# SOLVING WORD PROBLEMS WITH MIXED FRACTIONS

Problem 1 :

A can contains 10 kg of oil. 2 3/4 kg and 5 1/3 kg are poured into two vessels. How much is left in the can ?

Solution :

Quantity of oil originally  =  10 kg

Quantity of oil poured into the vessels  = 2  3/4 + 5  1/3

Remaining quantity of oil

=  Original quantity of oil - Quantity of oil poured

=  10 - (2  3/4 + 5  1/3)

=  10 - [(11/4) + (16/3)]

First let us combine the fractions inside the parentheses.

=  10 - { (33/12 ) + (64/12) }

=  10 - { (33 + 64)/12 }

=  10 - (97/12)

=  (120 - 97)/12

=   23/12

Therefore, quantity of oil left in the can is 23/12 kg.

Problem 2 :

A steel rod is 12 7/8 meters long. From this two pieces, one 3 1/4 meters long and another 4 2/3 meters long are cut off. What is the length of the remaining part of the rod?

Solution :

Original length of steel rod  =  12 7/8 meter

Length of rod which had been cut  =  3  1/4 +  4  2/3

=  (13/4) + (14/3)

L.C.M of 4 and 3 is 12

=  (13/4)  (3/3) + (14/3)  (4/4)

=  (39/12) + (56/12)

=  (39 + 56) /12

=  95/12

Length of remaining rod = length of original rod - Length of rod which had been cut

=  12  7/8 - (95/12)

=  (96 + 7)/8  - (95/12)

=  (103/8)  - 95/12

L.C.M of 8 and 12 is 24.

=  (103/8)  (3/3) - (95/12) ⋅ (2/2)

=  (309/24) - (190/24)

=  (390 - 190)/24

=  200/24

=  25/3

Therefore, the remaining length of rod is 25/3 m.

Problem 3 :

A man's monthly salary is \$800. From this he spends 305 3/4 for food and \$100 1/2 for children's education. How much will remaining with him ?

Solution :

Monthly salary of a man = 800

Amount spent for food = \$305  3/4

Amount spent for education = \$100  1/2

Remaining money  =  800 - (305  3/4 +  100  1/2)

=  800 - [ (1220+3)/4 + (200+1)/2 ]

=  800 - { (1223/4)  + (201/2) }

=  800 - { (1223/4)  + (201/2) ⋅ (2/2) }

=  800 - { (1223/4) + (402/4) }

=  800 - { (1223 + 402) /4 }

=  800 - 1625/4

=  (800/1) - (1625/4)

=  (3200/4) - (1625/4)

=   1575/4

So, the remaining amount he has is \$393.75.

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