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CHAPTER 3 — A PEEK BEYOND THE POINT
Class VII Mathematics — NCERT Ganita Prakash 2024
Preeti Kushwah Classes — Unit Test — Set 4 (Challenge)
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.
7. ★ questions require deeper reasoning — think before you write!
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.
7. ★ questions require deeper reasoning — think before you write!
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
A decimal number has: 5 in the tenths place, 3 in the hundredths place, 7 in the ones place, and 2 in the thousandths place. Write the number.
Q2.1REASON
Without calculating, predict: Is 0.999 × 0.999 greater than, less than, or equal to 1? Give one mathematical reason.
Q3.1
How many decimal numbers lie between 0.1 and 0.2? Give a mathematical reason for your answer.
Q4.1
Find the value of x: 3x + 0.5 = 2x + 3.5
Q5.1
A decimal d satisfies: d × 4 = d + 9. Find d.
Q6.1EXPLAIN
Using place value, explain why 0.25 × 4 = 1, given that 25 × 4 = 100.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
The perimeter of a square is 13.6 cm.
(a) Find the length of one side.
(b) Find the area of the square.
(a) Find the length of one side.
(b) Find the area of the square.
Q8.2
The temperature at 6 am is 18.5°C. It rises by 3.75°C by noon, then falls by 5.2°C by evening. Find the evening temperature.
Q9.2
A decimal number multiplied by itself gives 0.0625. What is the decimal number? Show your working and verify your answer.
Q10.2
Three friends owe ₹45.75, ₹38.50 and ₹52.25 respectively. They pool the total amount and split it equally. Find each person’s share.
Q11.2PATTERN
Arrange in descending order: 0.9, 0.99, 0.999, 0.9999. What value do these numbers get closer and closer to as you keep adding more 9s?
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3★
A shopkeeper earns a profit of ₹8.50 per kg of rice. He sells 12.5 kg on Day 1 and 8.75 kg on Day 2 at this profit. On Day 3, he reduces profit to ₹5.25 per kg and sells 15 kg.
(a) Find the total profit from Day 1 and Day 2 combined. [1]
(b) Find the profit on Day 3. [1]
(c) Find the total profit over all 3 days. [1]
(a) Find the total profit from Day 1 and Day 2 combined. [1]
(b) Find the profit on Day 3. [1]
(c) Find the total profit over all 3 days. [1]
Q13.3
(a) Find: 10 − {4.5 + (3.75 − 1.25) × 2} Show all steps. [1]
(b) The sum of 5 consecutive decimals, each differing by 0.1, is 6.5. Find all five decimals. [1]
(c) Starting from 3.5, a number is decreased by 0.1 five times. List all six values and find their sum. [1]
(b) The sum of 5 consecutive decimals, each differing by 0.1, is 6.5. Find all five decimals. [1]
(c) Starting from 3.5, a number is decreased by 0.1 five times. List all six values and find their sum. [1]
Q14.3★
Petrol costs ₹102.75 per litre. A car’s tank can hold 45.5 litres and it is currently 0.4 full.
(a) How many litres are currently in the tank? [1]
(b) How many more litres are needed to fill it completely? [1]
(c) Find the total cost to fill the remaining tank. [1]
(a) How many litres are currently in the tank? [1]
(b) How many more litres are needed to fill it completely? [1]
(c) Find the total cost to fill the remaining tank. [1]
Q15.3PATTERN ★
Observe the following pattern:
1/9 = 0.111… 2/9 = 0.222… 3/9 = 0.333…
(a) Using the pattern, write 7/9 as a decimal. [1]
(b) What is 9/9 as a decimal? What familiar value does this equal? Explain. [1]
(c) Multiply 0.333… by 3. What do you get? What does this tell you about the fraction 1/3? [1]
1/9 = 0.111… 2/9 = 0.222… 3/9 = 0.333…
(a) Using the pattern, write 7/9 as a decimal. [1]
(b) What is 9/9 as a decimal? What familiar value does this equal? Explain. [1]
(c) Multiply 0.333… by 3. What do you get? What does this tell you about the fraction 1/3? [1]
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5★
A company’s quarterly sales data (Q1):
(a) Find the total Q1 (Jan–Mar) sales. [1]
(b) Find the average monthly sales for Q1. [1]
(c) April’s target is 1.1 times January’s sales. Find the April target. [1]
(d) In February, expenses were ₹76,430.25. Find the profit in February. [1]
(e) The company offers a discount of 0.05 of March sales. Find the discounted March sales figure. [1]
| Month | Sales (₹) |
|---|---|
| January | 1,25,430.50 |
| February | 98,750.75 |
| March | 1,08,960.25 |
(b) Find the average monthly sales for Q1. [1]
(c) April’s target is 1.1 times January’s sales. Find the April target. [1]
(d) In February, expenses were ₹76,430.25. Find the profit in February. [1]
(e) The company offers a discount of 0.05 of March sales. Find the discounted March sales figure. [1]
Q17.5CHALLENGE ★
(a) The average of 3.5, x, and 4.5 is 4.0. Find the value of x. [1]
(b) Sahil spends 0.4 of his money. Then he spends 0.25 of the remainder. He is left with ₹270. How much money did he start with? Show all working. [2]
(c) A decimal d has 2 decimal places and d × d = 0.5625. Find d. Verify your answer. [1]
(d) Find: (3.5 + 1.5) × (3.5 − 1.5). Also predict the answer using the identity (a + b)(a − b) = a² − b². [1]
(b) Sahil spends 0.4 of his money. Then he spends 0.25 of the remainder. He is left with ₹270. How much money did he start with? Show all working. [2]
(c) A decimal d has 2 decimal places and d × d = 0.5625. Find d. Verify your answer. [1]
(d) Find: (3.5 + 1.5) × (3.5 − 1.5). Also predict the answer using the identity (a + b)(a − b) = a² − b². [1]
Bonus Question (Optional) [2 Marks]
Q18.2★★ MASTER CHALLENGE
A mysterious decimal is built from clues:
— Whole part = product of 2 and 3
— Tenths digit = (largest single-digit even number) ÷ 4
— Hundredths digit = 3² − 8
— Thousandths digit = √4 − 1
(a) Write the complete decimal number. [1]
(b) Multiply this number by 1000. Find the remainder when that result is divided by 7. [1]
— Whole part = product of 2 and 3
— Tenths digit = (largest single-digit even number) ÷ 4
— Hundredths digit = 3² − 8
— Thousandths digit = √4 − 1
(a) Write the complete decimal number. [1]
(b) Multiply this number by 1000. Find the remainder when that result is divided by 7. [1]
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Answer Key & Detailed Solutions
Q1. [1 Mark]
Ones = 7, tenths = 5, hundredths = 3, thousandths = 2
Number = 7.532
Number = 7.532
Q2. [1 Mark]
Less than 1.
Since 0.999 < 1, when we multiply two numbers that are each less than 1, the product is always less than either number. So 0.999 × 0.999 < 0.999 < 1.
Since 0.999 < 1, when we multiply two numbers that are each less than 1, the product is always less than either number. So 0.999 × 0.999 < 0.999 < 1.
Q3. [1 Mark]
Infinitely many.
Between any two distinct decimal numbers, there are always infinitely many more decimals. For example: 0.11, 0.12, …, 0.19 — that’s already 9 decimals, and we can further find 0.101, 0.102, 0.1001, etc., without end.
Between any two distinct decimal numbers, there are always infinitely many more decimals. For example: 0.11, 0.12, …, 0.19 — that’s already 9 decimals, and we can further find 0.101, 0.102, 0.1001, etc., without end.
Q4. [1 Mark]
3x + 0.5 = 2x + 3.5
3x − 2x = 3.5 − 0.5
x = 3
3x − 2x = 3.5 − 0.5
x = 3
Q5. [1 Mark]
d × 4 = d + 9
4d − d = 9
3d = 9
d = 3
4d − d = 9
3d = 9
d = 3
Q6. [1 Mark]
0.25 = 25/100 (25 hundredths).
We know 25 × 4 = 100.
So 25/100 × 4 = 100/100 = 1.
Alternatively: moving the decimal two places left divides by 100, so 0.25 = 25 ÷ 100; then (25 ÷ 100) × 4 = 100 ÷ 100 = 1.
We know 25 × 4 = 100.
So 25/100 × 4 = 100/100 = 1.
Alternatively: moving the decimal two places left divides by 100, so 0.25 = 25 ÷ 100; then (25 ÷ 100) × 4 = 100 ÷ 100 = 1.
Q7. [2 Marks]
(a) Side = perimeter ÷ 4 = 13.6 ÷ 4 = 3.4 cm
(b) Area = side × side = 3.4 × 3.4
34 × 34 = 1156; 2 decimal places → 11.56 cm²
(b) Area = side × side = 3.4 × 3.4
34 × 34 = 1156; 2 decimal places → 11.56 cm²
Q8. [2 Marks]
Temperature at 6 am = 18.5°C
After rising: 18.5 + 3.75 = 22.25°C (at noon)
After falling: 22.25 − 5.20 = 17.05°C (evening)
After rising: 18.5 + 3.75 = 22.25°C (at noon)
After falling: 22.25 − 5.20 = 17.05°C (evening)
Q9. [2 Marks]
d² = 0.0625 = 625/10000
√(625/10000) = 25/100 = 0.25
Verification: 0.25 × 0.25 = 25 × 25 ÷ 10000 = 625/10000 = 0.0625 ✓
√(625/10000) = 25/100 = 0.25
Verification: 0.25 × 0.25 = 25 × 25 ÷ 10000 = 625/10000 = 0.0625 ✓
Q10. [2 Marks]
Total = 45.75 + 38.50 + 52.25 = ₹136.50
Each person’s share = 136.50 ÷ 3 = ₹45.50
Each person’s share = 136.50 ÷ 3 = ₹45.50
Q11. [2 Marks]
Descending order: 0.9999 > 0.999 > 0.99 > 0.9
As we keep adding more 9s (0.9, 0.99, 0.999, 0.9999, …), the numbers get closer and closer to 1. In fact, 0.999… (infinitely repeating) equals exactly 1.
As we keep adding more 9s (0.9, 0.99, 0.999, 0.9999, …), the numbers get closer and closer to 1. In fact, 0.999… (infinitely repeating) equals exactly 1.
Q12. [3 Marks]
(a) Day 1 profit = 12.5 × 8.50 = ₹106.25
Day 2 profit = 8.75 × 8.50 = ₹74.375
Day 1 + Day 2 = 106.25 + 74.375 = ₹180.625
(b) Day 3 profit = 15 × 5.25 = ₹78.75
(c) Total profit = 180.625 + 78.75 = ₹259.375 (≈ ₹259.38)
Day 2 profit = 8.75 × 8.50 = ₹74.375
Day 1 + Day 2 = 106.25 + 74.375 = ₹180.625
(b) Day 3 profit = 15 × 5.25 = ₹78.75
(c) Total profit = 180.625 + 78.75 = ₹259.375 (≈ ₹259.38)
Q13. [3 Marks]
(a) 10 − {4.5 + (3.75 − 1.25) × 2}
Step 1 — Round brackets: (3.75 − 1.25) = 2.5
Step 2 — Multiply: 2.5 × 2 = 5
Step 3 — Curly brackets: {4.5 + 5} = 9.5
Step 4: 10 − 9.5 = 0.5
(b) Let middle decimal = x. The five: x−0.2, x−0.1, x, x+0.1, x+0.2
Sum = 5x = 6.5 → x = 1.3
Decimals: 1.1, 1.2, 1.3, 1.4, 1.5
(c) Values: 3.5, 3.4, 3.3, 3.2, 3.1, 3.0
Sum = 3.5+3.4+3.3+3.2+3.1+3.0 = 19.5
Step 1 — Round brackets: (3.75 − 1.25) = 2.5
Step 2 — Multiply: 2.5 × 2 = 5
Step 3 — Curly brackets: {4.5 + 5} = 9.5
Step 4: 10 − 9.5 = 0.5
(b) Let middle decimal = x. The five: x−0.2, x−0.1, x, x+0.1, x+0.2
Sum = 5x = 6.5 → x = 1.3
Decimals: 1.1, 1.2, 1.3, 1.4, 1.5
(c) Values: 3.5, 3.4, 3.3, 3.2, 3.1, 3.0
Sum = 3.5+3.4+3.3+3.2+3.1+3.0 = 19.5
Q14. [3 Marks]
(a) Current petrol = 45.5 × 0.4 = 18.2 litres
(b) Petrol needed = 45.5 − 18.2 = 27.3 litres
(c) Cost = 27.3 × 102.75
27.3 × 100 = 2730; 27.3 × 2.75 = 75.075
Total = 2730 + 75.075 = ₹2,805.08 (approx)
(b) Petrol needed = 45.5 − 18.2 = 27.3 litres
(c) Cost = 27.3 × 102.75
27.3 × 100 = 2730; 27.3 × 2.75 = 75.075
Total = 2730 + 75.075 = ₹2,805.08 (approx)
Q15. [3 Marks]
(a) Following the pattern (n/9 = 0.nnn…): 7/9 = 0.777…
(b) 9/9 = 1 (since 9 ÷ 9 = 1).
But from the pattern: 9/9 = 0.999…
Therefore 0.999… = 1. This is a famous mathematical fact — an infinitely repeating 9 is equal to 1.
(c) 0.333… × 3 = 0.999… = 1
This tells us 1/3 × 3 = 1, which confirms that 1/3 is the correct fraction — three thirds make a whole.
(b) 9/9 = 1 (since 9 ÷ 9 = 1).
But from the pattern: 9/9 = 0.999…
Therefore 0.999… = 1. This is a famous mathematical fact — an infinitely repeating 9 is equal to 1.
(c) 0.333… × 3 = 0.999… = 1
This tells us 1/3 × 3 = 1, which confirms that 1/3 is the correct fraction — three thirds make a whole.
Q16. [5 Marks]
(a) Total = 1,25,430.50 + 98,750.75 + 1,08,960.25 = ₹3,33,141.50
(b) Average = 3,33,141.50 ÷ 3 = ₹1,11,047.17 (approx)
(c) April target = 1,25,430.50 × 1.1 = ₹1,37,973.55
(d) February profit = 98,750.75 − 76,430.25 = ₹22,320.50
(e) Discount = 0.05 × 1,08,960.25 = ₹5,448.01
Discounted March sales = 1,08,960.25 × 0.95 = ₹1,03,512.24
(b) Average = 3,33,141.50 ÷ 3 = ₹1,11,047.17 (approx)
(c) April target = 1,25,430.50 × 1.1 = ₹1,37,973.55
(d) February profit = 98,750.75 − 76,430.25 = ₹22,320.50
(e) Discount = 0.05 × 1,08,960.25 = ₹5,448.01
Discounted March sales = 1,08,960.25 × 0.95 = ₹1,03,512.24
Q17. [5 Marks]
(a) (3.5 + x + 4.5) ÷ 3 = 4
8 + x = 12
x = 4
(b) After spending 0.4, Sahil has (1 − 0.4) = 0.6 of his money left.
After spending 0.25 of that: remaining = 0.6 × (1 − 0.25) = 0.6 × 0.75 = 0.45 of original.
0.45 × M = 270 → M = 270 ÷ 0.45 = ₹600
(c) d² = 0.5625 = 5625/10000; d = 75/100 = 0.75
Verify: 0.75 × 0.75 = 0.5625 ✓
(d) (3.5 + 1.5) × (3.5 − 1.5) = 5 × 2 = 10
Using identity: a² − b² = 3.5² − 1.5² = 12.25 − 2.25 = 10 ✓
8 + x = 12
x = 4
(b) After spending 0.4, Sahil has (1 − 0.4) = 0.6 of his money left.
After spending 0.25 of that: remaining = 0.6 × (1 − 0.25) = 0.6 × 0.75 = 0.45 of original.
0.45 × M = 270 → M = 270 ÷ 0.45 = ₹600
(c) d² = 0.5625 = 5625/10000; d = 75/100 = 0.75
Verify: 0.75 × 0.75 = 0.5625 ✓
(d) (3.5 + 1.5) × (3.5 − 1.5) = 5 × 2 = 10
Using identity: a² − b² = 3.5² − 1.5² = 12.25 − 2.25 = 10 ✓
Q18. [Bonus — 2 Marks]
(a) Clues decoded:
• Whole part = 2 × 3 = 6
• Tenths = largest single-digit even number ÷ 4 = 8 ÷ 4 = 2
• Hundredths = 3² − 8 = 9 − 8 = 1
• Thousandths = √4 − 1 = 2 − 1 = 1
Decimal number = 6.211
(b) 6.211 × 1000 = 6211
6211 ÷ 7: 7 × 887 = 6209; remainder = 6211 − 6209 = 2
• Whole part = 2 × 3 = 6
• Tenths = largest single-digit even number ÷ 4 = 8 ÷ 4 = 2
• Hundredths = 3² − 8 = 9 − 8 = 1
• Thousandths = √4 − 1 = 2 − 1 = 1
Decimal number = 6.211
(b) 6.211 × 1000 = 6211
6211 ÷ 7: 7 × 887 = 6209; remainder = 6211 − 6209 = 2