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The average price of 10 books is Rs.12 while the average price of 8 of these books is Rs.11.75. Of the remaining two books, if the price of one book is $60\%$ more than the price of the other, what is the price of each of these two books?


Rs. 5, Rs.8


Rs. 8, Rs. 12.8


Rs. 10, Rs. 16


Rs. 12, Rs. 19.2

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Option(C) is correct

Total cost of 10 books = Rs. 120

Total cost of 8 books = Rs. 94 

$⇒$ The cost of 2 books = Rs. 26

Let the price of each book be $x$ and $y$.

⇒ x + y = 26 ---------------- (1)

Given that the price of 1 book is $60\%$ more than the other price


\(y=\dfrac{26\times 100}{260}\)


Substituting $Y = 10$ in (1) we get,

$x + 10 = 26$

$x$ = 16

Edit: Option choices have been changed after has been pointed out by Parag.

(6) Comment(s)

Lakshmi Nath

check by option method


$y=\dfrac{26 \times 100}{260}$

How 260 came there?


If you solve the given equation, $\dfrac{160y}{100}+y=26$, you will see the term 260 appearing.

Here is how.

$\because \dfrac{160y}{100}+y=26$

$\Rightarrow y\left(\dfrac{160}{100}+1\right)=26$

$\Rightarrow y\left(\dfrac{160+100}{100}\right)=26$

$\Rightarrow y\left(\dfrac{260}{100}\right)=26$

$\Rightarrow y=\dfrac{26}{260/100}$

$\Rightarrow y=\dfrac{26\times 100}{260}$


you must give better options. only 16 is 60% more than 10. This question can be solved my just at options.


Thank you Parag for notifying it, modified the options.


All options mentioned above are sixteen percent more than other