Percentage problems
- 1. The value of a machine depreciates 10% annually. If its present value is Rs. 4000. Its value 2 years later will be?
Options- A. Rs. 3200
- B. Rs. 3240
- C. Rs. 3260
- D. Rs. 3280 Discuss
Correct Answer: Rs. 3240
Explanation:
Value of machine 2 years later = Rs. [ 4000 x {(1 - 10)/10)}2]
= Rs. (4000 x 9/10 x 9/10)
= Rs. 3240
- 2. The population of a town increase 4% annually but is decreases by emigration annually to the extent of 1/2%. What will be the increase percent in three years?
Options- A. 9.8
- B. 10
- C. 10.5
- D. 10.8 Discuss
Correct Answer: 10.8
Explanation:
Increase in 3 years over 100 = 100 x {1 + 7/(2 x 100)}3
= (100 x 207/200 x 207/200 x 207/200)
= (200 + 7)3 / 80000
= 88869743 / 80000 = 110.8718
Required increase % = (110.8 - 100)% = 10.8%
- 3. A man's wages were decreased by 50%. Again the reduced wages were increased by 50%. He has a loss of?
Options- A. 0 %
- B. 0.25 %
- C. 2.5 %
- D. 25 % Discuss
Correct Answer: 25 %
Explanation:
Let original wages = Rs. 100
Reduced wages = Rs. 50
Increased wages = 150% of Rs. 50 = (150 x 50)/100 = Rs. 75
? Loss = 25%
- 4. P is six times as large as q. The per cent that q is less than p,is?
Options- A. 831/3
- B. 162/3
- C. 90
- D. 60 Discuss
Correct Answer: 831/3
Explanation:
P = 6q. Thus q is less than p by 5q
? q is less than p by = {(6q - q)/6q} x 100%
= ( 5q/6q) x 100 % = 831/3 %
- 5. If A's salary is 30% more than B's then how much per cent is B's salary less than A's?
Options- A. 30%
- B. 25%
- C. 231/13
- D. 331/3 Discuss
Correct Answer: 231/13
Explanation:
B's salary is less than A's by (30 x 100)/130 % = 231/13%
- 6. A man spends 75% of his income. This income is increased by 20% and he increased his expenditure by 10%. His savings are increased by?
Options- A. 10%
- B. 25%
- C. 371/2%
- D. 50% Discuss
Correct Answer: 50%
Explanation:
Let income = Rs. 100
Then expenditure = Rs. 75 and saving = Rs. 25
New expenditure = 110% of Rs. 75 = Rs. 165 / 2
New saving = Rs. (120 - 165/2) = Rs. 75/2
Increase in saving = Rs. (75/2 - 25) = Rs. 25/2
? Increase% = (25/2) x (1/25) x 100% = 50%
- 7. The price of an article has been reduced by 25%. In order to restore the original price, the new price must be increased by?
Options- A. 331/3%
- B. 111/9%
- C. 91/11%
- D. 662/3% Discuss
Correct Answer: 331/3%
Explanation:
Let original price = Rs. 100
Reduced price = Rs. 75
Increase on Rs. 75 = Rs. 25
Increase on Rs. 100 = (25 x 100)/75 % = 331/3 %
- 8. If the numerator of a fraction is increased by 20% and he denominator be diminished by 10%, the value of the fraction is 16/21. The original fraction is?
Options- A. 3/5
- B. 4/7
- C. 2/3
- D. 5/7 Discuss
Correct Answer: 4/7
Explanation:
Let the original fraction be p/q
Now, (120% of p) / (90% of q) = 16 / 21
? 4p / 3q = 16/21
? p / q = (16/21) x (3/4) = 4 / 7
- 9. A mixture of 40 litres of milk and water contains 10% water. How much water should be added to this so that water may be 20% in the new mixture?
Options- A. 4 litres
- B. 5 litres
- C. 6.5 litres
- D. 7.5 litres Discuss
Correct Answer: 5 litres
Explanation:
? Milk = 90% 0f 40 = 36 litres and water = 4 litres.
Let L litre water to be added in order to make 20% water in new mixture.
? {(4 + L) / (40 + L)} x 100 = 20
? 20 x (40 + L) = 100 x (4 + L)
? 80 x L = 400
? L = 5 litres .
- 10. The boys and girls in a collage are in ratio 3 : 2. If 20% of the boys and 25% of the girls are adults, the percentage of students who are not adults is?
Options- A. 58%
- B. 67.5%
- C. 78%
- D. 82.5% Discuss
Correct Answer: 78%
Explanation:
Suppose boys = 3N and girls = 2N
Not adults = (80% of 3N) + (75% of 2N) = (12N/5 + 3N/2) = 39N/10
? Required percentage = (39N/10 / 1/5N) x 100 % = 78%
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