Problems On Numbers - Aptitude MCQ Questions and Solutions with Explanations

Let the fraction be $$\frac{x}{y}$$
Then,
$$\eqalign{ & \Leftrightarrow x + y = 11.....(\text{i}) \cr & \Leftrightarrow \frac{{x + 1}}{{y - 2}} = \frac{2}{3} \cr & \Leftrightarrow 3\left( {x + 1} \right) = 2\left( {y - 2} \right) \cr & \Leftrightarrow 3x - 2y = - 7.....(\text{ii}) \cr} $$
Solving (i) and (ii), we get:
x = 3 and y = 8
So, the fraction is $$\frac{3}{8}$$

Let the five successive odd number be,
x, x + 2, x + 4, x + 6, x + 8
Then, according to given information,
185 = x + x + 2 + x + 4 + x + 6 + x + 8
$$\eqalign{ & \Leftrightarrow 185 = 5x + 20 \cr & \Leftrightarrow 5x = 185 - 20 \cr & \Leftrightarrow 5x = 165 \cr & \Leftrightarrow x = 33 \cr} $$
Highest number = 33 + 8 = 41

Let the numbers be 3x, 4x and 6x
Then,
$$\eqalign{ & \Leftrightarrow 3x \times 4x \times 6x = 1944 \cr & \Leftrightarrow 72{x^3} = 1944 \cr & \Leftrightarrow {x^3} = 27 \cr & \Leftrightarrow x = 3 \cr} $$
∴ Largest number = 6x = 18

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

Let the ten's digit be x
Then, unit's digit = x + 2
∴ Number :
= 10x + (x + 2)
= 11x + 2
Sum of digits :
= x + (x + 2)
= 2x + 2
$$\eqalign{ & \therefore \left( {11x + 2} \right)\left( {2x + 2} \right) = 144 \cr & \Leftrightarrow 22{x^2} + 26x - 140 = 0 \cr & \Leftrightarrow 11{x^2} + 13x - 70 = 0 \cr & \Leftrightarrow \left( {x - 2} \right)\left( {11x + 35} \right) = 0 \cr & \Leftrightarrow x = 2 \cr} $$
Hence, required number:
= 11x + 2
= 11× 2 + 2
= 24

Let the denominator be x
Then, numerator = x + 5
Now,
$$\eqalign{ & \Leftrightarrow \frac{{x + 5}}{x} - \frac{{x + 5}}{{x + 5}} = \frac{5}{4} \cr & \Leftrightarrow \frac{{x + 5}}{x} = \frac{5}{4} + 1 \cr & \Leftrightarrow \frac{{x + 5}}{x} = \frac{9}{4} \cr & \Leftrightarrow \frac{{x + 5}}{x} = 2\frac{1}{4} \cr} $$
So, the fraction is $$2\frac{1}{4}$$

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow \left( {\frac{1}{5}x + 4} \right) = \left( {\frac{1}{4}x - 10} \right) \cr & \Leftrightarrow \frac{x}{{20}} = 14 \cr & \Leftrightarrow x = 14 \times 20 \cr & \Leftrightarrow x = 280 \cr} $$

Let the ten's digit be x
Then, units digit = x + 3
Number = 10x + (x + 3)
              = 11x + 3
Sum of digits = x + (x + 3)
                      = 2x + 3
$$\eqalign{ & \therefore \frac{{11x + 3}}{{2x + 3}} = \frac{4}{1} \cr & \Leftrightarrow 11x + 3 = 8x + 12 \cr & \Leftrightarrow 3x = 9 \cr & \Leftrightarrow x = 3 \cr} $$
Hence, Required number
= 11x + 3
= 11 × 3 + 3
= 36

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

Let the ten's digit be x and unit's digit be y
Then, number = 10x + y
Number obtained by interchanging the digits = 10y + x
∴ (10x + y) + (10y + x)
= 11(x + y), which is divisible by 11