Question

A bookshelf holds paperback and hardcover books. The ratio of paperback books to hardcover books is 22 to 3. How many paperback books are on the shelf? If 18 paperback books were removed from the shelf and replaced with 18 hardcover books, the resulting ratio of paperback books to hardcover books on the shelf would be 4 to 1.
(a) 158
(b) 200
(c) 198
(d) 164

Answer 1

Hint: To solve the given question, we will assume that the total number of paperback books is x and the total number of hardcover books are y. Now, we will take their ratio and equate it to 22:3. Then we will assume that the new number of paperback books are x’ where x’ = x – 18 and the new number of hardcover books are y’ where y’ = y + 18. We will take their ratio and equate it to 4:1. From here, we will get two equations in two variables which we will solve with the substitution method and find the value of x.

Complete step by step solution:
To start with, we will assume that the total number of paperback books is x and the total number of hardcover books are y. Now, according to the question, their ratio is 22:3. Thus, we will get the following equation
\[\dfrac{x}{y}=\dfrac{22}{3}\]
\[\Rightarrow 3x=22y......\left( i \right)\]
Now, 18 paperback books are removed, so we will assume the new number of paperback books = x’. hence, we can say that,
\[{{x}^{'}}=x-18.....\left( ii \right)\]
Now, 18 hardcover books are added so we will assume the new number of hardcover books as y’. Hence, we can say that,
\[{{y}^{'}}=y+18.....\left( iii \right)\]
It is given that the ratio of x’ to y’ is 4:1. Thus, we will get the following equation.
\[\dfrac{{{x}^{'}}}{{{y}^{'}}}=\dfrac{4}{1}\]
\[\Rightarrow {{x}^{'}}=4{{y}^{'}}.....\left( iv \right)\]
Now, we will substitute the values of x’ and y’ from (ii) and (iii) to (iv). Thus, we will get,
\[x-18=4\left( y+18 \right)\]
\[\Rightarrow x-18=4y+72\]
\[\Rightarrow x-4y=90.....\left( v \right)\]
Now, (i) and (v) are a pair of linear equations in two variables. We will solve these equations by the substitution method. From (i), we can say that,
\[\Rightarrow 3x=22y\]
\[\Rightarrow y=\dfrac{3x}{22}......\left( vi \right)\]
Now, we will put the value of y from (vi) to (v). Thus, we will get,
\[\Rightarrow x-4\left( \dfrac{3x}{22} \right)=90\]
\[\Rightarrow x-\dfrac{12x}{22}=90\]
On taking the LCM, we will get,
\[\Rightarrow \dfrac{22x-12x}{22}=90\]
\[\Rightarrow \dfrac{10x}{22}=90\]
\[\Rightarrow x=\dfrac{90\times 22}{10}\]
\[\Rightarrow x=198\]
Thus, there are 198 paperback books on the shelf.
Hence, option (c) is the correct answer.

Note: One thing to remember is that if two equations are of the form
\[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\]
\[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]
and we are solving these equations to find the single values of x and y using the substitution method then the ratio of \[{{a}_{1}}\] to \[{{a}_{2}}\] should not be equal to \[{{b}_{1}}\] to \[{{b}_{2}}\] i.e.
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\]
If they will be equal, either will get no solution or infinite solutions.