option (D)

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In $2010$, a library contained a total of $11500$ books in two categories -fiction and nonfiction. In $2015$, the library contained a total of $12760$ books in these two categories. During this period, there was $10\%$ increase in the fiction category while there was $12\%$ increase in the non-fiction category. How many fiction books were in the library in $2015?$

- $6000$
- $6160$
- $5500$
- $6600$

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Given that, in $2010$ total books of two categories (fiction and non-fiction) are $11500.$

Let be $x$ fiction categories books, then non-fiction books will be $11500 – x.$

And, in $2015,$ the library contained a total of $12760$ books in these two categories.

In $2015,$ there was $10\%$ increase in the fiction category while there was $12\%$ increase in the non-fiction category. Then,

$\Rightarrow x\times\frac{110}{100}+(11500-x)\times\frac{112}{100}=12760$

$\Rightarrow \frac{11}{10}x+(11500-x)\frac{28}{25}=12760$

$\Rightarrow \frac{55x+(11500-x)56}{50}=12760$

$\Rightarrow 55x+644000-56x=638000$

$\Rightarrow 644000-x=638000$

$\Rightarrow -x=63800-644000$

$\Rightarrow -x=-6000 $

$\Rightarrow x=6000$

$\therefore$ The number of fiction books in the library in $2015 = x \times \left(\frac{110}{100}\right)$

$\qquad = 6000 \times \left( \frac{110}{100} \right)$

$ \qquad = 6600$

Correct answer : D

Let be $x$ fiction categories books, then non-fiction books will be $11500 – x.$

And, in $2015,$ the library contained a total of $12760$ books in these two categories.

In $2015,$ there was $10\%$ increase in the fiction category while there was $12\%$ increase in the non-fiction category. Then,

$\Rightarrow x\times\frac{110}{100}+(11500-x)\times\frac{112}{100}=12760$

$\Rightarrow \frac{11}{10}x+(11500-x)\frac{28}{25}=12760$

$\Rightarrow \frac{55x+(11500-x)56}{50}=12760$

$\Rightarrow 55x+644000-56x=638000$

$\Rightarrow 644000-x=638000$

$\Rightarrow -x=63800-644000$

$\Rightarrow -x=-6000 $

$\Rightarrow x=6000$

$\therefore$ The number of fiction books in the library in $2015 = x \times \left(\frac{110}{100}\right)$

$\qquad = 6000 \times \left( \frac{110}{100} \right)$

$ \qquad = 6600$

Correct answer : D