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

Question

If x, y, and k are positive numbers such that  \frac{x}{x + y}*10 + \frac{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

Bunuel · Accepted Answer

If x, y, and k are positive numbers such that  \frac{x}{x + y}*10 + \frac{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

10*\frac{x}{x+y}+20*\frac{y}{x+y}=k

10*\frac{x+2y}{x+y}=k

10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k

Finally we get:  10*(1+\frac{y}{x+y})=k

We know that  x<y

Hence  \frac{y}{x+y}  is more than  0.5  and less than  1

0.5<\frac{y}{x+y}<1

So,  15<10*(1+\frac{y}{x+y})<20

Only answer between  15  and  20  is  18 .

Answer: D (18)

There can be another approach:

We have:  \frac{10x+20y}{x+y}=k , if you look at this equation you'll notice that it's a weighted average.

There are  x  red boxes and  y  blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?

k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.

Answer: D (18)

mbaiseasy · Answer

\frac{10x}{x+y}+ \frac{20y}{x+y}=k 
 10x + 20y = kx + ky 
 (10-k)x=(k-20)y 
 \frac{x}{y}=\frac{k-20}{10-k}

(A)  \frac{-10}{0}  Out! x and y are non-zeroes.
(B)  \frac{x}{y}=\frac{-8}{-2}=\frac{4}{1}  Out! x should be less than y.
(C)  \frac{-5}{-5}  Out! x should not be equal to y.
(D)  \frac{-2}{-8}=\frac{1}{4}  Bingo!

Answer: D

kraizada84 · Answer

main funda is y/(x+y) is equal to 1/2 if x=y but when x1/2 (x,y>0) also the maximum value is x tends to 0 i.e. as x>0 say x be 0.0000...(infinite times )1 then the expression x+y tends to y i.e. y/ (x+y) <1 hence 0.5 < y/(x+y) < 1 => 1+0.5 < 1+y/(x+y) < 1+1 => 10*1.5 < 10*(1+(y/x+y) <2 *10 => 15<10*(1+(y/x+y)<20 hope this helps..!!

mymbadreamz · Answer

Bunuel, I didn't understand how you got the 15 and 20 at the extreme ends in the equation. Could you please explain? thanks!
15<10*(1+(y/x+y)<20

Bunuel · Answer

Since  x<y  then  \frac{y}{x+y}  is more than  0.5  and less than  1 :  0.5<\frac{y}{x+y}<1 .

In this case for  1+\frac{y}{x+y}  is more than 1.5 and less than 2, so  10*(1+\frac{y}{x+y})  is more than 15 and less than 20:  15<10*(1+\frac{y}{x+y})<20 .

Hope it's clear.

EvaJager · Answer

\frac{10x}{x+y}+\frac{20y}{x+y}=\frac{10(x+y)+10y}{x+y}=10+\frac{10y}{x+y} .

\frac{10y}{x+y}>\frac{10y}{2y}=5  and  \,\,\frac{10y}{x+y}<\frac{10y}{y}=10 , therefore  10+5<k<10+10  or  15<k<20 .

Answer D.

kanusha · Answer

Hi Bunnel ,
Can you pls Explain how these Steps Came in your Solution

10*\frac{x}{x+y}+20*\frac{y}{x+y}=k

10*\frac{x+2y}{x+y}=k

10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k

Finally we get: 10*(1+\frac{y}{x+y})=k

I Didn't Understood how The Steps 3 and 4 Came?

Bunuel · Answer

\frac{x+2y}{x+y}=\frac{x+y}{x+y}+\frac{y}{x+y}=1+\frac{y}{x+y} .

Hope it's clear.

EMPOWERgmatRichC · Answer

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

anik19890 · Answer

what will happen when x>y in this question? since we dont know the value of x and y they can be 1:100 or something like that. what will be the ans then?

EMPOWERgmatRichC · Answer

Hi anik19890,

You ask a couple of different questions.

First, the prompt tells us that X < Y, so we CAN'T have a situation in which X > Y (or even X = Y). THAT specific inequality helps to define the correct answer.

Second, the prompt asks for which of the following COULD be the value of K, meaning that there is more than one possible answer (but only one of the 5 answer choices is correct). Based on the inequality we're given, we COULD have X = 1, Y = 100 (as you described), which would give us...

(1/101)(10) + (100/101)(20) = K

10/101 + 2000/101 = K
2010/101 = K

But that answer is NOT among the answer choices.

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

ScottTargetTestPrep · Answer

We are given:
 
(x/(x+y))(10) + (y/(x+y))(20) = k
 
  +   = k
 
We can combine the two fractions on the left side of the equation because they have the same denominator, (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.
 
Answer: D

uvee · Answer

Hi, Bunuel! I have a silly question. Can you please explain how you arrived at the following?
We know that  x<y

Hence  \frac{y}{x+y}  is more than  0.5  and less than  1

Bunuel · Answer

\frac{y}{x+y}

Since both x and y are positive, then the denominator is greater than numerator. So,  \frac{y}{x+y}<1 .

Next, if x were equal to y, then we'd have y/(2y) = 1/2 but x < y, so denominator is less than 2y, which makes the fraction greater than 1/2.

Hope it's clear.

Testluv · Answer

This one's all about pattern recognition.

Notice that the equation is giving us a weighted average. Because y is bigger than x, the average must be weighted towards y. Visually:

x---------avg---y
10--------avg---20

Thus, the answer must be between 15 and 20,   exclusive  .

Among the answer choices, only 18 fits the bill.

Choose D.

GMATGuruNY · Answer

I suspect that many test-takers would not recognize that this is a weighted average problem. For those test-takers, a straightforward and efficient approach would be to plug in the answer choices, which represent the value of k.

Answer choice C: k= 15  
(10x + 20y)/(x+y) = 15 
10x + 20y = 15x + 15y 
5y = 5x 
y = x 
Doesn't work because the problem states that x<y.

We need y to be larger, so let's try answer choice D: k=18  
10x + 20y = 18x + 18y 
2y = 8x 
y/x = 8/2 
Success! x<y.

The correct answer is D.

The process would be very quick if you recognized right away that since k = one of the answer choices, you could simply plug them in for k.

swadhakamal · Answer

Bunuel

We know that x<y

Hence y/(x+y) is more than 0.5 and less than 1

how we know its between 0.5 and less than 1? x can be 3 and y can be 4..there will be many possibilities.

please clear my confusion.

Thanks.

EMPOWERgmatRichC · Answer

Hi swadhakamal,

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

This means that if you take the sum of X and Y, then the Y will make up MORE than 1/2 of that sum. The reason why can be explained with a comparison:

(X + X) vs. (X + Y)

If we add two Xs together, then each variable represents EXACTLY HALF of the total. However, since Y is GREATER than X, if we take (X + Y), then the Y would have to be MORE than half of that total.

If you don't immediately recognize that pattern, then that's okay - you might find it helpful to come up with a few examples, and you'll see that that is ALWAYS the case. We can use your first example (X=3 and Y=4):

X+Y = 7 and 4 is more than half of 7.

Thus, when we're thinking about Y/(X+Y), we know that we'll have a positive fraction AND since Y is MORE than half of (X+Y), that the fraction must be GREATER than 1/2 (but less than 1).

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

BrentGMATPrepNow · Answer

Given:  \frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k

Rewrite as follows:  \frac{10x}{x + y} + \frac{20y}{x + y} = k

Rewrite as follows:  (\frac{10x}{x + y} + \frac{10y}{x + y}) + \frac{10y}{x + y}= k

Simplify:  (\frac{10x + 10y}{x + y}) + \frac{10y}{x + y}= k

Simplify:  10 + \frac{10y}{x + y}= k

Rewrite as follows:  10 + 10(\frac{y}{x + y})= k

IMPORTANT: If   x = y , then  \frac{y}{x + y} = \frac{y}{y + y} = \frac{y}{2y} = \frac{1}{2}

Since it's actually the case that   x < y , then we know that  \frac{1}{2} < \frac{y}{x + y} < 1  , which means  10(\frac{y}{x + y})  is greater than 5 but less than 10.

So, we know that  15 < k < 20

Answer: D

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

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GMAT Problem Solving (PS) Questions

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

Prep Toolkit

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