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. 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}

2. Sum of three numbers 264, If the first number be twice then second and third number be one third of the first, then the second number is

Answer: Option C

Explanation:

Let the second number is x, then first is 2x, and third is 1/3(2x)
\begin{aligned}
=>2x + x + \frac{2x}{3} = 264 <=> \frac{11x}{3} = 264
\end{aligned}

\begin{aligned}
=> x = 72
\end{aligned}

3. 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)

4. 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

5. 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