Question Detail

Two numbers differ by 5. If their product is 336, then sum of two number is

Answer: Option D

Explanation:

Friends you remember,
\begin{aligned}
=> (x+y)^2 = (x-y)^2 + 4xy
\end{aligned}

\begin{aligned}
=> (x+y)^2 = (5)^2 + 4(336)
\end{aligned}

\begin{aligned}
=> (x+y) = \sqrt{1369} = 37
\end{aligned}

1. A father is twice as old as his son. 20 years ago, the age of the father was 12 times the age of the son. The present age of the father (in years) is

Answer: Option C

Explanation:

Let sons age = x. Then, fathers age = 2x.
12(x—2O) = (2x - 20) so x = 22
Fathers present age = 44 years

2. The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is

Answer: Option B

Explanation:

We know
\begin{aligned}
(x + y)^2 = x^2 + y^2 + 2xy
\end{aligned}

\begin{aligned}
=> (x + y)^2 = 289 + 2(120)
\end{aligned}

\begin{aligned}
=> (x + y) = \sqrt{529} = 23
\end{aligned}

3. 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}

4. 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)

5. 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}