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- Question
- 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
- Correct Answer
- 200
ExplanationLet maximum marks be N.
According to the question,
40N/100 = 40 + 40
? 40N/100 = 80
? N = 200 - 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
- 2. The prices of sugar is incerased by 25%. If a family wants to keep its expenses on sugar unaltered, then the family will have to reduce the consumption of sugar by
Options- A. 20%
- B. 21%
- C. 22%
- D. 25% Discuss
- 3. Mangulal a shopkeeper, marks the prices of his goods at 20% higher than the original price. After that he allows a discount of 5%. What profit or loss did he get?
Options- A. 14% (profit)
- B. 14% (loss)
- C. 16% (loss)
- D. 16% (profit) Discuss
- 4. Because of scarcity of rainfall, the price of a land decrease by 12% and its production also decrease by 4%. What is the total effect on revenue?
Options- A. Loss of 16%
- B. Gain of 15%
- C. Loss of 15.48%
- D. Gain of 15.48% Discuss
- 5. In an examination, the marks obtained by Shantanu is 40% less than the marks obtained by Kamal, then marks obtained by Kamal is how much per cent more than the marks obtained by Shantanu?
Options- A. 552/3 %
- B. 443/5 %
- C. 331/3 %
- D. 662/3 % Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. 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
Percentage problems
Search Results
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%
Correct Answer: 20%
Explanation:
Let original price be ? 100
Then, increased price = ? 125
? Reduction in consumption
= [(125 - 100)/125 ] x 100%
= (25/125) x 100 %
= 20%
Correct Answer: 14% (profit)
Explanation:
Given that, a = 20% and b = 5%
According to the formula
Required percentage = [20 - 5 - {20 x (5/100)}% = 14%
Correct Answer:
Explanation:
Net effect = [- 12 - 4 + {(-12) (-4)/100}]%
= (-16 + 0.48)%
= -15.52%
Correct Answer: 662/3 %
Explanation:
Let kamal's marks be 100
? Shantanu's marks = 60
Kamal has 40 marks more than Shantanu
? Required percentage = (40/60) x 100%
= 662/3 %
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
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
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
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
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
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