Q17:
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
Tough algebra Equation
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
The trick is to recognize that the equation just gives a weighted average of x and y. The equation above is the same as you'd use if you were asked "If x pounds of peanuts and y pounds of cashews are mixed together, and peanuts cost $10/pound and cashews cost $20/pound, what is the price per pound of the resulting mixture?" The answer must be between 10 and 20, and if y > x, the answer must be closer to 20 than to 10. So 18 is the only possible answer.iikarthik wrote:Q17:
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
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com
-
- Senior | Next Rank: 100 Posts
- Posts: 74
- Joined: Sun Aug 02, 2009 9:10 pm
- Thanked: 1 times
- Followed by:1 members
-
- Master | Next Rank: 500 Posts
- Posts: 324
- Joined: Thu Dec 24, 2009 6:29 am
- Thanked: 17 times
- Followed by:1 members
Ian,Ian Stewart wrote:The trick is to recognize that the equation just gives a weighted average of x and y. The equation above is the same as you'd use if you were asked "If x pounds of peanuts and y pounds of cashews are mixed together, and peanuts cost $10/pound and cashews cost $20/pound, what is the price per pound of the resulting mixture?" The answer must be between 10 and 20, and if y > x, the answer must be closer to 20 than to 10. So 18 is the only possible answer.iikarthik wrote:Q17:
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 such a problem, where we need to experiment with numbers, how would we know which are the right numbers to choose?
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
Well, you don't need to experiment with numbers in the question above; I didn't in my post above. That said, if you don't see a conceptual or algebraic solution fairly quickly, plugging in numbers is a good fallback option. It would be a bit lucky to find numbers that give the exact answer to this question, but if you plug in a few very simple sets of numbers (you don't want to waste any time on complicated numbers), making sure that x is less than y, you'll always find that k is between 15 and 20, which may lead you to the correct answer here.rahul.s wrote: Ian,
In such a problem, where we need to experiment with numbers, how would we know which are the right numbers to choose?
There are also algebraic solutions to the question:
(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
and since 0 < x < y, then 0 < x/y < 1, and it must be that 0 < (20 - k) / (k - 10) < 1. From the answer choices we can be sure k - 10 isn't negative, so we can multiply through this inequality by k-10 to find that 0 < 20 - k < k - 10, or that 15 < k < 20.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com
-
- Master | Next Rank: 500 Posts
- Posts: 117
- Joined: Wed Jun 09, 2010 7:02 am
I think that knowing the quick weighted average formula using a number line is one of the fastest, nearly foolproof ways to get a seemingly difficult question right. AFter reading this, I'm inclined to ask you, Ian, are there any other "disguised" weighted average problems you've noted?Ian Stewart wrote:The trick is to recognize that the equation just gives a weighted average of x and y. The equation above is the same as you'd use if you were asked "If x pounds of peanuts and y pounds of cashews are mixed together, and peanuts cost $10/pound and cashews cost $20/pound, what is the price per pound of the resulting mixture?" The answer must be between 10 and 20, and if y > x, the answer must be closer to 20 than to 10. So 18 is the only possible answer.iikarthik wrote:Q17:
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
- 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
For those who worry that they wouldn't be able to recognize that this is a weighted average problem, an easy and efficient approach is to plug in the answer choices, which represent the value of k:iikarthik wrote:Q17:
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
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.
Eliminate C.
The value of y needs to be larger.
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 is very quick if you recognize right away that since k = one of the answer choices, you can simply plug them in for k.
Last edited by GMATGuruNY on Wed Jun 15, 2011 2:16 pm, edited 1 time in total.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
-
- Newbie | Next Rank: 10 Posts
- Posts: 7
- Joined: Fri Oct 08, 2010 3:12 pm
I think this will be a long method.. since you still need to check with option E. That makes it solving equation 3 times.GMATGuruNY wrote:For those who worry that they wouldn't be able to recognize that this is a weighted average problem, an easy and efficient approach would be to plug in the answer choices, which represent the value of k:iikarthik wrote:Q17:
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
Answer choice C:
(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 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.
I prefer simply put smallest value that satisfy the given condition of x and y. Preferably 1 and 2.
-
- Master | Next Rank: 500 Posts
- Posts: 370
- Joined: Sat Jun 11, 2011 8:50 pm
- Location: Arlington, MA.
- Thanked: 27 times
- Followed by:2 members
Hussi, no real need to do that , as only 1 condition will be an answer for the question!
As for K=30, gives you y = -2x, which indeed is not the solution!
As for K=30, gives you y = -2x, which indeed is not the solution!
Ian,Ian Stewart wrote:Well, you don't need to experiment with numbers in the question above; I didn't in my post above. That said, if you don't see a conceptual or algebraic solution fairly quickly, plugging in numbers is a good fallback option. It would be a bit lucky to find numbers that give the exact answer to this question, but if you plug in a few very simple sets of numbers (you don't want to waste any time on complicated numbers), making sure that x is less than y, you'll always find that k is between 15 and 20, which may lead you to the correct answer here.rahul.s wrote: Ian,
In such a problem, where we need to experiment with numbers, how would we know which are the right numbers to choose?
There are also algebraic solutions to the question:
(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
and since 0 < x < y, then 0 < x/y < 1, and it must be that 0 < (20 - k) / (k - 10) < 1. From the answer choices we can be sure k - 10 isn't negative, so we can multiply through this inequality by k-10 to find that 0 < 20 - k < k - 10, or that 15 < k < 20.
Can I ask you, how do I get from 0 < 20 - k < k - 10 to 15 < k < 20?
Sorry, I don't get it..
Thanks!
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
You can see this by separating the three-part inequality 0 < 20 - k < k - 10 into two inequalities:wieke13 wrote:[
Ian,
Can I ask you, how do I get from 0 < 20 - k < k - 10 to 15 < k < 20?
Sorry, I don't get it..
0 < 20 - k, and adding k to both sides, k < 20
20 - k < k - 10, and adding k and 10 to both sides, we find that 30 < 2k, or 15 < k
Putting those two inequalities together we have 15 < k < 20.
As a side note, looking back over that algebraic solution, it does seem a bit awkward. I certainly prefer the weighted average approach, but if you don't notice that here, an approach like backsolving might be more practical for many test takers than direct algebra.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com