CBSE Class 7 Mathematics A Peek Beyond the Point Practice Paper Assignment Set 02 with Answers

School Assignments for Class 7 Mathematics: Chapter 03 A Peek Beyond the Point

Access comprehensive school assignments for Chapter 03 A Peek Beyond the Point using the CBSE Class 7 Mathematics A Peek Beyond the Point Practice Paper Assignment Set 02 with Answers. Designed to align with the 2026-27 CBSE academic guidelines, these practice sets help Class 7 Mathematics students reinforce core concepts and improve their problem-solving accuracy.

Practice Class 7 Mathematics Assignments: Chapter 03 A Peek Beyond the Point

Access the complete assignment PDF for Class 7 Mathematics below. Regular practice with these targeted academic tasks builds familiarity with standard question patterns and helps secure higher marks in final school evaluations.

This Class 7 Maths Ganita Prakash Chapter 3 A Peek Beyond the Point Practice Paper 08 has 20 questions worth 40 marks, arranged in Sections A to E, and each one is solved step by step below. The questions practise building decimals from ones, tenths and hundredths, finding the largest or closest decimal, changing paise to rupees and mm or cm to metres, comparing decimals with >, < and =, and multi-step word problems on water in beakers, the weight of gold chains and filling a swimming pool.

Section A: Multiple Choice Questions (1 Mark Each)

Question 1: Which of the following shows 7 ones, 3 tenths, and 5 hundredths?
(a) 7.35
(b) 7.53
(c) 7.305
(d) 7.053
Show Answer & Explanation

Answer: (a) 7.35

Explanation:
1. 7 ones = 7, 3 tenths = 0.3 and 5 hundredths = 0.05.
2. 7 + 0.3 + 0.05 = 7.35.
3. In 7.35 the 3 is in the tenths place and the 5 is in the hundredths place.

Teacher's Note:
1. Tenths go in the first place after the point and hundredths in the second.
2. 7.305 would mean 3 tenths, 0 hundredths and 5 thousandths.

Question 2: The length of a pencil is 0.06 m. This is equal to:
(a) 6 mm
(b) 60 mm
(c) 0.6 cm
(d) 60 cm
Show Answer & Explanation

Answer: (b) 60 mm

Explanation:
1. 1 m = 100 cm, so 0.06 m = 0.06 × 100 cm = 6 cm.
2. 1 cm = 10 mm, so 6 cm = 6 × 10 mm = 60 mm.
3. Check: 1 m = 1000 mm, and 0.06 × 1000 = 60.

Teacher's Note:
1. Convert step by step: m to cm, then cm to mm.
2. 6 cm is the same as 60 mm, so watch the units in the options.

Question 3: Which number is the largest?
(a) 5.009
(b) 5.09
(c) 5.9
(d) 5.099
Show Answer & Explanation

Answer: (c) 5.9

Explanation:
1. Write all numbers with three decimal places: 5.009, 5.090, 5.900, 5.099.
2. All have 5 ones, so compare the tenths digits: 0, 0, 9, 0.
3. 5.9 has the largest tenths digit, so 5.9 is the largest.

Teacher's Note:
1. Compare place by place from the left; the first difference decides.
2. More digits after the point does not mean a bigger number.

Question 4: 250 paise = ______ rupees.
(a) ₹0.25
(b) ₹2.50
(c) ₹25
(d) ₹0.025
Show Answer & Explanation

Answer: (b) ₹2.50

Explanation:
1. 1 rupee = 100 paise, so divide the paise by 100.
2. 250 paise = \( \frac{250}{100} \) rupees = 2.50 rupees.
3. So 250 paise = ₹2.50.

Teacher's Note:
1. 200 paise make ₹2, and the remaining 50 paise make ₹0.50.
2. Rupee amounts are written with two decimal places for paise.

Question 5: Arrange in descending order: 3.25, 3.205, 3.52, 3.502.
(a) 3.52 > 3.502 > 3.25 > 3.205
(b) 3.502 > 3.52 > 3.25 > 3.205
(c) 3.52 > 3.25 > 3.502 > 3.205
(d) 3.52 > 3.502 > 3.205 > 3.25
Show Answer & Explanation

Answer: (a) 3.52 > 3.502 > 3.25 > 3.205

Explanation:
1. Write with three decimals: 3.250, 3.205, 3.520, 3.502.
2. Compare the decimal parts as thousandths: 520 > 502 > 250 > 205.
3. So the order is 3.52 > 3.502 > 3.25 > 3.205.

Teacher's Note:
1. Descending means largest first.
2. 3.52 is larger than 3.502 because 520 thousandths is more than 502 thousandths.

Question 6: Which is closest to 7?
(a) 7.08
(b) 6.99
(c) 7.005
(d) 7.1
Show Answer & Explanation

Answer: (c) 7.005

Explanation:
1. Distance from 7: 7.08 − 7 = 0.08; 7 − 6.99 = 0.01; 7.005 − 7 = 0.005; 7.1 − 7 = 0.1.
2. Compare the gaps: 0.005 < 0.01 < 0.08 < 0.1.
3. The smallest gap is 0.005, so 7.005 is closest to 7.

Teacher's Note:
1. 0.005 (5 thousandths) is smaller than 0.01 (10 thousandths).
2. 6.99 is very close too, but 7.005 is closer still.

Question 7: If 1.5 kg of sugar costs ₹45, then 1 kg costs:
(a) ₹30
(b) ₹67.50
(c) ₹45
(d) ₹33.33
Show Answer & Explanation

Answer: (a) ₹30

Explanation:
1. 1.5 kg is 3 half-kilograms.
2. So one half-kilogram costs ₹45 ÷ 3 = ₹15.
3. 1 kg = 2 half-kilograms costs 2 × ₹15 = ₹30.

Teacher's Note:
1. Breaking 1.5 kg into halves avoids dividing by a decimal.
2. Check: 1.5 × 30 = 45.

Question 8: Convert 34.5 cm into meters.
(a) 3.45 m
(b) 0.345 m
(c) 0.0345 m
(d) 34.05 m
Show Answer & Explanation

Answer: (b) 0.345 m

Explanation:
1. 1 m = 100 cm, so divide by 100.
2. 34.5 ÷ 100: move the decimal point two places to the left.
3. 34.5 cm = 0.345 m.

Teacher's Note:
1. Dividing by 100 moves the point two places left.
2. A sensible check: 34.5 cm is about one-third of a metre.

Question 9:
Assertion (A): 0.5 hours = 50 minutes.
Reason (R): In decimals, 0.5 means 50 parts out of 100.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. 1 hour = 60 minutes, not 100 minutes.
2. 0.5 hour = \( \frac{1}{2} \) of 60 minutes = 30 minutes, so the Assertion is false.
3. 0.5 = 0.50 = \( \frac{50}{100} \), so the Reason is true.
4. So the answer is (d) A is false but R is true.

Teacher's Note:
1. Time is not a base-10 measure: an hour has 60 minutes.
2. Half an hour is 30 minutes; 0.5 means a half, not 50 minutes.

Question 10:
Assertion (A): 1 mm = 0.1 cm.
Reason (R): 1 cm has 10 mm.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Since 1 cm = 10 mm, 1 mm = \( \frac{1}{10} \) cm = 0.1 cm.
2. So the Assertion is true, and the Reason is true.
3. The Reason is exactly the fact used to get the Assertion, so the answer is (a).

Teacher's Note:
1. A millimetre is one tenth of a centimetre.
2. R explains A when A follows directly from R.

Section B: Short Answer Questions (2 Marks Each)

Question 11. Express 4 ones, 2 tenths, and 8 hundredths in decimal form.
Answer:
4 ones = 4, 2 tenths = \( \frac{2}{10} \) = 0.2 and 8 hundredths = \( \frac{8}{100} \) = 0.08.
4 + 0.2 + 0.08 = 4.28.
In simple words: Put 4 in the ones place, 2 in the tenths place and 8 in the hundredths place: 4.28.

Teacher's Note:
1. Read the answer back: 4.28 is 4 ones, 2 tenths and 8 hundredths.
2. Each place to the right of the point is one tenth of the place before it.

 

Question 12. The weight of a packet is 2.075 kg. If 0.5 kg is removed, what is the new weight?
Answer:
Write 0.5 as 0.500 so both numbers have three decimal places.
  2.075
− 0.500
1.575

New weight = 1.575 kg.
In simple words: 2.075 kg − 0.500 kg = 1.575 kg.

Teacher's Note:
1. Subtract 5 tenths from the tenths place, not from the thousandths place.
2. Check: 1.575 + 0.5 = 2.075.

 

Question 13. Write 375 mm in meters.
Answer:
1 m = 1000 mm, so 1 mm = \( \frac{1}{1000} \) m.
375 mm = \( \frac{375}{1000} \) m = 0.375 m.
In simple words: Divide by 1000: 375 mm = 0.375 m.

Teacher's Note:
1. Three zeros in 1000 means the point moves three places left.
2. 375 mm is a bit more than one-third of a metre.

 

Question 14. Arrange the following in ascending order: 8.09, 8.19, 8.109, 8.9.
Answer:
Write all with three decimal places: 8.090, 8.190, 8.109, 8.900.
Compare the decimal parts as thousandths: 090 < 109 < 190 < 900.
Ascending order: 8.09 < 8.109 < 8.19 < 8.9.
In simple words: As thousandths the numbers are 8.090, 8.109, 8.190, 8.900, from smallest to largest.

Teacher's Note:
1. Ascending means smallest first.
2. 8.109 is less than 8.19 because 109 thousandths is less than 190 thousandths.

 

Section C: Short Answer Questions (3 Marks Each)

Question 15. A shopkeeper bought vegetables: 3.45 kg potatoes, 2.8 kg onions, and 5.075 kg tomatoes.
(a) Find the total weight of vegetables.
(b) Convert the total into grams.
Answer:
(a) Line up the decimal points (3.45 = 3.450, 2.8 = 2.800):
  3.450
+ 2.800
+ 5.075
11.325

Total weight = 11.325 kg.
(b) 1 kg = 1000 g, so 11.325 kg = 11.325 × 1000 g = 11325 g.
In simple words: The vegetables weigh 11.325 kg, which is 11325 g.

Teacher's Note:
1. Fill empty decimal places with zeros so the columns line up.
2. Multiplying by 1000 moves the decimal point three places right.

 

Question 16. Convert the following to tenths and then add: 12.45 m + 6.085 m.
Answer:
One tenth of a metre = 0.1 m, and every 1 m has 10 tenths.
12.45 m = 12.45 × 10 = 124.5 tenths of a metre.
6.085 m = 6.085 × 10 = 60.85 tenths of a metre.
Sum = 124.5 + 60.85 = 185.35 tenths of a metre.
Back in metres: 185.35 ÷ 10 = 18.535 m.
Check by direct addition: 12.450 + 6.085 = 18.535 m.
In simple words: Count both lengths in tenths of a metre, add them, then change back: 18.535 m.

Teacher's Note:
1. To find how many tenths are in a number, multiply it by 10.
2. The direct addition gives the same answer, which confirms the working.

 

Question 17. Fill in the blanks with >, <, or = :
(a) 4.05 ___ 4.5
(b) 0.7 ___ 0.70
(c) 3.408 ___ 3.48
Answer:
(a) 4.05 has 0 tenths and 4.5 has 5 tenths, so 4.05 < 4.5.
(b) 0.70 is 7 tenths and 0 hundredths, the same as 0.7, so 0.7 = 0.70.
(c) Both have 3 ones and 4 tenths; hundredths: 0 in 3.408 and 8 in 3.48, so 3.408 < 3.48.
In simple words: Compare digit by digit from the left: 4.05 < 4.5, 0.7 = 0.70 and 3.408 < 3.48.

Teacher's Note:
1. The first place where the digits differ decides which number is bigger.
2. Writing 4.5 as 4.50 and 3.48 as 3.480 makes the comparison clearer.

 

Section D: Long Answer Question (5 Marks)

Question 18. In a science experiment, water was poured into two beakers:
Beaker 1: 2.465 litres
Beaker 2: 5.875 litres
(a) Find the total water in both beakers.
(b) If 1.425 litres evaporated, how much water remained?
(c) Convert the remaining water into millilitres.
(d) If each mL of water weighs 1 gram, find the total weight of the remaining water.
(e) Convert this weight into kilograms.
Answer:
(a) Total water:
  2.465
+ 5.875
8.340

Total = 8.34 L.
(b) Remaining = 8.340 − 1.425 = 6.915 L.
(c) 1 L = 1000 mL, so 6.915 L = 6.915 × 1000 = 6915 mL.
(d) Each mL weighs 1 g, so 6915 mL weighs 6915 g.
(e) 1 kg = 1000 g, so 6915 g = 6915 ÷ 1000 = 6.915 kg.
In simple words: 8.34 L poured, 6.915 L left, which is 6915 mL weighing 6915 g = 6.915 kg.

Teacher's Note:
1. In (a), 5 + 5 thousandths = 10 thousandths, so carry 1 to the hundredths.
2. Litres of water and kilograms of water have the same number here because 1 mL weighs 1 g.
3. Multiply by 1000 to go from L to mL, divide by 1000 to go from g to kg.

 

Section E: Case Study Based Questions (4 Marks Each)

Question 19. A well-known jeweller in the city was approached by a regular customer who brought in three gold chains for repair. Each chain was of a different weight — Chain A weighed 45.75 g, Chain B weighed 32.8 g, and Chain C weighed 38.925 g. To restore their shine and strength, the jeweller planned to use additional gold dust during the repairs. He wanted to know the total gold he had received, the final weight after repairs, and its total value in the market.
(a) Find the total weight of gold received.
(b) If 5.5 g of gold dust is used in repairs and added to the chains, what is the final weight?
(c) Convert the final weight into kilograms.
(d) If 1 g of gold costs ₹5,400, find the total value of the final gold.
Answer:
(a) Total = 45.750 + 32.800 + 38.925:
  45.750
+ 32.800
+ 38.925
117.475

Gold received = 117.475 g.
(b) Final weight = 117.475 + 5.500 = 122.975 g.
(c) 1 kg = 1000 g, so 122.975 g = 122.975 ÷ 1000 = 0.122975 kg.
(d) Value = 122.975 × ₹5,400.
Split it: 122 × 5400 = 6,58,800 and 0.975 × 5400 = 5,265.
Total value = 6,58,800 + 5,265 = ₹6,64,065.
In simple words: The chains weigh 117.475 g, 122.975 g after repair (0.122975 kg), worth ₹6,64,065.

Teacher's Note:
1. Splitting 122.975 into 122 and 0.975 makes the multiplication easier.
2. 0.975 × 5400 = 975 × 5.4 = 5265 is a quick check.
3. Indian place-value commas: ₹6,64,065 is six lakh sixty-four thousand and sixty-five rupees.

 

Question 20. A local sports club was preparing its swimming pool for a summer training camp. The filling was done in two stages: 12.5 kilolitres on the first day and 8.375 kilolitres on the second day. However, due to the hot weather, some water evaporated before the pool was fully covered. The club manager wanted to calculate the total water initially filled, the remaining amount after evaporation, and the cost of the remaining water to manage the club’s budget for upcoming swimming sessions.
(a) Find the total water filled in litres.
(b) If 1500 litres evaporated, how much water remains?
(c) Convert the remaining water into kilolitres.
(d) If each kilolitre costs ₹35 to supply, find the total cost of the remaining water.
Answer:
(a) Total = 12.500 + 8.375 = 20.875 kL.
1 kL = 1000 L, so 20.875 kL = 20.875 × 1000 = 20875 litres.
(b) Water left = 20875 − 1500 = 19375 litres.
(c) 19375 ÷ 1000 = 19.375 kL.
(d) Cost = 19.375 × ₹35 = (19 × 35) + (0.375 × 35) = 665 + 13.125 = ₹678.125.
Money is written to the nearest paisa, so the cost is ₹678.13 (about ₹678).
In simple words: 20875 L filled, 19375 L (19.375 kL) left, costing about ₹678.13.

Teacher's Note:
1. 0.375 is three-eighths, so 0.375 × 35 = 105 ÷ 8 = 13.125.
2. Rupee amounts have only two decimal places (paise), so round ₹678.125 to ₹678.13.
3. Keep units consistent: change kL to L before subtracting litres.

 

Free study material for Mathematics

CBSE Class 7 Mathematics Assignments for Chapter 03 A Peek Beyond the Point

Download Assignment: Chapter 03 A Peek Beyond the Point (Class 7 Mathematics)

Access structured practice assignments for Chapter 03 A Peek Beyond the Point designed in alignment with the latest CBSE curriculum for Class 7 Mathematics. These printable sets cover objective and descriptive problem types to support thorough revision.

Key Advantages of Solving Chapter 03 A Peek Beyond the Point Assignments

  • Curriculum Standards: Assignments match modern CBSE sample formats to ensure relevant preparation.
  • Thorough Revision: Detailed problem sets reinforce core concepts and eliminate conceptual weak spots.
  • Execution Speed: Timed practice with assignment sets sharpens overall response timing.

How to Approach Mathematics Chapter 03 A Peek Beyond the Point Assignments

  1. Concept Foundation: Review the NCERT book for Class 7 Mathematics thoroughly before diving into assignment tasks.
  2. Self-Evaluation: Solve exercises independently before inspecting professional answer guides.
  3. Progress Monitoring: Note down complex formulas or concepts, clearing them up using available online practice aids.

FAQs

Where can I download the latest CBSE Class 7 Mathematics Chapter 03 A Peek Beyond the Point assignments?

You can download free PDF assignments for Class 7 Mathematics Chapter 03 A Peek Beyond the Point from StudiesToday.com. These practice sheets have been updated for the 2026-27 session covering all concepts from latest NCERT textbook.

Do these Mathematics Chapter 03 A Peek Beyond the Point assignments include solved questions?

Yes, our teachers have given solutions for all questions in the Class 7 Mathematics Chapter 03 A Peek Beyond the Point assignments. This will help you to understand step-by-step methodology to get full marks in school tests and exams.

Are the assignments for Class 7 Mathematics Chapter 03 A Peek Beyond the Point based on the 2026 exam pattern?

Yes. These assignments are designed as per the latest CBSE syllabus for 2026. We have included huge variety of question formats such as MCQs, Case-study based questions and important diagram-based problems found in Chapter 03 A Peek Beyond the Point.

How can practicing Chapter 03 A Peek Beyond the Point assignments help in Mathematics preparation?

Practicing topicw wise assignments will help Class 7 students understand every sub-topic of Chapter 03 A Peek Beyond the Point. Daily practice will improve speed, accuracy and answering competency-based questions.

Can I download Mathematics Chapter 03 A Peek Beyond the Point assignments for free on mobile?

Yes, all printable assignments for Class 7 Mathematics Chapter 03 A Peek Beyond the Point are available for free download in mobile-friendly PDF format.