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Bob and Cindy are playing a game in which Bob has to give

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Bob and Cindy are playing a game in which Bob has to give  [#permalink]

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New post 26 Mar 2014, 09:08
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Difficulty:

  55% (hard)

Question Stats:

70% (03:07) correct 30% (03:36) wrong based on 175 sessions

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Bob and Cindy are playing a game in which Bob has to give 1/4th of his toffees to Cindy after which they will have same number of toffees. Instead if Cindy gives 1/4th of her toffees to Bob then Bob has to have thrice as many as Cindy has. Find the original number of toffees with Bob before the beginning of the game?

A) 12
B) 18
C) 20
D) 24
E) 30
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Re: Bob and Cindy are playing a game in which Bob has to give  [#permalink]

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New post 26 Mar 2014, 09:55
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rrsnathan wrote:
Bob and Cindy are playing a game in which Bob has to give 1/4th of his toffees to Cindy after which they will have same number of toffees. Instead if Cindy gives 1/4th of her toffees to Bob then Bob has to have thrice as many as Cindy has. Find the original number of toffees with Bob before the beginning of the game?

A) 12
B) 18
C) 20
D) 24
E) 30


First of all notice that in order Bob to be able to give 1/4th of his toffees to Cindy, must have the number of toffees which is a multiple of 4. Eliminate B, and E. (Similarly, the number of toffees Cindy has must also be a multiple of 4).

Backsolve:

A. If originally Bob had 12 toffees, then after he gave 3 (1/4th) to Cindy he would be left with 9 toffees, so originally Cindy had 9-3=6. Eliminate because 6 is not a multiple of 4.

C. If originally Bob had 20 toffees, then after he gave 5 (1/4th) to Cindy he would be left with 15 toffees, so originally Cindy had 15-5=10. Eliminate because 10 is not a multiple of 4.

Only option D is left. Just to check:

D. If originally Bob had 24 toffees, then after he gave 6 (1/4th) to Cindy he would be left with 18 toffees, so originally Cindy had 18-6=12. The second condition also holds: if Cindy gives 1/4th of her toffees to Bob then Bob has to have thrice as many as Cindy has --> 3(12 - 3) = 24 + 3.

Answer: D.

Hope it's clear.
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Re: Bob and Cindy are playing a game in which Bob has to give  [#permalink]

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New post 26 Mar 2014, 10:08
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Bunuel wrote:
rrsnathan wrote:
Bob and Cindy are playing a game in which Bob has to give 1/4th of his toffees to Cindy after which they will have same number of toffees. Instead if Cindy gives 1/4th of her toffees to Bob then Bob has to have thrice as many as Cindy has. Find the original number of toffees with Bob before the beginning of the game?

A) 12
B) 18
C) 20
D) 24
E) 30


First of all notice that in order Bob to be able to give 1/4th of his toffees to Cindy, must have the number of toffees which is a multiple of 4. Eliminate B, and E. (Similarly, the number of toffees Cindy has must also be a multiple of 4).

Backsolve:

A. If originally Bob had 12 toffees, then after he gave 3 (1/4th) to Cindy he would be left with 9 toffees, so originally Cindy had 9-3=6. Eliminate because 6 is not a multiple of 4.

C. If originally Bob had 20 toffees, then after he gave 5 (1/4th) to Cindy he would be left with 15 toffees, so originally Cindy had 15-5=10. Eliminate because 10 is not a multiple of 4.

Only option D is left. Just to check:

D. If originally Bob had 24 toffees, then after he gave 6 (1/4th) to Cindy he would be left with 18 toffees, so originally Cindy had 18-6=12. The second condition also holds: if Cindy gives 1/4th of her toffees to Bob then Bob has to have thrice as many as Cindy has --> 3(12 - 3) = 24 + 3.

Answer: D.

Hope it's clear.


Thanks for the quick reply.Ur answers are really awesome.

Is there any other way to solve this? Using taking variables for unknowns?

Thanks in Advance,
Rrsnathan.
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Re: Bob and Cindy are playing a game in which Bob has to give  [#permalink]

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New post Updated on: 26 Mar 2014, 20:29
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are questions like this fair game for the gmat problem solving q's (i.e. questions with no concrete solution - Bob could have just as easily had 48, 72 - that can only be solved through POE with answer choices). thx
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Originally posted by m3equals333 on 26 Mar 2014, 20:27.
Last edited by m3equals333 on 26 Mar 2014, 20:29, edited 1 time in total.
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Re: Bob and Cindy are playing a game in which Bob has to give  [#permalink]

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New post 26 Mar 2014, 22:36
2
rrsnathan wrote:
Bunuel wrote:
rrsnathan wrote:
Bob and Cindy are playing a game in which Bob has to give 1/4th of his toffees to Cindy after which they will have same number of toffees. Instead if Cindy gives 1/4th of her toffees to Bob then Bob has to have thrice as many as Cindy has. Find the original number of toffees with Bob before the beginning of the game?

A) 12
B) 18
C) 20
D) 24
E) 30


First of all notice that in order Bob to be able to give 1/4th of his toffees to Cindy, must have the number of toffees which is a multiple of 4. Eliminate B, and E. (Similarly, the number of toffees Cindy has must also be a multiple of 4).

Backsolve:

A. If originally Bob had 12 toffees, then after he gave 3 (1/4th) to Cindy he would be left with 9 toffees, so originally Cindy had 9-3=6. Eliminate because 6 is not a multiple of 4.

C. If originally Bob had 20 toffees, then after he gave 5 (1/4th) to Cindy he would be left with 15 toffees, so originally Cindy had 15-5=10. Eliminate because 10 is not a multiple of 4.

Only option D is left. Just to check:

D. If originally Bob had 24 toffees, then after he gave 6 (1/4th) to Cindy he would be left with 18 toffees, so originally Cindy had 18-6=12. The second condition also holds: if Cindy gives 1/4th of her toffees to Bob then Bob has to have thrice as many as Cindy has --> 3(12 - 3) = 24 + 3.

Answer: D.

Hope it's clear.


Thanks for the quick reply.Ur answers are really awesome.

Is there any other way to solve this? Using taking variables for unknowns?

Thanks in Advance,
Rrsnathan.



Using variable, we can reach upto a certain point after which plugin would require. Infact, I tried to solve using the same.

Initial sweets with Bob = x

initial sweets with Candy = y

By the condition given,

\(x - \frac{x}{4} = y + \frac{x}{4}\)

\(\frac{x}{2} = y\) OR x = 2y ....... (1)


As Bunuel pointed out, B & E can be ignored (not divisible by 4)

We should also look for that value of y which is divisible by 4 (as given in the next part)

Putting values

A. 12 & 6 ....... (6 not divisibly by 4; ignore it)

C: 20 & 10 ....... (10 not divisibly by 4; ignore it)

D: 24 & 12 ......... (12 is divisibly by 4)

Answer = D


One more method; but after reaching equation (1)

Look out for the option which is divisible by 8; because we require to divide by 4 twice

Only option D stands out;
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Re: Bob and Cindy are playing a game in which Bob has to give  [#permalink]

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New post 27 Mar 2014, 00:40
m3equals333 wrote:
are questions like this fair game for the gmat problem solving q's (i.e. questions with no concrete solution - Bob could have just as easily had 48, 72 - that can only be solved through POE with answer choices). thx


Since it is a PS question, there is only one correct answer. If it would have been a DS question then may be we could have ended up in multiple options and hence chose E

Using variables, You end up with same equation so invariably you have to pluggin

For Plugin always start with Option C and then decide whether to move up or down.

If originally Let there B candies and C candies
Then after 1st we have C=1/2B
From St 2 we have 1/4C+B=9/4C or C=1/2B
Only D fits

All other options will give you C candies as non-integer values.
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Re: Bob and Cindy are playing a game in which Bob has to give  [#permalink]

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New post 27 Mar 2014, 03:24
I tried to solve it algebraically but trick is that there is not any absolute values to solve it in such way. The only information we can get is that initially B=2C. The option should be multiple of four after halving, so it is 24. Answer is D
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Re: Bob and Cindy are playing a game in which Bob has to give  [#permalink]

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New post 25 May 2014, 08:32
Yeah, I also backsolved this one.

Actually, when doing algebraically I got the same equation.

First C + 1/4B = 3/4B

B=2C

Then if B+1/4C = (3)(3/4C)
B=2C again

Backup approach? Backsolve

Then if B=20, C=10

Since C not divisible by 4, not valid.
Try next

B=24, C=12

C gives 3, is left with 9

Bob now was 27, which is three times C

BINGO

Answer: D

Hope this helps
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Re: Bob and Cindy are playing a game in which Bob has to give  [#permalink]

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New post 19 Apr 2017, 12:21
Nice question.
It took me more than 3 min to realize that checking the answers choice was faster.
If you go with algebra, you find only one equation x=2y , where x:number of toffees of Bob and Y: numbers of toffees of Cindy.
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Re: Bob and Cindy are playing a game in which Bob has to give   [#permalink] 19 Apr 2017, 12:21
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