14
Q:

Five years ago, the average age of A, B, C and D was 45 years. With E Joining them now, the average of all the five is 49 years. The age of E is?

Q:

The ratio of Karthik’s and Kalyan’s ages is 4: 5. If the difference between the present age of Kalyan and the age of Karthik 5 years hence is 3 years, then what is the total of present ages of Karthik and Kalyan?

 A) 56 years B) 62 years C) 69 years D) 72 years

Answer & Explanation Answer: D) 72 years

Explanation:

Let us consider Karthik as K1 and Kalyan as K2.
Given $\inline&space;\small&space;\frac{K1}{K2}&space;=&space;\frac{4}{5}$

and K2 - (K1 + 5) = 3
K2 - K1 = 8
K2 = 8 + K1
Now,

$\inline&space;\small&space;\frac{K1}{(8+K1)}$ $\inline&space;\small&space;=&space;\frac{4}{5}$
K1 = 32 years
Therefore K2 = 40 years.
K1 + K2 = 72 years.

4 72
Q:

Ten years ago, Kumar was thrice as old as Sailesh was but 10 years hence, he will be only twice as old. Find Kumar’s present age ?

 A) 50 years B) 70 years C) 60 years D) 40 years

Answer & Explanation Answer: B) 70 years

Explanation:

Let Kumar’s present age be "x" years and Sailesh’s present age be "y" years.
Then, according to the first condition,
x - 10 = 3(y - 10) $\inline&space;\small&space;\Rightarrow$ x - 3y = - 20 ........(1)
Now, Kumar’s age after 10 years = (x + 10) years.
Sailesh’s age after 10 years = (y + 10) years.
(x + 10) = 2 (y + 10) $\inline&space;\small&space;\Rightarrow$ x - 2y = 10 ......(2)

Solving (1) and (2),

we get x = 70 and y = 30

Kumar’s present age = 70 years    and

Sailesh’s present age = 30 years.

4 63
Q:

Ten years ago, A was half of B in age. If the ratio of their present ages is 3:4, then what will be the total of their present ages?

Let A's age 10 years ago be x years . Then B's age 10 years ago be 2x years

$\inline \fn_cm \Rightarrow \frac{x+10}{2x+10}=\frac{3}{4}$

$\inline \fn_cm \Rightarrow4(x+10)=3(2x+10)$

$\inline \fn_cm \Rightarrow x=5$

$\inline \fn_cm \therefore$ Total of their present ages = x + 10 + 2x + 10

= 3x+20

= 3 * 5 + 20 = 35 years

299
Q:

The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ratio of number of boys to the number of girls in the class is

Suppose the number of boys and girls are x and y respectively

$\inline \fn_cm \Rightarrow \frac{x\times 16.4 + y\times 15.4}{x+y}=15.8$

$\inline \fn_cm \Rightarrow 16.4x + 15.4y = 15.8x+15.8y$

$\inline \fn_cm \Rightarrow0.6x=0.4y$

$\inline \fn_cm \therefore \frac{x}{y}=\frac{0.4}{0.6}=\frac{2}{3}=2:3$

999
Q:

The ratio of the father's age to the son's age is 3 : 1. The product of their ages is 147. The ratio of their ages after 5 years will be?

Let father's age be 3x and son's age be x years

$\inline \fn_cm \Rightarrow 3x\times x=147$

$\inline \fn_cm \Rightarrow x^{2}=49\Rightarrow x=7$

Father's age after 5 years = 3x + 5 = 26 years

Son's age after 5 years = x + 5 = 12 years

$\inline \fn_cm \therefore$ Ratio of father's and son's ages = 26 : 12 = 13 : 6