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cfc198
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$3000 is distributed among A, B and C such that A gets 2 3 rd of w 21 Feb 2019, 10:13
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45% (medium)

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75% (02:45) correct 25% (02:55) wrong based on 337 sessions

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Rs. 3000 is distributed among A, B and C such that A gets 2/3 rd of what B and C together get and C gets 1/2 of what A and B together get. Find B’s share.
(1) 800
(2) 1000
(3) 1200
(4) 1500
(5) 1800
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$3000 is distributed among A, B and C such that A gets 2 3 rd of w 21 Feb 2019, 14:37
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The ratio of A's share to the rest is 2 to 3, so A gets 2/5 of the total, or $1200.

The ratio of C's share to the rest is 1 to 2, so C gets 1/3 of the total, or $1000.

So B must get the remaining $800.
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$3000 is distributed among A, B and C such that A gets 2 3 rd of w 21 Feb 2019, 13:33
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cfc198
Rs. 3000 is distributed among A, B and C such that A gets 2/3 rd of what B and C together get and C gets 1/2 of what A and B together get. Find B’s share.
(1) 800
(2) 1000
(3) 1200
(4) 1500
(5) 1800
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From the question we know:

1) A gets 2/3 rd of what B and C together get
Lets assume B+C get x
=> 2/3x + x = 3000
=> x = 1800
A gets 2/3x = 1200

2) C gets 1/2 of what A and B together get
Lets assume A+B get y
=> 1/2y + y = 3000
=> y = 2000
C gets 1/2y = 1000

Hence B gets = 3000 - 1200 - 1000 = 800

Hence A is the correct answer.
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jfranciscocuencag
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$3000 is distributed among A, B and C such that A gets 2 3 rd of w 01 Jul 2019, 18:12
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IanStewart
The ratio of A's share to the rest is 2 to 3, so A gets 2/5 of the total, or $1200.

The ratio of C's share to the rest is 1 to 2, so C gets 1/3 of the total, or $1000.

So B must get the remaining $800.
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Hello IanStewart!

Is it possible to apply this shortcutt to most of the problems related with ratios?

Kins regards!
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$3000 is distributed among A, B and C such that A gets 2 3 rd of w 02 Jul 2019, 05:27
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jfranciscocuencag

Is it possible to apply this shortcutt to most of the problems related with ratios?
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I'm not sure precisely what shortcut you mean, but there is so much variety in GMAT ratio problems that there could never be one shortcut that would solve 'most' problems.
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$3000 is distributed among A, B and C such that A gets 2 3 rd of w 12 Jul 2021, 10:31
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jfranciscocuencag
IanStewart
The ratio of A's share to the rest is 2 to 3, so A gets 2/5 of the total, or $1200.

The ratio of C's share to the rest is 1 to 2, so C gets 1/3 of the total, or $1000.

So B must get the remaining $800.

Hello IanStewart!

Is it possible to apply this shortcutt to most of the problems related with ratios?

Kins regards!
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Hi ,
By observing Qs and POE ,answer can easily be derived without solving.

Sum of A+B+C =3000
C= (A+B)/2
So C is mean of A & B.
Numbers could be arrange like A/B, C, B/A

now, A is 2/3 of (B+C) ..i.e A is 66% of sum of rest of numbers.

We can derive that , numbers will be arrange as B,C,A.
Value of C is 1000 (From equn. A+B/2). Therefore , only option 800 is feasible out of given options.
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$3000 is distributed among A, B and C such that A gets 2 3 rd of w 13 Jul 2021, 04:58
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A = (3000-A)*2/3

A = 1200

2C = 3000-C

C = 1000

B = 800
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$3000 is distributed among A, B and C such that A gets 2 3 rd of w 04 Feb 2026, 08:50
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Deconstructing the Question

Total = 3000 distributed among A, B, C.

Given:
\(A = \frac{2}{3}(B+C)\)
\(C = \frac{1}{2}(A+B)\)

Find B.

Step-by-step

From \(A = (2/3)(B+C)\):
\(3A = 2B + 2C\)

From \(C = (1/2)(A+B)\):
\(2C = A + B \Rightarrow A = 2C - B\)

Substitute into \(3A = 2B + 2C\):
\(3(2C - B) = 2B + 2C\)
\(6C - 3B = 2B + 2C\)
\(4C = 5B \Rightarrow B = \frac{4}{5}C\)

Then:
\(A = 2C - B = 2C - \frac{4}{5}C = \frac{6}{5}C\)

So:
\(A:B:C = 6:4:5\)

Total parts:
\(6+4+5=15\)
1 part:
\(3000/15 = 200\)

B has 4 parts:
\(B = 4 \cdot 200 = 800\)

Answer: (1) 800
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