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Aptitude

Percentage Quicknotes

How To Solve Percentange Questions Quickly

Percentage Questions & Definition

  • A percentage is a ratio expressed as a fraction whose denominator (bottom) is 100.Thus, x percent means x hundredths, written as x%.
  • A percentage is a fraction of an amount expressed as a particular number of hundredths of that amount.

What is P percent of X?

  • Written as an equation: Y = P% × X.
  • The ‘what’ is Y that we want to solve for
  • Remember to first convert percentage to decimal, dividing by 100
  • Solution: Solve for Y using the percentage formula Y = P% × X.
Type 1: Percentage Problems- based on Mixtures and Alligation

1. A shopkeeper mixes two varieties of Tea, one costing Rs. 40/kg and another Rs. 50/kg in the ratio 3:2. if he sells the mixed variety of Tea at Rs. 48/kg, his gain or loss percent is

Options:

A. 25%

B. 30%

C. 20%

D. 10%

Solution:

Average CP of Mixed Tea when Mixed in Ratio 3:2

$= \frac{40 \times 3 + 50 \times 2}{3 + 2} = \frac{220}{5} = Rs.44$

When SP = Rs. 48/kg

Therefore, CP of 1 Kg

$= \frac{4}{44} \times 100 = 9.90 \approx{10} %$

Correct option: D

2.A bar tender served a jar full of vodka containing 50% alcohol to a customer. After few minutes, the customer asked the bar tender to replace the vodka by another drink containing 19% alcohol and now the percentage of alcohol was found to be 25%. Find out the quantity of vodka replaced?

Options:

A.$ \frac{25}{31}$

B.$ \frac{20}{31}$

C.$ \frac{1}{2}$

D. $\frac{31}{20}$

Solution:

By the rule of alligation, we have:

Ratio of 1st and 2nd quantities = 6: 25

By the rule of ratio, if x:y is the ratio, to get the quantity of x, the formula is $\frac{x}{x+y}$,and to get the quantity of y, the formula is $\frac{y}{x+y}$

Therefore, required quantity replaced = $\frac{25}{25 + 6} = \frac{25}{31}$

Correct option: A

3. One type of liquid contains 25% of milk, the other contains 30% of milk. A container is filled with 6 parts of the first liquid and 4 parts of the second liquid. The percentage of milk in the mixture is:

Options:

A. 27%

B. 32%

C. 19%

D. 30%

Solution:

Let the quantity of a container is 10

So, milk from the first liquid $ 6 \times \frac{25}{100} = 1.5$

And, milk from 2nd liquid $ 4 \times \frac{30}{100} = 1.2$

Total milk = 1.5 + 1.2 = 2.7

In terms of percentage = $\frac{2.7}{10} ×100 = 27%$

Correct Option: A

Type 2: Problems based on Ratios and Fractions

1.If 40% of a number is equal to ${\frac{2}{3}}^{rd}$ of another number, what is the ratio of first number to the second number?

Options:

A. 1: 2

B. 2: 3

C. 3: 5

D. 5: 3

Solution:

Let the two numbers be x and y

40% of x = $\frac{2}{3}y$

Then, $\frac{40x}{100} = \frac{2}{3}$

$\frac{2}{5} = \frac{2}{3}$

$\frac{x}{y} = \frac{2}{3} × \frac{5}{2} = \frac{5}{3}$

Therefore the ratio = 5: 3

Correct option: D

2. In a box of 10 cakes ,5 cakes have Choco chip sprinkle on them. Find out how many percent of the cakes in the box has Choco chip sprinkle?

Options:

A. 50%

B. 40%

C. 25%

D. 20%

Solution:

$\frac{5}{10} = \frac{x}{100}$

2x = 100

x = \frac{100}{2}$

x = 50%

Correct option: A

3.Two numbers are respectively 40% and 80% more than a third number. The ratio of the two numbers is:

Options:

A. 6: 5

B. 7: 9

C. 2: 5

D. 1: 2

Solution:

Let the third number x

First number = 40 % more than $x = \frac{140}{100}x$ =$ \frac{7}{5}x$

Second number = 80% more than x =$ \frac{180}{100}x$ =$ \frac{9}{5}x$

Therefore their ratio = $\frac{7}{5}x: \frac{9}{5} x = 7: 9$

Correct Option: B

Type 3: Percentage- Problems based on Income, Salary, Expenditure

1.Radhika spends 40% of her salary on food, 20% on house rent, 10% on entertainment and 10% on conveyance. If her savings at the end of a month are Rs 1500, then what is her monthly salary?

Options:

A. Rs. 5500

B. Rs. 8500

C. Rs. 7500

D. Rs. 6000

Solution:

Let Radha's monthly salary be Rs x.

Then, from the given data we can write

40% of x+20% of x+10% of x+10% of x+ Rs 1500=x

⇒$ \frac{40}{100} \times x + \frac{20}{100} \times x + \frac{10}{100} \times x + \frac{10}{100} \times x + Rs 1500 = x$

$\implies \frac{80x}{100} + 1500 = x$

$\implies x - \frac{4x}{5} = Rs. 1500$

$\frac{x}{5} = Rs. 1500$

$\implies x = 1500 \times 5 = Rs 7500 $

Correct option: C

2. A man spends 60% of his income. If his income increases by 15% and his expenditure increases by 15%. Find the percentage change in his saving.

Options:

A. 15%

B. 18%

C. 25%

D. 45%

Solution:

Income - Expenditure = Saving

Let the income be 100x

A man spends 60% of his income

Expenditure = 60x

⇒ 100x - 60x = 40x

Savings = 40x

Income increases by 15% 

⇒ 100x + 15x = 115x

Expenditure increases by 15%

⇒ 15% = 3/20

Increase of 15% = 23/20

⇒ 60x × 23/20

⇒ 69x

New savings = 115x - 69x = 46x

Change in savings 

⇒ 46x - 40x = 6x

Percentage increase in saving = (6x/40x) × 100 = 15%

∴ Percentage increase in saving is 15%

Trick :

When income and expenditure are increased/decreased respectively by the same percentage of the amount i.e. x%, then the percentage change in (income - expenditure) i.e. saving would also be x%. 

So, here percentage change in saving = 15%

Correct option: A

3. Suresh salary was decreased by 30% and subsequently increased by 40%. How much percent does she loose?

Options:

A. 2%

B. 7%

C. 5%

D. 4%

Solution:

cording to the formula, Suresh loose percentage =$ -30 + 40 + \frac{(-30)(+40)}{100} = 10 - 2 = -2 %$

It means He lost Total 2% of his salary.

Correct Option: A

Type 4: Percentage- Problems based on Population

1.The population of Udaipur increased from 100000 to 250000 in 10 years. The average percent increase of population per year is:

Options:

A. 5%

B. 15%

C. 25%

D. 40%

Solution:

Increase of population in 10 years = 250000 – 100000 = 150000

Increase % = $\frac{150000}{100000 }× 100 = 150%$

Monthly Expenditure =

Required average $= \frac{150}{10}% = 15%$

Correct option: B

2.Population of a town increases by 10% in 1st year. Again, it increased by 20% in next year. Calculate the equivalent net % increase?

Options:

A. 35%

B. 25%

C. 23%

D. 32%

Solution:

Final Population= Initial Population $\frac{150}{10}$

increase in first Year × increase in second Year = Initial Population × $(1+ \frac{10}{100})(1+ \frac{20}{100})$

Final Population= Initial Population × 1.1 × 1.2

Final Population= Initial Population × 1.32

So, here net Multiplying Factor =1.32 = which means 32% increase

Correct option: D

3.The current population of a village is 12000. If it increases at the rate of 5% p.a. then at the end of 2 years, it will be:

Options:

A. 13230

B. 13320

C. 13200

D. 13323

Solution:

Current population of thevillage = 12000

Increase rate = R = 5%

We know that,

The population after 2 years = P *(1 + R/100)n = 12000 * (1 + 5/100)²

$= 12000 ×(\frac{105}{100})^2$

$= 12000 × \frac{105}{100} × \frac{105}{100}$

Correct Option: A

Type 5: Percentage- Problems based on profit and loss

1.A fruit seller had some oranges. He sells 40% oranges and still has 540 oranges. Find out the total oranges he had?

Options:

A. 1000

B. 420

C. 460

D. 900

Solution:

Let the total oranges be x

So, 1(00- 40)% of x = 540

$\frac{60}{100}$ of x = 540

= $540 × \frac{100}{60}$ = 900

Correct option: D

2.Rakesh brought 100 apples at the rate of Rs. 250. He sold them at the rate of Rs. 50 per dozen. Find the percentage of profit or loss?

Options:

A. 66.4% gain

B. 66.4% loss

C. 60.4% gain

D. 66.2% gain

Solution:

CP of one apple = $ \frac{250}{100}$ = Rs. 2.50

SP of one apple = $\frac{50}{12} = Rs 4.16 $

It is clear, that S.P. > C. P, therefore, there is gain.

Gain % = $\frac{SP – CP}{CP}× 100$

Gain % = 66.4%

Correct option: A

3. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:

Options:

A. 8%

B. 5%

C. 10%

D. 12%

Solution:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

Gain $ = \frac{80}{1600} \times 100$

Correct Option: B

Type 6: How To Solve Percentage Questions Quickly.

1.Ram was multiplying a number. By mistake he multiplied $\frac{2}{5}$ instead of $\frac{5}{2}$. What is the percentage error in the calculation?

Options:

A. 84%

B. 89%

C. 78%

D. 94%

Solution:

Let the number be x

Error = $\frac{5}{2} – \frac{2}{5}x = \frac{21x}{10}$

Error % = $\frac{21x}{10}x × \frac{5}{2x}× 100 = 84%$

Correct option: A

2.If P = x% of y and Q = y% of x, then which of the following is true?

Options:

A. P is smaller than Q

B. P is greater than Q

C. P is equal to Q

D. Cannot be determined

Solution:

P = x% of y = $\frac{x}{100}$ x y

Q = y% of $x = \frac{y}{100}y × x$

$\frac{x}{100}x × y = \frac{y}{100}$

$\frac{xy}{100} xy= \frac{yx}{100}$

Therefore, P = Q

Correct option: C

3. What percentage of numbers from 1 to 50 have 5 in the unit’s digit?

Options:

A. 10%

B. 20%

C. 30%

D. 25%

Solution:

Clearly, the numbers which have 0 or 5 in the unit’s digit from 1 to 50 are 5, 15, 25, 35 and 45

Total numbers are = 5

Required percentage $ = \frac{5}{50}× 100 = 10\%$

Correct Option: A