Oct 16 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions)
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 18 Jan 2012
Posts: 9

If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
Show Tags
Updated on: 23 Feb 2014, 06:43
Question Stats:
78% (02:05) correct 22% (02:27) wrong based on 169 sessions
HideShow timer Statistics
If x and y are nonnegative integers and x + y = 25 what is x? (1) 20x + 10y < 300 (2) 20x + 10y > 280 hey guys and gals, on p52 of the strategy guide 3 of the MGMAT they give an example if x and y are nonnegative integers and x+y=25 what is x 1) 20x+10y<300 2) 20x+10y>280 the solution they provide makes sense but 2 things are confusing a) they say the maximum for 20x + 10y is 250 when x=0 and y=25, why didn't they make the maximum value using x=25 , y=0 since that would give you a larger number? b) why didn't they use option (1) and solve for Y , then plug that into X=Y=25 from the original problem? wouldn't that solve for x right away? its probably something simple that I'm missing but any clarity would be appreciated its really bugging me! thanks in advance!
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by c210 on 12 Feb 2012, 02:21.
Last edited by Bunuel on 23 Feb 2014, 06:43, edited 2 times in total.
Edited the question and added the OA




Math Expert
Joined: 02 Sep 2009
Posts: 58381

Re: question on example given on mgmat strategy guide 3 p 152
[#permalink]
Show Tags
12 Feb 2012, 02:44
If x and y are nonnegative integers and x+y=25 what is xGiven: x+y=25 > y=25x. Question: x=? (1) 20x+10y<300 > 2x+y<30 > 2x+(25x)<30 > x<5. Not sufficient. (2) 20x+10y>280 > 2x+y>28 > 2x+(25x)>28 > x>3. Not sufficient. (1)+(2) From (1) and (2) 3<x<5 > since given that x is an integer then x=4. Sufficient. Answer: C. Hope it's clear.
_________________




Intern
Joined: 18 Jan 2012
Posts: 9

Re: If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
Show Tags
12 Feb 2012, 02:47
ooh that makes sense!
not that important but what about for :
a) they say the maximum for 20x + 10y is 250 when x=0 and y=25, why didn't they make the maximum value using x=25 , y=0 since that would give you a larger number?



Math Expert
Joined: 02 Sep 2009
Posts: 58381

Re: If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
Show Tags
12 Feb 2012, 03:01
c210 wrote: a) they say the maximum for 20x + 10y is 250 when x=0 and y=25, why didn't they make the maximum value using x=25 , y=0 since that would give you a larger number?
b) why didn't they use option (1) and solve for Y , then plug that into X=Y=25 from the original problem? wouldn't that solve for x right away?
its probably something simple that I'm missing but any clarity would be appreciated its really bugging me!
thanks in advance! As for your questions (notice that x and y are nonnegative integers): A. 250 is the smallest possible value of 20x+10y: minimize x, as it has greater multiple (20) and maximize y, as it has smaller multiple (10) > x=0 and y=25 > 20x +10y=250. The same way the largest possible value of 20x+10y is for x=25 and y=0 > 20x +10y=500. Or another way: 20x+10y=20x+10(25x)=10x+250 > to get the smallest possible value minimize x, so make it 0: 10x+250=250, and to get the largest possible value maximize x, so make it 25: 10x+250=500. B. You can not solve for x or y as you get only the ranges for them from (1), as well as from (2), and not the unique values (refer to the solution above). Hope it helps.
_________________



Manager
Joined: 12 Feb 2012
Posts: 114

Re: If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
Show Tags
12 Aug 2012, 18:06
Bunuel wrote: c210 wrote: a) they say the maximum for 20x + 10y is 250 when x=0 and y=25, why didn't they make the maximum value using x=25 , y=0 since that would give you a larger number?
b) why didn't they use option (1) and solve for Y , then plug that into X=Y=25 from the original problem? wouldn't that solve for x right away?
its probably something simple that I'm missing but any clarity would be appreciated its really bugging me!
thanks in advance! As for your questions (notice that x and y are nonnegative integers): A. 250 is the smallest possible value of 20x+10y: minimize x, as it has greater multiple (20) and maximize y, as it has smaller multiple (10) > x=0 and y=25 > 20x +10y=250. The same way the largest possible value of 20x+10y is for x=25 and y=0 > 20x +10y=500. Or another way: 20x+10y=20x+10(25x)=10x+250 > to get the smallest possible value minimize x, so make it 0: 10x+250=250, and to get the largest possible value maximize x, so make it 25: 10x+250=500. B. You can not solve for x or y as you get only the ranges for them from (1), as well as from (2), and not the unique values (refer to the solution above). Hope it helps. Hey Bunuel, You taught me one hard rule, I'll never forget with inequilities. You can only add two inequalities in the same direction, never subtract!! Ex a<b x<y a+x<b+y is ok but ax<by is WRONG!! Now, does this rule differ if I compare an inequality to an equation? Can I subtract an inequality and an equation in the same direction? (1) 2x+y<30 (2) x+y=25 Can I simply subtract with confidence to get: x<5? I know that I can solve for y in equation (2) and substitute in equation (1). I would appreciate if you could confirm Many thanks!!



Director
Joined: 22 Mar 2011
Posts: 590
WE: Science (Education)

Re: If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
Show Tags
12 Aug 2012, 23:19
alphabeta1234 wrote: Bunuel wrote: c210 wrote: a) they say the maximum for 20x + 10y is 250 when x=0 and y=25, why didn't they make the maximum value using x=25 , y=0 since that would give you a larger number?
b) why didn't they use option (1) and solve for Y , then plug that into X=Y=25 from the original problem? wouldn't that solve for x right away?
its probably something simple that I'm missing but any clarity would be appreciated its really bugging me!
thanks in advance! As for your questions (notice that x and y are nonnegative integers): A. 250 is the smallest possible value of 20x+10y: minimize x, as it has greater multiple (20) and maximize y, as it has smaller multiple (10) > x=0 and y=25 > 20x +10y=250. The same way the largest possible value of 20x+10y is for x=25 and y=0 > 20x +10y=500. Or another way: 20x+10y=20x+10(25x)=10x+250 > to get the smallest possible value minimize x, so make it 0: 10x+250=250, and to get the largest possible value maximize x, so make it 25: 10x+250=500. B. You can not solve for x or y as you get only the ranges for them from (1), as well as from (2), and not the unique values (refer to the solution above). Hope it helps. Hey Bunuel, You taught me one hard rule, I'll never forget with inequilities. You can only add two inequalities in the same direction, never subtract!! Ex a<b x<y a+x<b+y is ok but ax<by is WRONG!! Now, does this rule differ if I compare an inequality to an equation? Can I subtract an inequality and an equation in the same direction? (1) 2x+y<30 (2) x+y=25 Can I simply subtract with confidence to get: x<5? I know that I can solve for y in equation (2) and substitute in equation (1). I would appreciate if you could confirm Many thanks!! YES, you can subtract with confidence, because this operation is equivalent to subtracting the same quantity from both sides, operation allowed when working with inequalities. In the above case, you subtract 25, only that on the left you express it as x + y, and on the right you just simply write it as 25. Therefore, 2x + y  (x + y) < 30  25 or x < 5. If you use substitution, replacing x + y in the inequality by 25, you obtain x + 25 < 30, from which by now subtracting 25 from both sides, you get the same conclusion x < 5. So, you do the same thing, just in slightly different forms.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Current Student
Joined: 12 Aug 2015
Posts: 2568

Re: If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
Show Tags
13 Mar 2016, 09:32
Awesome question.. Inequality mixed with word problems here x=4 and y=21 is the only value value that will work SO C
_________________



Intern
Joined: 30 Mar 2015
Posts: 10

Re: If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
Show Tags
03 Dec 2016, 07:00
Bunuel wrote: If x and y are nonnegative integers and x+y=25 what is x
Given: x+y=25 > y=25x. Question: x=?
(1) 20x+10y<300 > 2x+y<30 > 2x+(25x)<30 > x<5. Not sufficient. (2) 20x+10y>280 > 2x+y>28 > 2x+(25x)>28 > x>3. Not sufficient.
(1)+(2) From (1) and (2) 3<x<5 > since given that x is an integer then x=4. Sufficient.
Answer: C.
Hope it's clear. for me ,the right answer is E because the question says that x and y are nonnegative integer means that x and y can be negative fraction and could be positive integer and fraction Therefore when we combine statments we find that x is between 3 and 4 means that x can be 3.5 3.6 3.2 so is insufficient can you please tell me where my reasoning is false ? Thankyou



Math Expert
Joined: 02 Sep 2009
Posts: 58381

Re: If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
Show Tags
03 Dec 2016, 07:47
hichem wrote: Bunuel wrote: If x and y are nonnegative integers and x+y=25 what is x
Given: x+y=25 > y=25x. Question: x=?
(1) 20x+10y<300 > 2x+y<30 > 2x+(25x)<30 > x<5. Not sufficient. (2) 20x+10y>280 > 2x+y>28 > 2x+(25x)>28 > x>3. Not sufficient.
(1)+(2) From (1) and (2) 3<x<5 > since given that x is an integer then x=4. Sufficient.
Answer: C.
Hope it's clear. for me ,the right answer is E because the question says that x and y are nonnegative integer means that x and y c an be negative fraction and could be positive integer and fraction Therefore when we combine statments we find that x is between 3 and 4 means that x can be 3.5 3.6 3.2 so is insufficient can you please tell me where my reasoning is false ? Thankyou x and y are nonnegative integers means that x and y are integers greater than or equal to 0: 0, 1, 2, 3, ... This is the only reading of this statement.
_________________



Manager
Joined: 02 Nov 2013
Posts: 72
Location: India

Re: If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
Show Tags
03 Dec 2016, 09:17
I hope, I can think the way Banuel does. To solve this simple question I used substitution method.
Yes, only one value of X can satisfy the equations, x=4 and y=21.
Answer C.



NonHuman User
Joined: 09 Sep 2013
Posts: 13210

Re: If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
Show Tags
15 Mar 2018, 05:37
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: If x and y are nonnegative integers and x+y=25 what is x?
[#permalink]
15 Mar 2018, 05:37






