Percentages 2/1

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1. A's salary is 20% more than B's, B's salary is 10% less than C's. If A's salary is Rs. 1080, find C's salary.
a. 1000
b. 1500
c. 1800
d. 2500

Answer: A

Explanation:
If B's salary is Rs. 100, then A's salary = Rs. 100 + Rs. 20 = Rs. 120
Therefore, B's salary = Rs. 1080 × $\displaystyle\frac{{{\rm{100}}}}{{{\rm{120}}}}$
If C's salary is Rs. 100, then B's salary = Rs. 100 - Rs. 10 = Rs. 90.
Therefore, C's salary = Rs. 1080 × $\displaystyle\frac{{{\rm{100}}}}{{{\rm{120}}}}{\rm{ \times }}\frac{{{\rm{100}}}}{{{\rm{90}}}}$ = Rs. 1000

2. In an examination, 50% students failed in English and 40% in Math and 15% students failed in both subjects. If 200 students passed in both the subjects, find the number of students appeared in the exam.
a. 500
b. 600
c. 700
d. 800

Answer: D

Explanation:
Students failed in English only = (50 - 15)% = 35%
Students failed in Math only = (40 - 15)% = 25%
Students failed in both subjects = 15%
Therefore, Students failed in either or both subjects = 35 + 25 + 15 = 75%
Therefore, Students passed in both subjects = (100 - 75)% = 25%
But students passed = 200 (i.e. 8 times of 25)
Therefore, Students appeared = 8 × 100 = 800

3. The price of petrol increased by 25% and so a person reduced his consumption by 25%. What percentage is the rise or fall in the expenditure incurred by him on petrol ?
a. No change
b. 5% rise
c. 6.25% rise
d. 6.25% reduction

Answer: D

Explanation:
Use above formula $\left( {25 - 25 - \displaystyle\frac{{25 \times 25}}{{100}}} \right)\% \Rightarrow - \displaystyle\frac{{625}}{{100}}\% \Rightarrow - 6.25\% $
So reduction of 6.25% in his expenditure.

Alternate Method:
Assume he consumes 100 liters of petrol,and petrol costs him Rs.100/-
His total expenditure = Rs.10000/-
Now price got increased by 25%. If a value got increased by 25% then the new value is 125% of the previous value. so 125%(100) = Rs.125
But he reduced his consumption by 25%, so his new consumption is 75%(100) = 75
His total expenditure = 125 x 75 = Rs.9375
Percentage change in the expenditures = $\displaystyle\frac{{10000 - 9375}}{{10000}} \times 100 = 6.25\% $

4. Radius of the base of a right circular cylinder is increased by 10% and height is decreased by 10%. What is the percentage increase or decrease in
I. Area of the base ?
II. Volume of a right circular cylinder ?
a. 10% rise , 8.9% reduction
b. 20% rise , No change
c. 21% rise , 8.9% reduction
d. 10% rise , No change

Answer: C

Explanation:
I. As the base of the cylider is a two dimensional figure so if radius got increased by 10% then the change in the base of area = $\left( {10 + 10 + \displaystyle\frac{{10 \times 10}}{{100}}} \right)\% \Rightarrow 21\% $
II. To calculate the change in the volume we can substitute A = 21 and B = -10, as the height got decreased by 10%
$\left( {21 - 10 - \displaystyle\frac{{21 \times 10}}{{100}}} \right)\% \Rightarrow 8.9\% $

5. A man bought some apples of which 13% of then were rotten. He sold 75% of the balance and was left with 261 apples. How many apples did he have originally?
a. Rs.1000
b. Rs.1200
c. Rs.1500
d. Rs.2000

Answer: B

Explanation:
Look at the diagrammatic representation of the problem

Now take initial number of apples x.
Then $x \times (100 - 13)\% \times (100 - 75)\% = 261$
$ \Rightarrow x = \displaystyle\frac{{261}}{{(87)\% \times (25)\% }} \Rightarrow \frac{{261 \times 100 \times 100}}{{87 \times 25}} \Rightarrow 1200$

6. In printing test papers for Excel, Gayatri found that if she used Arial font size 10 instead of Times Roman font size 10 there was a reduction of 18% in the number of pages required for test papers. Further, if she reduced the font size from 10 to 9, the savings were 14% and 10% in Times Roman and Arial fonts respectively. If an test papers printing in Times Roman font size 9 is converted to Arial font size 9, what is the percentage reduction in the number of pages?
a. 12.7
b. 14.1
c. 16.8
d. 17.5

Answer: B

Explanation:

From the above diagram it is clear that from TR-10 to A - 10 there is a reduction of 18% in the number of pages. and From TR-10 to TR-9 14% reduction, A - 10 to A - 9 10% reduction.
Assume Test paper in TR- 10 is 100 pages. So in TR- 9 the number of pages are 100 × 86% = 86
Now in A-10 total pages are 100 × 82% = 82
Now in A-9 total pages are 82 × 90% = 73.8
So total change in the number of pages = $\displaystyle\frac{{\left( {86 - 73.8} \right)}}{{86}} \times 100$ = 14.1

7. The total weight of six people of equal weights inside a car is 12% of the total weight of the car (inclusive of the six people inside). If three people get down from the car, what will be the ratio of the weight of people inside the car to the weight of the empty car?
a. 3 : 44
b. 5 : 44
c. 7 : 44
d. 3 : 53

Answer: A

Explanation:
Assume total weight of the car along with people is 100kgs.
Then Total weight of 6 people is 12% of 100 = 12 kgs. So empty car weight = 88 kgs
That is, the weight of a single person = 12/6 = 2 kgs.
If three people get down from the car, then there is a six kgs reduction in the total weight and the weight of the remaining people is only 6 kgs.
The required ratio = 6 : 88 = 3 : 44

8. Fresh grapes contains 90% water, and dry grapes contains 85% matter. To get 25 kgs of dry grapes who many kgs of fresh grapes have to be processed?
a. 200
b. 212.5
c. 225
d. 250

Answer: A

Explanation:
To solve problems which involves 2 variables, concentrate on the variable that won't change. In this problem we concentrate on the variable matter rather than water as it changes.
Let us take K kgs of fresh grapes. Now total mater in these grapes = K × 10%
In 25 kgs dry grapes total mater = 25 × 85%
We know that these two equation are equal so
K × 10% = 25 × 85%
Solving K = 212.5

Formulas Percentages Exercise - 1 Percentages Exercise - 2