# HOW TO SOLVE PERCENTAGE PROBLEMS IN APTITUDE EASILY

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⁄ₒᵣᵢgnl 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 -

= 1.2(1.2x) -

= 1.44x - 2°

sum of the angles of a triangle = 180°

x + 1.2x + 1.44x - 2° = 180°

3.64x - 2° = 180°

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

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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