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kevch25: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sat Jan 20, 2007 9:20 pm; Followed by:1 members

PS: Can you help?

by kevch25 » Thu Feb 08, 2007 2:27 pm

If x, y, and k are positive numbers 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

I will post the OA later tonight. Don't know how to hide the answer.

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Source: Beat The GMAT — Problem Solving | Thu Feb 08, 2007 2:27 pm

kevch25: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sat Jan 20, 2007 9:20 pm; Followed by:1 members

by kevch25 » Thu Feb 08, 2007 4:05 pm

Thanks for your help Mark

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BTGmoderatorRO: Moderator; Posts: 772; Joined: Wed Aug 30, 2017 6:29 pm; Followed by:6 members

by BTGmoderatorRO » Sun Sep 03, 2017 12:58 pm

(x/(x+y)) *10 +( y/ (x+y)) *20 = k
let us rearrange this question
10 ( x/ (x+y) + 10 ( y/ (x+y)+ 10 ( y/ (x+y) = k
(10x + 10y) / (x+y)) + 10 ( y/ (x+y) = k
(10(x+y) / (x+y)) + 10 ( y/ (x+y) = k
(10 + 10 ( y/ (x+y) = k
x < y ( i.e y>x )
x and y are positive +ve
0.5 < y/ (x+y) < 1 as well.
let us take the limit,
when x=0, y=max
y/ (0+y)=1
10 + 10 ( 1) = k
k=20

when x=y, y=min
y/ (y+y)=0.5
10 + 10 ( 0.5) = k
k=15

therefore, 15 < k < 20
form the option given K=18 is the option that satisfy this condition. hence , option D is correct.

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Admin1: Site Admin; Posts: 28; Joined: Wed Aug 30, 2017 10:11 pm

Answer

by Admin1 » Sun Sep 03, 2017 4:20 pm

We have that [x/(x+y)]*10+[y/(x+y)]*20=k, i.e. we have (10x+20y)/(x+y)=k.

The left side of the equation can be rewritten as (10x+10y)/(x+y) + 10y/(x+y) = 10(x+y)/(x+y) + 10y/(x+y) = 10 + 10y/(x+y).

Replacing this on the equation above we will get:
10+10y/(x+y)=k.

So, k>10.

On the other hand, x<y, so
x+y <2y
1/2 < y/(x+y)
10*1/2 < 10*y/(x+y)
5<10y/(x+y)

In addition we have that y /(x+y) < 1 which implies 10y/(x+y) < 10.

From the red and the orange equations we get 5 < 10y/(x+y) < 10. Adding 10 to each side of the inequality we get 15 < 10 + 10y/(x+y) < 20, i.e. 15 < k < 20.

So, the value of k is 18. The correct answer is D.

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 am; Location: Grand Central / New York; Thanked: 470 times; Followed by:34 members

by Jay@ManhattanReview » Sun Sep 03, 2017 9:30 pm

Nice solutions by all.

Here's for someone who cannot think of what possible maximum and minimum values x/y can take.

So, I pick up the solution from 10+10y/(x+y) = k.

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

Since we know that y > x, let's assume a convenient value. Say x = 1 and y = 2

Thus, @ x = 1 and y =2, we have 10+10y/(x+y) = k => 10+10*2/(1+2) = k

10+20/3 = k

k = 16.66

We see that among the options there is no such value as 16.66. So the correct answer would be either k = 15 (< 16.66) or k = 18 (> 16.66). This is because if k = 12 then any values of k within the range of 12 to 16.66 are correct, thus 15 is also correct; however, there must be only one correct answer. Thus, k = 12 must be the incorrect value of k. Similarly, if k = 30 then any values of k within the range of 16.66 to 30 are correct, thus 18 is also correct; however, there must be only one correct answer. Thus, k = 30 must be the incorrect value of k.

Let's try with a large value of k. Say y = 10 and x = 1.

Thus, @ x = 1 and y =10, we have 10+10y/(x+y) = k => 10+10*10/(1+10) = k

10+100/11 = k

k = 19.09

Thus, the value of k must least be correct in the range of 16.66 to 19.09. Luckily, k= 18 (option D) falls in this range, thus this must be the correct answer.

Had you opted to choose the smallest possible value of y, i.e., closest to x, you may choose, for example, y = 1.001; however, dealing with 1.001 is cumbersome. Let's assume that x = y = 1. This will certainly give an incorrect value of k; however, we can be sure that the correct value of k must be greater than that value.

Thus, @ x = 1 and y =1, we have 10+10y/(x+y) > k => 10+10*1/(1+1) > k

10+10/2 > k

k > 15

Since k = 15 is incorrect, the correct value is k = 18.

The correct answer: D

Hope this helps!

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-Jay
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Admin1: Site Admin; Posts: 28; Joined: Wed Aug 30, 2017 10:11 pm

by Admin1 » Mon Sep 04, 2017 12:25 am

Solution:
Given x, y, k > 0 and the equation,

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

we obtain, upon factoring process,

[10 / (x + y)] (x + 2y) = k.

Dividing (x + y) by (x + 2y) would simplify further the equation to obtain

10 (1 + [y / (x + y)] ) = k.

Now, since x, y, are positive numbers, and that y > x, hence, the expression, [y / (x + y)] is a positive fraction. Thus, the choices 10 & 30 are eliminated from possible answers.

We are left with the following choices:

12 = 10 (1 + 0.2) = 10 (1 + (2 / 10)) = 10 (1 + (1/5))
15 = 10 (1 + 0.5) = 10 (1 + (5 / 10)) = 10 (1 + (1/2))
18 = 10 (1 + 0.8) = 10 (1 + (8 / 10)) = 10 (1 + (4/5)).

Keeping in mind, that, x < y, and by looking at these three fractions, namely, (1/5), (1/2), and (4/5), the one matching our expression, [y / (x + y)], is (4/5). In that case, y = 4 and x = 1.

Therefore, the answer is D. 18.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Mon Sep 04, 2017 2:54 am

If x, y and k are positive numbers such that 10*x/(x+y)+20*y/(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

Putting the sum over a common denominator, we get:
(10x + 20y) / (x+y) = k.

Let x = the number of $10 shirts purchased at a certain store.
Let y = the number of $20 shirts purchased at a certain store.
Total cost of the $10 shirts = 10x.
Total cost of the $20 shirts = 20y.
Total number of shirts purchased = x+y.
Thus, the AVERAGE cost per shirt is equal to the following:

(10x + 20y) / (x+y).

In the problem above, the value of k is equal to the AVERAGE cost per shirt.
Since each shirt costs either $10 or $20, the average cost per shirt must be BETWEEN 10 and 20.
Since y>x, the number of $20 shirts purchased is GREATER than the number of $10 shirts purchased, with the result that the average cost per shirt must be CLOSER TO 20 than to 10.
Of the answer choices, the only viable option is k=18.

The correct answer is D.

An alternate approach is to PLUG IN THE ANSWERS, which represent the value of k.
Let x=1.
When we plug in the correct answer choice for k, x<y.

Answer choice B: k=12
(10*1 + 20y)/(1+y) = 12
10 + 20y = 12 + 12y
8y = 2
y = 2/8 = 1/4.
Since x>y, eliminate B.

Answer choice D: k=18
(10*1 + 20y)/(1+y) = 18
10 + 20y = 18 + 18y
2y = 8
y = 4.
Since x<y, success!

The correct answer is D.

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[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pm; Location: Palo Alto, CA; Thanked: 2867 times; Followed by:511 members; GMAT Score:800

by [email protected] » Mon Sep 04, 2017 1:02 pm

Hi All,

You can take advantage of the answer choices and some math "logic" to get to the correct answer. Let's TEST THE ANSWERS....

We're told that X, Y and K are POSITIVE and that X < Y.

We're also told that 10X/(X+Y) + 20Y/(X+Y) = K. We're asked what COULD be the value of K, which means that there's more than one possible answer. Since the answers are NUMBERS, one of the them MUST be a possible answer.

We can manipulate the given equation into:

10X + 20Y = K(X+Y)

Since all of the variables are POSITIVE, K MUST be between 10 and 20. Here's the proof:

IF.....K=10, then the equation becomes...
10X + 20Y = 10X + 10Y
20Y = 10Y
Since Y is positive, 20Y = 10Y is NOT possible.
Eliminate Answer A

In that same way, K can't be 20 (or greater) because the end equation would be an impossibility.

With the remaining 3 answers, we can TEST the possibilities...

IF...K = 12, then the equation becomes...
10X + 20Y = 12X + 12Y
8Y = 2X
4Y = X
In this scenario, X > Y which is the OPPOSITE of what we were told. This is NOT the answer.
Eliminate B.

IF....K=15, then the equation becomes...
10X + 20Y = 15X + 15Y
5Y = 5X
Y = X
In this scenario, X = Y, which is NOT a match for what we were told. This is NOT the answer.
Eliminate C.

IF....K=18, then the equation becomes....
10X + 20Y = 18X + 18Y
2Y = 8X
Y = 4X
Here, X < Y, which IS a match for what we were told. This IS a POSSIBLE answer.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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Re: PS: Can you help?

by Scott@TargetTestPrep » Sun Feb 09, 2020 4:40 am

kevch25 wrote: ↑
Thu Feb 08, 2007 2:27 pm
If x, y, and k are positive numbers 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

I will post the OA later tonight. Don't know how to hide the answer.

Solution:

We are given:

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

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

We can combine the two fractions on the left side of the equation because they have the same denominator, (x + y).

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

We see that we have a weighted average equation in which x items have an average of 10, and another y items have an average of 20 and a weighted average of k. In this case, the value of k must be between 10 and 20. However, since x is less than y, the weighted average (or k) must be closer to 20 than to 10. Thus k must be 18.

Alternate Solution:

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

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

We can combine the two fractions on the left side of the equation because they have the same denominator, (x + y).

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

10x + 20y = kx + ky

20y - ky = kx - 10x

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

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

Since y > x, x/y is less than 1. Since both x and y are positive, x/y is also positive. Trying each answer choice for k, we see that k = 10, 12 and 15 produce a value of x/y that is greater than or equal to 1 while k = 30 produces a negative value of x/y. k = 18 is the only value that results in a positive proper fraction.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]