Profit/Loss Numerical Questions

EXERCISE 13.1 PAGE NO: 13.11

1. A student buys an office table for  900 euro and sells it for  1000. Find his gain and gain percent.

Solution:

We know that the cost price of the table =  90

Selling price = 100

By using the formula,

Gain = selling price – cost price

= 1000 – 900

= 100

By using the formula,

Gain % = (gain/cost price) × 100

= (100/900) × 100

= 100/9

= 11 1/9 %

2. Mary bought a car for 12400 euro and sold it for 11470. Find her loss and loss percent.

Solution:

We know that the cost price of car = 12400

Selling price of car = 11470

By using the formula,

Loss = cost price – selling price

= 12400 – 11470

= 930

By using the formula,

Loss % = (loss/cost price) × 100

= (930/12400) × 100

= 7.5 %

3. The cost price of 10 kgs of a certain variety of apples  is equal to the selling price of 9 kgs of the same apples. Find the profit percent.

Solution:

We know that the cost price of 10 kgs = selling price of 9 kgs

Let us consider CP of 1 kg as  X euro

Selling price of 9 kgs = 10X

Selling price of 1 kg = 10x/9

Profit = 10x/9 – x

= x/9

Profit % = Gain % = (gain/cost price) × 100

= (x/9)/x × 100

= 100/9

= 11 1/9%

4. A retailer buys a bicycle for 225 euro. His overhead expenses are  15 euro. If he sells it for  300 euro, determine his profit percent.

Solution:

The cost price of a bicycle = 225

Overhead expenses =  15

Total cost = cost price + overhead expenses = 225+15 =  240

Selling price  = 300

By using the formula,

Gain = selling price – cost price

= 300 – 240 =  60

By using the formula,

Gain % = (gain/cost price) × 100

= 60/240 × 100

= 25 %

5. A merchant bought two cars for  $13000. He sold one car at a profit of 20% and the other at a loss of 12%. If the selling price of both cars is the same, find the cost price of each car.

Solution:

We know the cost price of two cars = $13000

So let us consider cost price of one car be $ x

Cost price of other car = $( 13000 – x )

Selling price of first car = x + x × 20/100

= x + x/5

=  6x/5

Selling price of second car = (13000 – x) – (13000 – x) × 12/100

=  (13000 – x ) × 88/100

By equating Selling Price of first and second car we get,

6x/5 = (13000 – x ) × 88/100

1.2x = 13000 × 0.88 – 0.88x

1.2x + 0.88x   = 11440

2.08x = 11400

x = 11400 / 2.08

= 5480

∴ Cost price of first car = $5480

Cost price of second car = 13000 – 5480 = $7520

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