Percentage problems
- 1. If the prices of tea falls down by 6% by how much per cent must a householder increased its consumption, so as not to decreased expenditure?
Options- A. 516/47%
- B. 418/67%
- C. 618/47%
- D. 617/47% Discuss
Correct Answer: 618/47%
Explanation:
Let original price be ? 100
Then, reduced price = ? 94
? Increase in consumption
= [(100 - 94)/94] x 100 %
= (6/94) x 100 %
= 618/47%
- 2. A student has to score 40% marks to get through. if he gets 40 marks and fails by 40 marks, then find the maximum marks set for the examination.?
Options- A. 200
- B. 250
- C. 300
- D. 150 Discuss
Correct Answer: 200
Explanation:
Let maximum marks be N.
According to the question,
40N/100 = 40 + 40
? 40N/100 = 80
? N = 200
- 3. A candidate scores 20% and fails by 50 marks, while another candidate who score 40% marks, gests 30 marks more than the minimun required marks to pass the examination. Find the maximum marks for the examination.
Options- A. 500
- B. 450
- C. 300
- D. 400 Discuss
Correct Answer: 400
Explanation:
Let the maximum marks be N.
According to the question,
(20N/100) + 50 = (40N/100) - 30
? 20N/100 = 80
? N = (80 x 100)/20 = 400
- 4. The population of a town is 126800. It increase by 15% in the 1st year and decrease by 20% in the 2nd year. What is the population of the town at the end of 2 yr.
Options- A. 174984
- B. 135996
- C. 116656
- D. 145820 Discuss
Correct Answer: 116656
Explanation:
Given, R = 15% and R2 = 20%
Required population
= P(1 + R1/100) (1 - R2/100)
= 126800(1 + 15/100) (1- 20/100)
= 126800 (1 + 3/20 ) (1- 1/5)
= 126800 x (23/20) x (4/5) = 116656
- 5. The population of a town is 705600. If it increases at the rate of 5% per annum, then what will be its population 2yr hence?
Options- A. 777924
- B. 777881
- C. 778781
- D. 797724 Discuss
Correct Answer: 777924
Explanation:
Given that,
P = 705600, R = 5% and n = 2
According to the formula,
Population after n yr = p(1 + R/100)n
? Population after 2 yr.
= 705600 x (1+ 5/100)2
= 705600 x (105/100)2
= 705600 x (21/20 x 21/20) = 777924
- 6. The population of a town is 1058400. if it increase at the rate of 5% per annum, then find the population of the town 2 yr ago.
Options- A. 949000
- B. 930000
- C. 960000
- D. 950000 Discuss
Correct Answer: 960000
Explanation:
Given that,
p = 1058400, R = 5% and n = 2
According the formula,
Population n yr ago = P/(1 + R/100) n
? Population 2yr ago = 1058400/(1+ 5/100)2
= 1058400 x 20/21 x 20/21 = 960000
- 7. The population of a city is 250000. It is increasing at the rate of 2% every year. The growth in the population after 2 yr is?
Options- A. 2500
- B. 10000
- C. 252000
- D. 10100 Discuss
Correct Answer: 10100
Explanation:
Population after 2 yr
= P (1 + R/100)2
= 250000 (1 + 2/100)2
? 250000 x (51/50) x (51/50) = 260100
? Growth = 260100 - 250000 = 10100
- 8. In a school, 10% of boys are equal to the one - fourth of the girls. What is the ration of boys and girls in that school?
Options- A. 3 : 2
- B. 5 : 2
- C. 2 : 1
- D. 4 : 3 Discuss
Correct Answer: 5 : 2
Explanation:
Let the number of boys = B
and number of girls = G
Then, 10% of B = 1/4 of G
? B/10 = G/4
? B/G = 10/4 = 5/2
? B : G = 5 : 2
- 9. Income of Suman is first increased by 7% and then it is decreased by 7%. What is the change in her income?
Options- A. 0.49% (increase)
- B. 0.39% (decrease)
- C. 0.39% (increase)
- D. 0.49% (decrease) Discuss
Correct Answer: 0.49% (decrease)
Explanation:
In such case, there is always decrease.
Given that, a = (common increase or decreased) = 7%
According to the formula,
Decreased Percentage = a2/100 %
= 72/100 %
= 0.49%
- 10. 48% of the 1st number is 60% of the 2nd number. What is the ratio of the 1st number to the 2 nd number?
Options- A. 4 : 7
- B. 3 : 4
- C. 5 : 4
- D. Could't be determined Discuss
Correct Answer: 5 : 4
Explanation:
Let 1st number be M and 2nd number be M.
According to the question,
48% of M = 60% N
? (M x 48)/100 = (N x 60)/100
? M/N = (60/100) x (100/48) = 5/4
? M : N = 5 : 4
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