Problems On Numbers - Aptitude MCQ Questions and Solutions with Explanations

21

The digit in the unit's place of a numbers is equal to the digit in the ten's place of half of that number and the digit in the ten's place of that number is less than the digit in unit's place of half of the number by 1. If the sum of the digits of the number is 7, then what is the number ?

A) 34 B) 52 C) 162 D) Data inadequate E) None of these

Discuss

Answer & Solution

Answer: Option B

Solution:

Let the ten's digit be x and unit's digit be y
Then,
$$\eqalign{ & \Leftrightarrow \frac{{10x + y}}{2} = 10y + \left( {x + 1} \right) \cr & \Leftrightarrow 10x + y = 20y + 2x + 2 \cr & \Leftrightarrow 8x - 19y = 2.....(i) \cr & {\text{And}} \Leftrightarrow x + y = 7.....(ii) \cr} $$
Solving (i) and (ii), we get :
x = 5, y = 2
Hence, required number = 52

22

Of the three number, the sum of the first two is 73, the sum of the second and the third is 77 and the sum of the third and thrice the first is 104. The third number is ?

A) 25 B) 39 C) 48 D) Cannot be determined E) None of these

Discuss

Answer & Solution

Answer: Option E

Solution:

Let the numbers be x, y and z
Then,
$$\eqalign{ & {\bf{x + y = 73}} \cr & \Leftrightarrow y = 73 - x \cr & {\bf{y + z = 77 }} \cr & \Leftrightarrow z = 77 - y \cr & \Leftrightarrow z = 77 - \left( {73 - x} \right) \cr & \Leftrightarrow z = 4 + x \cr & {\bf{3x + z = 104}} \cr & \Leftrightarrow 3x + 4 + x = 104 \cr & \Leftrightarrow 4x = 100 \cr & \Leftrightarrow x = 25 \cr & \because y = \left( {73 - 25} \right) \cr & \Leftrightarrow y = 48 \cr & \because z = \left( {4 + 25} \right) \cr & \Leftrightarrow z = 29 \cr} $$
∴ Third number = 29

23

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

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow x - \frac{3}{5}x = 50 \cr & \Leftrightarrow \frac{2}{5}x = 50 \cr & \Leftrightarrow x = \left( {\frac{{50 \times 5}}{2}} \right) \cr & \Leftrightarrow x = 125 \cr} $$

24

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) $$\frac{144}{25}$$

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the number be x
Then,
$$\eqalign{ & \frac{2}{3}x = \frac{25}{216} \times \frac{1}{x} \cr & \Leftrightarrow {x^2} = \frac{{25}}{{216}} \times \frac{3}{2} \cr & \Leftrightarrow {x^2} = \frac{{25}}{{144}} \cr & \Leftrightarrow x = \sqrt {\frac{{25}}{{144}}} \cr & \Leftrightarrow x = \frac{5}{{12}} \cr} $$

25

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

Discuss

Answer & Solution

Answer: Option B

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

26

What is the sum of two consecutive even numbers, the difference of whose squares is 84 ?

A) 34 B) 38 C) 42 D) 46

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the numbers be x and x + 2
Then,
(x + 2)² - x² = 84
⇔ 4x + 4 = 84
⇔ 4x = 80
⇔ x = 20
∴ Required sum
= x + (x + 2)
= 2x + 2
= 42

27

A certain number of two digits is three times the sum of its digits and if 45 be added to it, the digits are reversed. The number is ?

A) 23 B) 27 C) 32 D) 72

Discuss

Answer & Solution

Answer: Option B

Solution:

Let the ten's digit be x and unit's digit be y
Then,
$$\eqalign{ & \Rightarrow 10x + y = 3\left( {x + y} \right) \cr & \Rightarrow 7x - 2y = 0.....(i) \cr & 10x + y + 45 = 10y + x \cr & \Rightarrow y - x = 5.....(ii) \cr} $$
Solving (i) and (ii), we get :
x = 2 and y = 7
∴ Required number = 27

28

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) $$5\frac{1}{6}$$

Discuss

Answer & Solution

Answer: Option C

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 = $$4\frac{{5}}{6}$$

29

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 E) None of these

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow x + \frac{2}{5}x = 455 \cr & \Leftrightarrow \frac{7}{5}x = 455 \cr & \Leftrightarrow x = \left( {\frac{{455 \times 5}}{7}} \right) \cr & \Leftrightarrow x = 325 \cr} $$

30

Find the whole number which when increased by 20 is equal to 69 times the reciprocal of the number.

A) 2.5 B) 3 C) 5 D) 7

Discuss

Answer & Solution

Answer: Option B

Solution:

Let the required number be x
Then,
$$\eqalign{ & \Leftrightarrow x + 20 = \frac{{69}}{x} \cr & \Leftrightarrow {x^2} + 20x - 69 = 0 \cr & \Leftrightarrow {x^2} + 23x - 3x - 69 = 0 \cr & \Leftrightarrow x\left( {x + 23} \right) - 3\left( {x + 23} \right) = 0 \cr & \Leftrightarrow \left( {x + 23} \right)\left( {x - 3} \right) = 0 \cr & \Leftrightarrow x = 3\,\,\,\,\,\,\,\,\,\left[ {\because x \ne - 23} \right] \cr} $$