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CHAPTER 3 — A PEEK BEYOND THE POINT
Class VII Mathematics — NCERT Ganita Prakash 2024
Preeti Kushwah Classes — Unit Test — Set 3 (Advanced)
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 a decimal that lies between 0.4 and 0.5. Is there only one such decimal? Give a reason.
Q2.1
The product of a decimal and 1000 is 47. What is the decimal?
Q3.1
Express 5 kg 375 g as a decimal in kilograms.
Q4.1
Find: 2.5 × 2.5
Q5.1
Convert 11/8 to a decimal. Is it a terminating decimal?
Q6.1
Round 7.845 to: (a) 1 decimal place (b) 2 decimal places.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
A water tank holds 45.6 litres. 12.75 L is used in the morning and 8.5 L in the evening. Write a single expression for the water remaining and evaluate it.
Q8.2
Find: 4.5 × 6 − 2.75 × 4 + 1.5 Show each step clearly.
Q9.2
Ramesh earns ₹12,500.50 per month. He saves ₹3,250.75. Find his monthly expenses.
Q10.2
Write in ascending order: 1.1, 1.01, 1.001, 1.0001. Describe the pattern you notice in these numbers.
Q11.2
A rope 15.5 m long is cut into pieces of 2.5 m each. How many complete pieces can be cut? What length of rope is left over?
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
A train travels at 65.5 km/h.
(a) Find the distance covered in 2.5 hours.
(b) At the same speed, how long does it take to travel 196.5 km?
(c) If the speed increases to 82 km/h, find the distance covered in 2.5 hours.
(a) Find the distance covered in 2.5 hours.
(b) At the same speed, how long does it take to travel 196.5 km?
(c) If the speed increases to 82 km/h, find the distance covered in 2.5 hours.
Q13.3
Using only + and × with the decimals 1.5, 2.5 and 3.5 (each used exactly once):
(a) Write the expression that gives the largest possible value and evaluate it.
(b) Write the expression that gives the smallest possible value and evaluate it.
(c) Write one expression where the result is a whole number and verify.
(a) Write the expression that gives the largest possible value and evaluate it.
(b) Write the expression that gives the smallest possible value and evaluate it.
(c) Write one expression where the result is a whole number and verify.
Q14.3
A farmer has a field of 100 m². He grows wheat on 0.6 of it and vegetables on 0.25 of it.
(a) Find the area under wheat (in m²).
(b) Find the area under vegetables (in m²).
(c) Find the fallow (unused) area. Show all working.
(a) Find the area under wheat (in m²).
(b) Find the area under vegetables (in m²).
(c) Find the fallow (unused) area. Show all working.
Q15.3
Petrol costs ₹102.50 per litre. A car’s tank holds 45 litres. It is currently 0.4 full.
(a) How much petrol is currently in the tank?
(b) How much more petrol is needed to fill the tank completely?
(c) Find the cost to fill the remaining tank.
(a) How much petrol is currently in the tank?
(b) How much more petrol is needed to fill the tank completely?
(c) Find the cost to fill the remaining tank.
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
Monthly expenses of a student’s household (₹):
(a) Find the total monthly expenses. [1]
(b) Monthly income is ₹20,000. Find the monthly savings. [1]
(c) If rent rises by ₹750.50, find the new savings. [1]
(d) What decimal fraction of income (to 2 decimal places) is spent on education? [1]
(e) Express savings (from part b) as a decimal fraction of income (to 3 decimal places). [1]
| Category | Amount (₹) |
|---|---|
| Food | 4,250.75 |
| Rent | 8,000.00 |
| Transport | 1,320.50 |
| Education | 2,500.25 |
| Others | 875.60 |
(b) Monthly income is ₹20,000. Find the monthly savings. [1]
(c) If rent rises by ₹750.50, find the new savings. [1]
(d) What decimal fraction of income (to 2 decimal places) is spent on education? [1]
(e) Express savings (from part b) as a decimal fraction of income (to 3 decimal places). [1]
Q17.5
(a) Find: 10 − [2.5 + {3.75 − (1.25 + 0.5)}] Show all steps. [2]
(b) A decimal is 0.03 more than 0.7. Multiply it by 25. What do you get? [1]
(c) The sum of three decimals is 10. Two of them are 3.45 and 2.8. Find the third decimal. [1]
(d) Find the product: 1.2 × 2.5 × 4 [1]
(b) A decimal is 0.03 more than 0.7. Multiply it by 25. What do you get? [1]
(c) The sum of three decimals is 10. Two of them are 3.45 and 2.8. Find the third decimal. [1]
(d) Find the product: 1.2 × 2.5 × 4 [1]
Bonus Question (Optional) [2 Marks]
Q18.2
★ CHALLENGE
(a) Find a 2-decimal number such that when 3 is added to it, the result is 8.25. [1]
(b) Find a 2-decimal number such that when multiplied by 8, the result is 6. [1]
(a) Find a 2-decimal number such that when 3 is added to it, the result is 8.25. [1]
(b) Find a 2-decimal number such that when multiplied by 8, the result is 6. [1]
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Answer Key & Detailed Solutions
Q1. [1 Mark]
Any decimal between 0.4 and 0.5, e.g. 0.42 or 0.47 or 0.456.
No, there is not only one — there are infinitely many decimals between 0.4 and 0.5, because between any two distinct decimal numbers there are always more decimals.
No, there is not only one — there are infinitely many decimals between 0.4 and 0.5, because between any two distinct decimal numbers there are always more decimals.
Q2. [1 Mark]
Let the decimal be d.
d × 1000 = 47
d = 47 ÷ 1000 = 0.047
d × 1000 = 47
d = 47 ÷ 1000 = 0.047
Q3. [1 Mark]
1 kg = 1000 g, so 375 g = 375/1000 kg = 0.375 kg
5 kg 375 g = 5 + 0.375 = 5.375 kg
5 kg 375 g = 5 + 0.375 = 5.375 kg
Q4. [1 Mark]
Ignore decimal: 25 × 25 = 625
2.5 has 1 decimal place, 2.5 has 1 decimal place → total 2 decimal places
Answer: 6.25
2.5 has 1 decimal place, 2.5 has 1 decimal place → total 2 decimal places
Answer: 6.25
Q5. [1 Mark]
To get denominator 1000: 8 × 125 = 1000; so 11 × 125 = 1375
11/8 = 1375/1000 = 1.375
Yes, it is a terminating decimal (it ends after 3 decimal places).
11/8 = 1375/1000 = 1.375
Yes, it is a terminating decimal (it ends after 3 decimal places).
Q6. [1 Mark]
(a) Round 7.845 to 1 d.p.: look at 2nd decimal place digit = 4 (< 5), round down → 7.8
(b) Round 7.845 to 2 d.p.: look at 3rd decimal place digit = 5 (≥ 5), round up → 7.85
(b) Round 7.845 to 2 d.p.: look at 3rd decimal place digit = 5 (≥ 5), round up → 7.85
Q7. [2 Marks]
Expression: 45.6 − 12.75 − 8.5
= 45.60 − 12.75 − 8.50
Step 1: 45.60 − 12.75 = 32.85
Step 2: 32.85 − 8.50 = 24.35 litres
= 45.60 − 12.75 − 8.50
Step 1: 45.60 − 12.75 = 32.85
Step 2: 32.85 − 8.50 = 24.35 litres
Q8. [2 Marks]
4.5 × 6 − 2.75 × 4 + 1.5
Step 1 (multiply first): 4.5 × 6 = 27 and 2.75 × 4 = 11
Expression: 27 − 11 + 1.5
Step 2: 27 − 11 = 16
Step 3: 16 + 1.5 = 17.5
Step 1 (multiply first): 4.5 × 6 = 27 and 2.75 × 4 = 11
Expression: 27 − 11 + 1.5
Step 2: 27 − 11 = 16
Step 3: 16 + 1.5 = 17.5
Q9. [2 Marks]
Monthly expenses = Earnings − Savings
= 12,500.50 − 3,250.75
= ₹9,249.75
= 12,500.50 − 3,250.75
= ₹9,249.75
Q10. [2 Marks]
Ascending order: 1.0001 < 1.001 < 1.01 < 1.1
Pattern: Each number has one more decimal place than the previous. As we add more decimal places (more zeros before the last digit), the numbers get closer and closer to 1.
Pattern: Each number has one more decimal place than the previous. As we add more decimal places (more zeros before the last digit), the numbers get closer and closer to 1.
Q11. [2 Marks]
Number of pieces = 15.5 ÷ 2.5 = 155 ÷ 25 = 6.2
Complete pieces = 6
Length used = 6 × 2.5 = 15 m
Length left over = 15.5 − 15 = 0.5 m
Complete pieces = 6
Length used = 6 × 2.5 = 15 m
Length left over = 15.5 − 15 = 0.5 m
Q12. [3 Marks]
(a) Distance = speed × time = 65.5 × 2.5
655 × 25 = 16375; 2 decimal places → 163.75 km
(b) Time = distance ÷ speed = 196.5 ÷ 65.5 = 1965 ÷ 655 = 3 hours
(c) Distance = 82 × 2.5 = 820 ÷ 10 = 205 km
655 × 25 = 16375; 2 decimal places → 163.75 km
(b) Time = distance ÷ speed = 196.5 ÷ 65.5 = 1965 ÷ 655 = 3 hours
(c) Distance = 82 × 2.5 = 820 ÷ 10 = 205 km
Q13. [3 Marks]
(a) Largest: (1.5 + 2.5) × 3.5 = 4 × 3.5 = 14
(Grouping the two smaller decimals for addition maximises the product)
(b) Smallest: 1.5 + 2.5 + 3.5 = 7.5
(Adding all three is smallest since no multiplication)
(c) Whole number result: (1.5 + 2.5) × 3.5 = 4 × 3.5 = 14 ✓ (whole number)
(Grouping the two smaller decimals for addition maximises the product)
(b) Smallest: 1.5 + 2.5 + 3.5 = 7.5
(Adding all three is smallest since no multiplication)
(c) Whole number result: (1.5 + 2.5) × 3.5 = 4 × 3.5 = 14 ✓ (whole number)
Q14. [3 Marks]
(a) Wheat area = 100 × 0.6 = 60 m²
(b) Vegetable area = 100 × 0.25 = 25 m²
(c) Fallow area = 100 − 60 − 25 = 15 m²
(b) Vegetable area = 100 × 0.25 = 25 m²
(c) Fallow area = 100 − 60 − 25 = 15 m²
Q15. [3 Marks]
(a) Petrol currently in tank = 45 × 0.4 = 18 litres
(b) Petrol needed = 45 − 18 = 27 litres
(c) Cost = 27 × 102.50 = 27 × 102.50
27 × 100 = 2700; 27 × 2.50 = 67.50
Total = ₹2,767.50
(b) Petrol needed = 45 − 18 = 27 litres
(c) Cost = 27 × 102.50 = 27 × 102.50
27 × 100 = 2700; 27 × 2.50 = 67.50
Total = ₹2,767.50
Q16. [5 Marks]
(a) Total = 4250.75 + 8000.00 + 1320.50 + 2500.25 + 875.60 = ₹16,947.10
(b) Savings = 20,000 − 16,947.10 = ₹3,052.90
(c) New expenses = 16,947.10 + 750.50 = 17,697.60
New savings = 20,000 − 17,697.60 = ₹2,302.40
(d) Education fraction = 2500.25 ÷ 20000 = 0.125025 ≈ 0.13 (to 2 d.p.)
(e) Savings fraction = 3052.90 ÷ 20000 = 0.153 (to 3 d.p.)
(b) Savings = 20,000 − 16,947.10 = ₹3,052.90
(c) New expenses = 16,947.10 + 750.50 = 17,697.60
New savings = 20,000 − 17,697.60 = ₹2,302.40
(d) Education fraction = 2500.25 ÷ 20000 = 0.125025 ≈ 0.13 (to 2 d.p.)
(e) Savings fraction = 3052.90 ÷ 20000 = 0.153 (to 3 d.p.)
Q17. [5 Marks]
(a) 10 − [2.5 + {3.75 − (1.25 + 0.5)}]
Step 1 — Round brackets: (1.25 + 0.5) = 1.75
Step 2 — Curly brackets: {3.75 − 1.75} = 2
Step 3 — Square brackets: [2.5 + 2] = 4.5
Step 4: 10 − 4.5 = 5.5
(b) 0.7 + 0.03 = 0.73; 0.73 × 25 = 73 × 25 ÷ 100 = 1825 ÷ 100 = 18.25
(c) Third decimal = 10 − 3.45 − 2.80 = 10 − 6.25 = 3.75
(d) 1.2 × 2.5 = 3.0; 3.0 × 4 = 12
Step 1 — Round brackets: (1.25 + 0.5) = 1.75
Step 2 — Curly brackets: {3.75 − 1.75} = 2
Step 3 — Square brackets: [2.5 + 2] = 4.5
Step 4: 10 − 4.5 = 5.5
(b) 0.7 + 0.03 = 0.73; 0.73 × 25 = 73 × 25 ÷ 100 = 1825 ÷ 100 = 18.25
(c) Third decimal = 10 − 3.45 − 2.80 = 10 − 6.25 = 3.75
(d) 1.2 × 2.5 = 3.0; 3.0 × 4 = 12
Q18. [Bonus — 2 Marks]
(a) Let the number be x.
x + 3 = 8.25 → x = 8.25 − 3 = 5.25
(b) Let the number be x.
x × 8 = 6 → x = 6 ÷ 8 = 0.75
Answer: 0.75
x + 3 = 8.25 → x = 8.25 − 3 = 5.25
(b) Let the number be x.
x × 8 = 6 → x = 6 ÷ 8 = 0.75
Answer: 0.75