Percentages Quiz Set 007

Question 1

The sum of two numbers is 792. One of them is 164% of the other. The smaller number is?

A

300.

B

310.

C

290.

D

320.

Soln.
Ans: a

Let the numbers be x and \${164x}/100\$. The sum is x + \${164x}/100\$ which is same as \${264x}/100 = 792\$. Solving we get the smaller number x = 300.

Question 2

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.

Question 3

In a sample there are 750 items having a value of 50°C. 50% 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

2500 items.

B

2510 items.

C

2490 items.

D

2520 items.

Soln.
Ans: a

The %age ≥ 50°C = (100 - 50) = 50. If x is the size of the sample, 50% of x = 750 + \$2/3 × 750\$ = \${5 × 750}/3\$, which is same as \${50 × x}/100\$ = \${5 × 750}/3\$. Solving, we get x = 2500.

Question 4

In a sample there are 1620 items having a value of 50°C. 10% 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 - 10) = 90. If x is the size of the sample, 90% of x = 1620 + \$2/3 × 1620\$ = \${5 × 1620}/3\$, which is same as \${90 × x}/100\$ = \${5 × 1620}/3\$. Solving, we get x = 3000.

Question 5

A number is first increased by 60%. By what percent must it be reduced so as to restore it to its previous value?

A

\$37{1/2}\$ %.

B

\$38{1/2}\$ %.

C

\$36{1/2}\$ %.

D

\$20{1/4}\$ %.

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. Let us suppose that it has to be reduced by p% to restore it to 100. Then, (100 + x) - p × \$(100 + x)/100\$ = 100. We can cancel away 100, and easily simplify it to p = \${100 × x}/{100 + x}\$ = \${100 × 60}/{100 + 60}\$ = \${75/2}\$, which is same as: \$37{1/2}\$%.