A Milk dealer has 125 liters, 150 liters and 275 liters of three different kinds of milk. Find the l

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29 Nov 2021 Milk Dealer

A Milk dealer has 125 liters, 150 liters and 275 liters of three different kinds of milk. Find the least number of containers of equal size required to store all the milk without mixing.

OPtion

1. 22
2. 28
3. 51
4. 48
5. 35
6. 37
7. 53
8. 43
9. 42
10. None of the above

Solution

Since minimum number of containers are required, the size of the containers is greatest. Also, the containers in three cases are of equal size.
The size of the containers is the H.C.F. of 125, 150 and 275 which is 25.
Now, the number of containers required for storing the milk = (125+150+275) ÷ 25 = 22

Correct Option: 1 Best Solution (4)

G.S.Kantharao   3 years AGO

H.C.F of 125,150,275 is 25
So required =(125/25)+(150/25)+(275/25)=5+6+11=22
My option is 1👈

Like? Yes (2) | No   

Varun   3 years AGO

Number per type of milk / HCF
= 25 * ((125+150+275)/25) = 22

Like? Yes (1) | No   

Devendra Marghade   3 years AGO

Maximum size of container = HCF (125, 150, 275) = 25
Therefore minimum number of containers required = (125/25) + (150/25) + (275/25) = 5+6+11 = 22
Correct option 1)

Like? Yes | No   

RAMADURAI NATESAN   3 years AGO

Total quantity = 550. HDF of different types = 25.
ANS = 550/25 = 22

Like? Yes | No