WORD PROBLEMS WITH MIXED FRACTIONS
Problem 1 :
Peter had $75 3/4 in his purse. He spent $40 1/2 for his text books and $20 3/4 for his note books. How much will remain with him ?
Solution :
Money had by Peter = 75 3/4
Remaining money = 75 3/4 - ( 40 1/2 + 20 3/4)
= 303/4 - (81/2 + 83/4)
= 303/4 - { (81/2) ⋅ (2/2) + 83/4 }
= 303/4 - { (162/4 + 83/4 }
= 303/4 - { (162 + 83)/4 }
= (303/4) - (245/4)
= 58/4
= 29/2
Problem 2 :
The total weight of 4 girls is 155 1/4. If three of them weighs 20 1/5 kg 44 1/2 kg and 30 1/3 kg. Find the weight o the remaining girl.
Solution :
Total weight of 4 girls = 155 1/4
Let x be the weight of required girl.
Sum of the weight of three girls
= (20 1/5 + 44 1/2 + 30 1/3)
= (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 = 155 1/4 - 2851 /30
= (621/4) - (2851/30)
L.C.M = 60
= (621/4) ⋅ (15/15) - (2851/30)
= (9315 - 2851)/30
= 6464/30
= 215 7/15
Problem 3 :
In three packets there were $13 2/3, $25 1/3 and $30 1/6. The total amount in all the four packets is $95. What is the amount in the fourth packet ?
Solution :
Sum of amount of three packets = 13 2/3 + 25 1/3 + 30 1/6
= (41/3) + (76/3) + (181/6)
= (82 + 152 + 181)/6
= 415/6
Let x be the amount of fourth packet
Remaining amount of fourth packet = 95 - (415/6)
= (570 - 415)/6
= 155/6
Problem 4 :
Maria worked 5 1/2 hours on Tuesday and 6 1/4 hours on Wednesday. John worked 2 1/2 on Tuesday and 5 1/4 hours on Wednesday. How many fewer hours did John work ?
Solution :
Time taken by Maria = 5 1/2 + 6 1/4
= 11/2 + 25/4
= 22/4 + 25/4
= (22+25)/4
= 47/4
Time taken by John = 2 1/2 + 5 1/4
= (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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