Practice MCQ Questions and Answer on Problems on Numbers

1.

A number when multiplied by 13 is increased by 180. The number is :

• (A) 5
• (B) 12
• (C) 15
• (D) 45

Show Answer

 Let the number be x Then, $$\eqalign{ & \Rightarrow 13x = x + 180 \cr & \Rightarrow 12x = 180 \cr & \Rightarrow x = \frac{{180}}{{12}} \cr & \Rightarrow x = 15 \cr} $$

2.

If a number is multiplied by two-third of itself the value so obtained is 864. What is the number ?

• (A) 34
• (B) 36
• (C) 38
• (D) 44

Show Answer

 Let the number be x Then, $$\eqalign{ & \Leftrightarrow x \times \frac{2}{3}x = 864 \cr & \Leftrightarrow \frac{2}{3}{x^2} = 864 \cr & \Leftrightarrow {x^2} = \left( {\frac{{864 \times 3}}{2}} \right) \cr & \Leftrightarrow {x^2} = 1296 \cr & \Leftrightarrow x = \sqrt {1296} \cr & \Leftrightarrow x = 36 \cr} $$

3.

The product of two numbers is 120 and the sum of their square is 289. The sum of the numbers is :

• (A) 20
• (B) 23
• (C) 169
• (D) None of these

Show Answer

 Let the numbers be x and y Then, xy = 120 and x2 + y2 = 289 $$\eqalign{ & \therefore {\left( {x + y} \right)^2} \cr & = {x^2} + {y^2} + 2xy \cr & = 289 + 240 \cr & = 529 \cr & \therefore x + y \cr & = \sqrt {529} \cr & = 23 \cr} $$

4.

The sum of two numbers is 40 and their product is 375. What will be the sum of their reciprocals ?

• (A) $$\frac{1}{40}$$
• (B) $$\frac{8}{75}$$
• (C) $$\frac{75}{4}$$
• (D) $$\frac{75}{8}$$

Show Answer

 Let the numbers be x and y Then, x + y = 40 and xy = 375 $$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{y} = \frac{{x + y}}{{xy}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{40}}{{275}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{75}} \cr} $$

5.

The sum of two numbers is 75 and their difference is 25. The product of the two numbers is :

• (A) 1350
• (B) 1250
• (C) 125
• (D) 1000

Show Answer

 Let the numbers be a and b According to the question, $$\eqalign{ & a + b = 75 \cr & a - b = 25 \cr & \because {\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} = 4ab \cr & \Rightarrow {75^2} - {25^2} = 4ab \cr & \Rightarrow 4ab = \left( {75 + 25} \right)\left( {75 - 25} \right) \cr & \left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \cr & \Rightarrow 4ab = 100 \times 50 \cr & \Rightarrow ab = \frac{{100 \times 50}}{4} \cr & \Rightarrow ab = 1250 \cr} $$

6.

If the digit in the unit's place of a two-digit number is halved and the digit in the ten's place is doubled, the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true ?

• (A) Sum of the digits is a two-digit number
• (B) Digit in the unit's place is half of the digit in the ten's place
• (C) Digit in the unit's place and the ten's place are equal
• (D) Digit in the unit's place is twice the digit in the ten's place

Show Answer

 Let the ten's digit be x and the unit's digit be y Then, number = 10x + y New number : $$\eqalign{ & = 10 \times 2x + \frac{y}{2} \cr & = 20x + \frac{y}{2} \cr} $$ $$\eqalign{ & \therefore 20x + \frac{y}{2} = 10y + x \cr & \Leftrightarrow 40x + y = 20y + 2x \cr & \Leftrightarrow 38x = 19y \cr & \Leftrightarrow y = 2x \cr} $$ So, the unit's digit is twice the ten's digit.

7.

A fraction is such that if the double of the numerator and the triple pf the denominator is changed by +10 percent and -30 percent respectively, then we get 11 percent of $$\frac{16}{21}$$. Find the fraction :

• (A) $$\frac{2}{25}$$
• (B) $$\frac{3}{25}$$
• (C) $$\frac{4}{25}$$
• (D) None of these

Show Answer

 Let the fraction be $$\frac{x}{y}$$ Then, $$\eqalign{ & \Leftrightarrow \frac{{110\% {\text{ of 2}}x}}{{70\% {\text{ of 3}}y}} = 11\% {\text{ of }}\frac{{16}}{{21}} \cr & \Leftrightarrow \frac{{22x}}{{21y}} = \frac{{11}}{{100}} \times \frac{{16}}{{21}} \cr & \Leftrightarrow \frac{x}{y} = \left( {\frac{{11}}{{100}} \times \frac{{16}}{{21}} \times \frac{{21}}{{22}}} \right) \cr & \Leftrightarrow \frac{x}{y} = \frac{2}{{25}} \cr} $$

8.

If the product of three consecutive integers is 120, then the sum of the integers is :

• (A) 9
• (B) 12
• (C) 14
• (D) 15

Show Answer

 $$\eqalign{ & 120: \cr & = 2 \times 2 \times 2 \times 3 \times 5 \cr & = \left( {2 \times 2} \right) \times 5 \times \left( {2 \times 3} \right) \cr & = 4 \times 5 \times 6 \cr} $$ Clearly, the three consecutive integers whose product is 120 are 4, 5 and 6 Required sum : = 4 + 5 + 6 = 15

9.

The sum and product of two numbers are 12 and 35 respectively. The sum of their reciprocals will be :

• (A) $$\frac{12}{35}$$
• (B) $$\frac{1}{35}$$
• (C) $$\frac{35}{8}$$
• (D) $$\frac{7}{32}$$

Show Answer

 Let the numbers be x and y Then, $$\eqalign{ & x + y = 12\,\, \& \,\, xy = 35 \cr & \therefore \frac{1}{x} + \frac{1}{y} = \frac{{x + y}}{{xy}} = \frac{{12}}{{35}} \cr} $$

10.

The sum of two numbers is 40 and their difference is 4. The ratio of the numbers is :

• (A) 11 : 9
• (B) 11 : 18
• (C) 21 : 19
• (D) 22 : 19

Show Answer

 Let the numbers be x and y Then, $$\eqalign{ & \Leftrightarrow \frac{{x + y}}{{x - y}} = \frac{{40}}{4} \cr & \Leftrightarrow \frac{{x + y}}{{x - y}} = 10 \cr & \Leftrightarrow \left( {x + y} \right) = 10\left( {x - y} \right) \cr & \Leftrightarrow 9x = 11y \cr & \Leftrightarrow \frac{x}{y} = \frac{{11}}{9} \cr & \Leftrightarrow x:y = 11:9 \cr} $$