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ICSE Class X Math 2013 Paper

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Question : 46 of 46
Marks: +1, -0
A shopkeeper purchases a certain number of books for ₹ 960. If the cost per book was ₹ 8 less, the number of books that could be purchased for ₹ 960 would be 4 more. Write an equation, taking the original cost of each book to be ₹x\text{₹} x, and solve it to find the original cost of the books.
Solution:  
Given the original cost of each book be ₹x\text{₹} x.
Total cost =₹960=\text{₹} 960 (Given)
∴    \therefore \;\; Number of books for 960=  960x960=\;\frac{960}{x}
If the cost per book was ₹8\text{₹} 8 less, (i.e., x−8x-8 ) then
   Number of books   =  960x−8\;\text{ Number of books }\;=\;\frac{960}{x-8}
According to question,
  960x−8  =  960x+4\;\frac{960}{x-8}\;=\;\frac{960}{x}+4
  960x−8−  960x  =4\;\frac{960}{x-8}-\;\frac{960}{x}\;=4
960[  x−x+8x(x−8)]  =4960 \left[ \;\frac{x-x+8}{x(x-8)} \right] \;=4
  8x2−8x  =  1240\;\frac{8}{x^2-8 x}\;=\;\frac{1}{240}
⇒    x2−8x=1,920\Rightarrow \;\; x^2-8x = 1,920
⇒    x2−8x−1,920=0\Rightarrow \;\; x^2-8x-1,920 = 0
  ⇒x2−48x+40x−1,920=0\;\Rightarrow x^2-48x+40x-1,920=0
  ⇒x(x−48)+40(x−48)=0\;\Rightarrow x(x-48)+40(x-48)=0
  ⇒    (x−48)(x+40)=0\;\Rightarrow \;\; (x-48)(x+40)=0
  x=48   or   x=−40\;x=48 \;\text{ or }\; x=-40
∵−40\because -40 is not possible.
Hence, the original cost of each book =₹48=\text{₹} 48 .
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