \sqrt{784} = 2 \times 2 \times 7\)
\( = 28\)
(ii) 225

\( \therefore 225 = 3 \times 3 \times 5 \times 5\)
\( \therefore \sqrt{225} = 3 \times 5\)
\( = 15\)
(iii) 1296

\( \therefore 1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3\)
\( \therefore \sqrt{1296} = 2 \times 2 \times 3 \times 3\)
\( = 36\)
(iv) 2025

\( \therefore 2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5\)
\( \therefore \sqrt{2025} = 3 \times 3 \times 5\)
\( = 45\)
(v) 256

\( \therefore 256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\)
\( \therefore \sqrt{256} = 2 \times 2 \times 2 \times 2\)
\( = 16\)
In simple words: This question requires finding the square root of given numbers using the prime factorization method. By breaking down each number into its prime factors, we can identify pairs of identical factors; one factor from each pair is then multiplied to find the square root.

🎯 Exam Tip: When finding square roots by prime factorization, ensure you completely factorize the number into its smallest primes. Group the factors into pairs; for each pair, take one number. The product of these single numbers is the square root. Neatly organize your prime factorization steps.

Question 5. There are four polling booths for a certain election. The numbers of men and women who cast their vote at each booth is given in the table below. Draw a joint bar graph for this data.
Answer:
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक संयुक्त बार ग्राफ है जो चार अलग-अलग मतदान केंद्रों - नवोदय विद्यालय, विद्यानिकेतन स्कूल, सिटी हाई स्कूल और एकलव्य स्कूल में पुरुषों और महिलाओं द्वारा डाले गए वोटों की संख्या को दर्शाता है। Y-अक्ष पर 'व्यक्तियों की संख्या' दिखाई गई है, जहाँ 1 सेमी 100 व्यक्तियों के बराबर है, और X-अक्ष पर 'मतदान केंद्र' दिखाए गए हैं। प्रत्येक मतदान केंद्र के लिए, महिलाओं (हल्के भूरे रंग का बार) और पुरुषों (गहरे भूरे रंग का बार) के वोटों की संख्या को बगल-बगल में दर्शाया गया है, जिससे विभिन्न मतदान केंद्रों पर मतदान करने वालों की तुलना करना आसान हो जाता है।
In simple words: This question asks to represent the provided data about voting by men and women in different polling booths using a joint bar graph. A joint bar graph uses adjacent bars for each category (men and women) at each location (polling booth) to allow for easy comparison.

🎯 Exam Tip: When drawing a joint bar graph, clearly label both axes (X-axis: Polling Booths, Y-axis: Number of Persons) and specify the scale used on the Y-axis. Use distinct colors or patterns for different categories (men and women) and include a legend. Ensure the bars are of equal width and the spacing between groups of bars is consistent.

Question 6. Simplify the expressions:
(i) 45 ÷ 5 + 120 × 4 - 12
(ii) (38 - 8) × 2 ÷ 5 + 13
(iii) \(\frac{5}{3} + \frac{4}{7} \div \frac{32}{21}\)
(iv) 3 × {4 [85 + 5 - (15 - 3)] + 2}
Answer:
Solution:
(i) 45 ÷ 5 + 120 × 4 - 12
= 9 + 80 - 12
= 89 - 12
= 77
(ii) (38 - 8) × 2 ÷ 5 + 13
= 30 × 2 ÷ 5 + 13
= 60 ÷ 5 + 13
= 12 + 13
= 25
(iii) \(\frac{5}{3} + \frac{4}{7} \div \frac{32}{21}\)
\( = \frac{5}{3} + \frac{4}{7} \times \frac{21}{32}\)
\( = \frac{5}{3} + \frac{1}{1} \times \frac{3}{8}\)
\( = \frac{5}{3} + \frac{3}{8}\)
\( = \frac{40}{24} + \frac{9}{24}\)
\( = \frac{49}{24}\)
(iv) 3 × {4 [85 + 5 - (15 - 3)] + 2}
= 3 × {4[90 - 5] + 2}
= 3 × {4 × 85 + 2}
= 3 × (340 + 2)
= 3 x 342
= 1026
In simple words: This question requires simplifying mathematical expressions by following the order of operations (BODMAS/PEMDAS). This means performing operations inside brackets first, then multiplication and division from left to right, and finally addition and subtraction from left to right.

🎯 Exam Tip: Always strictly adhere to the BODMAS/PEMDAS rule. Misinterpreting the order of operations is a common error. Show each step of your simplification clearly, especially when dealing with multiple operations and nested brackets, to avoid calculation mistakes.

Question 7. Solve:
(i) \(\frac{5}{12} + \frac{7}{16}\)
(ii) \(3\frac{2}{5} - 2\frac{1}{4}\)
(iii) \(\frac{12}{5} \times (\frac{-10}{3})\)
(iv) \(4\frac{3}{8} \div \frac{25}{18}\)
Answer:
Solution:
(i) \(\frac{5}{12} + \frac{7}{16}\)
\( = \frac{5 \times 4}{12 \times 4} + \frac{7 \times 3}{16 \times 3}\)
\( = \frac{20}{48} + \frac{21}{48}\)
\( = \frac{20+21}{48}\)
\( = \frac{41}{48}\)
(ii) \(3\frac{2}{5} - 2\frac{1}{4}\)
\( = \frac{17}{5} - \frac{9}{4}\)
\( = \frac{17 \times 4}{5 \times 4} - \frac{9 \times 5}{4 \times 5}\)
\( = \frac{68}{20} - \frac{45}{20}\)
\( = \frac{68-45}{20}\)
\( = \frac{23}{20}\)
(iii) \(\frac{12}{5} \times (\frac{-10}{3})\)
\( = \frac{4}{1} \times (\frac{-2}{1})\)
\( = 4 \times (-2)\)
\( = -8\)
(iv) \(4\frac{3}{8} \div \frac{25}{18}\)
\( = \frac{35}{8} \div \frac{25}{18}\)
\( = \frac{35}{8} \times \frac{18}{25}\)
\( = \frac{7}{4} \times \frac{9}{5}\)
\( = \frac{63}{20}\)
In simple words: This question involves performing arithmetic operations (addition, subtraction, multiplication, and division) with fractions, including mixed numbers and negative fractions. Key steps include finding common denominators for addition/subtraction, converting mixed numbers to improper fractions, and inverting the second fraction for division.

🎯 Exam Tip: Remember to convert mixed numbers to improper fractions before performing any operations. For addition/subtraction, always find the Least Common Multiple (LCM) of the denominators. For division, multiply by the reciprocal of the second fraction. Simplify fractions at each possible step to manage calculations more easily.

Question 8. Construct \( \Delta ABC \) such that m\( \angle A \) = 55°, m\( \angle B \) = 60° and l(AB) = 5.9 cm.
Answer:
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिकोण ABC का निर्माण दर्शाता है जहाँ आधार AB की लंबाई 5.9 सेमी है। कोण A को 55 डिग्री पर और कोण B को 60 डिग्री पर मापा गया है। बिंदु C उन दो कोणों की भुजाओं के प्रतिच्छेदन पर स्थित है।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह त्रिभुज ABC का एक रफ स्केच है। इसमें भुजा AB की लंबाई 5.9 सेमी है, और कोण A 55° और कोण B 60° पर दिखाए गए हैं। यह स्केच छात्रों को वास्तविक निर्माण से पहले त्रिभुज के अनुमानित आकार और कोणों की कल्पना करने में मदद करता है।
In simple words: To construct triangle ABC, first draw the base segment AB with the given length. Then, use a protractor to draw angles at points A and B according to their given measures. The point where the rays of these angles intersect will be vertex C, completing the triangle.

🎯 Exam Tip: When constructing triangles with Angle-Side-Angle (ASA) criteria, accurately measure the base length first. Use a protractor carefully to draw the given angles at the endpoints of the base. Precision in angle and length measurements is crucial for an accurate construction.

Question 9. Construct \( \Delta XYZ \) such that, l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.
Answer:
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिकोण XYZ का निर्माण दिखाता है जिसकी सभी तीन भुजाओं की लंबाई दी गई है: XY 3.7 सेमी, YZ 7.7 सेमी और XZ 6.3 सेमी। निर्माण में एक आधार खींचना और फिर कम्पास का उपयोग करके अन्य दो भुजाओं को उनके प्रतिच्छेदन बिंदु पर चाप लगाकर बनाना शामिल है।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह त्रिभुज XYZ का एक रफ स्केच है। इसमें भुजाओं XY, YZ, और XZ की लंबाई क्रमशः 3.7 सेमी, 7.7 सेमी और 6.3 सेमी के रूप में चिह्नित की गई है। यह स्केच छात्रों को वास्तविक निर्माण से पहले त्रिभुज के अनुमानित आकार की कल्पना करने में मदद करता है।
In simple words: To construct triangle XYZ when all three side lengths are given, first draw one side as the base. Then, using a compass, draw arcs from the endpoints of the base with radii equal to the lengths of the other two sides. The intersection point of these arcs will be the third vertex.

🎯 Exam Tip: For SSS (Side-Side-Side) triangle construction, ensure your compass measurements for the arcs are precise. Drawing one side as the base, then using the compass from its endpoints to locate the third vertex, is the standard method. Accuracy in compass settings is key to a correct construction.

Question 10. Construct \( \Delta PQR \) such that, m\( \angle P \) = 80°, m\( \angle Q \) = 70°, l(QR) = 5.7 cm.
Answer:
Ans:
In \( \Delta PQR \),
m\( \angle P \) + m\( \angle Q \) + m\( \angle R \) = 180° .... (Sum of the measures of the angles of a triangle is 180°)
\( \therefore \) 80 + 70 + m\( \angle R \) = 180
\( \therefore \) 150 + m\( \angle R \) = 180
\( \therefore \) m\( \angle R \) = 180 - 150
\( \therefore \) m\( \angle R \) = 30°
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR के निर्माण को दिखाता है जहाँ भुजा QR की लंबाई 5.7 सेमी है, कोण Q 70° है और कोण R 30° है। बिंदु P उन दो कोणों की भुजाओं के प्रतिच्छेदन पर स्थित है। कोण P का मान 80° है जो त्रिभुज के आंतरिक कोणों के योग से प्राप्त होता है।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह त्रिभुज PQR का एक रफ स्केच है। इसमें भुजा QR की लंबाई 5.7 सेमी है, कोण Q 70° और कोण P 80° पर चिह्नित हैं। कोण R को 30° के रूप में दर्शाया गया है, जिसे त्रिभुज के कोणों के योग नियम से गणना करके प्राप्त किया गया है। यह स्केच छात्रों को वास्तविक निर्माण से पहले त्रिभुज के अनुमानित आकार और कोणों की कल्पना करने में मदद करता है।
In simple words: First, calculate the third angle (angle R) using the angle sum property of a triangle. Then, draw the side QR with the given length. At points Q and R, draw the calculated angles (angle Q and angle R) using a protractor. The intersection of these two rays will give you vertex P.

🎯 Exam Tip: When given two angles and a non-included side (AAS criterion), first find the third angle using the angle sum property (180 degrees). Then, use the Angle-Side-Angle (ASA) construction method with the given side and its adjacent angles. Ensure accurate protractor usage for angles and ruler for length.

Question 11. Construct \( \Delta EFG \) from the given measures. l(FG) = 5 cm, m\( \angle EFG \) = 90°, l(EG) = 7 cm.
Answer:
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज EFG के निर्माण को दर्शाता है। भुजा FG की लंबाई 5 सेमी है, कोण EFG 90° का है (जो कि एक समकोण है), और कर्ण EG की लंबाई 7 सेमी है। इस निर्माण में एक भुजा और एक कर्ण का उपयोग करके एक समकोण त्रिभुज बनाया गया है।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह त्रिभुज EFG का एक रफ स्केच है। इसमें भुजा FG 5 सेमी और कर्ण EG 7 सेमी के रूप में चिह्नित है। कोण F को 90° पर दर्शाया गया है, जिससे यह एक समकोण त्रिभुज बन जाता है। यह स्केच छात्रों को वास्तविक निर्माण से पहले त्रिभुज के अनुमानित आकार की कल्पना करने में मदद करता है।
In simple words: To construct triangle EFG, draw the base segment FG with length 5 cm. At point F, construct a 90° angle. With G as the center and a radius of 7 cm (length of EG), draw an arc that intersects the ray from F. This intersection point is E, completing the right-angled triangle.

🎯 Exam Tip: For constructing a right-angled triangle with one side and the hypotenuse (RHS criterion), draw the given side first. Construct the right angle at the appropriate vertex. Then, use a compass with the hypotenuse length from the other known vertex to intersect the perpendicular ray, thus locating the third vertex. Accuracy in constructing the 90° angle is vital.

Question 12. In \( \Delta LMN \), l(LM) = 6.2 cm, m\( \angle LMN \) = 60°, l(MN) 4 cm. Construct \( \Delta LMN \).
Answer:
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज LMN के निर्माण को दर्शाता है। भुजा LM की लंबाई 6.2 सेमी है, भुजा MN की लंबाई 4 सेमी है, और इन दोनों भुजाओं के बीच का कोण, कोण LMN, 60° है। निर्माण में दो भुजाओं और उनके बीच के कोण का उपयोग किया जाता है।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह त्रिभुज LMN का एक रफ स्केच है। इसमें भुजा LM की लंबाई 6.2 सेमी, भुजा MN की लंबाई 4 सेमी और कोण M 60° के रूप में चिह्नित है। यह स्केच छात्रों को वास्तविक निर्माण से पहले त्रिभुज के अनुमानित आकार और कोणों की कल्पना करने में मदद करता है।
In simple words: To construct triangle LMN, first draw the segment LM with length 6.2 cm. At point M, use a protractor to draw a 60° angle. Along the ray of this angle, mark point N such that MN is 4 cm. Finally, join L and N to complete the triangle.

🎯 Exam Tip: When constructing a triangle with Side-Angle-Side (SAS) criteria, always draw one of the given sides first. Then, construct the given angle at the common vertex of the two sides. Finally, measure the second given side along the ray of the angle. Precision in measuring the angle and both lengths is key.

Question 13. Find the measures of the complementary angles of the following angles:
(i) 35°
(ii) a°
(iii) 22°
(iv) (40 - x)°
Answer:
Solution:
(i) Let the measure of the complementary angle be x°.
35 + x = 90
\( \therefore \)35 + x - 35 = 90 - 35
....(Subtracting 35 from both sides)
\( \therefore \)x = 55
The complementary angle of 35° is 55°.
(ii) Let the measure of the complementary angle be x°.
a + x = 90
\( \therefore \)a + x - a = 90 - a
....(Subtracting a from both sides)
\( \therefore \)x = (90 - a)
The complementary angle of a° is (90 - a)°.
(iii) Let the measure of the complementary angle be x°.
22 + x = 90
\( \therefore \)22 + x - 22 = 90 - 22
....(Subtracting 22 from both sides)
\( \therefore \)x = 68
The complementary angle of 22° is 68°.
(iv) Let the measure of the complementary angle be a°.
40 - x + a = 90
\( \therefore \)40 - x + a - 40 + x = 90 - 40 + x
....(Subtracting 40 and adding x on both sides)
\( \therefore \)a = (50 + x)
The complementary angle of (40 - x)° is (50 + x)°.
In simple words: Complementary angles are two angles whose sum is exactly 90 degrees. To find the complementary angle of a given angle, you simply subtract the given angle's measure from 90 degrees.

🎯 Exam Tip: Clearly state the definition of complementary angles (sum is 90°) at the start. For algebraic expressions, be careful with signs when rearranging the equation to solve for the unknown angle. Always ensure your final answer is in degrees.

Question 14. Find the measures of the supplements of the following angles:
(i) 111°
(ii) 47°
(iii) 180°
(iv) (90 - x)°
Answer:
Solution:
(i) Let the measure of the supplementary angle be x°.
111 + x = 180
\( \therefore \) 111 + x - 111 = 180 - 111
....(Subtracting 111 from both sides)
\( \therefore \) x = 69
The supplementary angle of 111° is 69°.
(ii) Let the measure of the supplementary angle be x°.
47 + x = 180
\( \therefore \) 47 + x - 47 = 180 - 47
....(Subtracting 47 from both sides)
\( \therefore \)x = 133
The supplementary angle of 47° is 133°.
(iii) Let the measure of the supplementary angle be x°.
180 + x = 180
\( \therefore \)180 + x - 180 = 180 - 180
....(Subtracting 180 from both sides)
\( \therefore \)x = 0
The supplementary angle of 180° is 0°.
(iv) Let the measure of the supplementary angle be a°.
90 - x + a = 180
\( \therefore \) 90 - x + a - 90 + x = 180 - 90 + x
....(Subtracting 90 and adding x on both sides)
\( \therefore \)a = 180 - 90 + x
\( \therefore \)a = (90 + x)
The supplementary angle of (90 - x)° is (90 + x)°.
In simple words: Supplementary angles are two angles whose sum is 180 degrees. To find the supplementary angle of a given angle, you simply subtract the given angle's measure from 180 degrees.

🎯 Exam Tip: Define supplementary angles (sum is 180°) clearly. For algebraic expressions, handle negative signs carefully when transposing terms across the equals sign. Always ensure the units (degrees) are included in your final answer.

Question 15. Construct the following figures:
(i) A pair of adjacent angles
(ii) Two supplementary angles which are not adjacent angles.
(iii) A pair of adjacent complementary angles.
Answer:
Solution:
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो आसन्न कोणों (adjacent angles) को दर्शाता है, कोण ABC और कोण CBD। ये कोण एक उभयनिष्ठ शीर्ष B और एक उभयनिष्ठ भुजा BC साझा करते हैं, और उनकी आंतरिक भुजाएँ अतिव्यापी नहीं होती हैं।
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो ऐसे कोणों को दर्शाता है जो आसन्न नहीं हैं लेकिन पूरक हैं, अर्थात उनका योग 180° है। एक कोण 120° और दूसरा 60° है। ये कोण एक-दूसरे से अलग-अलग दिखाए गए हैं, जो आसन्न न होने की स्थिति को उजागर करता है।
(iii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो आसन्न पूरक कोणों (adjacent complementary angles) को दर्शाता है, कोण VST और कोण VSU, जिनका योग 90° है। ये कोण एक उभयनिष्ठ शीर्ष S और एक उभयनिष्ठ भुजा VS साझा करते हैं। कोण VST 30° है और कोण VSU 60° है, जो एक साथ एक समकोण बनाते हैं।
In simple words: This question requires drawing different pairs of angles based on their properties. Adjacent angles share a common vertex and arm. Supplementary angles add up to 180 degrees, while complementary angles add up to 90 degrees. These pairs can either be adjacent or non-adjacent.

🎯 Exam Tip: When constructing angles, ensure you understand the definitions: adjacent angles share a vertex and a ray; supplementary angles sum to 180°; complementary angles sum to 90°. Use a protractor for accurate angle measurements. Clearly label vertices and rays in your constructions.

Question 16. In \( \Delta PQR \) the measures of \( \angle P \) and \( \angle Q \) are equal and m\( \angle PRQ \) = 70°, Find the measures of the following angles.
1. m\( \angle PRT \)
2. m\( \angle P \)
3. m\( \angle Q \)
Answer:
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR और उसकी एक विस्तारित भुजा RT को दर्शाता है। त्रिभुज के आंतरिक कोण P, Q और R दिखाए गए हैं, जहाँ कोण PRQ 70° है। कोण PRT, जो कोण PRQ का एक रैखिक युग्म है, को बाहरी कोण के रूप में दिखाया गया है।
Here, \( \angle PRQ \) and \( \angle PRT \) are angles in a linear pair.
m\( \angle PRQ \) + m\( \angle PRT \) = 180°
\( \therefore \)70 + m\( \angle PRT \) = 180
\( \therefore \)m\( \angle PRT \) = 180 - 70
\( \therefore \)m\( \angle PRT \) = 110°
Now, \( \angle PRT \) is the exterior angle of \( \Delta PQR \).
\( \therefore \)m\( \angle P \) + m\( \angle Q \) = m\( \angle PRT \)
\( \therefore \)m\( \angle P \) + m\( \angle P \) = m\( \angle PRT \) ....(The measures of \( \angle P \) and \( \angle Q \) is same)
\( \therefore \)2m\( \angle P \) = 110
\( \therefore \)m\( \angle P \) = \(\frac{110}{2}\)
\( \therefore \)m\( \angle P \) = 55°
\( \therefore \)m\( \angle Q \) = 55°
In simple words: This problem uses the properties of linear pairs and exterior angles of a triangle. First, find the exterior angle PRT by subtracting the interior angle PRQ from 180°. Then, use the property that an exterior angle is equal to the sum of the two opposite interior angles (P and Q). Since angles P and Q are equal, you can solve for their individual measures.

🎯 Exam Tip: Clearly state the geometric reasons for each step (e.g., "angles in a linear pair," "exterior angle property"). Remember that angles in a linear pair sum to 180° and an exterior angle of a triangle equals the sum of its two interior opposite angles. Be methodical in your calculations, especially when dealing with algebraic expressions for angles.

Question 17. Simplify
(i) 5\(^4\) x 5\(^3\)
(ii) \((\frac{2}{3})^6 \div (\frac{2}{3})^9\)
(iii) \((\frac{7}{2})^8 \times (\frac{7}{2})^{-6}\)
(iv) \((\frac{4}{5})^2 \div (\frac{5}{4})\)
Answer:
i. \(5^4 \times 5^3 = 5^{4+3} = 5^7\)
ii. \((\frac{2}{3})^6 \div (\frac{2}{3})^9 = (\frac{2}{3})^{6-9} = (\frac{2}{3})^{-3} = (\frac{3}{2})^3\)
iii. \((\frac{7}{2})^8 \times (\frac{7}{2})^{-6} = (\frac{7}{2})^{8+(-6)} = (\frac{7}{2})^2\)
iv. \((\frac{4}{5})^2 \div (\frac{5}{4}) = (\frac{4}{5})^2 \times (\frac{4}{5})^1 = (\frac{4}{5})^{2+1} = (\frac{4}{5})^3\)
In simple words: This question demonstrates the rules of exponents for multiplication and division, showing how to combine powers with the same base and handle negative exponents.

🎯 Exam Tip: Remember to apply the correct exponent rules: \(a^m \times a^n = a^{m+n}\), \(a^m \div a^n = a^{m-n}\), \( (a^m)^n = a^{mn} \), and \(a^{-n} = \frac{1}{a^n} \).

Question 18. Find the value:
(i) 17\(^{16}\) ÷ 17\(^{16}\)
(ii) 10\(^{-3}\)
(iii) (2\(^3\))\(^2\)
(iv) 4\(^6\) x 4\(^{-4}\)
Answer:
i. \(17^{16} \div 17^{16} = 17^0 = 1\)
ii. \(10^{-3} = \frac{1}{10^3} = \frac{1}{1000}\)
iii. \((2^3)^2 = 2^{3 \times 2} = 2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64\)
iv. \(4^6 \times 4^{-4} = 4^{6+(-4)} = 4^2 = 4 \times 4 = 16\)
In simple words: This question tests understanding of basic exponent rules, including zero exponent, negative exponents, power of a power, and multiplication with different exponents.

🎯 Exam Tip: Pay close attention to the base and the operation. A common mistake is miscalculating negative exponents or power of a power. Simplify step-by-step.

Question 19. Solve:
(i) (6a - 5b - 8c) + (15b + 2a - 5c)
(ii) (3x + 2y) (7x - 8y)
(iii) (7m - 5n) - (-4n - 11m)
(iv) (11m - 12n + 3p) - (9m + 7n - 8p)
Answer:
i. \((6a - 5b - 8c) + (15b + 2a - 5c) = (6a + 2a) + (-5b + 15b) + (-8c - 5c) = 8a + 10b - 13c\)
ii. \((3x + 2y) (7x - 8y) = 3x \times (7x - 8y) + 2y \times (7x - 8y)\)
\( = 21x^2 - 24xy + 14xy - 16y^2 = 21x^2 - 10xy - 16y^2\)
iii. \((7m - 5n) - (-4n - 11m) = 7m - 5n + 4n + 11m = (7m + 11m) + (-5n + 4n) = 18m - n\)
iv. \((11m - 12n + 3p) - (9m + 7n - 8p) = 11m - 12n + 3p - 9m - 7n + 8p = (11m - 9m) + (-12n - 7n) + (3p + 8p) = 2m - 19n + 11p\)
In simple words: This question involves simplifying algebraic expressions by combining like terms, applying distributive property for multiplication, and handling subtraction of polynomials.

🎯 Exam Tip: When simplifying expressions, always group like terms together. Be careful with signs, especially when subtracting negative terms or distributing negative numbers.

Question 20. Solve the following equations:
(i) 4(x + 12) = 8
(ii) 3y + 4 = 5y - 6
Answer:
i. \(4(x + 12) = 8\)

\( \implies 4x + 48 = 8\)

\( \implies 4x + 48 - 48 = 8 - 48\)
....(Subtracting 48 from both sides)

\( \implies 4x = -40\)

\( \implies x = \frac{-40}{4}\)

\( \implies x = -10\)
ii. \(3y + 4 = 5y - 6\)

\( \implies 3y + 4 + 6 = 5y - 6 + 6\)
....(Adding 6 on both sides)

\( \implies 3y + 10 = 5y\)

\( \implies 3y + 10 - 3y = 5y - 3y\)
....(Subtracting 3y from both sides)

\( \implies 10 = 2y\)

\( \implies 2y = 10\)

\( \implies y = \frac{10}{2}\)

\( \implies y = 5\)
In simple words: This question requires solving linear equations by isolating the variable using inverse operations, such as distributing, adding, subtracting, multiplying, and dividing terms on both sides.

🎯 Exam Tip: Always perform inverse operations to both sides of the equation to maintain equality. Distribute terms correctly and be meticulous with arithmetic, especially involving negative numbers.