You-have-a-3-litre-jug-and-a-5-litre-jug-Make-4-litres-

Question Number 2603 by Yozzis last updated on 23/Nov/15

$${You}\:{have}\:{a}\:\mathrm{3}\:{litre}\:{jug}\:{and}\:{a}\:\mathrm{5}\:{litre}\:{jug}.\:{Make}\:\mathrm{4}\:{litres}. \\ $$

Answered by RasheedAhmad last updated on 23/Nov/15

2×(5 litre jug)−2×(3 litre jug) =4 litres

$$\mathrm{2}×\left(\mathrm{5}\:{litre}\:{jug}\right)−\mathrm{2}×\left(\mathrm{3}\:{litre}\:{jug}\right) \\ $$$$=\mathrm{4}\:{litres} \\ $$

Commented by Rasheed Soomro last updated on 23/Nov/15

^(RasheedAhmad) Pour two times 5 litre jug in avessel. Now take two times 3 litre jug out of vessel. Remaining liquid in the vessel =4 litres

$$\:^{{RasheedAhmad}} \\ $$$${Pour}\:{two}\:{times}\:\mathrm{5}\:{litre}\:{jug}\:{in}\:{avessel}. \\ $$$${Now}\:{take}\:{two}\:{times}\:\mathrm{3}\:{litre}\:{jug}\:{out}\:{of} \\ $$$${vessel}. \\ $$$${Remaining}\:{liquid}\:{in}\:{the}\:{vessel}\:=\mathrm{4}\:{litres} \\ $$

Commented by Yozzis last updated on 23/Nov/15

Practically speaking, explain to me how you′d get 4 litres.

$${Practically}\:{speaking},\:{explain}\:{to}\:{me}\:{how} \\ $$$${you}'{d}\:{get}\:\mathrm{4}\:{litres}. \\ $$

Commented by prakash jain last updated on 23/Nov/15

I think you are right. The question did not specify that you don′t have any other vessel.

$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{are}\:\mathrm{right}.\:\mathrm{The}\:\mathrm{question}\:\mathrm{did}\:\mathrm{not}\:\mathrm{specify} \\ $$$$\mathrm{that}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{any}\:\mathrm{other}\:\mathrm{vessel}. \\ $$

Commented by Yozzis last updated on 23/Nov/15

$${Thank}\:{you}! \\ $$

Answered by prakash jain last updated on 23/Nov/15

3 litre − empty into 5 litres 3 litre − fill up 5 litre. 2 l used, 1l remains in 3l jug. Empty 5l jug. put 1l remaining in 3l jug into 5l jug. fill up 3l jug and empty into 5l jug. 5l jug now has 4l. 2×3−5+3=4 Another option fill up 5l jug. empty into 3l jug−2l remain in 5l jug. empty 3l jug. fill up 3l jug with remain 2l from 5l. 3l jug now has 2l. fill up 5l jug. empty 5l jug into 3l jug. so 3l jug becomes full 1l used and 4l remains in 5l jug. 2×5−2×3=4

$$\mathrm{3}\:\mathrm{litre}\:−\:\mathrm{empty}\:\mathrm{into}\:\mathrm{5}\:\mathrm{litres} \\ $$$$\mathrm{3}\:\mathrm{litre}\:−\:\mathrm{fill}\:\mathrm{up}\:\mathrm{5}\:\mathrm{litre}.\:\mathrm{2}\:\mathrm{l}\:\mathrm{used},\:\mathrm{1l}\:\mathrm{remains} \\ $$$$\mathrm{in}\:\mathrm{3l}\:\mathrm{jug}. \\ $$$$\mathrm{Empty}\:\mathrm{5l}\:\mathrm{jug}. \\ $$$$\mathrm{put}\:\mathrm{1l}\:\mathrm{remaining}\:\mathrm{in}\:\mathrm{3l}\:\mathrm{jug}\:\mathrm{into}\:\mathrm{5l}\:\mathrm{jug}. \\ $$$$\mathrm{fill}\:\mathrm{up}\:\mathrm{3l}\:\mathrm{jug}\:\mathrm{and}\:\mathrm{empty}\:\mathrm{into}\:\mathrm{5l}\:\mathrm{jug}. \\ $$$$\mathrm{5l}\:\mathrm{jug}\:\mathrm{now}\:\mathrm{has}\:\mathrm{4l}. \\ $$$$\mathrm{2}×\mathrm{3}−\mathrm{5}+\mathrm{3}=\mathrm{4} \\ $$$$ \\ $$$$\mathrm{Another}\:\mathrm{option} \\ $$$$\mathrm{fill}\:\mathrm{up}\:\mathrm{5l}\:\mathrm{jug}. \\ $$$$\mathrm{empty}\:\mathrm{into}\:\mathrm{3l}\:\mathrm{jug}−\mathrm{2l}\:\mathrm{remain}\:\mathrm{in}\:\mathrm{5l}\:\mathrm{jug}. \\ $$$$\mathrm{empty}\:\mathrm{3l}\:\mathrm{jug}. \\ $$$$\mathrm{fill}\:\mathrm{up}\:\mathrm{3l}\:\mathrm{jug}\:\mathrm{with}\:\mathrm{remain}\:\mathrm{2l}\:\mathrm{from}\:\mathrm{5l}. \\ $$$$\mathrm{3l}\:\mathrm{jug}\:\mathrm{now}\:\mathrm{has}\:\mathrm{2l}. \\ $$$$\mathrm{fill}\:\mathrm{up}\:\mathrm{5l}\:\mathrm{jug}. \\ $$$$\mathrm{empty}\:\mathrm{5l}\:\mathrm{jug}\:\mathrm{into}\:\mathrm{3l}\:\mathrm{jug}.\:\mathrm{so}\:\mathrm{3l}\:\mathrm{jug}\:\mathrm{becomes}\:\mathrm{full} \\ $$$$\mathrm{1l}\:\mathrm{used}\:\mathrm{and}\:\mathrm{4l}\:\mathrm{remains}\:\mathrm{in}\:\mathrm{5l}\:\mathrm{jug}. \\ $$$$\mathrm{2}×\mathrm{5}−\mathrm{2}×\mathrm{3}=\mathrm{4} \\ $$

Commented by Yozzis last updated on 23/Nov/15

$${Thank}\:{you}! \\ $$

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