Problem 1 :
A can contains 10 kg of oil. 2¾ kg and 5⅓ 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 + 5⅓
Remaining quantity of oil :
= Original quantity of oil - Quantity of oil poured
= 10 - (2¾ + 5⅓)
= 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 12⅞ meters long. From this two pieces, one 3¼ meters long and another 4⅔ meters long are cut off. What is the length of the remaining part of the rod?
Solution :
Original length of steel rod = 12⅞ meter.
Length of rod which had been cut :
= 3¼ + 4⅔
= 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
= 12⅞ - 95/12
= 103/8 - 95/12
LCM of 8 and 12 is 24.
= 309/24 - 190/24
= (390 - 190)/24
= 200/24
= 25/3
= 8⅓
Therefore, the remaining length of rod is 25⅓ 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 = $305¾.
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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