TWO CONTAINER OF MILK CONTAINS 398 L AND 436 L OF MILK THE MILK IS TO BE TRANSFERRED TO ANOTHER CONTAINER WITH THE HELP OF DRUM WLE TRANSFERRING TO ANOTHER CONTAINER 7 L AND 11 L OF MILK ARE LEFT IN BOTH THE CONTAINER RESPECTIVELY WHAT WILL BE THE MAXIMUM CAPACITY OF DRUM
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HEY DEAR .
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Volume V1 = 398 liters
V2 = 436 Litres
a drum of capacity D litres is used to transfer the milk to other containers. When we transfer milk, we transfer D litres, 2D litres, 3 D , 4D etc. amounts. If there is any milk left out, that will be a fraction of the capacity of the drum.
let m and n be integers.
V1 = 398 = m * D + 7 => m * D = 391 = 17 * 23
V2 = 436 = n * D + 11 => n * D = 425 = 5 * 5 * 17
Looking at both equations , D = 17 , as only 17 occurs in both.
answer is 17 liters.
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___________________________
Volume V1 = 398 liters
V2 = 436 Litres
a drum of capacity D litres is used to transfer the milk to other containers. When we transfer milk, we transfer D litres, 2D litres, 3 D , 4D etc. amounts. If there is any milk left out, that will be a fraction of the capacity of the drum.
let m and n be integers.
V1 = 398 = m * D + 7 => m * D = 391 = 17 * 23
V2 = 436 = n * D + 11 => n * D = 425 = 5 * 5 * 17
Looking at both equations , D = 17 , as only 17 occurs in both.
answer is 17 liters.
HOPE , IT HELPS.
FOLLOW ME. ✌
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