Problem 1 :
John bought a car for $18000 and sold it to his friend for $18900. Find the profit percentage.
Solution :
Profit = Selling price - Cost price
Profit = 18900 - 18000
Profit = 900
Profit % = (Profit/Cost price) ⋅ 100%
Profit % = (900/18000) ⋅ 100%
Profit % = 5%
Problem 2 :
Peter bought an item at $300 and sold it for $240. Find the loss percentage.
Solution :
Loss = Cost price - Selling price
Loss = 300 - 240
Loss = 60
Loss % = (Loss/Cost price) ⋅ 100%
Loss % = (60/300) ⋅ 100%
Loss % = 20 %
Problem 3 :
A sold a laptop to B at 15% profit. B sold the same laptop to C at 20% profit. If C had paid $1656 for it, find the price at which A bought the laptop.
Solution :
Let x be the price at which A bought the laptop.
Cost price of B = 115% of x = 1.15x
Cost price of C = 120% of 1.15x = 1.38x
It is given that C paid $1656.
Then, we have
1.38x = 1656
Divide both sides by 1.38
x = 1656/1.38
x = 1200
So, the price at which A bought the laptop is $1200.
Problem 4 :
Michael sold an item at a loss of 5%. If he had sold at a profit of 20%, he would have got $100 more. Find the cost is the cost price of the item.
Solution :
Let x be the cost price of the chair
S.P (-5%) = 95% of x
S.P (-5%) = 0.95x -----(1)
S.P (+10%) = 120% of x
S.P (+10%) = 1.2x -----(2)
In (2), he got $100 more than (1).
Then, we have
(2) - (1) = 100
1.2x - 0.95x = 100
0.25x = 100
Divide both sides by 0.25
x = 100/0.25
x = 10000/25
x = 400
So, the cost price of the item is $400.
Problem 5 :
A man bought 20 units of a product at the cost of $400 and sold 5 units at the cost of $125. Find the profit or loss percentage.
Solution :
Cost price for 20 units = $400
Cost price for one unit = 400/20 = $20
Selling price for 5 units = $125
Selling price for one unit = 125/5 = $25
Because selling price is more than cost price, there is profit.
Profit = Selling price - Cost price
Profit = 25 - 20
Profit = 5
Profit % = (Profit/Cost price) ⋅ 100%
Profit % = (5/20) ⋅ 100%
Profit % = 25%
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Nov 21, 24 06:23 AM
Nov 21, 24 06:13 AM
Nov 20, 24 08:12 AM