# Percentages Quiz Set 015

### 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. 