Question

A shopkeeper earns a profit of ₹10.by selling one geometry box and incurs a loss of ₹5 on lunch box .which are defective 1)in a perticular month he incur a loss of ₹50 in this period. If he sold 25 geometry boxes how many lunch boxes did he sell.2)in the next month he earns neither profit not loss if he sold 50 lunch boxes how many geometry boxes did he sell?

Answer 1

Given:

A Shopkeeper earns a profit of ₹10 on the geometry box and loss ₹5 on the lunch box.

To find:

1) Number of lunch boxes sold.

2) Number of geometry boxes sold.

Solution:

1) Let the shopkeeper sold X lunch boxes.

Total loss occurred in this case is ₹50.

So according to the question,

• 25×10 + X×(-5) = -50

• 250 - 5X = -50

• 5X = 300

• X = 60.

2) Let the shopkeeper sold Y geometry boxes.

Now there is no loss no gain.

So according to the question,

• 10×Y + 50×(-5) = 0

• 10Y -250 = 0

• 10Y = 250

• Y = 25

The shopkeeper sold 60 lunch boxes and 25 geometry boxes.

Answer 2

First month: 25 geometry box and 60 lunch box  = loss of Rs. 50

Second month: 25 geometry box and 50 lunch box = breakeven

Step-by-step explanation:

Given: A shopkeeper earns a profit of ₹10 by selling one geometry box and incurs a loss of ₹5 on lunch box which are defective.

Find: 1) In a particular month, he incurs a loss of ₹50 in this period. If he sold 25 geometry boxes, how many lunch boxes did he sell?

2) In the next month, he earns neither profit nor loss. If he sold 50 lunch boxes, how many geometry boxes did he sell?

Solution:

Let x be the geometry box and y be the lunch box.

1) In the first month, shopkeeper incurs a loss of Rs. 50.

   25(10) -5y = -50

   250 - 5y = -50

   5y = 300

   So y = 300/5 = 60

Shopkeeper sold 60 lunch boxes in the first month.

2) Shopkeeper earned neither profit nor loss.

  10x + 50(-5) = 0

  10x = 250

   So x = 250 / 10 = 25

Shopkeeper sold 25 geometry boxes in the second month.