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. 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. if the sum of \begin{aligned} \frac{1}{2} \end{aligned} and \begin{aligned} \frac{1}{5} \end{aligned} of a number exceeds \begin{aligned} \frac{1}{3} \end{aligned} of the number by \begin{aligned} 7\frac {1}{3} \end{aligned}, then number is

Answer: Option B

Explanation:

Seems a bit complicated, isnt'it, but trust me if we think on this question with a cool mind then it is quite simple...
Let the number is x,
then, \begin{aligned} (\frac{1}{2}x + \frac{1}{5}x) - \frac{1}{3}x = \frac{22}{3} \end{aligned}

\begin{aligned}
=> \frac{11x}{30} = \frac{22}{3}
\end{aligned}

\begin{aligned}
=> x = 20
\end{aligned}

4. Which among following is not a prime number ?

Answer: Option A

5. Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number

Answer: Option D

Explanation:

If the number is x,

Then, x + 17 = 60/x

x2 + 17x - 60 = 0

(x + 20)(x - 3) = 0

x = 3, -20, so x = 3 (as 3 is positive)