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Variations

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Variations

by Aman verma » Sun Aug 21, 2011 8:58 am
Q: Suppose y varies as the sum of two quantities of which one varies directly as x and the other inversely as x. If y = 6 when x = 4 and y = 10/3 when x = 3 , then the relation between x and y is :

a) x = y + 4

b) y = 2x + 8/x

c) y = 2x - 8/x

d) y = 2x - 4/x

e) y = 3x + 4


P.S- Now this problem can be very easily solved by plugging the options. But can this be solved algebraically. This problem has something to do with proportionality constant. I am yet unable to solve this algebraically.Looking for some tips !!
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by Frankenstein » Sun Aug 21, 2011 9:29 am
Hi,

y varies as the sum of two quantities. Let the two quantities be q1,q2.
So, y = k(q1+q2), where k is a constant.
q1 varies directly as x.
So, q1 = ax, a is a constant
q2 varies inversely as x
So, q2 = b/x
y = k(ax + b/x) is the equation.
We will A,B such that A = ka and B = kb.
So, y = Ax + B/x
We have two sets of values of x and y
6 = 4A+B/4 => 16A + B=24
10/3 = 3A + B/3 => 9A + B = 10
So, we get two equations in terms on A and B. Solve for A,B. We get A = 2,B=-8

Hence, C
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by Aman verma » Mon Aug 22, 2011 5:48 am
Awesome !! OA[spoiler]c)[/spoiler]
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by Whitney Garner » Mon Aug 22, 2011 10:45 am
Received PM asking for clarification. Frankenstein is right on but let me see if a slight variation in the notation might clear it up for anyone that isn't 100% with us!

If Y moves directly with x, that means that there exists a constant (let's call it j) where y=jx. If Y were to move inversely with x, then that means there exists a constant (let's call this one k) where y=k/x.

This problem says that the value of Y is the SUM of two values - one that varies directly with x and one that varies inversely with x, so we can use what we learned above.

Y = jX + k/X.

Now, we are given 2 cases {y=6,x=4} and {y=10/3, x=3}. We can plug these each into our equation above to get the following 2 equations:

(a) 6 = j(4) + k/(4)
(b) 10/3 = j(3) + k/(3)

Let's multiply both equations through with their denominators to simplify:

(a) 24 = 16j + k
(b) 10 = 9j + k

Now we have 2 equations in 2 unknowns, the constants j and k. We can subtract equation (b) from equation (a):

24 = 16j + k
-10 = -9j - k
---------------
14 = 7j
2 = j

Now plug back in to either to solve for k:

10 = 9(2) + k
-8 = k

We can then plug these into our original equation:

Y = jX + k/X
Y = 2X - 8/X

So the answer is C[/spoiler]

:)
Whit
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