Question

In a library, 50 percent of the total number of books is of Marathi. The books of English are \[\dfrac{1}{3}\] of Marathi books, the books on mathematics are 25 percent on English books, the remaining 560 books are of other subjects. What's the total number of books in the library?

Answer 1

Hint:
We will assume the total number of books to be x, after that we will calculate all the total number of books for the different subjects in terms of x. Once we have calculated all the books present in the library for each subject in terms of x, we will add all the books and equate it to x because both represent the total number of books present in the library. Hence, we will get the answer.

Complete step by step solution:
Let us assume the total number of books to be x,
Now according to the question,
 \[50\% \] of the total number of books is of Marathi, we get
 $ \Rightarrow {B_{marathi}} = (50\% ) \times x$
We know that
 $a\% = \dfrac{a}{{100}}$
Hence, we get
 $ \Rightarrow {B_{marathi}} = \dfrac{{50}}{{100}} \times x$
So, we have,
 $ \Rightarrow {B_{marathi}} = \dfrac{1}{2} \times x$
Hence, the number of Marathi books in the library are
 $ \Rightarrow {B_{marathi}} = \dfrac{x}{2}$ … (1)
Now, according to the question,
The books of English are $\dfrac{1}{3}$ of Marathi books, that is
 $ \Rightarrow {B_{english}} = {B_{marathi}} \times \dfrac{1}{3}$

On substituting, ${B_{marathi}} = \dfrac{x}{2}$ , we get
 $ \Rightarrow {B_{english}} = \dfrac{x}{2} \times \dfrac{1}{3}$
Hence, the number of English books in the library are
 $ \Rightarrow {B_{english}} = \dfrac{x}{6}$ … (2)
Now, according to the question,
The books of Maths are $\dfrac{1}{3}$ of English books, that is
 $ \Rightarrow {B_{Maths}} = (25)\% \times {B_{english}}$
We know that
 $a\% = \dfrac{a}{{100}}$
Hence, we get
 $ \Rightarrow {B_{maths}} = \dfrac{{25}}{{100}} \times {B_{english}}$
On substituting ${B_{english}} = \dfrac{x}{6}$ , we get
 $ \Rightarrow {B_{maths}} = \dfrac{{25}}{{100}} \times \dfrac{x}{6}$
Hence on simplification we get,
 $ \Rightarrow {B_{maths}} = \dfrac{1}{4} \times \dfrac{x}{6}$
Hence, we have,
 $ \Rightarrow {B_{maths}} = \dfrac{x}{{24}}$ … (3)
Now, the remaining 560 books present in the library are of other subjects.
Hence, the total number of books present in the library x can also be given as the sum of Marathi, English, Maths, and the other 560 books present in the library. Hence, we get
Total books ${B_{total}}$ as
 $ \Rightarrow {B_{total}} = {B_{Marathi}} + {B_{English}} + {B_{English}} + 560$ … (4)

We also assumed the total number of books to be x, that is
 $ \Rightarrow {B_{total}} = x$ .. (5)
Using (4) and (5), we get
 $ \Rightarrow x = {B_{Marathi + }}{B_{English}} + {B_{English}} + 560$
Using (1), (2), and (3) we get
 $ \Rightarrow x = \dfrac{x}{2} + \dfrac{x}{6} + \dfrac{x}{{24}} + 560$
Now taking LCM at RHS, we get
 $ \Rightarrow x = \dfrac{{12x}}{{24}} + \dfrac{{4x}}{{24}} + \dfrac{x}{{24}} + \dfrac{{24 \times 560}}{{24}}$
On simplification we get,
 $ \Rightarrow x = \dfrac{{12x + 4x + x + 24 \times 560}}{{24}}$
Hence adding the terms in numerator, we get,
 $ \Rightarrow x = \dfrac{{17x + 13440}}{{24}}$
By cross-multiplying, we get
 $ \Rightarrow 24 \times x = 17x + 13440$
Taking all the terms with x to LHS, we get
 $ \Rightarrow 7x = 13440$
On dividing the equation by 7 we get,
 $ \Rightarrow x = \dfrac{{13440}}{7}$
On simplification we get,
 $ \Rightarrow x = 1920$

Hence, the total number of books present in the library are 1920.

Note:
In the above question, the number which we calculated MUST be a positive integer, if we get anything else, our answer is wrong somewhere. Also, be very careful about what the question is saying, here students may get confused and calculate all the number of books from each subject with respect to the total number of books whereas according to the question different subject’s number of books are related to different subject’s number of books, that is, all subject’s total number of books are not directly related to the total number of books.