Problems On Numbers - Aptitude MCQ Questions and Solutions with Explanations

1

A number consists of two digits such that the digit in the ten's place is less by 2 than the digit in the unit's place. Three times the number added to $$\frac{6}{7}$$ times the number obtained by reversing the digits equals 108. The sum of the digits in the number is :

A) 6 B) 7 C) 8 D) 9

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the unit's digit be x
Then, ten's digit = (x - 2)
$$\therefore 3\left[ {10\left( {x - 2} \right) + x} \right] + \frac{6}{7}$$ $$\left[ {10x + \left( {x - 2} \right)} \right]$$ $$ = 108$$
⇔ 231x - 420 + 66x - 12 = 756
⇔ 297x = 1188
⇔ x = 4
Hence, sum of the digits :
= x + (x - 2)
= 2x - 2
= 6

2

If (73)² is subtracted from the square of a number, the answer so obtained is 5075. What is the number ?

A) 96 B) 98 C) 102 D) 106

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow {x^2} - {\left( {73} \right)^2} = 5075 \cr & \Leftrightarrow {x^2} - 5329 = 5075 \cr & \Leftrightarrow {x^2} = 5075 + 5329 \cr & \Leftrightarrow {x^2} = 10404 \cr & \Leftrightarrow x = \sqrt {10404} \cr & \Leftrightarrow x = 102 \cr} $$

3

The sum of three consecutive odd numbers is 20 more than the first of these numbers. What is the middle number ?

A) 7 B) 9 C) 11 D) Data inadequate E) None of these

Discuss

Answer & Solution

Answer: Option B

Solution:

Let the numbers be x, (x + 2) and (x + 4)
Then,
$$\eqalign{ & \Leftrightarrow x + \left( {x + 2} \right) + \left( {x + 4} \right) = x + 20 \cr & \Leftrightarrow 2x = 14 \cr & \Leftrightarrow x = 7 \cr} $$
∴ Middle number :
= x + 2
= 9

4

The product of two numbers is 192 and the sum of these two numbers is 28. What is the smaller these two numbers ?

A) 12 B) 14 C) 16 D) 18 E) None of these

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the numbers be x and (28 - x)
Then,
$$\eqalign{ & \Leftrightarrow x\left( {28 - x} \right) = 192 \cr & \Leftrightarrow {x^2} - 28x + 192 = 0 \cr & \Leftrightarrow \left( {x - 16} \right)\left( {x - 12} \right) = 0 \cr & \Leftrightarrow x = 16{\text{ or }}x = 12 \cr} $$
So, the numbers are 16 and 12

5

If the square of a two-digit number is reduced by the square of the number formed by reversing the digits of the number, the final result is :

A) Divisible by 11 B) Divisible by 9 C) Necessarily irrational D) Both (A) and (B)

Discuss

Answer & Solution

Answer: Option D

Solution:

Let the two-digit number be 10x + y
Then, number formed by reversing the digits = 10y + x
Difference of square of the numbers :
$$ = {\left( {10x + y} \right)^2} - {\left( {10y + x} \right)^2}$$
$$ = \left( {100{x^2} + {y^2} + 20xy} \right) - $$ $$\left( {100{y^2} + {x^2} + 20xy} \right)$$
$$ = 99\left( {{x^2} - {y^2}} \right)$$ which is divisible by both 9 and 11

6

Three times the first of three consecutive odd integers is 3 more than twice the third. The integer is :

A) 9 B) 11 C) 13 D) 15

Discuss

Answer & Solution

Answer: Option D

Solution:

Let the three integers be x, (x + 2) and (x + 4)
Then,
Three times the first of three consecutive odd integers is 3 more than twice the third.
⇔ 3x = 2(x + 4) + 3
⇔ x = 11
∴ Third integer :
= x + 4
= 11 + 4
= 15