Question Detail

Sum of two numbers is 25 and their difference is 13. Find their product.

Answer: Option C

Explanation:

Friends, this sort of question is quite important in competitive exams, whenever any question come which have relation between sum, product and difference, this formula do the magic:
$\begin{aligned} => (x+y)^2 = (x-y)^2 + 4xy \end{aligned}$

$\begin{aligned} <=> (25)^2 = (13)^2 + 4xy \end{aligned}$

$\begin{aligned} <=> 4xy = (25)^2 - (13)^2 \end{aligned}$
$\begin{aligned} <=> xy = \frac{456}{4} = 114 \end{aligned}$

1. find the number, If 50 is subtracted from two-third of number, the result is equal to sum of 40 and one-fourth of that number.

Answer: Option B

Explanation:

Let the number is x,
$\begin{aligned} => \frac{2}{3}x-50 = \frac{1}{4}x + 40 \end{aligned}$

$\begin{aligned} <=> \frac{2}{3}x-\frac{1}{4}x = 90 \end{aligned}$
$\begin{aligned} <=> \frac{5x}{12} = 90 \end{aligned}$
$\begin{aligned} <=> x = 216 \end{aligned}$

2. Find the number which when multiplied by 15 is increased by 196

Answer: Option C

Explanation:

Let the number be x.
Then, 15x = x + 196
=› 14 x= 196
=› x = 14.

3. If one third of one fourth of number is 15, then three tenth of number is

Answer: Option C

Explanation:

Let the number is x,
$\begin{aligned} \frac{1}{3} of\frac{1}{4} * x = 15 \end{aligned}$

$\begin{aligned} => x = 15\times 12 = 180 \end{aligned}$

$\begin{aligned} => so \frac{3}{10} \times x = \frac{3}{10} \times 180 = 54 \end{aligned}$

4. A father is twice as old as his son. 20 years ago, the age of the father was 12 times the age of the son. The present age of the father (in years) is

Answer: Option C

Explanation:

Let sons age = x. Then, fathers age = 2x.
12(x—2O) = (2x - 20) so x = 22
Fathers present age = 44 years

5. Product of two natural numbers is 17. Then, the sum of reciprocals of their squares is

$\begin{aligned} \frac{290}{289} \end{aligned}$
$\begin{aligned} \frac{1}{289} \end{aligned}$
$\begin{aligned} \frac{290}{90} \end{aligned}$
$\begin{aligned} \frac{290}{19} \end{aligned}$

Answer: Option A

Explanation:

If the numbers are a, b, then ab = 17,
as 17 is a prime number, so a = 1, b = 17.

$\begin{aligned} \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{1^2} + \frac{1}{17^2} \end{aligned}$
$\begin{aligned} = \frac{290}{289} \end{aligned}$