Arithmetic SNAP 2024 Slot 2 Question paper with answers

Percentages (2 Questions)

Q1. A shopkeeper increases the price of an item by 20% and then offers a discount of 10%. What is the net percentage change in the price of the item?

1. 8% increase
2. 8% decrease
3. 10% increase
4. 12% decrease

Q2. A man spends 60% of his income and saves ₹12,000. What is his total income?

1. ₹20,000
2. ₹25,000
3. ₹30,000
4. ₹40,000

Ratios (2 Questions)

Q3. The ratio of boys to girls in a school is 3:5. If there are 120 more girls than boys, how many boys are there in the school?

1. 180
2. 240
3. 300
4. 360

Q4. The present age of A and B are in the ratio 4:5. After 8 years, their ages will be in the ratio 5:6. What is A’s present age?

1. 24
2. 28
3. 32
4. 36

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Interest (1 Question)

Q5. A sum of ₹10,000 is invested at a compound interest rate of 10% per annum for 2 years. What is the compound interest accrued?

1. ₹1,000
2. ₹2,000
3. ₹2,100
4. ₹2,210

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Mixtures (1 Question)

Q6. A mixture contains milk and water in the ratio 3:1. If 20 liters of the mixture is replaced with water, the ratio becomes 2:3. What is the initial quantity of milk in the mixture?

1. 24 liters
2. 30 liters
3. 36 liters
4. 40 liters

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Averages (1 Question)

Q7. The average of 7 consecutive integers is 23. What is the largest integer in the set?

1. 25
2. 26
3. 27
4. 28

Percentages

Q1 Solution: Net change formula: \( \text{Net Change} = x + y + \frac{xy}{100} \), where \(x = 20\), \(y = -10\). \[ \text{Net Change} = 20 – 10 + \frac{(20 \times -10)}{100} = 8\% \text{ increase.} \]

Q2 Solution: Savings = \(40\%\), so income \( = \frac{12,000}{0.4} = 30,000\).

Ratios

Q3 Solution: Girls – Boys = 120. Let boys = \(3x\), girls = \(5x\). \[ 5x – 3x = 120 \implies x = 60, \text{ Boys } = 3x = 180. \]

Q4 Solution: Let present ages be \(4x\) and \(5x\). After 8 years: \[ \frac{4x + 8}{5x + 8} = \frac{5}{6} \implies 24x + 48 = 25x + 40 \implies x = 8. \] Present age of A = \(4x = 32\).

Interest

Q5 Solution: Compound interest: \( A = P \left(1 + \frac{r}{100}\right)^n \) \[ A = 10,000 \times \left(1 + \frac{10}{100}\right)^2 = 10,000 \times 1.1 \times 1.1 = 12,100. \] Compound Interest = \(12,100 – 10,000 = 2,100\).

Mixtures

Q6 Solution: Let initial quantity of mixture = \(x\). Milk = \( \frac{3}{4}x \). After replacing 20 liters: \[ \frac{\frac{3}{4}x}{x} = \frac{2}{5} \implies 3x = 2(x + 20) \implies x = 40. \] Milk = \( \frac{3}{4} \times 40 = 30 \).

Averages

Q7 Solution: Sum of 7 integers = \( 7 \times 23 = 161 \). Integers are: \( 20, 21, 22, 23, 24, 25, 26 \). Largest = \(26\).