Question Detail

Find the number, when 15 is subtracted from 7 times the number, the result is 10 more than twice of the number

Answer: Option A

Explanation:

Let the number be x.
7x -15 = 2x + 10 => 5x = 25 => x = 5

1. 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

2. 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}

3. Find the number which when multiplied by 15 is increased by 196

Answer: Option C

Explanation:

Let the number be x.
Then, 15x = x + 196
=› 14 x= 196
=› x = 14.

4. 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}

5. Sum of two numbers is 25 and their difference is 13. Find their product.

Answer: Option C

Explanation:

Friends, this sort of question is quite important in competitive exams, whenever any question come which have relation between sum, product and difference, this formula do the magic:
\begin{aligned}
=> (x+y)^2 = (x-y)^2 + 4xy
\end{aligned}

\begin{aligned}
<=> (25)^2 = (13)^2 + 4xy
\end{aligned}

\begin{aligned}
<=> 4xy = (25)^2 - (13)^2
\end{aligned}
\begin{aligned}
<=> xy = \frac{456}{4} = 114
\end{aligned}