Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59020

Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
20 Jun 2017, 22:01
uvee wrote: Bunuel wrote: BANON wrote: 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 \(10*\frac{x}{x+y}+20*\frac{y}{x+y}=k\) \(10*\frac{x+2y}{x+y}=k\) \(10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k\) 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)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 \(1\) \(\frac{y}{x+y}\) 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.
_________________



Intern
Joined: 26 Aug 2017
Posts: 3

Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
19 Jan 2018, 08:31
Could you guys please explain if the answer holds true when x<0 and y>0 ? Thanks !



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15446
Location: United States (CA)

Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
19 Jan 2018, 20:26
Hi Phlaryu, The prompt tells us that X, Y and K are all POSITIVE, so you shouldn't be wasting time on the outcome if X < 0. GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★



Kaplan GMAT Instructor
Joined: 21 Jun 2010
Posts: 66
Location: Toronto

Re: If x, y, and k are positive numbers such that 10*x/(x+y)+ 20
[#permalink]
Show Tags
04 Dec 2018, 00:58
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:
xavgy 10avg20
Thus, the answer must be between 15 and 20, exclusive.
Among the answer choices, only 18 fits the bill.
Choose D.



Senior Manager
Joined: 04 Aug 2010
Posts: 492
Schools: Dartmouth College

Re: If x, y, and k are positive numbers such that 10*x/(x+y)+ 20
[#permalink]
Show Tags
04 Dec 2018, 00:59
pzazz12 wrote: 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 suspect that many testtakers would not recognize that this is a weighted average problem. For those testtakers, 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.
_________________
GMAT and GRE Tutor Over 1800 followers GMATGuruNY@gmail.com New York, NY If you find one of my posts helpful, please take a moment to click on the "Kudos" icon. Available for tutoring in NYC and longdistance. For more information, please email me at GMATGuruNY@gmail.com.



VP
Joined: 23 Feb 2015
Posts: 1297

Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
27 May 2019, 13:24
Bunuel wrote: 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 \(18\).
BunuelWhich values confirm the highlighted part? Also, could you help me to figure out the highlighted part without putting the value? Thanks__
_________________
“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.”― Henry Wadsworth LongfellowSEARCH FOR ALL TAGS



VP
Joined: 23 Feb 2015
Posts: 1297

If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
27 May 2019, 13:33
GMATGuruNY wrote: pzazz12 wrote: 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 suspect that many testtakers would not recognize that this is a weighted average problem. For those testtakers, 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. GMATGuruNYFrom your explanation it seems that y=x in choice C. Next step, you choose choice D (where the value is more than 15). So, how did you confirm that the value which is more than 15 will make the condition legit (x<y)? I'll appreciate your help. Thanks__
_________________
“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.”― Henry Wadsworth LongfellowSEARCH FOR ALL TAGS



VP
Joined: 23 Feb 2015
Posts: 1297

Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
27 May 2019, 14:07
EvaJager wrote: ausadj18 wrote: Can someone show me the shortcut to solving number 148 from the OG: If x,y and k are positive numbers such that (x/x+y)(10)+(y/x+y)(20)=k and x <y, which of following could be value for k? a) 10 b) 12 c)15 d)18 e) 30 \(\frac{10x}{x+y}+\frac{20y}{x+y}=\frac{10(x+y)+10y}{x+y}=10+\frac{10y}{x+y}\). \(\frac{10y}{x+y}>\frac{10y}{2y}=5\) and \(\,\,\frac{10y}{x+y}<\frac{10y}{y}=10\), therefore \(10+5<k<10+10\) or \(15<k<20\).Answer D. Hi Bunuel, IanStewartCould you help me to comprehend the highlighted part?
_________________
“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.”― Henry Wadsworth LongfellowSEARCH FOR ALL TAGS



Math Expert
Joined: 02 Sep 2009
Posts: 59020

Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
27 May 2019, 21:29
Asad wrote: Bunuel wrote: 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 \(18\).
BunuelWhich values confirm the highlighted part? Also, could you help me to figure out the highlighted part without putting the value? Thanks__ Check here: https://gmatclub.com/forum/ifxyandk ... 3120.html
_________________



SVP
Joined: 03 Jun 2019
Posts: 1838
Location: India

Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
01 Sep 2019, 09:08
BANON wrote: 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 Given: 1. x, y, and k are positive numbers such that \(\frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k\) 2. x < y Asked: Which of the following could be the value of k? \(\frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k\) \(\frac{x+y}{x + y}*10 + \frac{y}{x + y}*10 = k\) \(10 + \frac{y}{x + y}*10 = k\) SInce y>x \(y> \frac{(x+y)}{2}\) \(\frac{y}{x+y}>\frac{1}{2}\) \(k > 10 + \frac{1}{2}*10\) \(k > 15\) \(10 + \frac{y}{x + y}*10 = k\) Since \(\frac{y}{x+y}< 1\) \(k< 10 + 1*10\) k<20 15<k<20 Only option D 18 satisfies these conditions IMO D
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts." Please provide kudos if you like my post. Kudos encourage active discussions. My GMAT Resources:  Efficient LearningAll you need to know about GMAT quantTele: +911140396815 Mobile : +919910661622 Email : kinshook.chaturvedi@gmail.com



SVP
Joined: 03 Jun 2019
Posts: 1838
Location: India

Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
01 Sep 2019, 09:11
Asad wrote: EvaJager wrote: ausadj18 wrote: Can someone show me the shortcut to solving number 148 from the OG: If x,y and k are positive numbers such that (x/x+y)(10)+(y/x+y)(20)=k and x <y, which of following could be value for k? a) 10 b) 12 c)15 d)18 e) 30 \(\frac{10x}{x+y}+\frac{20y}{x+y}=\frac{10(x+y)+10y}{x+y}=10+\frac{10y}{x+y}\). \(\frac{10y}{x+y}>\frac{10y}{2y}=5\) and \(\,\,\frac{10y}{x+y}<\frac{10y}{y}=10\), therefore \(10+5<k<10+10\) or \(15<k<20\).Answer D. Hi Bunuel, IanStewartCould you help me to comprehend the highlighted part? Asad Please see my solution for explanation of the highlighted part in Bunuel's solution.
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts." Please provide kudos if you like my post. Kudos encourage active discussions. My GMAT Resources:  Efficient LearningAll you need to know about GMAT quantTele: +911140396815 Mobile : +919910661622 Email : kinshook.chaturvedi@gmail.com



Intern
Joined: 11 Jun 2019
Posts: 33
Location: India

Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
01 Sep 2019, 21:58
10x/ (X+Y) + 20Y/(X+Y) = K
10*{ (X+2Y)/ (X+Y) } = K
(X+2Y) / (X+Y) = K/10 and we know that X<Y, so here we plug in values.
(X+2Y) / (X+Y) = 10/10......Can't be true as Y will have to be 0 for this. (X+2Y)/ (X+Y) = 12/10.... Gives us 2 equations. X+ 2Y = 12 and X+Y = 10...solving this we get X= 8 and Y =2, so incorrect. (X+2Y) / (X+Y) = 15/10...similar to above, we will get two equations and by solving them, we get X= 5 and Y= 5, so incorrect. (X+2Y) / (X+Y) = 18/10.......X+ 2Y= 18 and X+Y = 10........solving them we get, X= 2 and Y = 8...This is the correct solution. (X+2Y) / (X+Y) = 30/10......we get ve value of X. Hence the answer is (D)



Manager
Joined: 18 Jan 2017
Posts: 127

If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
09 Sep 2019, 10:52
uvee wrote: Bunuel wrote: BANON wrote: 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 \(10*\frac{x}{x+y}+20*\frac{y}{x+y}=k\) \(10*\frac{x+2y}{x+y}=k\) \(10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k\) 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)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 \(1\) Hey Bunuel Can you please calrify this Query? I would like an answer for this as well. Thanks in advance.



Math Expert
Joined: 02 Sep 2009
Posts: 59020

Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
Show Tags
09 Sep 2019, 10:57
Inten21 wrote: uvee wrote: Bunuel wrote: \(10*\frac{x}{x+y}+20*\frac{y}{x+y}=k\)
\(10*\frac{x+2y}{x+y}=k\)
\(10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k\)
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) 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 \(1\) Hey Bunuel Can you please calrify this Query? I would like an answer for this as well. Thanks in advance. Check here: https://gmatclub.com/forum/ifxyandk ... l#p1873381
_________________




Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)
[#permalink]
09 Sep 2019, 10:57



Go to page
Previous
1 2
[ 34 posts ]



