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Each of the four identical containers (A,B,C and D) contains ‘b’ balls

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Each of the four identical containers (A,B,C and D) contains ‘b’ balls [#permalink]

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New post 27 Aug 2014, 23:42
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Difficulty:

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Question Stats:

65% (01:52) correct 35% (01:44) wrong based on 144 sessions

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Each of the four identical containers (A,B,C and D) contains ‘b’ balls. When some of the balls from container A are moved to the other 3 containers, the ratio of the number of the balls in A,B,C and D are in the ratio 1:4:4:3. How many balls are moved from the first container in terms of ‘b’?

A) (5/6)b
B) (1/6)b
C) (2/6)b
D) (4/6)b
E) (3/6)b




Source: 4Gmat
[Reveal] Spoiler: OA

Kudos [?]: 320 [0], given: 197

Manager
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Re: Each of the four identical containers (A,B,C and D) contains ‘b’ balls [#permalink]

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New post 27 Aug 2014, 23:44
I got the answer be taking a number for 'b' .

But can someone tell me there is a better method to solve this. If deriving is better than taking values.



Also I am able to get answers for Quant but invariably it takes a lot of time sometimes to get the answer.

Would you suggest some method to improve speed :(

I don't want to miss out on questions because of speed.

Any suggestions to build speed? :(

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Each of the four identical containers (A,B,C and D) contains ‘b’ balls [#permalink]

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New post 28 Aug 2014, 00:00
alphonsa wrote:
I got the answer be taking a number for 'b' .

But can someone tell me there is a better method to solve this. If deriving is better than taking values.



Also I am able to get answers for Quant but invariably it takes a lot of time sometimes to get the answer.

Would you suggest some method to improve speed :(

I don't want to miss out on questions because of speed.

Any suggestions to build speed? :(



We have different methods to approach this problem,

1st look at the answer choices, all the values contain denominator 6. So B must be a multiple of 6, only then the value for all the answer choices will be an integer.

Since we cannot have the number of balls in fractions.

Now plugging a value. Clue here is answer choices. So take b=6.

If all the jars has same number of (b) balls initially, then the total number of balls will be 6+6+6+6= 4*6= 24.

We are removing some balls from jar A and moving it to other jars. The ratio is 1:4:4:3.

Notice that, we are not adding new balls from outside. So the sum will be always equal to 24.

Let x be the multiplier, then

1x+4x+4x+3x= 24.

12x=24

x=2.

Now, the number of balls in each jar will be,

A:2
B:4*2=8
C:4*2=8
D: 3*2=6

Initially we had 6 balls in jar A. Now we have 2 balls. So 4 balls removed from the jar A.

Check the answer choices which gives us the value of 4, when we substitute b=6.

Only one option does that.

So Answer is D.

Hope it helps.
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Re: Each of the four identical containers (A,B,C and D) contains ‘b’ [#permalink]

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New post 28 Aug 2014, 00:16
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alphonsa wrote:
Each of the four identical containers (A,B,C and D) contains ‘b’ balls. When some of the balls from container A are moved to the other 3 containers, the ratio of the number of the balls in A,B,C and D are in the ratio 1:4:4:3. How many balls are moved from the first container in terms of ‘b’?

A) (5/6)b
B) (1/6)b
C) (2/6)b
D) (4/6)b
E) (3/6)b

Source: 4Gmat


After ball movement, the ratio = 1:4:4:3

Before ball movement, all 4 containers had same "b" no. of balls = 3:3:3:3

Comparing before/after ratios.....

It means 2 balls (out of 3) were taken out from A; 1 went in B & 1 went in C (No change in D)

Equation \(= \frac{2b}{3} = \frac{4b}{6}\)

Answer = D
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Re: Each of the four identical containers (A,B,C and D) contains ‘b’ balls [#permalink]

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New post 28 Aug 2014, 00:39
Paresh's approach is the quickest that I know of. Although the numbers are relatively easy in this question, I'd choose the denominator of the solutions as b (ie 6 in this question), that way I can be sure of getting an integer.

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Re: Each of the four identical containers (A,B,C and D) contains ‘b’ balls [#permalink]

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New post 06 Mar 2016, 20:55
Gnpth wrote:
alphonsa wrote:
I got the answer be taking a number for 'b' .

But can someone tell me there is a better method to solve this. If deriving is better than taking values.



Also I am able to get answers for Quant but invariably it takes a lot of time sometimes to get the answer.

Would you suggest some method to improve speed :(

I don't want to miss out on questions because of speed.

Any suggestions to build speed? :(



We have different methods to approach this problem,

1st look at the answer choices, all the values contain denominator 6. So B must be a multiple of 6, only then the value for all the answer choices will be an integer.

Since we cannot have the number of balls in fractions.

Now plugging a value. Clue here is answer choices. So take b=6.

If all the jars has same number of (b) balls initially, then the total number of balls will be 6+6+6+6= 4*6= 24.

We are removing some balls from jar A and moving it to other jars. The ratio is 1:4:4:3.

Notice that, we are not adding new balls from outside. So the sum will be always equal to 24.

Let x be the multiplier, then

1x+4x+4x+3x= 24.

12x=24

x=2.

Now, the number of balls in each jar will be,

A:2
B:4*2=8
C:4*2=8
D: 3*2=6

Initially we had 6 balls in jar A. Now we have 2 balls. So 4 balls removed from the jar A.

Check the answer choices which gives us the value of 4, when we substitute b=6.

Only one option does that.

So Answer is D.

Hope it helps.



here is my approach
each jar has initially b balls
next after removing => x,4x,4x,3x
balls in A = x+4x-b+4x-b+3x-b = 12x-3b
b=12x-3b => x=b/3

now balls removed from A= b-b/3 = 2b/3 =4b/6 => option D..

is it wrong to think like this?
Anyways your solution is awesome and it takes less time..

thanks
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Re: Each of the four identical containers (A,B,C and D) contains ‘b’ balls [#permalink]

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New post 06 Apr 2016, 18:35
alphonsa wrote:
Each of the four identical containers (A,B,C and D) contains ‘b’ balls. When some of the balls from container A are moved to the other 3 containers, the ratio of the number of the balls in A,B,C and D are in the ratio 1:4:4:3. How many balls are moved from the first container in terms of ‘b’?

A) (5/6)b
B) (1/6)b
C) (2/6)b
D) (4/6)b
E) (3/6)b




Source: 4Gmat


i solved it algebraically and it took me under 1 minute to do so.

we have 1x:4x:4x:3x = total =12x.
we know that 12x=4b
b=3x

now, we are left with 1 x, then 2x must have been removed.
(2x/3x)*b
since we need 4 in denominator, multiply everything by 2:
(4/6)b

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Re: Each of the four identical containers (A,B,C and D) contains ‘b’ balls [#permalink]

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New post 27 Oct 2017, 21:28
alphonsa wrote:
Each of the four identical containers (A,B,C and D) contains ‘b’ balls. When some of the balls from container A are moved to the other 3 containers, the ratio of the number of the balls in A,B,C and D are in the ratio 1:4:4:3. How many balls are moved from the first container in terms of ‘b’?

A) (5/6)b
B) (1/6)b
C) (2/6)b
D) (4/6)b
E) (3/6)b

Source: 4Gmat


Let \(y\) balls be transferred from A to other three containers. we need to find the value of \(y\)

After the transfer number of ball in each container be \(x\), \(4x\), \(4x\) & \(3x\)

number of balls left in container A; \(x= b-y\)

And total number of balls in the remaining three containers after increment of \(y\); \((4x+4x+3x)=3b+y => 11x=3b+y\)

substitute the value of \(x\) in the above equation to get

\(11(b-y)=3b+y\), solve this to get

\(y= \frac{8b}{12} => \frac{4b}{6}\)

Option D

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Re: Each of the four identical containers (A,B,C and D) contains ‘b’ balls   [#permalink] 27 Oct 2017, 21:28
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Each of the four identical containers (A,B,C and D) contains ‘b’ balls

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