WORD PROBLEMS WITH MIXED FRACTIONS

Problem 1 :

Peter had $75¾ in his purse. He spent $40½ for his text books and $20¾ for his note books. How much will remain with him?

Solution :

Money had by Peter = 75¾.

Remaining money :

= 75¾ - ( 40½ +  20 ¾)

= 303/4 - (81/2 + 83/4)

= 303/4 - (162/4 + 83/4)

= 303/4 - (162 + 83)/4

= 303/4 - 245/4

= (303 - 245)/4

= 54/4

= 29/2

= $14½

Problem 2 :

The total weight of 4 girls is 155¼.  If three of them weigh 20 kg, 44½ kg and 30 kg. Find the weight o the remaining girl.

Solution :

Total weight of 4 girls = 155¼.

Let x be the weight of required girl.

Sum of the weight of three girls :

= (20 + 44½ + 30)

= 101/5 + 89/2 + 91/3

Here L.C.M is 5 ⋅ 2  3 = 30.

= (101/5)  (6/6) +  (89/2)  (15/15) + (91/3) ⋅ (10/10)

= 606/30 + 1335/30 + 910/30

= (606 + 1335 + 910)/30

= 2851/30

Weight of required girl :

= 155¼ - 2851/30

= 621/4 - 2851/30

L.C.M  =  60.

= (621/4) ⋅ (15/15) - (2851/30)

= (9315 - 2851)/30

= 6464/30

= 3232/15

= 215⁷⁄₁₅ kg

Problem 3 :

In three packets there were $13, $25 and $30. The total amount in all the four packets is $95. What is the amount in the fourth packet?

Solution :

Sum of amount in three  packets :

= 13⅔ + 25⅓ + 30

= 41/3 + 76/3 + 181/6

= 82/6 + 152/6 + 181/6

= (82 + 152 + 181)/6

= 415/6

Remaining amount of fourth packet :

= 95 - 415/6

= (570 - 415)/6

= 155/6

= $25

Problem 4 :

Maria worked 5½ hours on Tuesday and 6¼ hours on Wednesday. John worked 2½ on Tuesday and 5¼ hours on Wednesday. How many fewer hours did John work?

Solution :

Time taken by Maria :

= 5½ + 6¼

= 11/2 + 25/4

= 22/4 +  25/4

= (22 + 25)/4

= 47/4 

Time taken by John :

= 2½ + 5¼

= 5/2 + 21/4

= 10/4 + 21/4

= (10 + 21)/4

= 31/4

Difference of time taken between  Maria and John

= 47/4 - 31/4

= (47 - 31)/4

= 16/4

= 4 hours 

Maria had worked 4 hours fewer than John.

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