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Value of k (OG12 - PS-148)

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by karthikpandian19 » Tue Jan 03, 2012 10:04 pm
If x, y, and k are positive integers such that (x/(x+y))(10)+(y/(x+y))(20)=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

PS: i didnt understand the long explanation for the correct answer D
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by rijul007 » Tue Jan 03, 2012 10:25 pm
k = 10x/(x+y) + 20y/(x+y) = (10x+20y)/(x+y) = (10x+10y+10y)/(x+y) = 10 + 10y/(x+y)

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

Option given are

a.10
b.12
c.15
d.18
e.30


a.10

10 = 10 + 10y/(x+y)
10y/(x+y) = 0
this means y is +ve. whereas, ques states that both x and y are positive integers
eliminate


b.12

12 = 10 + 10y/(x+y)
10y/(x+y) = 2
y/(x+y) = 1/5
4y = x
x>y

which does not satisify the condition given in the question
Eliminate

c.15

15 = 10 + 10y/(x+y)
5 = 10y/(x+y)
1/2 = y/(x+y)
y = x

Eliminate


d.18

18 = 10 + 10y/(x+y)
8 = 10y/(x+y)
4x + 4y = 5y
y = 4x
X<y

This option is Correct

e.30

30 = 10 + 10y/(x+y)
2 = y/(x+y)
2x + 2y = y
2x = -y

This states that one of the numbers is -ve
Wrong


Option D is Correct.
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by Anurag@Gurome » Tue Jan 03, 2012 10:35 pm
karthikpandian19 wrote:If x, y, and k are positive integers such that (x/(x+y))(10)+(y/(x+y))(20)=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

PS: i didnt understand the long explanation for the correct answer D
(x/(x + y))(10) + (y/(x + y))(20) = k and x < y
10x + 20y = k(x + y)
x(10 - k) = y(k - 20)
y/x = (10 - k)/(k - 20) and we know y > x or y/x > 0

So, 15 < k < 20 implies k = 18

The correct answer is D.
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by karthikpandian19 » Tue Jan 03, 2012 10:43 pm
Really i got benefited with this explanation
rijul007 wrote:k = 10x/(x+y) + 20y/(x+y) = (10x+20y)/(x+y) = (10x+10y+10y)/(x+y) = 10 + 10y/(x+y)

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

Option given are

a.10
b.12
c.15
d.18
e.30


a.10

10 = 10 + 10y/(x+y)
10y/(x+y) = 0
this means y is +ve. whereas, ques states that both x and y are positive integers
eliminate


b.12

12 = 10 + 10y/(x+y)
10y/(x+y) = 2
y/(x+y) = 1/5
4y = x
x>y

which does not satisify the condition given in the question
Eliminate

c.15

15 = 10 + 10y/(x+y)
5 = 10y/(x+y)
1/2 = y/(x+y)
y = x

Eliminate


d.18

18 = 10 + 10y/(x+y)
8 = 10y/(x+y)
4x + 4y = 5y
y = 4x
X<y

This option is Correct

e.30

30 = 10 + 10y/(x+y)
2 = y/(x+y)
2x + 2y = y
2x = -y

This states that one of the numbers is -ve
Wrong


Option D is Correct.
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by karthikpandian19 » Tue Jan 03, 2012 10:50 pm
In this condition if i substitute the values of k in the finally derived equation of yours "y/x = (10 - k)/(k - 20) and we know y > x or y/x > 0
I am getting for 15 the value is 1 which is acceptable right???? y/x>0 and it is 1???

In this method both C & D are right? Am i right?
Anurag@Gurome wrote:
karthikpandian19 wrote:If x, y, and k are positive integers such that (x/(x+y))(10)+(y/(x+y))(20)=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

PS: i didnt understand the long explanation for the correct answer D
(x/(x + y))(10) + (y/(x + y))(20) = k and x < y
10x + 20y = k(x + y)
x(10 - k) = y(k - 20)
y/x = (10 - k)/(k - 20) and we know y > x or y/x > 0

So, 15 < k < 20 implies k = 18

The correct answer is D.
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by Anurag@Gurome » Tue Jan 03, 2012 10:56 pm
karthikpandian19 wrote:In this condition if i substitute the values of k in the finally derived equation of yours "y/x = (10 - k)/(k - 20) and we know y > x or y/x > 0
I am getting for 15 the value is 1 which is acceptable right???? y/x>0 and it is 1???

In this method both C & D are right? Am i right?
C cannot be the right answer as k lies between 15 ans 20, not inclusive. So, the only correct choice is 18.
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by chufus » Wed Jan 04, 2012 3:39 am
Anurag@Gurome wrote:
karthikpandian19 wrote:If x, y, and k are positive integers such that (x/(x+y))(10)+(y/(x+y))(20)=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

PS: i didnt understand the long explanation for the correct answer D
(x/(x + y))(10) + (y/(x + y))(20) = k and x < y
10x + 20y = k(x + y)
x(10 - k) = y(k - 20)
y/x = (10 - k)/(k - 20) and we know y > x or y/x > 0

So, 15 < k < 20 implies k = 18

The correct answer is D.
Dear Anurag I'm still lost on this one.......

When you say "we know y > x or y/x > 0"
shudn't it be y/x > 1 since x and y are positive and x < y ( but this is the least of my concerns)

Now how do you get from this to "15 < k < 20" . I'm completely lost on this one.. Don't see the connection.... Could you kindly elaborate on the thought process in-between?
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by GMATGuruNY » Wed Jan 04, 2012 5:14 am
karthikpandian19 wrote:If x, y, and k are positive integers such that (x/(x+y))(10)+(y/(x+y))(20)=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

PS: i didnt understand the long explanation for the correct answer D
I posted an explanation here:

https://www.beatthegmat.com/tough-algebr ... tml#292229.

As for the more theoretical approach:
Let x = the number of $10 towels purchased and y = the number of $20 towels purchased.
Average cost per towel = (total cost)/(total number of towels) = (10x + 20y)/(x+y).

This is the value of k in the problem above:
(x/(x+y))(10) + (y/(x+y))(20) = k
(10x + 20y)/(x+y) = k.

Since the price of each towel is either $10 or $20, the value of k -- the AVERAGE cost per towel -- must be between 10 and 20.
Since y>x -- implying that more $20 towels are purchased than $10 towels -- the average cost per towel must be CLOSER TO 20 than to 10.
Thus, the only viable answer choice is D.
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by leumas » Thu Jan 05, 2012 3:08 am
karthikpandian19 wrote:In this condition if i substitute the values of k in the finally derived equation of yours "y/x = (10 - k)/(k - 20) and we know y > x or y/x > 0
I am getting for 15 the value is 1 which is acceptable right???? y/x>0 and it is 1???

In this method both C & D are right? Am i right?
Anurag@Gurome wrote:
karthikpandian19 wrote:If x, y, and k are positive integers such that (x/(x+y))(10)+(y/(x+y))(20)=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

PS: i didnt understand the long explanation for the correct answer D
(x/(x + y))(10) + (y/(x + y))(20) = k and x < y
10x + 20y = k(x + y)
x(10 - k) = y(k - 20)
y/x = (10 - k)/(k - 20) and we know y > x or y/x > 0

So, 15 < k < 20 implies k = 18

The correct answer is D.
In Option C, x=y i.e both will be 5. But the question clearly says x,y, which holds true for 18, x=2, y =8.

Hope this helps.

Samuel
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