Practice MCQ Questions and Answer on Problems on Numbers
91.
A number consists of two digits. If the digits interchange place and the new number is added to the original number, then the resulting number will be divisible by :
(A) 3
(B) 5
(C) 9
(D) 11
Solution:
Let the ten's digit be x and unit's digit be y Then, number = 10x + y Number obtained by interchanging the digits = 10y + x ∴ (10x + y) + (10y + x) = 11(x + y), which is divisible by 11
92.
The sum of two numbers is 37 and the difference of their squares is 185, then the difference between the two numbers is :
(A) 10
(B) 4
(C) 5
(D) 3
Solution:
Let the numbers be a and b, where a > b According to the question, $$\eqalign{ & a + b = 37\& {a^2} - {b^2} = 185 \cr & \Rightarrow \left( {a + b} \right)\left( {a - b} \right) = 185 \cr & \Rightarrow 37\left( {a - b} \right) = 185 \cr & \Rightarrow a - b = \frac{{185}}{{37}} \cr & \Rightarrow a - b = 5 \cr} $$
93.
A student was asked to divide the half of a certain number by 6 and the other half by 4 and then to add the two quantities so obtained. Instead of doing so the student divided the number by 5 and the result fell short by 4. The given number was ?
(A) 240
(B) 288
(C) 384
(D) 480
Solution:
Let the number be x Then, $$\eqalign{ & \Leftrightarrow \left[ {\frac{{\left( {\frac{x}{2}} \right)}}{6} + \frac{{\left( {\frac{x}{2}} \right)}}{4}} \right] - \frac{x}{5} = 4 \cr & \Leftrightarrow \frac{x}{{12}} + \frac{x}{8} - \frac{x}{5} = 4 \cr & \Leftrightarrow \frac{{10x + 15x - 24x}}{{120}} = 4 \cr & \Leftrightarrow x = 4 \times 120 \cr & \Leftrightarrow x = 480 \cr} $$
94.
Two-third of a positive number and $$\frac{25}{216}$$ of its reciprocal are equal, The number is :
(A) $$\frac{5}{12}$$
(B) $$\frac{12}{5}$$
(C) $$\frac{25}{144}$$
(D) $$\frac{144}{25}$$
Solution:
Let the number be x Then, $$\eqalign{ & \frac{2}{3}x = \frac{25}{216} \times \frac{1}{x} \cr & \Leftrightarrow {x^2} = \frac{{25}}{{216}} \times \frac{3}{2} \cr & \Leftrightarrow {x^2} = \frac{{25}}{{144}} \cr & \Leftrightarrow x = \sqrt {\frac{{25}}{{144}}} \cr & \Leftrightarrow x = \frac{5}{{12}} \cr} $$
95.
A number when multiplied by 13 is increased by 180. The number is :
(A) 5
(B) 12
(C) 15
(D) 45
Solution:
Let the number be x Then, $$\eqalign{ & \Rightarrow 13x = x + 180 \cr & \Rightarrow 12x = 180 \cr & \Rightarrow x = \frac{{180}}{{12}} \cr & \Rightarrow x = 15 \cr} $$
96.
The product of three consecutive even numbers when divided by 8 is 720. The product of their square roots is :
If the product of two numbers is 5 and one of the number is $$\frac{3}{2}$$, then sum of two numbers is :
(A) $$4\frac{1}{3}$$
(B) $$4\frac{2}{3}$$
(C) $$4\frac{5}{6}$$
(D) $$5\frac{1}{6}$$
Solution:
Let two numbers be a and b Given ab = 5 and a = $$\frac{3}{2}$$ $$\eqalign{ & \Rightarrow b = \frac{5}{a} \cr & \Rightarrow b = \frac{5}{{\frac{3}{2}}} \cr & \Rightarrow b = \frac{{5 \times 2}}{3} \cr & \Rightarrow b = \frac{{10}}{3} \cr} $$ Required sum of : ⇒ a + b = $$\frac{{3}}{2}$$ + $$\frac{{10}}{3}$$ L.C.M. of 2 and 3 = 6 ⇒ a + b = $$\frac{{9 + 20}}{6}$$ ⇒ a + b = $$\frac{{29}}{6}$$ ⇒ a + b = $$4\frac{{5}}{6}$$
99.
The difference between a two-digit number and the number obtained by interchanging the two digits is 63. Which is the smaller of the two numbers ?
(A) 29
(B) 70
(C) 92
(D) Cannot be determined
Solution:
Let the ten's digit be x and unit's digit be y Then, $$\eqalign{ & \Leftrightarrow \left( {10x + y} \right) - \left( {10y + x} \right) = 63 \cr & \Leftrightarrow 9\left( {x - y} \right) = 63 \cr & \Leftrightarrow x - y = 7 \cr} $$
100.
Find the whole number which when increased by 20 is equal to 69 times the reciprocal of the number.