Question Detail

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)

1. 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}

2. 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

3. 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}

4. 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}

5. 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}