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Bert and Clair's coins (OG13)

This topic has 2 expert replies and 2 member replies
alex.gellatly Master | Next Rank: 500 Posts
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Bert and Clair's coins (OG13)

Thu Aug 02, 2012 10:02 pm
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.

OK, I got the correct answer by picking numbers and kind of guessing. Can someone show me a good algebraic approach?

Thanks!

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A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-website-with-og-video-explanations-t112985.html#475231

vinodsundaram Senior | Next Rank: 100 Posts
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Target GMAT Score:
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Mon Aug 06, 2012 1:36 am
there's catch in these kind of problems.

Once you combine both equations, you solve and get multiple answers.
It is important to ELIMINATE All possible wrong answers.

We know Coins can always be Positive numbers. Hence using the relation between B and C we can eliminate.

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GMATGuruNY GMAT Instructor
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Fri Aug 03, 2012 10:09 am
alex.gellatly 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.

OK, I got the correct answer by picking numbers and kind of guessing. Can someone show me a good algebraic approach?

Thanks!
This problem is restricted to POSITIVE INTEGERS.

Statement 1: Bert has 50 percent more coins than Claire.
Thus, if C=2, then B = (1.5)2 = 3, implying that B:C = 3:2.
Since the sum of the parts of this ratio = 3+2 = 5, the total number of coins must be a multiple of 5.
INSUFFICIENT.

Statement 2: The total number of coins that Bert and Claire have is between 21 and 28.
Thus, the total number of coins could be any integer between 21 and 28.
INSUFFICIENT.

Statements 1 and 2 combined:
The only multiple of 5 between 21 and 28 is 25.
SUFFICIENT.

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Anurag@Gurome GMAT Instructor
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Thu Aug 02, 2012 10:12 pm
alex.gellatly 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.

OK, I got the correct answer by picking numbers and kind of guessing. Can someone show me a good algebraic approach?

Thanks!
Let us assume that the no. of coins that Bert has = B, and
no. of coins that Claire has = C
We have to find the value of B + C.

(1) Bert has 50% more coins than Claire implies B = C + 0.50C or B = 1.5C
So, B + C = 1.5C + C = 2.5C, but we do not know the value of C; NOT sufficient.

(2) The total number of coins that Bert and Claire have is between 21 and 28 implies 21 < (B + C) < 28.
This means B + C can be 22, 23, 24, 25, 26, or 27. But there is no unique value; NOT sufficient.

Combining (1) and (2), B + C = 2.5C and 21 < (B + C) < 28
21 < 2.5C < 28
21/2.5 < C < 28/2.5
8.4 < C < 11.2 implies C can be 9 or 10.

When C = 9, B = 1.5C = 1.5 * 9 = 13.5, which is not possible as B is the no. of coins that Bert has, which should be an integer.
When C = 10, B = 1.5C = 1.5 * 10 = 15. Here B + C = 15 + 10 = 25; SUFFICIENT.

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sm3.friendz@gmail.com Junior | Next Rank: 30 Posts
Joined
24 Jan 2012
Posted:
12 messages
Fri Aug 03, 2012 9:02 am
I> Not Sufficient because total coins are
x+1.5x=2.5x (X can take any value)
II> Not Sufficient because total coins could be anything between
21 to 28
Taking
I+II> 2.5x will be between 21 and 28
25 could be 1 value
Lets search for other values(should be less than or more than in multiples of 2.5)
25-2.5*(1) = 22.5 (Not possible)
25-2.5*(2) = 20 (Less than 21)
Similarly,
25+2.5*(1) = 27.5 (Not possible)
25+2.5*(2) = 30 (More than 28)

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