Practice MCQ Questions and Answer on Problems on Numbers
71.
If doubling a number and adding 20 to the result gives the same answer as multiplying the number by 8 and taking away 4 from the product, the number is :
(A) 2
(B) 3
(C) 4
(D) 6
Solution:
Let the number be x Then, ⇔ 2x + 20 = 8x - 4 ⇔ 6x = 24 ⇔ x = 4
72.
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 ?
A positive number when decreased by 4 is equal to 21 times the reciprocal of the number. The number is ?
(A) 3
(B) 5
(C) 7
(D) 9
Solution:
Let the number be x Then, $$\eqalign{ & \Leftrightarrow x - 4 = \frac{{21}}{x} \cr & \Leftrightarrow {x^2} - 4x - 21 = 0 \cr & \Leftrightarrow \left( {x - 7} \right)\left( {x + 3} \right) = 0 \cr & \Leftrightarrow x = 7 \cr} $$
74.
The difference between two integers is 5. Their product is 500. Find the numbers.
(A) 15, 20
(B) 20, 25
(C) 30, 25
(D) 21, 26
Solution:
Let the integers be x and (x + 5) Then, $$\eqalign{ & \Leftrightarrow x\left( {x + 5} \right) = 500 \cr & \Leftrightarrow {x^2} + 5x - 500 = 0 \cr & \Leftrightarrow \left( {x + 25} \right)\left( {x - 20} \right) = 0 \cr & \Leftrightarrow x = 20 \cr} $$ So, the numbers are 20 and 25
75.
The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:
(A) 20
(B) 30
(C) 40
(D) None of these
Solution:
Let the numbers be a, b and c Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131 (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400 ⇒ (a + b + c) = $$\sqrt {400} $$ = 20
76.
The difference between two numbers is 16. If one-third of the smaller number is greater than one-seventh of the larger number by 4, then the two numbers are :
(A) 9 and 25
(B) 12 and 28
(C) 33 and 49
(D) 56 and 72
Solution:
Let the numbers be x and (x + 16) Then, $$\eqalign{ & \Leftrightarrow \frac{x}{3} - \frac{{\left( {x + 16} \right)}}{7} = 4 \cr & \Leftrightarrow 7x - 3\left( {x + 16} \right) = 84 \cr & \Leftrightarrow 4x = 84 + 48 \cr & \Leftrightarrow 4x = 132 \cr & \Leftrightarrow x = 33 \cr} $$ Hence, the numbers are 33 and 49
77.
Two-third of a positive number and of its reciprocal are equal, The number is :
(A)
(B)
(C)
(D)
Solution:
Let the number be x Then, $$\eqalign{ & \frac{2}{3}x = \frac{25}{216} \times \frac{1}{x} \cr & \Leftrightarrow {x^2} = \frac{{25}}{{216}} \times \frac{3}{2} \cr & \Leftrightarrow {x^2} = \frac{{25}}{{144}} \cr & \Leftrightarrow x = \sqrt {\frac{{25}}{{144}}} \cr & \Leftrightarrow x = \frac{5}{{12}} \cr} $$
78.
A man bought some eggs of which 10% are rotten. He gives 80% of the remainder to his neighbours. Now he is left out with 36 eggs. How many eggs he bought ?
(A) 40
(B) 100
(C) 200
(D) 72
Solution:
Let the total number of eggs bought be a 10% of eggs are rotten ∴ Remaining eggs : $$\eqalign{ & = a - 10\% {\text{ of }}a \cr & = a - \frac{{10a}}{{100}} \cr & = \frac{{100a - 10a}}{{100}} \cr & = \frac{{90a}}{{100}} \cr & = \frac{{9a}}{{10}} \cr} $$ Man gives 80% of $$\frac{{9a}}{{100}}$$ eggs to his neighbour $$\eqalign{ & = \frac{{80}}{{100}} \times \frac{{9a}}{{10}} \cr & = \frac{{72a}}{{100}} \cr} $$ Remaining eggs : $$\eqalign{ & = \frac{{9a}}{{10}} - \frac{{72a}}{{100}} \cr & = \frac{{90a - 72a}}{{100}} \cr & = \frac{{18a}}{{100}} \cr & = \frac{{9a}}{{50}} \cr} $$ According the question, $$\eqalign{ & \Rightarrow \frac{{9a}}{{50}} = 36 \cr & \Rightarrow 9a = 36 \times 50 \cr & \Rightarrow a = \frac{{36 \times 50}}{9} \cr & \Rightarrow a = 200 \cr} $$ Hence, the total number of eggs bought be = 200
79.
If a number is added to two-fifths of itself, the value so obtained is 455. What is the number ?
(A) 325
(B) 350
(C) 400
(D) 420
Solution:
Let the number be x Then, $$\eqalign{ & \Leftrightarrow x + \frac{2}{5}x = 455 \cr & \Leftrightarrow \frac{7}{5}x = 455 \cr & \Leftrightarrow x = \left( {\frac{{455 \times 5}}{7}} \right) \cr & \Leftrightarrow x = 325 \cr} $$
80.
If the sum of numbers is 33 and their difference is 15, the smaller number is ?
(A) 9
(B) 12
(C) 15
(D) 18
Solution:
Let the numbers be x and y Then, $$x + y = 33.....(i)$$And, $$x - y = 15.....(ii)$$ Solving (i) and (ii), we get : x = 24 , y = 9 ∴ Smaller number = 9