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# If x, y, and k are positive numbers such that

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Manager
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If x, y, and k are positive numbers such that [#permalink]

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19 Nov 2006, 14:13
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x, y, and k are positive numbers such that (10x+20y)/(x+y) = k and if x<y, which of the following could be the value of k?

a) 10
b) 12
c) 15
d) 18
e) 30
Director
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19 Nov 2006, 14:43
Just by substitution I get 18.

If all 3 are positive, and y >x, the minimum value of x is 1. then y can be 2, 3, 4 etc. When y = 4, k =18. For y=2, and y=3, k is not a positive integer.
SVP
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19 Nov 2006, 15:09
(D) as well

(10x+20y)/(x+y) = k

<=> (10x+20y) = k*(x+y)
<=> (k-10)*x = (20-k)*y

To make x and y positive, k have to be such that 10 < k < 20.

This rules out A and E.

Then, since x < y, k must be bigger than 15 (we must multiply a bigger number in front of x than in front of y).

Hence, k = 18.
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19 Nov 2006, 16:21
Fig,

Help me out a little. Firstly, I've done the first part of the question two ways and come up with two different answers:

First:

10x + 20y = k(x +y) ---> 10x - kx = ky - 20y ---> x (10 - k) = y (k - 20)

For x and y to be positive k<10 or k<20???? Where'd I go wrong????

The other way I arranged it was the same as you:

<=> (10x+20y) = k*(x+y)
<=> (k-10)*x = (20-k)*y

How did you go from here to working out it's 18?

I think I'm up too late!!
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19 Nov 2006, 16:56
Fig,

Help me out a little. Firstly, I've done the first part of the question two ways and come up with two different answers:

First:

10x + 20y = k(x +y) ---> 10x - kx = ky - 20y ---> x (10 - k) = y (k - 20)

For x and y to be positive k<10 or k<20???? Where'd I go wrong????

The other way I arranged it was the same as you:

<=> (10x+20y) = k*(x+y)
<=> (k-10)*x = (20-k)*y

How did you go from here to working out it's 18?

I think I'm up too late!!

By your first way, it's a little harder to view the elements ... It makes us deal with an equation with negative factors in front. But it's possible.

x (10 - k) = y (k - 20)

Imagine now:
o If k = 21 the equation is turned to be -11*x = y than x and y cannot both be positive. >>>> k < 20.
o If k = 9 the equation is turned to be x = -11*y than x and y cannot both be positive. >>>> k > 10.

Then, 10-k and k-20 suggest a sort of symetry with a center at 15. To confirm this feeling, we can take k=15.

x (10 - 15) = y (15 - 20) <=> x = y : Bingo there is a symetrical behaviour at 15.

So now we can try k = 12,

x (10 - 12) = y (12 - 20) <=> 2 x = 8 y >>>> x > y : the opposite of what we are looking for.

So k = 18

By the second way, we are using positive factors in front of x & y.

(k-10)*x = (20-k)*y

Here, we can observe without calculation that:
> at left side of the equation, k must be greater than 10 (otherwise we have a negative factor)
> at right side, k must be fewer than 20 (otherwise we have a negative factor)

One more time, k-10 and 20-k suggest a symetry at 15. Then, similar to the first way but without calculation, we can see that greater k is fewer x is.

Thus, as we want x < y and as k = 15 is a symetry point, k must be equal to 18.
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19 Nov 2006, 17:23
Thanks a lot Fig and hats off to you.
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19 Nov 2006, 18:00
(10x+20y)/(x+y) = 10(x+2y)/(x+y) = 10[(x+y)+y]/x+y = 10 + 10y/(x+y) = 10(1+y/x+y)

10(1+y/x+y) = k
1 + y/(x+y) = k/10
y/(x+y) = k/10 - 1

So k cannot be 10 or 30 since y/x+y is a proper fraction.

Testing:
k = 12, y/x+y = 1/5 --> would require x > y. Out.
k = 15, y/x+y = 1/2 --> would mean x = y. Out.
k = 18, y/x+y = 4/5 --> possible since y can be greater than x for this result.

Ans D
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19 Nov 2006, 18:07
plug in, D is right.
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19 Nov 2006, 20:34
(10x + 20y)/(x + y) = 10 + 10 * y / (x + y)
as y > x
1 > y / (x + y) > 1/2

Only possible ans is 18
19 Nov 2006, 20:34
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