SOLVING WORD PROBLEMS WITH MIXED FRACTIONS

Problem 1 :

A can contains 10 kg of oil. 2¾ kg and 5 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 :

= 2¾ +  5

Remaining quantity of oil :

= Original quantity of oil - Quantity of oil poured

= 10 - (2¾ +  5)

= 10 - (11/4 + 16/3)

LCM of 3 and 4 is 12.

= 10 - (33/12 + 64/12)

= 10 - (33 + 64)/12

= 10 - 97/12

= (120 - 97)/12

= 23/12

= 1¹¹⁄₁₂

Therefore, quantity of oil left in the can is 1¹¹⁄₁₂ kg.

Problem 2 :

A steel rod is 12 meters long. From this two pieces, one 3¼ meters long and another 4 meters long are cut off. What is the length of the remaining part of the rod?

Solution :

Original length of steel rod = 12 meter.


Length of rod which had been cut :

= 3¼ + 4

= 13/4 + 14/3

LCM of  4 and 3 is 12.

= 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 - 95/12

= 103/8 - 95/12

LCM of 8 and 12 is 24.

= 309/24 - 190/24

= (390 - 190)/24

= 200/24

= 25/3

= 8

Therefore, the remaining length of rod is 25 m.

Problem 3 :

A man's monthly salary is $800. From this he spends $305¾ for food and $100½ for children's education. How much will remaining with him?

Solution :

Monthly salary of a man = $800.

Amount spent for food = $305
¾.

Amount spent for education = $100½.

Remaining money :

= 800 - (305¾ +  100½)

= 800 - (1223/4 + 201/2)

= 800 - (1223/4 + 402/4)

= 800 - (1223 + 402)/4

= 800 - 1625/4

= 3200/4 - 1625/4

= 1575/4

= 393¾

So, the remaining amount he has is $393¾

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