**Fractions decimals and percentages word problems :**

Word problems on fractions decimals and percentages are much useful to the students who would like practice problems with fractions, decimals and percentages on real world situations.

**Problem 1 :**

The denominator of a fraction exceeds the numerator by 5. If 3 be added to both, the fraction becomes 3/4. Find the fraction.

**Solution :**

Let "x" be the numerator.

"The denominator of the fraction exceeds the numerator"

From the above information, fraction = x / (x+5) ------(1)

"If 3 be added to both, the fraction becomes 3/4"

From the above information, we have (x+3) / (x+5+3) = 3/4

(x+3)/(x+8) = 3/4 ---------> 4(x+3) = 3(x+8)

4x + 12 = 3x + 24 ---------> x = 12

(1)--------> x / (x+5) = 12 / (12+5) = 12/27

Hence, the required fraction is 12/27

**Problem 2 :**

In a school, there are 450 students in total. If 2/3 of the total strength are boys, find the number of girls in the school.

**Solution :**

No. of boys in the school = 450 x 2/3 = 300

Total no. of students = 450.

Then no. of girls = 450 - 300 = 150

Hence the numbers girls in the school = 150

**Example 3 :**

The width of the rectangle is 2/3 of its length. If the perimeter of the rectangle is 80 cm. Find its area.

**Solution :**

Let "x" be the length of the rectangle.

Then, width of the rectangle = (2/3)x

Perimeter = 80 cm -----> 2(l + w) = 80 -------> l + w = 40

l + w = 40 -------> x + (2/3)x = 40 -------> (3x+2x) / 3 = 40

(3x+2x) / 3 = 40 ------> 5x = 120 ------> x = 24

So, length = x = 24 cm

and width = (2/3)x = (2/3)24 = 16 cm

Area = l x w = 24x16 = 384 square cm.

Hence, area of the rectangle is 384 square cm

**Problem 4 :**

If good are purchased for $ 1500 and one fifth of them sold at a loss of 15%. Then at what profit percentage should the rest be sold to obtain a profit of 15%?

**Solution :**

As per the question, we need 15% profit on $1500.

Selling price for 15% on 1500

S.P =115% x 1500 = 1.15x1500 = 1725

When all the good sold, we must have received $1725 for 15% profit.

When we look at the above picture, in order to reach 15% profit overall, the rest of the goods ($1200) has to be sold for $1470.

That is,

C.P = $1200, S.P = $1470, Profit = $270

Profit percentage = (270/1200) x 100

Profit percentage = 22.5 %

Hence, the rest of the goods to be sold at 22.5% profit in order to obtain 15% profit overall.

**Problem 5 :**

I purchased 120 books at the rate of $3 each and sold 1/3 of them at the rate of $4 each. 1/2 of them at the rate of $ 5 each and rest at the cost price. Find my profit percentage.

**Solution :**

Total money invested = 120x3 = $360 -------(1)

Let us see, how 120 books are sold in different prices.

From the above picture,

Total money received = 160 + 300 +60 = $ 520 --------(2)

Profit = (2) - (1) = 520 - 360 = $160

Profit percentage = (160/360)x100 % = 44.44%

Hence the profit percentage is 44.44

**Problem 6 :**

If A's salary is 20% less than B's salary. By what percent is B's salary more than A's salary ?

**Solution :**

Let us assume B's salary = $ 100 ----------(1)

Then, A's salary = $ 80 --------(2)

Now we have to find the percentage increase from (2) to (1).

Difference between (1) and (2) = $ 20

Percentage increase from (2) to (1) = (20/80) x 100% = 25%

Hence, B's salary is 25% more than A's salary.

Let us look at the next problem on "Fractions decimals and percentages word problems"

**Problem 7 :**

In an election, a candidate who gets 84% of votes is elected by majority with 588 votes. What is the total number of votes polled ?

**Solution :**

Let "x" be the total number of votes polled.

Given : A candidate who gets 84% of votes is elected by majority of 476 votes

From the above information, we have

84% of x = 588 ---------> 0.84x = 588

x = 588 / 0.84 = 700

Hence, the total number of votes polled 700.

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**Problem 8 :**

When the price of a product was decreased by 10 % , the number sold increased by 30 %. What was the effect on the total revenue ?

**Solution :**

Before decrease in price and increase in sale,

Let us assume that price per unit = $ 100.

Let us assume that the number of units sold = 100

Then the total revenue = 100 x 100 = 10000 -----(1)

After decrease 10 % in price and increase 30 % in sale,

Price per unit = $ 90.

Number of units sold = 130

Then the total revenue = 90 x 130 = 11700 -----(2)

From (1) and (2), it is clear that the revenue is increased.

Difference between (1) and (2) = 1700

Percent increase in revenue

= (Actual increase / Original revenue) x 100 %

= (1700/10000) x 100 %

= 17 %

Hence, the net effect in the total revenue is 17 % increase.

Let us look at the next problem on "Fractions decimals and percentages word problems"

**Problem 9 :**

A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation ?

**Solution :**

In the given two fractions, the denominators are 5 and 3.

Let assume a number which is divisible by both 5 and 3.

Least common multiple of (5, 3) = 15.

So, let the number be 15.

15 x 3/5 = 9 ----------(1) ---------incorrect

15 x 5/3 = 25 ---------(2) --------correct

Difference between (1) and (2) is 16

Percentage error = (Actual error / Correct answer ) x 100 %

= (16 / 25) x 100 %

= 64 %

Hence, the percentage error in the calculation is 64 %.

Let us look at the next problem on "Fractions decimals and percentages word problems"

**Problem 10 :**

There are 15 boys and 12 girls in a section A of class 7. If 3 boys are transferred to section B of class 7,then find the percentage of boys in section A.

**Solution :**

Before transfer :

No. of boys in section A = 15

No. of boys in section B = 12

Given : 3 boys are transferred from section A to B

After transfer :

No. of boys in section A = 12

No. of boys in section B = 12

Hence, percentage of boys in section A is 50%

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**Problem 11 :**

A chemist mixed 6.35 grams of one compound with 2.45 grams of another compound. How many grams were there in the mixture ?

**Solution : **

Quantity of one compound = 6.35 grams

Quantity of another compound = 2.45 grams

Total quantity of mixture is

= Quantity of one comp + quantity of another comp

= 6.35 + 2.45

= 8.8 grams

Hence, the quantity of the mixture is 8.8 grams.

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**Problem 12 : **

David buys 3 pens where the price of each pen is $1.5 and the 4 pencils where the price of each pen is $0.75. If he gets a discount of 10% in the total bill, how much does he have to pay ?

**Solution : **

First let us find the total bill.

Total bill = 3x1.5 + 4x0.75

Total bill = 4.5 + 3

Total bill = $ 7.5

Given : discount 10%

The money that he has to pay is

= (100 - 10)% of total bill

= 90% x 7.5

= 0.9 x 7.5

= 6.75

Therefore, he has to pay $6.75

Let us look at the next problem on "Fractions decimals and percentages word problems"

**Problem 13 : **

Joseph earned $24.60 for working 6 hours. How much will earn, if he works for 7.5 hours ?

**Solution : **

Given : Money earned in 6 hours = $24.60

Money earned in 1 hour = $24.60 / 6

Money earned in 1 hour = $4.10

Therefore money earned in 7.5 hours is

= 7.5 x $4.10

= $30.75

Let us look at the next problem on "Fractions decimals and percentages word problems"

**Problem 14 : **

A pipe is 76.8 meters long. What will the greatest number of pieces of pipe each 8 meters long that can be cut from this pipe ?

**Solution :**

The original length of the pipe = 76.8 meters

The length of each pipe piece = 8 meters

No of pieces where each piece has length 8 meters is

= 76.8 / 8

= 9.6

Therefore, the greatest number of pieces of pipe each 8 meters long is 9.

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**Problem 15 :**

The cost of a toy is $22.5. If the profit is 20%, what is the selling price of the toy ?

**Solution :**

The cost of a toy is $22.5

The selling price = (100 + 20) % of cots price

= 120% x 22.5

= 120% x 22.5

= 1.2 x 22.5

= 27

Therefore the selling price of the toy is $27

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