# PROFIT AND LOSS WORKSHEET PDF

Problem 1 :

A man buy an article for \$27.50 and sells it for \$28.60. Find his profit percent.

Problem 2 :

If a camera is purchased for \$490 and sold it for \$465.50, find the loss percent.

Problem 3 :

By selling 33 meters of cloth, one gains the selling price of 11 meters. Find the profit percent.

Problem 4 :

A person incurs 5% loss by selling a laptop for \$1140. At what price should the laptop be sold to earn 5% profit ?

Problem 5 :

The cost price of 21 articles is equal to S.P of 18 articles. Find the profit or loss percent.

Problem 6 :

A man bought candies at 3 for a dollar. How many for a dollar must he sell to earn a profit of 50% ?

Problem 7 :

If the cost price is 96% of the selling price, then what is the profit percent ?

Problem 8 :

An article is sold at a certain price. By selling it at 2/3 of that price, one loses 10%. Find the profit percent at original price.

## Solutions

Problem 1 :

A man buy an article for \$27.50 and sells it for \$28.60. Find his profit percent.

Solution :

Given :

Cost price  =  \$27.50

Selling price  =  \$28.60

Finding Profit :

Profit  =  Selling price - Cost price

Profit  =  28.60 - 27.50

Profit  =  1.10

Finding Profit Percent :

Profit percent  =  (Profit/Cost price)  100 %

Profit percent  =  (1.10/27.50) ⋅ 100 %

Profit percent  =  4 %

Problem 2 :

If a camera is purchased for \$490 and sold it for \$465.50, find the loss percent.

Solution :

Given :

Cost price  =  \$490

Selling price  =  \$465.50

Finding Loss :

Loss  =  Cost price - Selling price

Loss  =  490 - 465.50

Loss  =  24.50

Finding Loss Percent :

Loss percent  =  (Loss/Cost price)  100 %

Loss percent  =  (24.50/490) ⋅ 100 %

Loss percent  =  5 %

Problem 3 :

By selling 33 meters of cloth, one gains the selling price of 11 meters. Find the profit percent.

Solution :

Let the selling price of 1 meter of cloth be \$1.

Then, the selling price of 33 meters of cloth  =  \$33

By fact, we have

S.P of 33 m - C.P of of 33 m  =  Profit

Given :

By selling 33 meters of cloth, the profit earned is the selling price of 11 meters.

So, we have

S.P of 33 m - C.P of of 33 m  =  S.P of 11 m

S.P of 33 m - S.P of of 11 m  =  C.P of 33 m

S.P of 22 m  =  C.P of 33 m

22 ⋅ 1  =  C.P of 33 m

22  =  C.P of 33 m

Finding Profit :

Profit  =  Selling price - Cost price

Profit  =  33 - 22

Profit  =  11

Finding Profit Percent :

Profit percent  =  (Profit/Cost price)  100 %

Profit percent  =  (11/22)  100 %

Profit percent  =  (1/2) ⋅ 100 %

Profit percent  =  50 %

Problem 4 :

A person incurs 5% loss by selling a laptop for \$1140. At what price should the laptop be sold to earn 5% profit ?

Solution :

Let "x" be the cost price of the laptop.

Given :

The laptop is sold for \$1140 at 5% loss.

So, we have

(100 - 5)% of x  =  1140

95% of x  =  1140

0.95x  =  1140

Divide both sides by 0.95

0.95x / 0.95  =  1140 / 0.95

x  =  1200

The cost price of the laptop  =  \$1200.

Selling price of the laptop at 5% proft is

=  (100 + 5)% of 1200

=  105% of 1200

=  1.05 ⋅ 1200

=  1260

Hence, the laptop should be sold at \$1260 to earn a profit of 5%.

Problem 5 :

The cost price of 21 articles is equal to S.P of 18 articles. Find the profit or loss percent.

Solution :

Let the cost of 1 article \$1.

Then, the cost price of 18 articles  =  \$18

Given :

Cost price of 21 articles  =  Selling price of 18 articles

21 ⋅ 1  =  Selling price of 18 articles

21  =  Selling price of 18 articles

Finding Profit :

Profit  =  S.P of 18 articles  - C.P of 18 articles

Profit  =  21 - 18

Profit  =  3

Finding Profit Percent :

Profit percent  =  (Profit/Cost price)  100 %

Profit percent  =  (3/18)  100 %

Profit percent  =  (1/6) ⋅ 100 %

Profit percent  =  100/6 %

Profit percent  =  16  %

Problem 6 :

A man bought candies at 3 for a dollar. How many for a dollar must he sell to earn a profit of 50% ?

Solution :

Cost price of 3 candies  =  \$1

Selling price of 3 candies at 50% is

=  (100 + 50)% of cost price of 3 candies

=  150% of 1

=  1.5 ⋅ 1

=  1.5

So, he has to sell 3 candies for \$1.50 to earn a profit of 50%.

Selling price of 1 candy  =  1.50/3

Selling price of 1 candy  =  \$0.50

No. of candies he has to sell for a dollar is

=  1/0.50

=  2

Hence, the person must sell 2 candies for a dollar to earn a profit of 50%.

Problem 7 :

If the cost price is 96% of the selling price, then what is the profit percent ?

Solution :

Let the selling price be \$100.

Then the cost price is \$96.

Finding Profit :

Profit  =  Selling price  - Cost price

Profit  =  100 - 96

Profit  =  4

Finding Profit Percent :

Profit percent  =  (Profit/Cost price)  100 %

Profit percent  =  (4/96)  100 %

Profit percent  =  (1/24) ⋅ 100 %

Profit percent  =  4.17 %

Problem 8 :

An article is sold at a certain price. By selling it at 2/3 of that price, one loses 10%. Find the profit percent at original price.

Solution :

Let the original selling price be \$3 and the cost price be x.

(Here, original selling price is assumed as \$3, because when it is multiplied by 2/3, the result will be an integer)

Then, 2/3 of the original selling price is

=  2/3 ⋅ 3

=  1

Selling price at 10% loss  =  2/3 of the original S.P

(100 - 10)% of cost price  =  2

90% of "x"  =  2

0.9 ⋅ x  =  2

0.9x  =  2

Divide both sides by 0.9

0.9x / 0.9  =  2 / 0.9

x  =  20/9

So, the cost price  =  \$20/9.

Finding Profit :

Profit  =  Selling price  - Cost price

Profit  =  3 - 20/9

Profit  =  27/9 - 20/9

Profit  =  (27 - 20)/9

Profit  =  7/9

Finding Profit Percent :

Profit percent  =  (Profit/Cost price)  100 %

Profit percent  =  [(7/9) / (20/9)]  100 %

Profit percent  =  (7/9) ⋅ (9/20) ⋅ 100 %

Profit percent  =  35 %

To learn profit and loss shortcuts,

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

## Recent Articles

1. ### SAT Math Videos

May 22, 24 06:32 AM

SAT Math Videos (Part 1 - No Calculator)

2. ### Simplifying Algebraic Expressions with Fractional Coefficients

May 17, 24 08:12 AM

Simplifying Algebraic Expressions with Fractional Coefficients