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Official Explanation

Answer: C - This is a good question on which to start with a made-up number. Recognizing that we'll have to divide by 2 and by 3, let's say the capacity of C is 30. If it's half full of water, then, it contains 15 of water (there's no unit, but we don't need one). D has twice the capacity, so its capacity is 60. It's 1/3 full of water, so it contains 20.

The next step has us pouring as much of what's in D as possible into C. C has a capacity of 30 and is currently holding 15, so it could hold 15 more. D currently holds 20, so 15 of that could be poured into C, leaving 5.

If D has a capacity of 60 and is currently holding 5, it contains 5/60 of it's capacity, or 1/12, choice (C).
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How do you suddenly jump to 6 litter and 12 litter
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How do you suddenly jump to 6 litter and 12 litter

We are told that canister canister D has twice the capacity of canister C.

Thus if we assume that C has a capacity of 6 liters, then the capacity of D must be twice of that, so 12 liters.

Hope it's clear.
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Hi all,

If we assume the total Volume of C to be P, then it is P/2 full, then volume of D would be 2P and it would be 2P/3 full

If we pour P/2 from D to C, we would be left with 2P/3-P/2 = P/6...

I'm not able to point out where the error is. Can someone assist me?
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appsy01
Hi all,

If we assume the total Volume of C to be P, then it is P/2 full, then volume of D would be 2P and it would be 2P/3 full

If we pour P/2 from D to C, we would be left with 2P/3-P/2 = P/6...

I'm not able to point out where the error is. Can someone assist me?

you have got right upto P/6
But ATQ --->canister D will still contain what fraction of its capacity of water
and the capacity of D is 2P...so the remaining fraction should be (p/6)/2P--1/12
Ans C
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Canister C is 1/2 full of water and canister D, which has twice the capacity of canister C, is 1/3 full of water. If the water in canister D is poured in canister C until canister C is completely full of water, canister D will still contain what fraction of its capacity of water?

(A) 0
(B) 1/36
(C) 1/12
(D) 1/6
(E) 1/4

C needs 1/4 of D to become full, leaving D filled to 1/3-1/4=1/12 of its capacity
1/12
C
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Let the capacities of the canisters be x and 2x then water in canisters will be (1/2) x and (2/3)x respectively.
(1/2)x more will go into 1st canister =>left behind water in 2nd canister=(2/3)x-(1/2)x=(1/6)x
So, (1/6)x/2x= (1/12)..C is the correct choice.
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Canister C, using smart numbers = 60 liters, 1/2 = 30

so 30 liters of 60 liters,

Canister D = 2x C so 120 liters, and it is 1/3 full = 40

It will take 40 liters of D to fill canister C so 10 liters remain = 10/120 = 1/12
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avohden
Canister C is 1/2 full of water and canister D, which has twice the capacity of canister C, is 1/3 full of water. If the water in canister D is poured in canister C until canister C is completely full of water, canister D will still contain what fraction of its capacity of water?

(A) 0
(B) 1/36
(C) 1/12
(D) 1/6
(E) 1/4



let capacity of C = 2
so full of water = 1
and capacity of D = 4
full of water = 4/3
now D is poured into C until it gets full and we are left with ; 4/3-1 ; 1/3
so now fraction of D water 1/3
which in D capacity 1/3/4 = 1/12
OPTION C
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I did this in kind of a round about.

Canister C is 1/2 full of water and canister D, which has twice the capacity of canister C, is 1/3 full of water. If the water in canister D is poured in canister C until canister C is completely full of water, canister D will still contain what fraction of its capacity of water?

(A) 0
(B) 1/36
(C) 1/12
(D) 1/6
(E) 1/4

Suppose the capacities are...
C = 100 --> so 50L
D = 200 --> ~66.6L

C has a deficit of 50.

66.6 - 50 = 16.6 <--- amount left in D.

16.6/200 = ?

15/100 x 100 = 15%
15/200 x 100 = 7.5%

So 16.6/200 is a little bit greater than 7.5%

1/12 = ~ 0.08

C.
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