# How To Solve Percentage Questions Quickly

## How to solve Percentages Questions Quickly

**In Aptitude , Percentage is the number expressed as a fraction of 100 Or we can say that a fraction with denominator 100 is called a per cent **

### How to Solve 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: How to Solve Percentage Problems- based on Mixtures and Alligation

**Question 1. A general store shopkeeper sells black pepper at some cost price but he mixes it with papaya’s seed and thereby gains 25%. Find the percentage of papaya’s seed mixed with black pepper?**

**Options:**

**A. 25%**

**B. 30%**

**C. 20%**

**D. 10%**

**Solution: **Let the CP of 1 kg black pepper = Re. 1

The SP of I Kg mixture = Re

1

Gain = 25%

Therefore, CP of 1 Kg

mixture = \frac{100}{125} × 1 = \frac{4}{5}

Ratio of black pepper: papaya seed = \frac{4}{5}: \frac{1}{5}

Hence, percentage of papaya seed in the mixture = \frac{1}{5} × 100 = 20%

**Correct option: C**

**Question 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**

**Question 3. There is one liquid A which contains 25% of water, the liquid B contains 30% of water. A vessel is filled with 6 parts of the liquid A and 4 parts of the liquid B. Find out the percentage of water in the mixture?**

**Options:**

**A. 27%**

**B. 25%**

**C. 20%**

**D. 30%**

**Solution: **Let the capacity of the vessel be 10

So, water from liquid A = 6 × \frac{25}{100} = 1.5

And, water from liquid B = 4 × \frac{30}{100} = 1.2

Total water = 1.5 + 1.2 = 2.7

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

**Correct option: A**

### Type 2: How To Solve- Problems based on Ratios and Fractions

**Question 1.If 40% of a number is equal to 2/3rd 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}y

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

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

Therefore the ratio = 5: 3

**Correct option: D**

**Question 2: In a box of 8 donuts 2 donuts have Choco chip sprinkle on them. Find out how many percent of the donuts in the box has Choco chip sprinkle?**

**Options: **

**A. 20%**

**B. 40%**

**C. 25%**

**D. 10%**

**Solution: ** \frac{2}{8}= \frac{x}{100}

8x = 200

x = \frac{200}{8}

x = 25%

**Correct option: C**

**Question 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: How To Solve Percentage- Problems based on Income, Salary, Expenditure

**Question 1. Luaa spends her monthly salary in the following manner: 20% on house rent, 20% on food, 5 % on transportation, 10% on the education, and 20% on other household expenses. She saves the remaining amount of ₨. 5000 at the end of the month. Find out her monthly salary?**

**Options:**

**A. Rs. 35000**

**B. Rs. 25000**

**C. Rs. 20000**

**D. Rs. 30000**

**Solution: **Let the

Monthly Salary = 100%

Monthly Expenditure =

20% + 20% + 5% + 10% + 20% = 75%

Monthly Savings = 100% – 75% = 25%

Now, 25% of salary saved = 5000

Let’s take 100% salary as x

25% of x = 5000

x = 5000 × \frac{100}{25}

x = \frac{500000}{25} x

x = 20000

Therefore, her monthly salary = Rs. 20000

**Correct option: C**

**Question 2.Aman spends 60% of his income. Suppose his income is increased by 21% and his expenditure increases by 5%, then what is the increase in his savings (in percentage)? **

**Options: **

**A. 60%**

**B. 18%**

**C. 40%**

**D. 45%**

**Solution: **Let Aman’s income = 100

Expenditure = 60 and

Savings = 40 (We get 40 by subtracting, 100 – 60)

New Income is increased by 21% = 100 + 21 = 121

Expenditure Increased by 5% = 5% more than 60 = 105% of 60 = 63

New Savings = 121 – 63 = 58

Therefore, increase in savings = 58 – 40 = 18

Increase in % = \frac{18}{40} × 100 = 45%

**Correct option: D**

**Question 3. Saloni’s salary was decreased by 30% and subsequently increased by 40%. How much percent does she loose?**

**Options:**

**A. 16%**

**B. 15%**

**C. 12%**

**D. 13%**

Solution: Let the salary be 100

100 × \frac{-30}{100} × \frac{40}{100} = -12%

It means she lost 12%

**Correct option: C**

### Type 4: How to Solve Percentage- Problems based on Population

**Question 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%

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

**Correct option: B**

**Question 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**

**Question 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}

= 13230

**Correct option: A**

### Type 5: How to Solve Percentage- Problems based on profit and loss

**Question 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

x = 540 × \frac{100}{60} = 900

**Correct option: D**

**Question 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. 66.4% 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 % = \frac{1.66}{2.50} × 100

Gain % = 66.4%

**Correct option: D**

**Question 3. A fruit seller mixes 25 kg apples of Rs. 20 per kg with 35 kg apples of other variety at Rs. 25 per kg and sells the mixture of apples at Rs. 30 per kg. Find his profit percentage**

**Options:**

**A. 39.9%**

**B. 30.9%**

**C. 32.9%**

**D. 31.9%**

**Solution: **CP of 60 kg apples = Rs. (25 × 20 + 35 × 25) = Rs. (500 + 875) = Rs. 1375

S.P. of 60 kg apple = Rs. (60 x 30) = Rs. 1800.

Gain% = \frac{1800-1375}{1375} × 100 = \frac{425}{1375} × 100 = 30.9%

**Correct ****option: B**

### Type 6: How To Solve Percentage Questions Quickly.

**Question 1.Ram was multiplying a number. By mistake he multiplied 2/5 instead of 5/2. What is the percentage error in the calculation? **

**Options:**

**A. 64%**

**B. 74%**

**C. 54%**

**D. 84%**

**Solution: **Let the number be x

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

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

**Correct option: D**

**Question 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}* y

Q = y% of x = \frac{y}{100} × x

\frac{x}{100}× y = \frac{y}{100} × x

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

Therefore, P = Q

**Correct option: C**

**Question 3. What percentage of numbers from 1 to 50 have 0 or 5 in the unit’s digit?**

**Options:**

**A. 50%**

**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, 10, 15, 20, 25, 30, 35, 40, 45, and 50

Total numbers are = 10

Required percentage = \frac{10}{50} × 100 = 20%

**Correct option: B**

**Read Also – Formulas to solve percentage questions **

in question 2 of type 5 option given is wrong but solution is right

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