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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. Which among following is not a prime number ?
- 97
- 33
- 45
- 72
Answer: Option A
2. find the number, difference between number and its 3/5 is 50.
- 120
- 123
- 124
- 125
Answer: Option D
Explanation:
Let the number = x,
Then, x-(3/5)x = 50,
=> (2/5)x = 50 => 2x = 50*5,
=> x = 125
3. 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}
4. If one third of one fourth of number is 15, then three tenth of number is
- 34
- 44
- 54
- 64
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}
5. Find the number which when multiplied by 15 is increased by 196
- 10
- 12
- 14
- 16
Answer: Option C
Explanation:
Let the number be x.
Then, 15x = x + 196
=› 14 x= 196
=› x = 14.
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