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What is the total number of coins that Bert and Claire have?

Question: \(B+C=?\)

(1) Bert has 50 percent more coins than Claire --> \(B=1.5C\). Not sufficient.

(2) The total number of coins that Bert and Claire have is between 21 and 28 --> \(21<B+C<28\). Not sufficient.

(1)+(2) Since from (1) \(B=1.5C\) and from (2) \(21<B+C<28\), then \(21<1.5C+C<28\) --> \(21<2.5C<28\). Three values of C satsify this inequality; 9, 10, and 11. But for 9 and 11, the value of B from \(B=1.5C.\) won't be an integer, thus C can only be 10 --> \(B=1.5C=15\) --> \(B+C=25\). Sufficient.

Re: What is the total number of coins that Bert and Claire have? [#permalink]

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16 Oct 2012, 05:04

2

This post received KUDOS

If Bert has B coins and Claire has C coins. Then total number of coins =B+C

From Statement 1 B =1.5 C or B/C=3/2 Ratio of B and C is 3:2 => Only inference from this we can make is, total number must be a multiple of 5. Not sufficient

From statement 2 21<=B+C<=28 Multiple values for B+C - not sufficient

Combining the information from statement 1 and 2. Total number of coins = 25

Hence C.

Bunuel wrote:

What is the total number of coins that Bert and Claire have?

(1) Bert has 50 percent more coins than Claire. (2) The total number of coins that Bert and Claire have is between 21 and 28.

Practice Questions Question: 65 Page: 280 Difficulty: 600

What is the total number of coins that Bert and Claire have?

Question: \(B+C=?\)

(1) Bert has 50 percent more coins than Claire --> \(B=1.5C\). Not sufficient.

(2) The total number of coins that Bert and Claire have is between 21 and 28 --> \(21<B+C<28\). Not sufficient.

(1)+(2) Since from (1) \(B=1.5C\) and from (2) \(21<B+C<28\), then \(21<1.5C+C<28\) --> \(21<2.5C<28\). Three values of C satsify this inequality; 9, 10, and 11. But for 9 and 11, the value of B from \(B=1.5C.\) won't be an integer, thus C can only be 10 --> \(B=1.5C=15\) --> \(B+C=25\). Sufficient.

Answer: C.

Kudos points given to everyone with correct solution. Let me know if I missed someone.
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Re: What is the total number of coins that Bert and Claire have? [#permalink]

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25 Nov 2013, 10:18

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Re: What is the total number of coins that Bert and Claire have? [#permalink]

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10 Jan 2014, 07:51

Bunuel wrote:

What is the total number of coins that Bert and Claire have?

(1) Bert has 50 percent more coins than Claire. (2) The total number of coins that Bert and Claire have is between 21 and 28.

Practice Questions Question: 65 Page: 280 Difficulty: 600

The stem asks us: B + C = TC, what is TC?

1) This tells us that B = 1.5 C, so now we have two unknowns: 2.5C = TC, insufficient 2) Gives us a range of possible values, so in itself 2 is insufficient because the value could be any of 6 values. Insufficient

1 + 2 : Look closely: given our equation from 1, TC must be a multiple of 5 (we cannot have fractions of coins), and given the restriction in 2, we only have ONE possible multiple of 5: 25...

Re: What is the total number of coins that Bert and Claire have? [#permalink]

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12 Mar 2015, 01:27

Hello from the GMAT Club BumpBot!

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On Test Day, you're going to face a few questions in the Quant section that are perfect for "brute force" - you don't have to be a genius to get the correct answer and you don't need to have any specialized knowledge either. You DO have to put pen to pad and do some basic work though.

Here, we're asked for the total number of coins that Bert and Clair have.

Fact 1: Bert has 50 percent more coins than Claire.

Since you can't have a "fraction" of a coin, this piece of information provides just a little bit of info about the relationship between the number of coins that Bert has and the number of coins that Claire has.

IF..... Claire has 1 coin, Bert has 1.5 coins, which does NOT make sense (he can't have half a coin). Claire has 2 coins, then Bert has 3 coins. THIS makes sense. The total is 5 coins. Claire has 4 coins, then bert has 6 coins. This also makes sense. The total is 10 coins. Fact 1 is INSUFFICIENT.

You may have noticed that the total number of coins is going to be a multiple of 5. You don't need to know that to answer the question (although it would likely save you some time later on).

Fact 2: The total number of coins that Bert and Claire have is between 21 and 28.

This gives us more direct information, but no exact value. Fact 2 is INSUFFICIENT.

Combined, we know.... Bert has 50 percent more coins than Claire. The total number of coins that Bert and Claire have is between 21 and 28.

From here, you can "map out" the possibilities.... We already know that Claire MUST have an EVEN number of coins....so let's start there....

IF... Claire = 6, then Bert = 9 and the total is 15. This does not fit the given range, so it can't be the answer. Claire = 8, then Bert = 12 and the total is 20. This does not fit the given range, so it can't be the answer. Claire = 10, then Bert = 15 and the total is 25. This DOES fit the given range, so it COULD be the answer. Claire = 12, then Bert = 18 and the total is 30. This does not fit the given range, so it can't be the answer.

As the number of coins Claire has increases, then the total will increase (and continue to be out of the given range). This means that there's JUST ONE answer that fits all of the given information. Combined, SUFFICIENT.

Re: What is the total number of coins that Bert and Claire have? [#permalink]

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16 May 2015, 08:34

B+C?

(1) B=1.5C. Clearly insufficient.

(2) 21<B+C<28

As we are working with integers, we can rule out 21 and 28. Thus: 22=<B+C=<27. This is still insufficient, though. As we still have 6 possibilities.

(1)+(2) B+C should be divisible by 5/2. Because B+C = 2,5C and as we are working with countable items, we want to find an integer.

As such. We will have: Tot/(5/2)=Tot*2/5=2*tot/5 = integer.

To be divisible by 5, the number has to end in 5 or 0. For this reason I will only look at the unit digits. (2->7) 4,6,8,0,2,4. Only one out of the seven possibilities ended in 0 or 5, and thus only one possibility exists. C is sufficient.

(This looks as if it takes a lot of time, but it's not very time consuming when done in your head)
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Re: What is the total number of coins that Bert and Claire have? [#permalink]

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31 May 2015, 02:43

B+C = ? 1) C + 1,5C = 2,5C Not sufficient 2) Clearly no sufficient. We need a relationship between B and C to make it work 1+2) 2,5C = 21-28 C = (21-28)*2/5, so we need a multiple of 5 --> 25*2 is the only mulstiple of 5 in the range 21 - 28
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Re: What is the total number of coins that Bert and Claire have? [#permalink]

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07 Jul 2015, 01:04

I tried the estimation strategy and I have 2 possible solutions that match the criterion. If they have 16 and 8 coins or they have 18 and 9 coins. Need help understanding the same.

I tried the estimation strategy and I have 2 possible solutions that match the criterion. If they have 16 and 8 coins or they have 18 and 9 coins. Need help understanding the same.

It says that Bert has 50 percent more coins than Claire (B = 1.5C), not twice as many (B = 2C). Please refer to the solutions above.
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Re: What is the total number of coins that Bert and Claire have? [#permalink]

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08 Mar 2016, 18:35

fameatop wrote:

1) Let of no of coins with Claire = 2x Let of no of coins with Bert = 3x 2x+3x = 5x --->x can take any value - Insufficient

2) No of coins can take any integer value from 21 to 28- Insufficient

1+2) Total no of coins must be a multiple of 5 & between 21 & 28. Only possible integer value is 25--->Sufficient Answer C

I did this question the same way that fametop did, can anyone comment on if this is a good way to think about it. In terms of total coins needing to be a multiple of 5?

1) Let of no of coins with Claire = 2x Let of no of coins with Bert = 3x 2x+3x = 5x --->x can take any value - Insufficient

2) No of coins can take any integer value from 21 to 28- Insufficient

1+2) Total no of coins must be a multiple of 5 & between 21 & 28. Only possible integer value is 25--->Sufficient Answer C

I did this question the same way that fametop did, can anyone comment on if this is a good way to think about it. In terms of total coins needing to be a multiple of 5?

Yes, this is a perfectly valid method for attacking this question.

Alternately, once you realize that B=1.5C and 21 < B+C < 28 ---> 21< 2.5C<28, remember that the total number of coins MUST be an integer.

Thus the final value must be a multiple of 2.5 ---> 25 is the ONLY value.

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