Question 97

A man purchased 40 fruits; Apples and oranges for Rs.17. Had he purchased as manyn oranges as apples and as many apples as oranges, he would have paid Rs.15/-. Find the cost of one pair of an apple and an orange.

Solution

Let us assume the number of apples purchased by the man to be x.

Then, the number of oranges will be (40-x).

Let the price of each apple be Rs. y and that of an orange be Rs. z

It is given that, xy+ (40-x)z= 17.

=> xy+40z-xz=17 ---------------(1)

Now, if the number of apples and oranges are interchanged, the apples will be (40-x) and the oranges will be x.

Given, the new total cost will be Rs. 15

So, (40-x)y + xz=15

=>40y-xy +xz= 15 ---------------(2)

Adding eqns 1 and 2, we get,

xy+40z-xz+40y-xy+xz=15+17

=> 40(y+z)=32

=>y+z= $$\ \frac{\ 32}{40}=\ \frac{\ 4}{5}$$

Or, the cost of one apple and one orange, x+y= Rs.$$\frac{\ 4}{5}=\ 80\ paise$$


Create a FREE account and get:

  • All Quant Formulas and shortcuts PDF
  • 170+ previous papers with solutions PDF
  • Top 5000+ MBA exam Solved Questions for Free

cracku

Boost your Prep!

Download App