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CHAPTER 3 — A PEEK BEYOND THE POINT
Class VII Mathematics — NCERT Ganita Prakash 2024
Preeti Kushwah Classes — Unit Test — Set 2 (Standard)
General Instructions:
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 5 marks each.
6. Show all working clearly. Marks are awarded for steps.
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 5 marks each.
6. Show all working clearly. Marks are awarded for steps.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
Write the expanded form of 20.705.
Q2.1
Compare using > or <: 3.09 _____ 3.90
Q3.1
Convert 125/1000 to a decimal.
Q4.1
Multiply: 0.35 × 100
Q5.1
True or False: “0.6 and 0.60 are unlike decimals.” Justify your answer.
Q6.1
How many times larger is 0.7 than 0.007?
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Convert 7/8 to a decimal. (Hint: make the denominator 1000.) Show all steps.
Q8.2
Subtract: 12.03 − 5.8
Show working by aligning decimal points clearly.
Show working by aligning decimal points clearly.
Q9.2
Convert 3.125 to a fraction in its simplest form. Show all steps.
Q10.2
A piece of wire is 8.4 m long. It is cut into 4 equal pieces. Find the length of each piece.
Q11.2
Find the value of: 6.5 × 100 − 2.5 × 10
Show each step clearly.
Show each step clearly.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Arrange in ascending order: 0.3, 0.33, 0.303, 0.033
Show your comparison at each step by converting to like decimals.
Show your comparison at each step by converting to like decimals.
Q13.3
The dimensions of a rectangle are: length = 7.5 cm and breadth = 4.25 cm.
(a) Find the perimeter of the rectangle.
(b) Find the area of the rectangle.
(c) If the length increases by 1.5 cm, find the new perimeter.
(a) Find the perimeter of the rectangle.
(b) Find the area of the rectangle.
(c) If the length increases by 1.5 cm, find the new perimeter.
Q14.3
A shopkeeper mixes 3.5 kg of one type of tea at ₹240/kg with 1.5 kg of another type at ₹320/kg.
(a) Find the total cost of both types of tea.
(b) Find the total weight of the mixture.
(c) Find the price per kg of the mixture.
(a) Find the total cost of both types of tea.
(b) Find the total weight of the mixture.
(c) Find the price per kg of the mixture.
Q15.3
Rainfall recorded (in cm) over 5 days: 3.5, 0.8, 2.75, 4.2, 1.05
(a) Find the total rainfall over 5 days.
(b) Find the average rainfall per day.
(c) On which day was the rainfall the least?
(a) Find the total rainfall over 5 days.
(b) Find the average rainfall per day.
(c) On which day was the rainfall the least?
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
(a) Find: 25.4 − 13.75 + 8.05 − 0.8. Show each step clearly. [2]
(b) The sum of two decimal numbers is 15.75. One of them is 8.4. Find the other number. [1]
(c) A decimal number is multiplied by 100 to give 347. What is the original decimal number? [1]
(d) What must be added to 9.25 to get the next whole number? [1]
(b) The sum of two decimal numbers is 15.75. One of them is 8.4. Find the other number. [1]
(c) A decimal number is multiplied by 100 to give 347. What is the original decimal number? [1]
(d) What must be added to 9.25 to get the next whole number? [1]
Q17.5
Monthly expenses of a family are given below:
(a) Find the total monthly expenses. [1]
(b) Monthly income is ₹20,000. Find the monthly savings. [1]
(c) If rent increases by ₹750.50, find the new total expenses. [1]
(d) What decimal fraction of income is spent on food? (to 2 decimal places) [1]
(e) Express savings as a decimal fraction of income. [1]
| Category | Amount (₹) |
|---|---|
| Food | 4250.75 |
| Rent | 8000.00 |
| Transport | 1320.50 |
| Education | 2500.25 |
| Others | 875.60 |
(b) Monthly income is ₹20,000. Find the monthly savings. [1]
(c) If rent increases by ₹750.50, find the new total expenses. [1]
(d) What decimal fraction of income is spent on food? (to 2 decimal places) [1]
(e) Express savings as a decimal fraction of income. [1]
Bonus Question (Optional) [2 Marks]
Q18.2
★ Three friends run a 100 m race. Their times are: Amit: 13.50 s, Beena: 13.05 s, Chintu: 13.5 s.
(a) Who finished first? [0.5]
(b) Are Amit’s and Chintu’s times identical? Explain using the concept of like decimals. [0.5]
(c) By how much time did Beena beat Chintu? [0.5]
(d) If Devika runs 0.38 s faster than Beena, what is Devika’s time? [0.5]
(a) Who finished first? [0.5]
(b) Are Amit’s and Chintu’s times identical? Explain using the concept of like decimals. [0.5]
(c) By how much time did Beena beat Chintu? [0.5]
(d) If Devika runs 0.38 s faster than Beena, what is Devika’s time? [0.5]
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Answer Key & Detailed Solutions
Q1. [1 Mark]
20.705 = 20 + 0.7 + 0.005
Or in fraction form: 20 + 7/10 + 5/1000
20.705 = 20 + 0.7 + 0.005
Or in fraction form: 20 + 7/10 + 5/1000
20.705 = 20 + 0.7 + 0.005
Q2. [1 Mark]
3.09 vs 3.90: whole parts equal (3). Compare tenths: 0 < 9.
3.09 < 3.90
3.09 < 3.90
Q3. [1 Mark]
125/1000: denominator is 1000, so 3 decimal places.
125/1000 = 0.125
125/1000 = 0.125
Q4. [1 Mark]
0.35 × 100: shift decimal 2 places right.
0.35 × 100 = 35
0.35 × 100 = 35
Q5. [1 Mark]
False.
0.6 and 0.60 are like decimals — both have one decimal place (0.60 = 0.6). Like decimals have the same number of decimal places. In fact they are equal: 0.6 = 0.60.
0.6 and 0.60 are like decimals — both have one decimal place (0.60 = 0.6). Like decimals have the same number of decimal places. In fact they are equal: 0.6 = 0.60.
Q6. [1 Mark]
0.7 ÷ 0.007 = 0.700 ÷ 0.007 = 700 ÷ 7 = 100
0.7 is 100 times larger than 0.007.
0.7 is 100 times larger than 0.007.
Q7. [2 Marks]
7/8 — make denominator 1000: multiply by 125/125
7/8 = (7 × 125)/(8 × 125) = 875/1000
7/8 = 0.875
7/8 = (7 × 125)/(8 × 125) = 875/1000
7/8 = 0.875
Q8. [2 Marks]
12.03
− 5.80 (add zero to align)
――――
= 6.23
− 5.80 (add zero to align)
――――
= 6.23
Q9. [2 Marks]
3.125 = 3125/1000
Simplify: HCF(3125, 1000) = 125
3125 ÷ 125 = 25; 1000 ÷ 125 = 8
3.125 = 25/8
Simplify: HCF(3125, 1000) = 125
3125 ÷ 125 = 25; 1000 ÷ 125 = 8
3.125 = 25/8
Q10. [2 Marks]
Length of each piece = 8.4 ÷ 4
= 84 ÷ 4 = 21, then place 1 decimal: 2.1 m
Each piece is 2.1 m long.
= 84 ÷ 4 = 21, then place 1 decimal: 2.1 m
Each piece is 2.1 m long.
Q11. [2 Marks]
Step 1: 6.5 × 100 = 650 (shift decimal 2 places right)
Step 2: 2.5 × 10 = 25 (shift decimal 1 place right)
Step 3: 650 − 25 = 625
Step 2: 2.5 × 10 = 25 (shift decimal 1 place right)
Step 3: 650 − 25 = 625
Q12. [3 Marks]
Convert to like decimals (3 places): 0.300, 0.330, 0.303, 0.033
Compare tenths: 0.033 has 0 tenths (smallest); others have 3 tenths.
Now compare 0.300, 0.330, 0.303 by hundredths: 0 < 0 < 3, so 0.300 = 0.303 hundredths comparison: 300 < 303 < 330.
Ascending: 0.033 < 0.3 < 0.303 < 0.33
Compare tenths: 0.033 has 0 tenths (smallest); others have 3 tenths.
Now compare 0.300, 0.330, 0.303 by hundredths: 0 < 0 < 3, so 0.300 = 0.303 hundredths comparison: 300 < 303 < 330.
Ascending: 0.033 < 0.3 < 0.303 < 0.33
Q13. [3 Marks]
(a) Perimeter = 2 × (l + b) = 2 × (7.5 + 4.25) = 2 × 11.75 = 23.50 cm
(b) Area = l × b = 7.5 × 4.25
= 75 × 425 ÷ 10000 = 31875 ÷ 1000 = 31.875 cm²
(c) New length = 7.5 + 1.5 = 9.0 cm
New perimeter = 2 × (9.0 + 4.25) = 2 × 13.25 = 26.50 cm
(b) Area = l × b = 7.5 × 4.25
= 75 × 425 ÷ 10000 = 31875 ÷ 1000 = 31.875 cm²
(c) New length = 7.5 + 1.5 = 9.0 cm
New perimeter = 2 × (9.0 + 4.25) = 2 × 13.25 = 26.50 cm
Q14. [3 Marks]
(a) Cost = 3.5 × 240 + 1.5 × 320 = 840 + 480 = ₹1320
(b) Total weight = 3.5 + 1.5 = 5 kg
(c) Price per kg = 1320 ÷ 5 = ₹264/kg
(b) Total weight = 3.5 + 1.5 = 5 kg
(c) Price per kg = 1320 ÷ 5 = ₹264/kg
Q15. [3 Marks]
(a) Total = 3.50 + 0.80 + 2.75 + 4.20 + 1.05 = 12.30 cm
(b) Average = 12.30 ÷ 5 = 2.46 cm/day
(c) Least rainfall: Day 2 = 0.8 cm
(b) Average = 12.30 ÷ 5 = 2.46 cm/day
(c) Least rainfall: Day 2 = 0.8 cm
Q16. [5 Marks]
(a) 25.40 − 13.75 = 11.65
11.65 + 8.05 = 19.70
19.70 − 0.80 = 18.90
(b) Other number = 15.75 − 8.4 = 15.75 − 8.40 = 7.35
(c) Original number = 347 ÷ 100 = 3.47
(d) Next whole number after 9.25 is 10. Amount to add = 10 − 9.25 = 0.75
11.65 + 8.05 = 19.70
19.70 − 0.80 = 18.90
(b) Other number = 15.75 − 8.4 = 15.75 − 8.40 = 7.35
(c) Original number = 347 ÷ 100 = 3.47
(d) Next whole number after 9.25 is 10. Amount to add = 10 − 9.25 = 0.75
Q17. [5 Marks]
(a) Total expenses = 4250.75 + 8000.00 + 1320.50 + 2500.25 + 875.60 = ₹16,947.10
(b) Savings = 20000 − 16947.10 = ₹3052.90
(c) New rent = 8000 + 750.50 = 8750.50
New total = 16947.10 + 750.50 = ₹17,697.60
(d) Food fraction = 4250.75 ÷ 20000 ≈ 0.2125 ≈ 0.21 (to 2 decimal places)
(e) Savings fraction = 3052.90 ÷ 20000 = 0.15264 ≈ 0.15
(b) Savings = 20000 − 16947.10 = ₹3052.90
(c) New rent = 8000 + 750.50 = 8750.50
New total = 16947.10 + 750.50 = ₹17,697.60
(d) Food fraction = 4250.75 ÷ 20000 ≈ 0.2125 ≈ 0.21 (to 2 decimal places)
(e) Savings fraction = 3052.90 ÷ 20000 = 0.15264 ≈ 0.15
Q18. [Bonus — 2 Marks]
(a) Lowest time = Beena (13.05 s). Beena finished first.
(b) Amit: 13.50 s = 13.5 s. Chintu: 13.5 s.
13.50 = 13.5 (trailing zero does not change value).
Yes, Amit’s and Chintu’s times are identical — like decimals, equal value.
(c) Chintu’s time − Beena’s time = 13.50 − 13.05 = 0.45 s
(d) Devika’s time = 13.05 − 0.38 = 12.67 s
(b) Amit: 13.50 s = 13.5 s. Chintu: 13.5 s.
13.50 = 13.5 (trailing zero does not change value).
Yes, Amit’s and Chintu’s times are identical — like decimals, equal value.
(c) Chintu’s time − Beena’s time = 13.50 − 13.05 = 0.45 s
(d) Devika’s time = 13.05 − 0.38 = 12.67 s