Practice MCQ Questions and Answer on Problems on Numbers
91.
The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is :
(A) 380
(B) 395
(C) 400
(D) 425
Solution:
Let the numbers be x and y Then, $$xy = 9375{\text{ and }}\frac{x}{y} = 15$$ $$\eqalign{ & \Leftrightarrow \frac{{xy}}{{\left( {\frac{x}{y}} \right)}} = \frac{{9375}}{{15}} \cr & \Leftrightarrow {y^2} = 625 \cr & \Leftrightarrow y = 25 \cr & \therefore x = 15y \cr & \Rightarrow x = 15 \times 25 \cr & \Rightarrow x = 375 \cr} $$ ∴ Sum of the numbers : = 375 + 25 = 400
92.
If 50 is subtracted from two-third of number, the result is equal to sum of 40 and one-fourth of that number. What is the number ?
(A) 174
(B) 216
(C) 246
(D) 336
Solution:
Let the number be x Then, $$\eqalign{ & \Leftrightarrow \frac{2}{3}x - 50 = \frac{1}{4}x + 40 \cr & \Leftrightarrow \frac{2}{3}x - \frac{1}{4}x = 90 \cr & \Leftrightarrow \frac{{5x}}{{12}} = 90 \cr & \Leftrightarrow x = \left( {\frac{{90 \times 12}}{5}} \right) \cr & \Leftrightarrow x = 216 \cr} $$
93.
The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?
(A) 4
(B) 8
(C) 16
(D) None of these
Solution:
Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit. Let ten's and unit's digits be 2x and x respectively. Then, (10 × 2x + x) - (10x + 2x) = 36 ⇒ 9x = 36 ⇒ x = 4 ∴ Required difference = (2x + x) - (2x - x) = 2x = 8
94.
The difference between a number and its three-fifths is 50, What is the number ?
(A) 75
(B) 100
(C) 125
(D) None of these
Solution:
Let the number be x Then, $$\eqalign{ & \Leftrightarrow x - \frac{3}{5}x = 50 \cr & \Leftrightarrow \frac{2}{5}x = 50 \cr & \Leftrightarrow x = \left( {\frac{{50 \times 5}}{2}} \right) \cr & \Leftrightarrow x = 125 \cr} $$
95.
The sum of the squares of two numbers is 3341 and the difference of their squares is 891. The numbers are :
(A) 25, 36
(B) 25, 46
(C) 35, 46
(D) None of these
Solution:
Let the numbers be x and y. Then, x2 + y2 = 3341..... (i) And, x2 - y2 = 891..... (ii) Adding (i) and (ii), we get : 2x2 = 4232 or x2 = 2116 or x = 46 Subtracting (ii) from (i), we get : 2y2 = 2450 or y2 = 1225 or y = 35 So, the numbers are 35 and 46
96.
The sum of two numbers is 25 and their difference is 13. Find their product :
(A) 104
(B) 114
(C) 315
(D) 325
Solution:
Let the numbers be x and y Then, x + y = 25 and x - y = 13 $$\eqalign{ & \Leftrightarrow 4xy = {\left( {x + y} \right)^2} - {\left( {x - y} \right)^2} \cr & \Leftrightarrow 4xy = {\left( {25} \right)^2} - {\left( {13} \right)^2} \cr & \Leftrightarrow 4xy = 625 - 169 \cr & \Leftrightarrow 4xy = 456 \cr & \Leftrightarrow xy = 114 \cr} $$
97.
The difference between two positive integers is 3. If the sum of their squares is 369, then the sum of the numbers is :
(A) 25
(B) 27
(C) 33
(D) 81
Solution:
Let the numbers be x and (x + 3) Then, $$\eqalign{ & \Leftrightarrow {x^2} + {\left( {x + 3} \right)^2} = 369 \cr & \Leftrightarrow {x^2} + {x^2} + 9 + 6x = 369 \cr & \Leftrightarrow 2{x^2} + 6x - 360 = 0 \cr & \Leftrightarrow {x^2} + 3x - 180 = 0 \cr & \Leftrightarrow \left( {x + 15} \right)\left( {x - 12} \right) = 0 \cr & \Leftrightarrow x = 12 \cr} $$ So, the numbers are 12 and 15 ∴ Required sum = (12 + 15) = 27
98.
Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.
In a three-digit number, the digit in the unit's place is 75% of the digit in the ten's place. The digit in the ten's place is greater than the digit in the hundred's place by 1. If the sum of the digits in the ten's place and the hundred's place is 15. What is the number ?