WORD PROBLEMS ON FRACTIONS

About "Word Problems on Fractions"

Word Problems on Fractions :

In this section, we are going to learn, how to solve word problems on fractions step by step. 

Let us look at some examples on "Word problems on fractions". 

Word Problems on Factions  - Examples

Example 1 :

A cookie factory uses 1/4 of a barrel of oatmeal in each batch of cookies. The factory used 1/2 of a barrel of oatmeal yesterday. How many batches of cookies did the factory make ?

Solution :

Amount of oatmeal used in each batch of cookies is

=  1/4 of a barrel

No. of cookies made yesterday is

=  Oatmeal used yesterday / Oatmeal used in each batch 

=  (1/2) / (1/4)

=  1/2 ⋅ 4/1

=  2

Hence, the factory made 2 batches of cookies yesterday. 

Example 2 :

Of the students in the band, 1/4 play the flute and another 1/10 play the clarinet. What fraction of the students in the band play either the flute or the clarinet ?

Solution :

Fraction of the students in the band play either the flute or the clarinet is 

=  1/4 + 1/10

L.C.M of (4, 10) is 20. 

So, make each denominator as 20 by multiplying the numerator and denominator of the first fraction by 5 and the second fraction by 2. 

Then we have, 

=  (1 ⋅ 5) / (4 ⋅ 5)  +  (1 ⋅ 2) / (10 ⋅ 2)

=  5/20 + 2/20

=  (5 + 2) / 20

=  7 / 20

Hence, the fraction of the students who play either the flute or the clarinet is 7/20.

Example 3 :

Lily rode her bicycle 5/4 miles from her house to the park. Then she rode 3/2 miles from the park to the library. How many miles did Lily ride in all ?

Solution :

Total number of miles that Lily rode in all is 

=  5/4 + 3/2

L.C.M of (4, 2) is 4. 

So, make each denominator as 4 by multiplying the numerator and denominator of the first fraction by 1 and the second fraction by 2. 

Then we have, 

=  (5 ⋅ 1) / (4 ⋅ 1)  +  (3 ⋅ 2) / (2 ⋅ 2)

=  5/4 + 6/4

=  (5 + 6) / 20

=  11 / 4

=  2¾

Hence, Lily rode 2¾ miles in all. 

Example 4 :

At the dealership where Mary works, she fulfilled 1/6 of her quarterly sales goal in January and another 3/10 of her sales goal in February. If her quarterly sales goal is $12,000, what amount of sales did she make in January and February ?

Solution :

Fraction of her sales done in January and February is

=  1/6 + 3/10

L.C.M of (6, 10) is 30. 

So, make each denominator as 30 by multiplying the numerator and denominator of the first fraction by 5 and the second fraction by 3. 

Then we have, 

=  (1 ⋅ 5) / (6 ⋅ 5)  +  (3 ⋅ 3) / (10 ⋅ 3)

=  5/30 + 9/30

=  (5 + 9) / 30

=  14 / 30

=  7/15

Amount of sales she made in January and February is

=  7/15 of quarterly sales goal 

=  7/15 ⋅ 12000

=  5600

Hence, amount of sales Mary made in January and February is $5,600. 

Example 5 :

Lucy made strawberry jam and raspberry jam. She made enough strawberry jam to fill 5/6 of a jar. If she made 2 times as much raspberry jam as strawberry jam, how many jars would the raspberry jam fill ?

Solution :

Amount of strawberry jam made by Lucy is 

=  5/6 of a jar

2 times as much raspberry jam as strawberry jam is 

=  2 ⋅ 5/6 of a jar

=  5/3 of a jar

=  1 ⅔ of a jar

Hence, Lucy would make enough raspberry jam to fill ⅔  jars.

Example 6 :

David's salary is $1800. David spent 2/3 of the total money for salary. He spent 1/2 of the remaining for his kids education and saved the rest. How much did he save ? 

Solution :

Money spent on food  is

=  1800 ⋅ 2/3 = 1200

Remaining  =  1800 - 1200  =  600 -------(1)

Money spent on kids education is

=  1/2 of remaining

=  1/2 ⋅ 600

=  300 --------(2)

Then, his savings is

=  (1) - (2)

=  600 - 300

=  300

Hence, his savings is $300. 

Example 7 :

A, B and C are friends. A has 1/3 of money that B has. C has 1/2 of money that A has. If they all together have  $450. How much money do A, B and C have separately ? 

Solution :

Let us assume that B has the money "x". 

Then, A  =  (1/3)x  =  x/3    

C  =  1/2 of A 

C  =  1/2  x/3

C  =  x/6

Given : A + B + C  =  300 -------> x/3 + x + x/6  =  450

4x/12 + 12x/12 + 2x/12  =  450

(4x+12x+2x)/12  =  450

18x/12  =  450

x  =  300 

So, we have

A  =  x/3  =  300/3  =  100

B  =  x  =  300

C  =  x/6  =  300/6  =  50

Hence, A has $100, B has $450 and C has $50. 

Example 8 :

John's present age is 1/3 of David's age  5 years back.If David is 20 years old now, find the present age  of John.  

Solution :

Present age of David  =  20 years

David's age 5 years back  =  20 - 5  =  15 years

John's present age is 1/3 of David's age  5 years back

John's present age is

=  1/3 of 15 years

=  (1/3) ⋅ 15

=  5 years. 

Hence, John's present age is 5 years. 

After having gone through the stuff given above, we hope that the students would have understood "Word problems on fractions". 

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