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Variations and Propotions - Difficult

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by gmatrant » Fri Oct 12, 2007 10:04 pm
A varies directly as the sum of the two quantities B and C. B in turn varies directly as x and C varies inversly as x.

When x=2, A=6 and when x=4 , A=9. Find the value of A when the value of x=16.

1)2.5 2)1 3)8.5 4)32.5

Answer [spoiler](4)[/spoiler]
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by Suyog » Sat Oct 13, 2007 12:13 am
When x=2, A=6
eqn: Bx + C/x = 6
i.e: 2B + C/2 = 6 ....... I

when x=4 , A=9
eqn: Bx + C/x = 9
i.e.: 4B + C/4 = 9 ....... II

Solving I and II we get B = 2, C = 4

Now B = 2, C = 4 and x = 16
eqn: Bx + C/x
= 2(16) + 4/16
= 32 + 0.25
= 32.25

Closest is option D
Can you check if its correct?
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by gmatrant » Sat Oct 13, 2007 3:15 am
Suyog wrote:When x=2, A=6
eqn: Bx + C/x = 6
i.e: 2B + C/2 = 6 ....... I

when x=4 , A=9
eqn: Bx + C/x = 9
i.e.: 4B + C/4 = 9 ....... II

Solving I and II we get B = 2, C = 4

Now B = 2, C = 4 and x = 16
eqn: Bx + C/x
= 2(16) + 4/16
= 32 + 0.25
= 32.25

Closest is option D
Can you check if its correct?
You are right the answer is 32.25.. Sorry for the typo
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