WORD PROBLEMS ON FRACTIONS WITH SOLUTIONS

About "word problems on fractions with solutions"

On this web page, word problems on fractions with solutions we are going to see practice questions on fractions.you can find the solution for this problems with detailed steps.

What are the points to be noticed when we add or subtract two or more fractions?

(i) Before adding or subtracting any two or three fractions first we need to check whether we have same denominator or not.

(ii) In case the denominators are same then we can put only one denominator and we can add or subtract the numerator (according to the symbol).

(iii) In case the denominators are different, we should take L.C.M.

(iv) For multiplying two fractions we don't have to consider the denominators. We should multiply numerator with numerator and denominator with denominator.

Practice questions word problems on fractions-with solutions

Question 1 :

 A fruit merchant bought mangoes in bulk. He sold 5/8 of the mangoes. 1/16 of the mangoes were spoiled. 300 mangoes remained with him. How many mangoes did he buy? 

Solution :

Number of mangoes sold = 5/8 parts

Number of mangoes spoiled = 1/16 parts

Number of mangoes remained = 300

Let "n" be the total number of mangoes

Number of mangoes sold = (5/8)  x n ==> 5n/8

Number of mangoes spoiled = 1/16 n ==> n/16

Total number of mangoes = Number of mangoes sold +             Number of mangoes spoiled + Number of mangoes remained  

=  5n/8 + n/16 + 300

Since the denominators are not same,we have to take L.C.M to make them same.

L.C.M = 16

16n  = 11n + 4800  ==> 5n = 4800 ==> n = 960

So, number of mangoes that he bought = 960 

Question 2 :

A family requires 2 ½ liters of milk per day.How much milk would family require in a month of 31 days?  

Solution :

Milk required for a day = ½ liters

Required quantity of milk for a day  =  5/2

Required quantity of milk for 31 days = (5/2) x 31

                                                      = (5 x 31)/2

                                                      =  155/2

 [Now we have to convert this improper fraction as mixed fraction.For that we need to divide 155 by 2.If we do so, we will get 77 as quotient,1 as remainder and 2 as divisor.  2

So, the final answer is  77 ½ liters

Question 3 :

A ream of paper weighs 12 ½ kg.  What is the weight per quire.

Solution :

Weight of 1 ream of paper  =  12 ½ kg

1 ream = 20 quires 

Weight of 1 quire =  12 ½  ÷ 20

Here we have to change this mixed fraction into improper fraction

=  25 / 2  ÷ 20 ==> (25/2) x 1/20 ==> 25/40 ==> 0.625

Weight of 1 quire of paper = 0.625 kg

Hence, weight of 1 quire of paper = 0.625 kg

Question 4 :

It was Richard's birthday.He distributed 6 kg of sweets among her friends.If he gave ⅛  kg of sweet to each.How many friends are there? 

Solution :

Total quantity of sweet = 6 kg

Share of each person = ⅛  kg

Let "X" be the number of friends that Richard has

So ,           X ⅛  = 6 ==> X =  68  ==> 48

Hence, there are 48 friends.   

Question 5 : 

6 students went on a picnic.One student agreed to bear half of the expenses. The remaining 5 students shared the remaining expenses equally. What fraction of the expenses does each of 5 students pay?

Solution :

Let "x" be the expense spent in a picnic

Share of one student = (x/2)

Share of remaining 5 students =  (x/2)  ÷ 5

 =  (x/2) ÷  (5/1) ==> (x/2) x (1/5) ==> x/10

Question 6 : 

I have 2 ½ times money that David has. If i have $100,how much money does David have? 

Solution :

Let "x" be the money that David has  

Value of money that i have =   $100 

So,  2 ½ x X   = 100 ==>  (5/2) x = 100 ==> (5x/2) = 100 ==> x  =  40

Hence, David had $40.

Question 7 :

In a basket there are two kinds of sweet packets. There are 7 packets of the first kind each weighing 1 ¼ kg and 9 packets of the second kind each weighing ¾ kg . What is the total weight of the sweets in the basket?

Solution :

From the above question we have the details as below:

Number of packets in Type A = 7

measurement of each packet in Type A = ¼ kg

Number of packets in Type B = 9

measurement of each packet in Type B = ¾ kg

Weight of packet A = 7 x ¼  ==> 7 x (5/4) ==> 35/4

weight of packet B = 9 x ¾ ==> 27/4

Total weight = Weight of packet A + Weight of packet B

=  (35/4) + (27/4) ==>  (35+27)/4 ==> 62/4 ==>  31/2 

Now we have to convert this improper fraction as mixed fraction.If we do so we will get 2 as divisor,15 as quotient and 1 as remainder.so the final answer be 15 ½

Question 8 :

How many half-liter bottles can be filled from a can containing 37 ½ liter of milk?   

Solution :

Total quantity of milk = 37 ½ liter

Number of half liters to be filled   = 37 ½ ÷ (1/2)

   = (74+1)/2 ÷ (1/2) ==> (75/2) ÷ (1/2) ==> (75/2) x (2/1)

Number of bottles required = 75

Hence, 75 bottles are required.

Question 9 :

A gentleman bought 200 liter of milk for a function. ⅘ of it was used for preparing candies. ¾ of the remaining milk was used for preparing coffee. How much of the milk remained.

Solution :

Total quantity of milk  =   200 liter  

Quantity of milk used for preparing candies  = 200 x (4/5)

=  (200x4)/5 ==> 800/5 ==> 160 liter

Quantity of milk after preparing candies = 200 - 160 ==> 40 liter

Quantity of milk used for preparing coffee   = 40 x (3/4)

 =  (40 x 3)/4 ==> 120/4 ==> 30 liter

Quantity of milk used for candies and coffee  = 160 + 30 ==> 190 liter

Remaining quantity of milk  = 200 - 190 ==> 10 liter

Hence, remaining quantity of milk is 10 liters.

Question 10 :

Two third of a tank can be filled in 18 minutes. How many minutes will it require to fill the whole tank?  

Let "X" be the time taken to fill the whole tank.

Time taken to fill  2/3 
part of the tank = 18 minutes                            

(2/3) x X  =  18 ==> 18 x (3/2) ==> 27 minutes

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