Question

Sixty – five percent of a number is 21 less than four – fifth of that number. What is the number?
A. 100
B. 120
C. 140
D. 160

Answer 1

Hint: We will first assume the number we require to find to be x and then apply the same conditions as given in the question to form an equation in one variable and thus solve it.

Complete step-by-step answer:
Let the number which we are required to find be x.
Now, since it is given to us that sixty – five percent of that number is 21 less than four – fifth of that number.
We will first find the sixty – five percent of the number x that is $\dfrac{{65}}{{100}}x$. Now, it is 21 less than four – fifth of the number. It means that $\dfrac{{65}}{{100}}x$ is 21 less than four – fifth of x.
We can write this as:-
$ \Rightarrow \dfrac{{65}}{{100}}x = \dfrac{4}{5}x - 21$
Now, we will club on one side, the terms with x and the constants on the other side of the equation and we will obtain the following expression:-
$ \Rightarrow \dfrac{4}{5}x - \dfrac{{65}}{{100}}x = 21$
Taking the least common multiple of the denominator on the left hand side of the above expression, we will then obtain the following expression:-
$ \Rightarrow \left( {\dfrac{{80 - 65}}{{100}}} \right)x = 21$
Now, if we simplify the above expression, we will then obtain the following equation:-
$ \Rightarrow \dfrac{{15}}{{100}}x = 21$
Simplifying the left hand side further, we will obtain:-
$ \Rightarrow \dfrac{3}{{20}}x = 21$
Cross – multiplying in above expression, we get:-
$ \Rightarrow x = \dfrac{{21 \times 20}}{3}$
On doing the calculations, we get: x = 140.

Hence, the required answer is (C) 140.

Note:
The students must note that we subtracted 21 from the four – fifth side of number not from sixty – five percent side because it is given that sixty – five percent is 21 less than four – fifth of that number, so when 65% is already less, if we subtract from this side only, we will create an imbalance.
The students must note that here we created an equation in one variable for which we required only one equation to solve. We require as many equations as many unknown variables we have in the solution.