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Interesting GMATFix Problem-35

Expert replies
by arora007 » Wed Sep 22, 2010 3:28 pm
A bowl contains 34 marbles colored red, black and green. The number of red marbles is fewer than a third of the number of black marbles. There are more than twice as many green marbles as red marbles. What is the least possible sum of black marbles and green marbles?
a)25
b)26
c)27
d)28
e)29
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Source: — Problem Solving |

by fatalityish » Wed Sep 22, 2010 3:49 pm
Answer for this one is (c) 27

Solution:-

let there be r red marbles, g green marbles and b black marbles.

hence
r + g + b = 34 ...........(I)

also as per given data

r < (b/3)
r < 2*g

or rearranging them

3*r<b
(r/2)<g

adding the two equations

3r + (r/2) < (b+g)
(7/2)r < (b+g)

put b+g = x for convenience
(7/2)r < x ........(II)

from (I)
r = 34 - x

substitute value of r in (II)
(7/2)*(34-x) < x

on solving this we get
x>26.4 (approx.)
or
b+g>26.4

since the number of marbles has to be an integral value hence least possible value of total of blacj and green marbles is 27.
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by diebeatsthegmat » Wed Sep 22, 2010 7:19 pm
arora007 wrote:A bowl contains 34 marbles colored red, black and green. The number of red marbles is fewer than a third of the number of black marbles. There are more than twice as many green marbles as red marbles. What is the least possible sum of black marbles and green marbles?
a)25
b)26
c)27
d)28
e)29
yeah, 27 should be the answer
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by arora007 » Wed Sep 22, 2010 8:28 pm
Yeah...27 it is...
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