Question

A shopkeeper buys some books at the the rate of 11 for rupees 10 and also same number of books of another Publication at the rate of 9 for rupees 10 if he sells the whole lot at the the rate of 1 Rupee per book find his gain or loss on the whole transaction

Answer 1

Answer:

shop keeper buys some books 11 =10rupees

same no.of books 9=10 rupees

so totally he buyed 20 books per 20 rupees

so each book cost is 1 rupees

if he sell each book per 1 rupees also he didn't loss money..

and he didn't gain any money..

he won't loss or gain any money by selling those book per 1 rupees..

how much he payed that much he gain thats all..

conclusion:

He didn't loss or gain any money by the whole transaction

Answer 2

Answer:

The profit % for (x + y) books is $\dfrac{9x - 11 y}{90 x + 110y}$  .

Step-by-step explanation:

Given as :

The cost of 11 books = Rs 10

Let the x books is bought from one publication

Applying unitary method

∵  The cost of 11 books = Rs 10

So, The cost of 1 book = Rs $\dfrac{10}{11}$

∴ The cost of x book = Rs $\dfrac{10}{11}$ × x

Again

The cost of 9 books = Rs 10

Let the y books is bought from other publication

Applying unitary method

∵  The cost of 9 books = Rs 10

So, The cost of 1 book = Rs $\dfrac{10}{9}$

∴ The cost of y book = Rs $\dfrac{10}{9}$ y

So, The total cost price of books = Rs ($\dfrac{10 x}{11}$  +  $\dfrac{10 y}{9}$ )

Or, The total cost price of books = Rs( $\dfrac{90 x + 110 y}{99}$ )

Again

The x + y books are sold at the rate Rs 1 per books

So, The selling price of 1 book = Rs 1

∴ The selling price of (x + y)  books = Rs 1  ×  (x + y)

i.e The selling price of (x + y)  books = Rs (x + y)

Again

profit % = $\dfrac{selling price - cost price}{cost price}$

profit % = $\dfrac{(x+y) - (\frac{90x+110y}{99})}{\frac{90x+110y}{99}}$

Or, profit % = $\dfrac{99 x + 99 y - 90 x - 110 y}{90 x + 110 y}$

Or, profit % = $\dfrac{9x - 11 y}{90 x + 110y}$

Hence So, The profit % for (x + y) books is $\dfrac{9x - 11 y}{90 x + 110y}$  . Answer