Problems On Numbers - Aptitude MCQ Questions and Solutions with Explanations

71

If the numerator of a fraction is increased by 2 and the denominator is increased by 3, the fraction becomes $$\frac{7}{9}$$ and if both the numerator as well as he denominator are decreased by 1, the fraction becomes $$\frac{4}{5}$$. What is the original fraction ?

A) $$\frac{5}{6}$$ B) $$\frac{9}{11}$$ C) $$\frac{13}{16}$$ D) $$\frac{17}{21}$$

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the fraction be $$\frac{x}{y}$$
Then,
$$\eqalign{ & \Leftrightarrow \frac{{x + 2}}{{y + 3}} = \frac{7}{9} \cr & \Leftrightarrow 9x - 7y = 3.....(i) \cr & {\text{And,}} \cr & \Leftrightarrow \frac{{x - 1}}{{y - 1}} = \frac{4}{5} \cr & \Leftrightarrow 5x - 4y = 1.....(ii) \cr} $$
Solving (i) and (ii), we get :
x = 5 and y = 6
Hence, the original fraction is $$\frac{5}{6}$$

72

The difference of two numbers is 20% of the larger number. If the smaller number is 12, the larger one is :

A) 15 B) 16 C) 18 D) 20

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow x - 12 = 20\% {\text{ of }}x \cr & \Leftrightarrow x - \frac{x}{5} = 12 \cr & \Leftrightarrow \frac{{4x}}{5} = 12 \cr & \Leftrightarrow x = \frac{{12 \times 5}}{4} \cr & \Leftrightarrow x = 15 \cr} $$

73

Two numbers are such that the ratio between them is 4 : 7. If each is increased by 4, the ratio becomes 3 : 5. The lager number is ?

A) 36 B) 48 C) 56 D) 64

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the numbers be 4x and 7x
Then,
$$\eqalign{ & \Leftrightarrow \frac{{4x + 4}}{{7x + 4}} = \frac{3}{5} \cr & \Leftrightarrow 5\left( {4x + 4} \right) = 3\left( {7x + 4} \right) \cr & \Leftrightarrow x = 8 \cr} $$
∴ Larger number :
= 7x
= 7 × 8
= 56

74

the difference between two numbers is 3 and the difference between their squares is 63. Which is the larger number ?

A) 9 B) 12 C) 15 D) Cannot be determined E) None of these

Discuss

Answer & Solution

Answer: Option B

Solution:

Let the number be x and y
Then,
$${x^2} - {y^2} = 63\,\,\,\,\& \,\,\,\,x - y = 3$$
On dividing, we get: x + y = 21
Solving x + y = 21 and x - y = 3,
We get: x = 12 and y = 9
∴ Larger number = 12

75

twenty times a positive integer is less than its square by 96. What is the integer ?

A) 20 B) 24 C) 30 D) Cannot be determined E) none of these

Discuss

Answer & Solution

Answer: Option B

Solution:

Let the integer be x
Them,
$$\eqalign{ & \Leftrightarrow {x^2} - 20x = 96 \cr & \Leftrightarrow {x^2} - 20x - 96 = 0 \cr & \Leftrightarrow \left( {x + 4} \right)\left( {x - 24} \right) = 0 \cr & \Leftrightarrow x = 24 \cr} $$

76

There are two numbers such that the sum of twice the first number and thrice the second number is 100 and the sum of thrice the first number and twice the second number is 120. Which is the larger number ?

A) 12 B) 14 C) 32 D) 35 E) None of these

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the numbers be x and y
Then,
2x + 3y = 100 ..... (i)
And,
3x + 2y = 120 ..... (ii)
Adding (i) and (ii), we get :
5x + 5y = 220
or x + y = 44 ..... (iii)
Subtracting (i) from (ii): we get :
x - y = 20 ..... (iv)
Adding (iii) and (iv), we get :
2x = 64
or x = 32
Putting x = 32 in (iii), we get :
y = 12
Hence, larger number = 32

77

If one-seventh of a number exceeds its eleventh part by 100, then the number is :

A) 770 B) 110 C) 1825 D) 1925

Discuss

Answer & Solution

Answer: Option D

Solution:

Let the number be x
Then,
$$\eqalign{ & \Rightarrow \frac{1}{7}x - \frac{1}{{11}}x = 100 \cr & \Rightarrow \frac{{4x}}{{77}} = 100 \cr & \Rightarrow x = \frac{{7700}}{4} \cr & \Rightarrow x = 1925 \cr} $$

78

The sum of three numbers is 264. If the first number be twice the second and third number be be one-third of the first, then the second number is :

A) 48 B) 54 C) 72 D) 84

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the second number be x
Then, first number = 2x,
And third number = $$\frac{2x}{3}$$
$$\eqalign{ & \therefore 2x + x + \frac{{2x}}{3} = 264 \cr & \Leftrightarrow \frac{{11x}}{3} = 264 \cr & \Leftrightarrow x = \left( {\frac{{264 \times 3}}{{11}}} \right) \cr & \Leftrightarrow x = 72 \cr} $$

79

A, B, C, D and E are five consecutive odd numbers. The sum of A and C is 146. What is the value of E ?

A) 71 B) 75 C) 79 D) 81 E) None of these

Discuss

Answer & Solution

Answer: Option C

Solution:

Let A = x
B = x + 2
C = x + 4
D = x + 6 and
E = x + 8
Then,
$$\eqalign{ & \Rightarrow A + C = 146 \cr & \Rightarrow x + \left( {x + 4} \right) = 146 \cr & \Rightarrow 2x = 142 \cr & \Rightarrow x = 71 \cr & \therefore E = x + 8 \cr & \,\,\,\,\,\,\,\,\,\,\, = 71 + 8 \cr & \,\,\,\,\,\,\,\,\,\,\, = 79 \cr} $$

80

The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of the number ?

A) 3 B) 4 C) 9 D) Cannot be determined 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,
$$\eqalign{ & \Leftrightarrow \left( {10x + y} \right) - \left( {10y + x} \right) = 36 \cr & \Leftrightarrow 9\left( {x - y} \right) = 36 \cr & \Leftrightarrow x - y = 4 \cr} $$