Question Detail

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}

1. Sum of a rational number and its reciprocal is 13/6. Find the number

Answer: Option B

Explanation:

\begin{aligned} => x + \frac{1}{x} = \frac{13}{6} \end{aligned}
\begin{aligned} => \frac{x^2+1}{x} = \frac{13}{6} \end{aligned}
\begin{aligned} => 6x^2-13x+6 = 0 \end{aligned}

\begin{aligned} => 6x^2-9x-4x+6 = 0 \end{aligned}
\begin{aligned} => (3x-2)(2x-3) \end{aligned}
\begin{aligned} => x = 2/3 or 3/2 \end{aligned}

2. 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.

3. 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}

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}