Question Detail

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}

1. if the sum of \begin{aligned} \frac{1}{2} \end{aligned} and \begin{aligned} \frac{1}{5} \end{aligned} of a number exceeds \begin{aligned} \frac{1}{3} \end{aligned} of the number by \begin{aligned} 7\frac {1}{3} \end{aligned}, then number is

Answer: Option B

Explanation:

Seems a bit complicated, isnt'it, but trust me if we think on this question with a cool mind then it is quite simple...
Let the number is x,
then, \begin{aligned} (\frac{1}{2}x + \frac{1}{5}x) - \frac{1}{3}x = \frac{22}{3} \end{aligned}

\begin{aligned}
=> \frac{11x}{30} = \frac{22}{3}
\end{aligned}

\begin{aligned}
=> x = 20
\end{aligned}

2. 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}

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 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}

5. A number is doubled and 9 is added. If resultant is trebled, it becomes 75. What is that number

Answer: Option A

Explanation:

=> 3(2x+9) = 75
=> 2x+9 = 25
=> x = 8