Percentage - Practice Ex.

1. In the first test of the semester, kiran scored 60. In the last test of the semester, kiran scored 75%. By what percent did kiran’s score improve?

Ans: 25%

Sol. In first test kiran got 60
In last test he got 75.
% increase in test ( 60(x+100))/100=75
0.6X+60=75
0.6X=15
X=15/0.6=25%
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2. Randy’s chain of used car dealership sold 16,400 cars in 1998. If the chain sold 15,744 cars in 1999, by what percent did the number of cars sold decrease?

Ans: 4%

Sol. Let percentage of decrease is x , then 16400(100-x)/100=15744
16400-15744=164x
x=656/164=4%
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3. A radio when sold at a certain price gives a gain of 20%. What will be the gain percent, if sold for thrice the price?
A) 260%
B) 150%
C) 100%
D) 50%
E) None of these

Ans: 260%

Sol. Let x be original cost of the radio.
The solding price = (100+20)x=120x
If , it is sold for thrice the price ,then 3*120x=360x
So, gain percent is (360-100)=260%.
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4. The boys and girls in a college are in the ratio 3:2. If 20% of the boys and 25% of the girls are adults, the percentage of students who are not adults is:??

Ans.78%

Sol: Suppose boys = 3x and girls = 2x
Not adults = (80*3x/100) + (75*2x/100) = 39x/10
Required percentage = (39x/10)*(1/5x)*100 = 78%
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5. 5/9 part of the population in a village are males. If 30% of the males are married, the percentage of unmarried females in the total population is:

Ans: (250/9)%

Sol: Let the population =x Males=(5/9)x
Married males = 30% of (5/9)x = x/6
Married females = x/6
Total females = (x-(5/9)x)=4x/9
Unmarried females = (4x/9 – x/6) = 5x/18
Required percentage = (5x/18 * 1/x * 100) = (250/9)%
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6. A man sells an article at a profit of 25%. If he had bought it at 20 % less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost price of the article?

Ans. Rs. 50.

Sol: Let the C.P be Rs.x.
1st S.P =125% of Rs.x.= 125*x/100= 5x/4.
2nd C.P=80% of x. = 80x/100 =4x/5.
2nd S.P =130% of 4x/5. = (130/100* 4x/5) = 26x/25.
Therefore, 5x/4-26x/25 = 10.50 or x = 10.50*100/21=50.
Hence, C.P = Rs. 50.
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7. A candidate who gets 20% marks fails by 10 marks but another candidate who gets 42% marks gets 12% more than the passing marks. Find the maximum marks.

Solution:
Let the maximum marks be x.
From the given statement pass percentage is 42% - 12% = 30%
By hypothesis, 30% of x – 20% of x = 10 (marks)
i.e., 10% of x = 10
Therefore, x = 100 marks.
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8. In an election contested by two parties, Party D secured 12% of the total votes more than Party R. If party R got 132,000 votes, by how many votes did it lose the election?

Solution:
Let the percentage of the total votes secured by Party D be x%
Then the percentage of total votes secured by Party R = (x – 12)%
As there are only two parties contesting in the election, the sum total of the votes secured by the two parties should total up to 100% i.e., x + x – 12 = 100
2x – 12 = 100 or 2x = 112 or x = 56%.
If Party D got 56% of the votes, then Party got (56 – 12) = 44% of the total votes.
44% of the total votes = 132,000
i.e.T = 300,000 votes.

The margin by which Party R lost the election = 12% of the total votes = 12% of 300,000 = 36,000.
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9. If the price of petrol increases by 25%, by how much must a user cut down his consumption so that his expenditure on petrol remains constant?

Solution:
Let the price of petrol be Rs.100 per litre. Let the user use 1 litre of petrol. Therefore, his expense on petrol = 100 * 1 = Rs.100
Now, the price of petrol increases by 25%. Therefore, the new price of petrol = Rs.125.
As he has to maintain his expenditure on petrol constant, he will be spending only Rs.100 on petrol.
Let ‘x’ be the number of litres of petrol he will use at the new price.

Therefore, 125*x = 100 => x = 0.8 litres.

He has cut down his petrol consumption by 0.2 litres = = 20% reduction.

**There is a short cut for solving this problem.

If the price of petrol has increased by 25%, it has gone up of its earlier price.
Therefore, the % of reduction in petrol that will maintain the amount of money spent on petrol constant = 20%

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