Competitive Exams : Quantitative Aptitude - Percentage (Set 4) | Concept of Maths

Question 1

1. In a multi-stage manufacturing process, the raw material weight decreases by 12.5% during refining, then increases by 20% during chemical treatment, and finally loses 10% of its weight during packaging. If the final weight of the packaged product is 151.2 kg, what was the initial weight of the raw material?

A) 150 kg

B) 155 kg

C) 160 kg

D) 165 kg

Answer

Correct Answer: C

Let initial weight be x. Refining: $x \\times (1 - 0.125) = 0.875x$ Chemical Treatment: $0.875x \\times 1.20 = 1.05x$ Packaging: $1.05x \\times 0.90 = 0.945x$ Given: $0.945x = 151.2 \\Rightarrow x = \\frac{151.2}{0.945} = 160$ kg.

Question 2

2. In an election between three candidates P, Q, and R, 15% of voters did not cast their votes. Out of the votes cast, 10% were declared invalid. Candidate P received 40% of the valid votes, while Candidate Q received 30% of the valid votes. If Candidate R received 27,540 votes, find the total number of voters on the electoral roll.

A) 1,00,000

B) 1,20,000

C) 1,15,000

D) 1,25,000

Answer

Correct Answer: B

Let total voters = V. Votes cast = 0.85V. Valid votes = $0.90 \\times 0.85V = 0.765V$. Candidate R's share = 100\% - (40\% + 30\%) = 30\% of valid votes. $0.30 \\times 0.765V = 27,540$ $0.2295V = 27,540 \\Rightarrow V = \\frac{27540}{0.2295} = 1,20,000$.

Question 3

3. The price of a rare earth metal increases by 25\%. By what percentage must a technologist reduce the consumption so that the total expenditure on the metal increases by only 5\%?

A) 12\%

B) 15\%

C) 18\%

Question 4

4. A mining company extracts ore containing 20\% copper. During the smelting process, 15\% of the copper is lost. If the company wants to produce 6,800 kg of pure copper, how much ore must be mined?

A) 40,000 kg

B) 34,000 kg

C) 42,500 kg

D) 38,000 kg

Answer

Correct Answer: A

Let Ore mined = x kg. Copper in ore = 0.20x. Copper after loss = $0.20x \\times (1 - 0.15) = 0.20x \\times 0.85 = 0.17x$. $0.17x = 6,800 \\Rightarrow x = \\frac{6800}{0.17} = 40,000$ kg.

Question 5

5. In a library, 30\% of the books are in English, 50\% of the remaining are in Hindi and the rest 1,400 are in Regional languages. How many books are in English?

A) 1,000

B) 1,500

C) 1,200

D) 800

Answer

Correct Answer: C

Let total books = T. English = 0.3T. Remaining = 0.7T. Hindi = $0.5 \\times 0.7T = 0.35T$. Regional = T - (0.3T + 0.35T) = 0.35T. 0.35T = 1,400 \\Rightarrow T = 4,000. English books = $0.3 \\times 4,000 = 1,200$.

Question 6

6. If the radius of a sphere is increased by 10\%, then what is the percentage increase in its volume?

A) 30\%

B) 33.1\%

C) 21\%

D) 35.2\%

Answer

Correct Answer: B

Volume $V \\propto r^3$. New radius r' = 1.1r. New Volume $V' \\propto (1.1r)^3 = 1.331r^3$. Percentage increase = $(1.331 - 1) \\times 100 = 33.1\%$.

Question 7

7. Mr. Sahay spends 20\% of his monthly income on household expenditure. Out of the remaining, he spends 25\% on children's education, 15\% on transport, and 10\% on entertainment. If he is left with Rs. 20,000, find his monthly income.

A) Rs. 40,000

B) Rs. 45,000

C) Rs. 60,000

D) Rs. 50,000

Answer

Correct Answer: D

Let income = I. Household = 0.20I. Remaining = 0.80I. Sum of percentages of remaining spent = 25\% + 15\% + 10\% = 50\%. Leftover = 50\% of 0.80I = 0.40I. 0.40I = 20,000 \\Rightarrow I = 50,000.

Question 8

8. A solution of salt and water contains 15\% salt by weight. Of this solution, 30 kg of water is evaporated and the concentration of salt becomes 20\%. Find the original weight of the solution.

A) 100 kg

B) 120 kg

C) 150 kg

D) 90 kg

Answer

Correct Answer: B

Let original weight = x. Salt = 0.15x. After evaporation, weight = x - 30. Salt = 0.20(x - 30). Since salt amount is constant: 0.15x = 0.20(x - 30) 0.15x = 0.20x - 6 \\Rightarrow 0.05x = 6 \\Rightarrow x = 120 kg.

Question 9

9. The population of a town increases by 10\% in the first year and decreases by 10\% in the second year. If the population at the end of the second year is 9,900, what was the population at the beginning of the first year?

A) 10,000

B) 9,800

C) 10,100

D) 11,000

Answer

Correct Answer: A

Let initial population = P. End of Year 1: 1.1P. End of Year 2: $1.1P \\times 0.9 = 0.99P$. 0.99P = 9,900 \\Rightarrow P = 10,000.

Question 10

10. Fresh fruit contains 68\% water and dry fruit contains 20\% water. How much dry fruit can be obtained from 100 kg of fresh fruits?

A) 32 kg

B) 50 kg

C) 40 kg

D) 48 kg

Answer

Correct Answer: C

Fruit pulp remains constant. Pulp in 100 kg fresh fruit = (100 - 68)\% of 100 = 32 kg. Let D be weight of dry fruit. Pulp in dry fruit = (100 - 20)\% of D = 0.8D. 0.8D = 32 \\Rightarrow D = 40 kg.

Question 11

11. In an exam, A got 10\% more marks than B. B got 20\% more marks than C. If A got 132 marks, how many marks did C get?

A) 90

B) 100

C) 110

D) 120

Answer

Correct Answer: B

Let C = x. B = 1.2x. $A = 1.1 \\times 1.2x = 1.32x$. 1.32x = 132 \\Rightarrow x = 100.

Question 12

12. If x\% of y is equal to z, then what percentage of z is x?

A) $\\frac{y^2}{100}$

B) $\\frac{x^2}{100}$

C) $\\frac{100^2}{y}$

D) $\\frac{10000}{y}$

Answer

Correct Answer: D

Given $\\frac{x}{100} \\times y = z$. We need to find k such that $\\frac{k}{100} \\times z = x$. $k = \\frac{100x}{z}$. Substitute $z = \\frac{xy}{100}$: $k = \\frac{100x}{xy/100} = \\frac{10000x}{xy} = \\frac{10000}{y}$.

Question 13

13. A worker's salary was increased by 40\% and then subsequently decreased by 25\%. What is the net percentage change in his salary?

A) 15\% increase

B) 10\% decrease

C) 5\% increase

D) No change

Answer

Correct Answer: C

Net change formula: $a + b + \\frac{ab}{100}$. a = 40, b = -25. $40 - 25 + \\frac{40 \\times (-25)}{100} = 15 - 10 = 5\%$. Positive sign indicates increase.

Question 14

14. In a class of 60 students, 40\% are girls. If 25\% of the girls and 10\% of the boys are absent on a particular day, what is the percentage of students present?

A) 84\%

B) 80\%

C) 82\%

D) 86\%

Answer

Correct Answer: A

Total = 60. Girls = $0.4 \\times 60 = 24$. Boys = 36. Absent Girls = $0.25 \\times 24 = 6$. Absent Boys = $0.10 \\times 36 = 3.6$. Total Absent = 9.6. Total Present = 60 - 9.6 = 50.4. Percentage Present = $\\frac{50.4}{60} \\times 100 = 84\%$.

Question 15

15. A salesman receives a basic salary of Rs. 10,000 per month plus a commission of 5\% on sales exceeding Rs. 50,000. If he earns Rs. 14,500 in a month, what were his total sales?

A) Rs. 1,20,000

B) Rs. 1,40,000

C) Rs. 1,50,000

D) Rs. 90,000

Answer

Correct Answer: B

Commission earned = 14,500 - 10,000 = 4,500. Let sales exceeding 50k be $S_{ex}$. $0.05 \\times S_{ex} = 4,500 \\Rightarrow S_{ex} = 90,000$. Total Sales = 50,000 + 90,000 = 1,40,000.

Question 16

16. If the length of a rectangle is increased by 20\% and the breadth is decreased by 20\%, what is the percentage change in the area?

A) No change

B) 2\% decrease

C) 4\% decrease

D) 4\% increase

Answer

Correct Answer: C

Successive change: $a + b + \\frac{ab}{100}$. $20 - 20 + \\frac{20 \\times (-20)}{100} = -4\%$. A 4\% decrease.

Question 17

17. A candidate scoring 25\% marks in an examination fails by 30 marks, while another candidate scoring 50\% marks gets 20 marks more than the minimum passing marks. Find the total marks of the exam.

A) 150

B) 180

C) 250

D) 200

Answer

Correct Answer: D

Let total marks be M. Pass Marks = 0.25M + 30 Pass Marks = 0.50M - 20 0.25M + 30 = 0.50M - 20 0.25M = 50 \\Rightarrow M = 200.

Question 18

18. If A is 25\% more than B, then B is how much percent less than A?

A) 20\%

B) 25\%

C) 16.67\%

D) 30\%

Answer

Correct Answer: A

A = 1.25B. Difference = A - B = 0.25B. Percentage less = $\\frac{0.25B}{A} \\times 100 = \\frac{0.25B}{1.25B} \\times 100 = \\frac{1}{5} \\times 100 = 20\%$.

Question 19

19. Due to a 20\% reduction in the price of sugar, a man can buy 5 kg more sugar for Rs. 600. What was the original price per kg?

A) Rs. 24

B) Rs. 30

C) Rs. 25

D) Rs. 32

Answer

Correct Answer: B

Let original price be P. Original quantity $Q = \\frac{600}{P}$. New price 0.8P. New quantity $Q' = \\frac{600}{0.8P}$. Q' - Q = 5 $\\frac{600}{0.8P} - \\frac{600}{P} = 5 \\Rightarrow \\frac{750}{P} - \\frac{600}{P} = 5$ $\\frac{150}{P} = 5 \\Rightarrow P = 30$.

Question 20

20. In an office, 70\% of employees like coffee, 40\% like tea, and each employee likes at least one of the two. If 15 employees like both coffee and tea, what is the total number of employees in the office?

A) 100

B) 120

C) 150

D) 200

Answer

Correct Answer: C

Using $n(A \\cup B) = n(A) + n(B) - n(A \\cap B)$. 100\% = 70\% + 40\% - \Both\%. \Both\% = 110\% - 100\% = 10\%. 10\% of Total = 15. Total = 150.

Accepted Answer

Correct Answer: D

Let Price = P, Consumption = C, Expenditure $E = P \\times C$. New Price P' = 1.25P, New Expenditure E' = 1.05E. $1.05E = 1.25P \\times C' \\Rightarrow 1.05(PC) = 1.25P \\times C'$ $C' = \\frac{1.05}{1.25}C = 0.84C$. Reduction = $(1 - 0.84) \\times 100 = 16\%$.