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Simplifying Algebraic Expressions

Problem Solving — algebra and arithmetic
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by bkearney31 » Thu Oct 18, 2012 8:08 am
I am new to the forum and was hoping someone could help with a quick question I had. It is straight from the Official Guide 12th Edition, question #148, page 233

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

In the book they show the steps but I am stuck on the progression of one step to the next.

1) 10x+20y/x+y (I understand the first step)

2) 10(x+y)+10y/x+y (This is the next step and I do not understand how they got the numerator)

Any help would be greatly appreciated!
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by Brent@GMATPrepNow » Fri Oct 19, 2012 7:46 am
bkearney31 wrote:I am new to the forum and was hoping someone could help with a quick question I had. It is straight from the Official Guide 12th Edition, question #148, page 233

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

In the book they show the steps but I am stuck on the progression of one step to the next.

1) 10x+20y/x+y (I understand the first step)

2) 10(x+y)+10y/x+y (This is the next step and I do not understand how they got the numerator)

Any help would be greatly appreciated!
Yeah, they could have added a step in the middle. First, however, I'll add some brackets to make it clear to others how the OG12 solution looks:

We have:
Step 1: (10x+20y)/(x+y)
Step 2: [10(x+y)+10y]/(x+y)

Okay, in between steps 1 and 2, we take 20y and rewrite it as 10y + 10y

Here we go:

Step 1: (10x+20y)/(x+y)
Step 1.5: (10x+10y+10y)/x+y
Then factor 10x+10y to get
Step 2: [10(x+y)+10y]/(x+y)

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Jim@StratusPrep » Fri Oct 19, 2012 8:07 am
Honestly, would take this problem away from the algebra...

This is a weighted average problem. If x < y then the number has to be closer to 20 than 10. 20 cannot be the answer because x and y both have positive values. Only if x = 0 could the answer be 20.

Hope that helps.
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by eaakbari » Thu Nov 08, 2012 6:58 am
Jim@StratusPrep wrote:Honestly, would take this problem away from the algebra...

This is a weighted average problem. If x < y then the number has to be closer to 20 than 10. 20 cannot be the answer because x and y both have positive values. Only if x = 0 could the answer be 20.

Hope that helps.
Jim,

I would appreciate if you could elaborate your answer

Thanks
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