Question Detail

Sum of three numbers 264, If the first number be twice then second and third number be one third of the first, then the second number is

Answer: Option C

Explanation:

Let the second number is x, then first is 2x, and third is 1/3(2x)
\begin{aligned}
=>2x + x + \frac{2x}{3} = 264 <=> \frac{11x}{3} = 264
\end{aligned}

\begin{aligned}
=> x = 72
\end{aligned}

1. 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

2. Sum of two numbers is 40 and their difference is 4. The ratio of the numbers is

10:3
5:9
11:9
13:9

Answer: Option C

Explanation:

\begin{aligned}
=> \frac{(x+y)}{(x-y)} = \frac{40}{4}
\end{aligned}

\begin{aligned}
=> (x+y)= 10(x-y)
\end{aligned}

\begin{aligned}
=> 9x = 11y => \frac{x}{y} = \frac{11}{9}
\end{aligned}

3. The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is

Answer: Option B

Explanation:

We know
\begin{aligned}
(x + y)^2 = x^2 + y^2 + 2xy
\end{aligned}

\begin{aligned}
=> (x + y)^2 = 289 + 2(120)
\end{aligned}

\begin{aligned}
=> (x + y) = \sqrt{529} = 23
\end{aligned}

4. 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}

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}