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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Variations and Propotions - Difficult
This topic has expert replies
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?
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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gmatrant
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You are right the answer is 32.25.. Sorry for the typoSuyog 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?
