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Current Student B
Joined: 22 Jul 2014
Posts: 120
Concentration: General Management, Finance
GMAT 1: 670 Q48 V34 WE: Engineering (Energy and Utilities)
Each of the four identical containers (A,B,C and D) contains ‘b’ balls  [#permalink]

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13 00:00

Difficulty:   65% (hard)

Question Stats: 63% (02:19) correct 37% (02:25) wrong based on 155 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
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Re: Each of the four identical containers (A,B,C and D) contains ‘b’  [#permalink]

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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 $$= \frac{2b}{3} = \frac{4b}{6}$$

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Current Student B
Joined: 22 Jul 2014
Posts: 120
Concentration: General Management, Finance
GMAT 1: 670 Q48 V34 WE: Engineering (Energy and Utilities)
Re: Each of the four identical containers (A,B,C and D) contains ‘b’ balls  [#permalink]

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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? Current Student P
Status: Chasing my MBB Dream!
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WE: General Management (Aerospace and Defense)
Each of the four identical containers (A,B,C and D) contains ‘b’ balls  [#permalink]

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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’ balls  [#permalink]

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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.
Current Student D
Joined: 12 Aug 2015
Posts: 2568
Schools: Boston U '20 (M)
GRE 1: Q169 V154 Re: Each of the four identical containers (A,B,C and D) contains ‘b’ balls  [#permalink]

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1
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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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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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
Intern  Joined: 09 May 2016
Posts: 2
Each of the four identical containers (A,B,C and D) contains ‘b’ balls  [#permalink]

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Another way to solve:

Total number of balls if all containers have 'b' balls = 4b

assume x balls were taken out of the first container and y, m, n balls were put in the rest three containers.

b-x: b+y: b+m: b+n = 1:4:4:3 ..... [1+4+4+3=12]

It can be said that:
(b-x) = (1/12)*(4b) [balls in first container are 1/12th of total balls]
(b+y) = (4/12)*(4b) [balls in second container are 4/12th of total balls]
(b+m) = (4/12)*(4b) [balls in third container are 4/12th of total balls]
(b+n) = (3/12)*(4b) [balls in fourth container are 3/12th of total balls]

pick any one of the above to solve:

b-x=(1/12)*(4b)
b-x=b/3
x=2b/3

2b/3 is the total number of balls removed from container 1.

which can also be written as 4b/6.
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Posts: 117
GMAT 1: 650 Q47 V33 Re: Each of the four identical containers (A,B,C and D) contains ‘b’ balls  [#permalink]

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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

A contains b-2x-y
B contains b+x
C contains b+x
D contains b+y

Now the ratio is 1:4:4:3

So, b+x = 4(b-2x-y)
=> 9x+4y = 3b .......................1

and b+y = 3(b-2x-y)
=> 6x + 4y = 2b ......................2

Solving (Subtracting 2 from 1)
3x = b
so y = 0

So, balls taken out from A = 2x-y = 2b/3 = 4b/6 Re: Each of the four identical containers (A,B,C and D) contains ‘b’ balls   [#permalink] 20 Aug 2019, 13:41
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