Question

A number plus two-third of itself, plus one-half of itself, plus one-seventh of itself equals 97. Find the number.

Answer 1

Hint: Let us assume the number be x. Here, two-third of the number is equivalent to \[\dfrac{2}{3}x\] and in the same manner we can find one-half of the number and one-seventh of the number in terms of x. We will sum up all the terms which are equal to 97.

Complete step by step answer:

In this question, we have to find the number which when added with two-third of itself, plus one-half of itself, plus one-seventh of itself equals 97.
So, let us assume that the number is x.
Then, two-third of the number equals to \[\dfrac{2}{3}x\]
And, one-half of the number equals to \[\dfrac{1}{2}x\]
Also, one-seventh of the number equals to \[\dfrac{1}{7}x\]
Now, as we have given in question that the number is added with two-third of itself, one-half of itself and one-seventh of itself equals 97.
So, we can write this as \[x+\dfrac{2}{3}x+\dfrac{1}{2}x+\dfrac{1}{7}x=97\]
Now, taking L.C.M. on left hand side
\[\Rightarrow \dfrac{42x+28x+21x+6x}{42}=97\]
Adding all the terms of numerator in L.H.S.
\[\Rightarrow \dfrac{97x}{42}=97\]
\[\Rightarrow x=\dfrac{97\times 42}{97}\]
Cancelling out 97 from numerator and denominator of right hand side
\[\Rightarrow x=42\]
Hence, the number is 42.

Note: Any ratio of the number itself is the ratio multiplied with number, that is, x by y^th of number is the same as \[\dfrac{x}{y}\left( \text{number} \right)\]. There are few possible mistakes which students can make in hurry are writing \[\dfrac{2}{3}x=\dfrac{3}{2}x\], also, sometimes in hurry they forget to add the number itself which leads to wrong answer.