Number System - Aptitude MCQ Questions and Solutions with Explanations

A + B + C + D = 60 -----------> (1)

Now According to Question,
A = $$\frac{1}{2}$$ (B + C + D)
2A = B + C + D
Put the value of B + C + D in Equation 1
A + 2A = 60
3A = 60
A = 20

Now again according to Question
B = $$\frac{1}{3}$$ (A + C + D)
A + C + D = 3B
By Putting the Value of A + C + D in Equation 1
we get,
B + 3B = 60
4B = 60
B = 15

Now again according to Question
C = $$\frac{1}{4}$$ (A + B + D)
A + B + D = 4C
By Putting the Value of A + B + D in equation 1
We get C + 4C = 60
5C = 60
C = 12

Since A + B + C + D = 60
20 + 15 + 12 + D = 60
D = 60 - 47 = 13
Amount of D = Rs. 13

$$ = \frac{1}{{15}} + \frac{1}{{35}} + \frac{1}{{63}} + \frac{1}{{99}} + \frac{1}{{143}}$$
$$ = \frac{1}{{3 \times 5}} + \frac{1}{{5 \times 7}} + \frac{1}{{7 \times 9}} + \frac{1}{{9 \times 11}} + \frac{1}{{11 \times 13}}$$
$$ = \frac{1}{2}\left[ {\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{13}} \right]$$
$$ = \frac{1}{2}\left[ {\frac{1}{3} - \frac{1}{{13}}} \right]$$
$$ = \frac{1}{2} \times \frac{{10}}{{39}}$$
$$ = \frac{5}{{39}}$$

Total students = z
Total boys = x
Total girls = z - x
⇒ $$\frac{{{\text{Number of girls}}}}{{{\text{Total students}}}}$$
= $$\frac{z - x}{{z}}$$
= 1 - $$\frac{x}{{z}}$$

Let the property is 'P' units
According to the question,
Given part,
$$\eqalign{ & = \frac{1}{4}P + \frac{1}{2}P + \frac{1}{5}P \cr & = \frac{{5P + 10P + 4P}}{{20}} \cr & = \frac{{19}}{{20}}P \cr} $$
Part of property given away $$\frac{19}{{20}}$$

38950 is completely divisible by 410 only.
Hence, Rs. 410 is the correct answer.

On dividing 5000 by 73, we get 36 remainder
Required number = 5000 + (73 - 36) = 5037

$$\eqalign{ & \Rightarrow 348 \times 265 \cr & = \left( {350 - 2} \right) \times 265 \cr & = \left( {350 \times 265} \right) - \left( {2 \times 265} \right) \cr & = \left\{ {\left( {300 + 50} \right) \times 265} \right\} - 530 \cr & = \left( {300 \times 265} \right) + \left( {50 \times 265} \right) - 530 \cr & = 79500 + 13250 - 530 \cr & = 92750 - 530 \cr & = 92220 \cr} $$

On dividing 534677 by 777 we get 101 as remainder.
∴ (Divisor) - (Remainder)
= (777 - 101)
= 676

Clearly, 19 is a prime number.

$$\eqalign{ & \Rightarrow \frac{{13}}{1} = \frac{{13w}}{{\left( {1 - w} \right)}} \cr & \Rightarrow \frac{w}{{\left( {1 - w} \right)}} = 1 \cr & \Rightarrow w = 1 - w \cr & \Rightarrow 2w = 1 \cr & \Rightarrow w = \frac{1}{2} \cr & \therefore {\left( {2w} \right)^2} \cr & = 4{w^2} \cr & = 4 \times \frac{1}{4} \cr & = 1 \cr} $$