Percentages Quiz Set 015

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Question 1

Price of sugar is increased by 100%. By what percentage should consumption be reduced so that the expenditure remains same?

 A

50 %.

 B

51 %.

 C

49 %.

 D

$17{2/3}$ %.

Soln.
Ans: a

Let price be Rs. 100/kg and the consumption be 1 kg. The current expenditure is 100 × 1 = Rs. 100. The new price of sugar is 100 + 100 = Rs. 200/kg. Let the new consumption be c kg. Now, if expenditure is to be same we should have 100 = 200 × c, which gives c = 100/200, so the decrease is 1 - $100/200$ = ${200 - 100}/200$ × 100% = 50%.


Question 2

If 13% of X + 2% of Y equals 2/3 of the sum of 11% of X and 15% of Y, then the ratio of X to Y is?

 A

$1{7/17}$.

 B

$2{9/16}$.

 C

${7/19}$.

 D

$3{18/19}$.

Soln.
Ans: a

We can drop the % from both sides and write the given equation as 13X + 2Y = $2/3$ × (11X + 15Y), which is same as 3 × (13X + 2Y) = 2 × (11X + 15Y). Let r = X/Y. Dividing both sides by Y and putting r = X/Y, we get, 3 × (13r + 2) = 2 × (11r + 15), solving for r we get r = ${24/17}$, which is same as: $1{7/17}$.


Question 3

A number is first increased by 80%, and thereafter it is reduced by 80% again. What is the overall reduction?

 A

64 %.

 B

66 %.

 C

62 %.

 D

68 %.

Soln.
Ans: a

Let us derive the shortcut formula first, so that you can remember it and use it when the need arises. Suppose the number is 100, and let the increase be x%. The number becomes 100 + x. When it is reduced by x%, we get 100 + x - $((100 + x) × x)/100$ = 100 + x - $(x + x^2/100)$ = 100 - $x^2/100$, hence a reduction of $x^2/100$ = 64%


Question 4

Price of sugar is increased by 40%. By what percentage should consumption be reduced so that the expenditure remains same?

 A

$28{4/7}$ %.

 B

$34{1/2}$ %.

 C

$21{4/9}$ %.

 D

$24{5/9}$ %.

Soln.
Ans: a

Let price be Rs. 100/kg and the consumption be 1 kg. The current expenditure is 100 × 1 = Rs. 100. The new price of sugar is 100 + 40 = Rs. 140/kg. Let the new consumption be c kg. Now, if expenditure is to be same we should have 100 = 140 × c, which gives c = 100/140, so the decrease is 1 - $100/140$ = ${140 - 100}/140$ × 100% = ${200/7}$, which is same as: $28{4/7}$%.


Question 5

In a sample there are 1260 items having a value of 50°C. 30% values are below 50°C, and the number of values above 50°C is $2/3$ of the items having a value of 50°C. What is the size of the sample?

 A

3000 items.

 B

3010 items.

 C

2990 items.

 D

3020 items.

Soln.
Ans: a

The %age ≥ 50°C = (100 - 30) = 70. If x is the size of the sample, 70% of x = 1260 + $2/3 × 1260$ = ${5 × 1260}/3$, which is same as ${70 × x}/100$ = ${5 × 1260}/3$. Solving, we get x = 3000.


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Creative Commons License
This Blog Post/Article "Percentages Quiz Set 015" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Updated on 2017-10-26.

Posted by Parveen(Hoven),
Aptitude Trainer


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