Percentage Questions for SSC CHSL set-3 PDF

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Percentage Questions for SSC CHSL set-3 PDF
Percentage Questions for SSC CHSL set-3 PDF

Percentage Questions for SSC CHSL set-3 PDF

Download SSC CHSL Percentage Questions with answers Set-3 PDF based on previous papers very useful for SSC CHSL Exams. Top-15 Very Important Questions for SSC Exam

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Question 1: The cost price of an article is 64% of the marked price. The gain percentage after allowing a discount of 12% on the market price is

a) 37.5%

b) 48%

c) 50.5%

d) 52%

Question 2: A box has 100 blue balls, 50 red balls and 50 black balls. 25% of blue balls and 50% of red balls are taken away. Then, percentage of black balls at present is

a) 25%

b) 100/3%

c) 40%

d) 50%

Question 3: The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is equal to

a) 32%

b) 34%

c) 42%

d) 44%

Question 4: The cost price of 100 books is equal to the selling price of 60 books. The gain or loss percentage will be :

a) $66$ %

b) $66\frac{2}{3}$ %

c) $66\frac{1}{4}$ %

d) $66\frac{3}{4}$ %

Question 5: A man spends 75% of his income. His income increases by 20% and his expenditure also increases by 10%. The percentage of increase in his savings is

a) 40%

b) 30%

c) 50%

d) 25%

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Question 6: A fruit-seller buys some oranges and by selling 40% of them he realises the cost price of all the oranges. As the oranges being to grow over-ripe, he reduces the price and sells 80% of the remaining oranges at half the previous rate of profit. The rest of the oranges being rotten are thrown away. The overall percentage of profit is

a) 80

b) 84

c) 94

d) 96

Question 7: The ratio of the number of boys and girls in a school is 2 : 3. If 25% of the boys and 30% of the girls are scholarship holders, the percentage of the school students who are not scholarship holders is

a) 72

b) 36

c) 54

d) 60

Question 8: A grain dealer cheats to the extent of 10% while buying as well as selling by using false weights. His total profit percentage is :

a) 20%

b) 22.22%

c) 23.23%

d) 25%

Question 9: If 30% of A is added to 40% of B, the answer is 80% of B. What percentage of A is B?

a) 30%

b) 40%

c) 70%

d) 75%

Question 10: The cost of an apple is twice that of a banana and the cost of a banana is 25% less than that of a guava. If the cost of each type of fruit increases by 10%, then the percentage increase in the cost of 4 bananas, 2 apples and 3 guavas is

a) 10%

b) 12%

c) 16%

d) 18%

Question 11: In a club, the average age of the members is 30 years, the average age of the male members is 34 years and that of the female members is 26 years. The percentage of the male members is

a) 50%

b) 60%

c) 30%

d) 40%

Question 12: A student multiplied a number by 5/7 instead of 7/5. What is the percentage error in the calculation?

a) 48.98 percent

b) 96 percent

c) 24.49 percent

d) 48 percent

Question 13: A student multiplied a number by 4/7 instead of 7/4. What is the percentage error in the calculation?

a) 206.25 percent

b) 67.35 percent

c) 33.67 percent

d) 103.13 percent

Question 14: A student multiplied a number by 6/13 instead of 13/6. What is the percentage error in the calculation?

a) 369.44 percent

b) 39.35 percent

c) 184.72 percent

d) 78.7 percent

Question 15: A student multiplied a number by 11/13 instead of 13/11. What is the percentage error in the calculation?

a) 39.67 percent

b) 28.4 percent

c) 14.2 percent

d) 19.83 percent

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Answers & Solutions:

1) Answer (A)

Let’s say marked price = 100
Then cost price = 64
and selling price = 88
discount = 24
%discount = $\frac{24}{64} \times 100$ = 37.5%

2) Answer (B)

Initially there were 100 blue balls, 50 red balls and 50 black balls.
After taking away there will remain 75 blue balls, 25 red balls and 50 black balls , a total of 150 balls
Hence percentage of black balls will be $\frac{50\times100}{150}$ = $\frac{100}{3}$

3) Answer (D)

let the length of rectangle be L units and breadth be B units.

So area of rectangle = L x B

As both length and breadth has been increased by 20%

So new length = 1.2L

New breadth = 1.2B

New area = 1.2L x 1.2B = 1.44LB

Percentage increase in area = $\frac{1.44LB – LB}{LB}$×100 = 44%

4) Answer (B)

Let the cost price of 1 book be Rs C

and the selling Price of 1 book be Rs S

It is given that cost price of 100 books is equal to the selling price of 60 books

So, 100xC = 60xS

S = $\frac{5}{3}$ C

So Profit % = $\frac{SellingPrice-CostPrice}{CostPrice}$x100

= $\frac{\frac{5C}{3}-C}{C}$x100

= $\frac{2}{3}$ * 100 = $66\frac{2}{3}$ %

 

5) Answer (C)

let the income of person be Rs y

So his expenditure be 75% of income which is = 0.75y

Saving = y- 0.75y = 0.25y

Now after increment of 20 % in income the new income becomes = 1.2y

Given that bew expenditure is 10% more than previous one.So , new expenditure is = 1.1 × 0.75y = 0.825y

New savings = 1.2y – 0.825y = 0.375y

Percentage increase in savings =$\frac {0.375y-0.25y}{0.25y}$×100 = 50%

6) Answer (B)

Let us assume that the fruit seller buys 100 oranges for Rs. 100
He sells 40 oranges for Rs. 100
Profit obtained = 100 – 40 = Rs. 60
% Profit = $\frac{60}{40} \times 100 = 150$%
Now, he sells 80% of the remaining oranges at half the profit
i.e., he sells 48 oranges at 75% profit.
Selling Price of 48 oranges = 48 + 75% of 48 = Rs. 84
Rest of them are thrown away.
Total SP = 100 + 84 = 184
Profit = 184 – 100 = 84
% Profit = 84%

7) Answer (A)

Let the number of boys be 2y and number of girls be 3y

25 % of boys = 0.25 x 2y = 0.5y

30% of girls = 0.3 x 3y = 0.9y

total scholarship holder = 0.5y + 0.9y = 1.4y

% of no scholarship holder = $\frac{3.6y}{5y}$ x 100 = 72%

8) Answer (B)

let the retailer buys 110 gram in Rs 100 because of 10 % cheating

so cost price = 1 gram = Rs 100/110 = Rs 0.909

and retailer sells 90 gram in Rs 100

and hence selling price = Rs 100 / 90 = Rs 1.111

Profit % = $\frac{1.111- 0.909}{0.909}$ = 22.22%

9) Answer (D)

it is given that

0.3 A + 0.4 B = 0.8 B

0.3 A = 0.4 B

A = 4/3 B

B = 3/4 A

hence B is 75 % of A

10) Answer (A)

let the cost of a banana be Rs b

and hence cost of an apple = 2b

and guava cost be = 4/3 b

now cost of every fruit is increased by 10%

and hence

cost of a banana be Rs 1.1b

and hence cost of an apple = 2.2 b

and guava cost be = (4.4/3)b

initially , 4 bananas + 2 apples + 3 guavas = 13/3 b

finally , 4 bananas + 2 apples + 3 guavas = 14.3/3 b

and hence percentage increment = $\frac{(14.3-13)}{13}$ x 100 = 10 %

11) Answer (A)

Let male members be $x$ and female members be $y$

=> Total members in the club = $(x+y)$

Now, total age of male members = $34x$

Total age of female members = $26y$

Also, total age of all the members = $30(x + y)$

Acc to ques :

=> $34x + 26y = 30x + 30y$

=> $4x = 4y$

=> $x = y$

Now, % of male members = $\frac{x}{x + y}*100$

= $\frac{x}{2x}*100 = 50$%

12) Answer (A)

Let the number be ‘x’.

This number was supposed to be multiplied by ‘7/5’ to get ‘7x/5’.

But it was multiplied by ‘5/7’ to get ‘5x/7’.

So, percentage error = $ \frac{\frac{7x}{5} – \frac{5x}{7}}{\frac{7x}{5}} * 100$

= $\frac{49-25}{49} * 100$

= $\frac{24}{49} *100 $

$\approx 48.98\%$

=> Ans – (A)

13) Answer (B)

Let the number be 28

When the student multiplied it by 4/7, => original result = $\frac{4}{7} \times 28 = 16$

When the student multiply it by 7/4, => New result = $\frac{7}{4} \times 28 = 49$

=> Percentage error in calculation = $\frac{(49 – 16)}{49} \times 100$

= $\frac{3300}{49} = 67.35\%$

=> Ans – (B)

14) Answer (D)

Let the number be 78

When the student multiplied it by 6/13, => original result = $\frac{6}{13} \times 78 = 36$

When the student multiply it by 13/6, => New result = $\frac{13}{6} \times 78 = 169$

=> Percentage error in calculation = $\frac{(169 – 36)}{169} \times 100$

= $\frac{13300}{169} = 78.69 \approx 78.7\%$

=> Ans – (D)

15) Answer (B)

Let the number be 143

When the student multiplied it by 11/13, => original result = $\frac{11}{13} \times 143 = 121$

When the student multiply it by 13/11, => New result = $\frac{13}{11} \times 143 = 169$

=> Percentage error in calculation = $\frac{(169 – 121)}{169} \times 100$

= $\frac{4800}{169} = 28.4\%$

=> Ans – (B)

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