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 20⅕ kg, 44½ kg and 30⅓ 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 :
= (20⅕ + 44½ + 30⅓)
= 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 $13⅔, $25⅓ and $30⅙. 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 + 30⅙
= 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
= $25⅚
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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