Home / Aptitude / Percentage / Change of Base or Line Method
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Its an important concept as far as percentage related topics.
Ex. If A is 25% more than B, B is what percent less than A.
Explanation: If B = 100, A = 125
B is 25 less than A
% less = 25 /125 × 100 = 20%.
Students come across many questions on this concept.
NOTE : TO SOLVE THIS QUESTIONS IN SECONDS, STUDENTS CAN WATCH THE VIDEO (LINE METHOD/CHANGE OF BASE).
Q1. |
If Y exceeds X by 14 2/7 %. By what percent X is less than Y? |
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A. 12.5% |
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B. 10% |
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C. 9.09% |
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D. 11.11% |
Apply the line method.
1/7 going down becomes 1/8= 12.5%
Q2. |
During semester break the weight of Raghav increases by 111/9%. What percent should he decrease his weight so as to keep his weight the same as before semester? |
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A. 11.11% |
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B. 10% |
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C. 9.09% |
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D. 12.5% |
Apply the line method.
1/9 going down becomes 1/10 = 10%
Q3. |
The price of cooking oil increases by 10%. By what percent must a family decrease its consumption so as to keep the same expenditure? |
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A. 12.5% |
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B. 11.11% |
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C. 9.09% |
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D. 10% |
Apply the line method.
1/10 going down becomes 1/11 = 9.09%
Q4. |
The marks scored by Kabita is 12.5% less than that of Sabita. By what percent is the mark of Sabita more than that of Kabita? |
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A. 9.09% |
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B. 11.11% |
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C. 12.5% |
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D. 14.28% |
Apply the line method.
1/8 going up becomes 1/7 = 14.28%
Q5. |
The height of Rajiv is 50% more than that of Rakesh. By what percent is the height of Rakesh less than that of Rajiv? |
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A. 33.33% |
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B. 50% |
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C. 20% |
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D. 25% |
Apply the line method.
1/2 going down becomes 1/3 = 33.33%
Q6. |
Ram's salary is 20% more than that of Shyam's. Shyam's salary is what percent less than Ram's? |
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A. 20% |
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B. 25% |
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C. 12.5% |
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D. 16.67% |
Apply the line method.
1/5 going down becomes 1/6 = 16.67%
Q7. |
A is 10% less than B. B is what percent more than A? |
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A. 9.09% |
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B. 10% |
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C. 11.11% |
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D. 12.5% |
Apply the line method.
1/10 going up becomes 1/9 = 11.11%
Q8. |
A is 10% more than B. B is what percent less than A? |
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A. 10% |
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B. 9.09% |
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C. 11.11% |
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D. 12.5% |
Apply Line Method.
1/10 when goes down becomes 1/11= 9.09%
Q1. |
Due to 20% reduction in the price of rice, a man is able to purchase 5 kg more rice for ₹800. What was the original price per kg of rice? |
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A. ₹ 32 |
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B. ₹ 40 |
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C. ₹ 50 |
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D. ₹20 |
1/5 reduction in the price of rice will increase 1/4 increase (Line Method) in the quantity.
1/4 = 5 kg
1 = 20 kgs (Original qtuantity purchased.)
Per kg = ₹800/20 = ₹40 per kg
Q2. |
The area of the base of a cylinder decreases by 60%. By what percent its height must be increased so as to maintain the volume of the cylinder the same as before? |
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A. 37.5% |
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B. 150% |
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C. 62.% |
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D. 50% |
Volume of cylinder = Area of base x height.
Base area decreases by 3/5
To find the increase in height of the cylinder, apply the line method to maintain the same volume.
3/5 going up becomes 3/2= 150%
Q3. |
The commission of a broker decreases from 10% to 8%. By what percent should he increase his business so as to keep same expenditure? |
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A. 20% |
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B. 33.33% |
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C. 50% |
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D. 25% |
The decrease in fraction = 2/10 = 1/5
Apply the line method.
1/5 going up becomes 1/4 = 25%
Q4. |
The price of tea powder per kg increased from ₹ 200 from kg to ₹ 300 per kg. By what percent must the consumption of tea powder be reduced to maintain its expenditure as before? |
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A. 50% |
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B. 33 1 / 3 % |
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C. 20% |
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D. 25% |
Increase = 100/200 = 1/2
Apply the Line method.
1/2 going down becomes 1/3 = 331/3%
Q5. |
The price of diesel increases from ₹ 60 per litre to ₹ 80 per litre. By what percent must a diesel car owner decrease the consumption so as to keep the same expenditure on diesel? |
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A. 50% |
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B. 20% |
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C. 33.33% |
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D. 25% |
Increase in price = ₹80 - ₹ 60 = ₹20
Increase in fraction = 20/60 = 1/3
Apply the line method.
1/3 going down becomes 1/4 =25%
Q1. |
Running at 70 km/hr a car consumes 10% more diesel than when it runs at 50 km/hr. The car runs 10km in 1 litre of diesel while running at 50 km/hr. How many km can the car cover in 50 litres of diesel while running at 70 km/hr? |
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A. 450 km |
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B. 550 km |
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C. 445.54 km |
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D. 454.54 km |
Running at 50 km/hr, the car can cover 50 x 10 = 500 km
1/10 more consumption will lead to 1/11 less distance.
Distance covered running at 70 km/hr = 500 x 10/11 = 454.54 km