Let A be the original quantity and it increases to a new quantity B by 15%.
Formula to find the new quantity B :
B = (100 + 15)% of A
B = 115% of A
B = ¹¹⁵⁄₁₀₀ ⋅ A
B = 1.15A
Let A be the original quantity and it decreases to a new quantity B by 15%.
Formula to find the new quantity B :
B = (100 - 15)% of A
B = 85% of A
B = ⁸⁵⁄₁₀₀ ⋅ A
B = 0.85A
Relationship between selling price and cost price when the profit/loss percentage is given :
When there is profit (say x%),
selling price = (100 + x)% of cost price
When there is loss (say y%),
selling price = (100 - y)% of cost price
Let X be the original quantity and it increaes to a new quantity Y.
Formula to find the percentage increase from X to Y :
= ⁽ʸ ⁻ ˣ⁾⁄ₓ ⋅ 100%
Let X be the original quantity and it decreases to a new quantity Y.
Formula to find the percentage decrease from X to Y :
= ⁽ˣ ⁻ ʸ⁾⁄ₓ ⋅ 100%
In general, percentage change :
= (ᶜʰᵃⁿᵍᵉ ⁱⁿ ⱽᵃˡᵘᵉ⁄ₒᵣᵢgᵢnₐl ᵥₐlᵤₑ) ⋅ 100%
Let X be the cost price of a product and Y be the selling price.
If there is profit on selling the product, formula to find the profit percentage :
= ⁽ʸ ⁻ ˣ⁾⁄ₓ ⋅ 100%
If there is loss on selling the product, formula to find the loss percentage :
= ⁽ˣ ⁻ ʸ⁾⁄ₓ ⋅ 100%
Let M and M be two quantities.
Formula to find M is what percentage of N :
= ᴹ⁄N ⋅ 100%
Formula to find N is what percentage of M :
= ᴺ⁄M ⋅ 100%
Problem 1 :
The price of a product is $80. If the price is increased by 25%, find the new price.
Solution :
The new price :
= (100 + 25)% of $80
= 125% of $80
= 1.25 ⋅ $80
= $100
Problem 2 :
The price of a product is $120. If the product is sold at 10% discount, find the price price of the product after discount.
Solution :
The price of the product after discount :
= (100 - 10)% of $120
= 90% of $120
= 0.90 ⋅ $120
= $108
Problem 3 :
A store buys a camara for $250. If the store wants to earns 12% profit, at what price should the store sell the camara?
Solution :
Selling price of the camara :
= (100 + 12)% of $250
= 112% of $250
= 1.12 ⋅ $250
= $280
Problem 4 :
A product was bought for $30 and sold for $36. Find the profit percentage.
Solution :
Profit percentage :
= ⁽³⁶ ⁻ ³⁰⁾⁄₃₀ ⋅ 100%
= ⁶⁄₃₀ ⋅ 100%
= ⁶⁄₃₀ ⋅ 100%
= ⅕ ⋅ 100%
= 20%
Problem 5 :
A product was bought for $90 and sold for $63. Find the loss percentage.
Solution :
Loss percentage :
= ⁽⁹⁰ ⁻ ⁶³⁾⁄₉₀ ⋅ 100%
= ²⁷⁄₉₀ ⋅ 100%
= ³⁄₁₀ ⋅ 100%
= 30%
Problem 6 :
The cost price of a product is $55. If the product is sold at a profit of 5%, find the selling price.
Solution :
selling price = (100 + 5)% of cost price
selling price = 105% of cost price
selling price = 1.05 ⋅ cost price
Substitute 55 for cost price.
selling price = 1.05 ⋅ 55
selling price = $57.75
Problem 7 :
The cost price of a product is $70. If the product is sold at a loss of 10%, find the selling price.
Solution :
selling price = (100 - 10)% of cost price
selling price = 90% of cost price
selling price = 0.9 ⋅ cost price
Substitute 70 for cost price.
selling price = 0.9 ⋅ 70
selling price = $63
Problem 8 :
The selling price of a product is $99. If there is 20% profit on selling the product, find the cost price.
Solution :
selling price = (100 + 20)% of cost price
selling price = 120% of cost price
selling price = 1.2 ⋅ cost price
Substitute 99 for selling price.
99 = 1.2 ⋅ cost price
Divide both sides by 1.2.
cost price = $82.50
Problem 9 :
The selling price of a product is $152. If there is 5% loss on selling the product, find the cost price.
Solution :
selling price = (100 - 5)% of cost price
selling price = 95% of cost price
selling price = 0.95 ⋅ cost price
Substitute 152 for selling price.
152 = 0.95 ⋅ cost price
Divide both sides by 0.95.
cost price = $160
Problem 10 :
The second angle of a triangle is 20% more than the first angle. The third angle is 2° less than 120% of the second angle. Find the angles of the triangle.
Solution :
Let x be the first angle.
second angle = (100 + 20)% of x
= 120% of x
= 1.2x
third angle = 120% of 1.2x - 2°
= 1.2(1.2x) - 2°
= 1.44x - 2°
sum of the angles of a triangle = 180°
x + 1.2x + 1.44x - 2° = 180°
3.64x - 2° = 180°
Add 2° to both sides.
3.64x = 182°
Divide both sides by 3.64.
x = 50°
fist angle = 50°
second angle = 1.2(50°)
= 60°
third angle = 1.44(50°) - 2°
= 72° - 2°
= 70°
The angles of the triangle are 50°, 60° and 70°.
Problem 11 :
There are 720 students in a school and 40% of the students are boys. If 62 new boys and 12 new girls are admitted, find the percentage of boys.
Solution :
Number of boys in the school :
= 40% of 720
= 0.4 ⋅ 720
= 288
Number of girls in the school :
= 720 - 288
= 432
After new admission,
number of boys = 288 + 62 = 350
number of girls = 432 + 12 = 450
total number of students = 350 + 450 = 800
Percentage of boys in the school after new admission :
= ³⁵⁰⁄₈₀₀ ⋅ 100%
= 0.4375 ⋅ 100%
= 43.75%
Problem 12 :
A is 20% of B and B is 40% of C. What percentage of A is C?
Solution :
Given : A is 20% of B and B is 40% of C.
A = 0.2B ----(1)
B = 0.4C
Substitute B = 0.4C in (1).
A = 0.2(0.4C)
A = 0.8C
Divide both sides by 0.8.
ᴬ⁄₀.₈ = C
(¹⁄₀.₈)A = C
1.25A = C
125% of A = C
Problem 13 :
A is 10% more than B and B is 20% more than C. A is what percentage more than of C?
Solution :
Given : A is 10% more than B and B is 20% more than C.
A = (100 + 10)% of B
A = 110% of B
A = 1.1B ----(1)
B = (100 + 20)% of C
B = 120% of C
B = 1.2C
Substitute B = 1.2C in (1).
A = 1.1(1.2C)
A = 1.32C
A = 132% of C
A = (100 + 32%) of C
A is 32% more than C
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