Question 1

1. Find the unit digit in the product $(2467)^{153} \times (341)^{72}$.

A) 1B) 3C) 7D) 9

Answer

✓ Correct Answer: Option C

Explanation: Unit digit of $7^{153}$: Cyclicity of 7 is 4. $153 \pmod 4 = 1$. So, $7^1 = 7$. Unit digit of $1^{72}$ is always 1. Product: $7 \times 1 = 7$.

Question 2

2. What is the remainder when $3^{21}$ is divided by 5?

A) 2B) 3C) 4D) 1

Question 3

3. The difference between the local value (place value) and the face value of 7 in the numeral 657321 is:

A) 6993B) 7000C) 7D) 6930

Answer

✓ Correct Answer: Option A

Explanation: Place value of 7 in 657321 is $7 \times 1000 = 7000$. Face value of 7 is 7. Difference: 7000 - 7 = 6993.

Question 4

4. Find the largest 4-digit number which is exactly divisible by 15, 25, 40 and 75.

A) 9000B) 9400C) 9800D) 9600

Answer

✓ Correct Answer: Option D

Explanation: LCM of 15, 25, 40, 75: $15 = 3 \times 5$, $25 = 5^2$, $40 = 2^3 \times 5$, $75 = 3 \times 5^2$. LCM = $2^3 \times 3 \times 5^2 = 8 \times 3 \times 25 = 600$. Largest 4-digit number = 9999. $9999 \div 600$ gives remainder 399. Number = 9999 - 399 = 9600.

Question 5

5. The LCM of $\frac{2}{3}, \frac{4}{9}, \frac{5}{6}$ is:

A) $\frac{8}{27}$B) $\frac{20}{3}$C) $\frac{10}{3}$D) $\frac{20}{9}$

Answer

✓ Correct Answer: Option B

Explanation: LCM of fractions = $\frac{LCM of Numerators}{HCF of Denominators}$. Numerators: 2, 4, 5. LCM = 20. Denominators: 3, 9, 6. HCF = 3. Result: $\frac{20}{3}$.

Question 6

6. The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other.

A) 88B) 121C) 99D) 66

Answer

✓ Correct Answer: Option C

Explanation: Formula: $Product of numbers = HCF \times LCM$. $77 \times x = 11 \times 693$. $x = \frac{11 \times 693}{77} = \frac{693}{7} = 99$.

Question 7

7. Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together?

A) 3 hoursB) 180 hoursC) 60 minutesD) 1.5 hours

Answer

✓ Correct Answer: Option A

Explanation: Required time = LCM of 9, 12, and 15. $9 = 3^2, 12 = 2^2 \times 3, 15 = 3 \times 5$. LCM = $2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$ minutes. $180 minutes = \frac{180}{60} = 3$ hours.

Question 8

8. Two numbers are in the ratio 3 : 4. If their HCF is 4, their LCM is:

A) 12B) 16C) 24D) 48

Answer

✓ Correct Answer: Option D

Explanation: Let the numbers be 3x and 4x. HCF of 3x and 4x is x. Given x = 4. The numbers are 3(4)=12 and 4(4)=16. LCM of 12 and 16 is 48. (Or simply LCM = $3 \times 4 \times HCF = 12 \times 4 = 48$).

Question 9

9. Simplify: $18 \div 3 \times (5 - 2) + 4^2$.

A) 22B) 34C) 24D) 40

Answer

✓ Correct Answer: Option B

Explanation: Using BODMAS: 1. Bracket: (5 - 2) = 3. 2. Orders (Square): $4^2 = 16$. 3. Division: $18 \div 3 = 6$. 4. Multiplication: $6 \times 3 = 18$. 5. Addition: 18 + 16 = 34.

Question 10

10. The value of $\sqrt{42 + \sqrt{42 + \sqrt{42 + \dots \infty}}}$ is:

A) 6B) 8C) 7D) 42

Answer

✓ Correct Answer: Option C

Explanation: Let $x = \sqrt{42 + x}$. Squaring: $x^2 = 42 + x ⇒ x^2 - x - 42 = 0$. (x - 7)(x + 6) = 0. Since x must be positive, x = 7. Shortcut: $42 = 6 \times 7$ (consecutive factors). For '+', the larger factor (7) is the answer.

Question 11

11. Find the value of $\frac{0.05 \times 0.05 \times 0.05 - 0.04 \times 0.04 \times 0.04}{0.05 \times 0.05 + 0.05 \times 0.04 + 0.04 \times 0.04}$.

A) 0.01B) 0.09C) 0.001D) 1

Answer

✓ Correct Answer: Option A

Explanation: This is in the form $\frac{a^3 - b^3}{a^2 + ab + b^2}$, where a = 0.05 and b = 0.04. We know $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. So the expression simplifies to a - b = 0.05 - 0.04 = 0.01.

Question 12

12. Find the value of $(1 - \frac{1}{3})(1 - \frac{1}{4})(1 - \frac{1}{5}) \dots (1 - \frac{1}{n})$.

A) $\frac{1}{n}$B) $\frac{2(n-1)}{n}$C) $\frac{2}{n(n+1)}$D) $\frac{2}{n}$

Answer

✓ Correct Answer: Option D

Explanation: Expression: $\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \dots \times \frac{n-1}{n}$. Observe the cancellation: the denominator of one term cancels the numerator of the next. Only the first numerator (2) and the last denominator (n) remain. Result: $\frac{2}{n}$.

Question 13

13. If a salary is first increased by 10% and then decreased by 10%, what is the net change?

A) No changeB) 1% decreaseC) 1% increaseD) 2% decrease

Answer

✓ Correct Answer: Option B

Explanation: Using the successive formula: $x + y + \frac{xy}{100}$. $10 + (-10) + \frac{10 \times (-10)}{100} = 0 - 1 = -1%$. A negative result indicates a 1% decrease.

Question 14

14. If 40% of (A + B) = 60% of (A - B), then $\frac{2A - 3B}{A + B}$ is:

A) $\frac{7}{6}$B) $\frac{6}{7}$C) $\frac{7}{12}$D) $\frac{5}{12}$

Answer

✓ Correct Answer: Option A

Explanation: 0.4(A+B) = 0.6(A-B) ⇒ 2(A+B) = 3(A-B) 2A + 2B = 3A - 3B ⇒ A = 5B. Substitute A = 5B into the expression: $\frac{2(5B) - 3B}{5B + B} = \frac{10B - 3B}{6B} = \frac{7B}{6B} = \frac{7}{6}$.

Question 15

15. A candidate who gets 30% marks fails by 50 marks. Another candidate who gets 45% marks gets 25 marks more than the minimum passing marks. Find the maximum marks.

A) 400B) 500C) 600D) 800

Answer

✓ Correct Answer: Option B

Explanation: Difference in percentage = 45% - 30% = 15%. Difference in marks = 25 - (-50) = 75. 15% of Total = 75. Total Marks = $\frac{75}{15} \times 100 = 5 \times 100 = 500$.

Question 16

16. The population of a town is 10,000. It increases by 10% in the 1st year and 20% in the 2nd year. What is the population after 2 years?

A) 13000B) 13100C) 13200D) 12000

Answer

✓ Correct Answer: Option C

Explanation: Population after 1st year = 10000 + 10% = 11000. Population after 2nd year = 11000 + 20% of 11000 = 11000 + 2200 = 13200. Alternatively: $10000 \times 1.1 \times 1.2 = 13200$.

Question 17

17. If the cost price of 15 articles is equal to the selling price of 12 articles, the gain percentage is:

A) 25%B) 20%C) 18%D) 30%

Answer

✓ Correct Answer: Option A

Explanation: Let CP of 1 article = x. CP of 15 articles = 15x. SP of 12 articles = 15x ⇒ SP of 1 article = $\frac{15x}{12} = 1.25x$. Gain per article = 1.25x - x = 0.25x. Gain% = $\frac{0.25x}{x} \times 100 = 25%$.

Question 18

18. A man buys a cycle for ₹450 and spends ₹50 on its repair. If he sells the cycle for ₹600, what is his gain percentage?

A) 10%B) 15%C) 25%D) 20%

Answer

✓ Correct Answer: Option D

Explanation: Total Cost Price (CP) = 450 + 50 = ₹500. Selling Price (SP) = ₹600. Profit = 600 - 500 = ₹100. Profit% = $\frac{100}{500} \times 100 = 20%$.

Question 19

19. Two items are sold for ₹990 each. On one, the merchant gains 10% and on the other, he loses 10%. What is the overall profit or loss?

A) No profit no lossB) 1% lossC) 1% gainD) ₹10 loss

Answer

✓ Correct Answer: Option B

Explanation: When two items are sold at the same price, one at a gain of x% and other at a loss of x%, there is always a loss. Loss% = $\frac{x^2}{100} = \frac{10^2}{100} = 1%$.

Question 20

20. A dishonest dealer professes to sell his goods at cost price but uses a weight of 900g for a kg weight. His gain percent is:

A) 10%B) 11%C) $11\frac{1}{9}%$D) 9%

Answer

✓ Correct Answer: Option C

Explanation: Gain% = $\frac{Error}{True Value - Error} \times 100$ Error = 1000g - 900g = 100g. Gain% = $\frac{100}{900} \times 100 = \frac{100}{9} = 11\frac{1}{9}%$.

Quantitative Aptitude Mix Quiz Set 4

1. Find the unit digit in the product $(2467)^{153} \times (341)^{72}$.

2. What is the remainder when $3^{21}$ is divided by 5?

3. The difference between the local value (place value) and the face value of 7 in the numeral 657321 is:

4. Find the largest 4-digit number which is exactly divisible by 15, 25, 40 and 75.

5. The LCM of $\frac{2}{3}, \frac{4}{9}, \frac{5}{6}$ is:

6. The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other.

7. Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together?

8. Two numbers are in the ratio 3 : 4. If their HCF is 4, their LCM is:

9. Simplify: $18 \div 3 \times (5 - 2) + 4^2$.

10. The value of $\sqrt{42 + \sqrt{42 + \sqrt{42 + \dots \infty}}}$ is:

11. Find the value of $\frac{0.05 \times 0.05 \times 0.05 - 0.04 \times 0.04 \times 0.04}{0.05 \times 0.05 + 0.05 \times 0.04 + 0.04 \times 0.04}$.

12. Find the value of $(1 - \frac{1}{3})(1 - \frac{1}{4})(1 - \frac{1}{5}) \dots (1 - \frac{1}{n})$.

13. If a salary is first increased by 10% and then decreased by 10%, what is the net change?

14. If 40% of (A + B) = 60% of (A - B), then $\frac{2A - 3B}{A + B}$ is:

15. A candidate who gets 30% marks fails by 50 marks. Another candidate who gets 45% marks gets 25 marks more than the minimum passing marks. Find the maximum marks.

16. The population of a town is 10,000. It increases by 10% in the 1st year and 20% in the 2nd year. What is the population after 2 years?

17. If the cost price of 15 articles is equal to the selling price of 12 articles, the gain percentage is:

18. A man buys a cycle for ₹450 and spends ₹50 on its repair. If he sells the cycle for ₹600, what is his gain percentage?

19. Two items are sold for ₹990 each. On one, the merchant gains 10% and on the other, he loses 10%. What is the overall profit or loss?

20. A dishonest dealer professes to sell his goods at cost price but uses a weight of 900g for a kg weight. His gain percent is: