Question Detail

find the number, difference between number and its 3/5 is 50.

Answer: Option D

Explanation:

Let the number = x,
Then, x-(3/5)x = 50,
=> (2/5)x = 50 => 2x = 50*5,
=> x = 125

1. 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}

2. 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}

3. Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is

Answer: Option C

Explanation:

Let the three integers be x, x+2 and x+4.
Then, 3x = 2(x+4)+3,
x= 11
Therefore, third integer x+4 = 15

4. The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is

Answer: Option B

Explanation:

Let the numbers be a, b and c.
Then,
\begin{aligned}
a^2 + b^2 + c^2 = 138
\end{aligned}
and (ab + bc + ca) = 131
\begin{aligned}
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
\end{aligned}
= 138 + 2 x 131 = 400

\begin{aligned}
=> (a + b + c) = \sqrt{400} = 20.
\end{aligned}

5. Difference between a two-digit number and the number obtained by interchanging the two digits is 36, what is the difference between two numbers

Answer: Option B

Explanation:

Let the ten digit be x, unit digit is y.
Then (10x + y) - (10y + x) = 36
=> 9x - 9y = 36
=> x - y = 4.