Question Detail

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}

1. A number is doubled and 9 is added. If resultant is trebled, it becomes 75. What is that number

Answer: Option A

Explanation:

=> 3(2x+9) = 75
=> 2x+9 = 25
=> x = 8

2. Which among following is not a prime number ?

Answer: Option A

3. 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}

4. 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}