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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!

Re: If x and y are nonnegative integers and x+y=25 what is x? [#permalink]

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12 Feb 2012, 01:47

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?

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(25-x)=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).

Re: If x and y are nonnegative integers and x+y=25 what is x? [#permalink]

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12 Aug 2012, 17:06

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(25-x)=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 a-x<b-y 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

Re: If x and y are nonnegative integers and x+y=25 what is x? [#permalink]

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12 Aug 2012, 22:19

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(25-x)=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 a-x<b-y 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.

Re: If x and y are nonnegative integers and x+y=25 what is x? [#permalink]

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31 Jul 2014, 06:48

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Re: If x and y are nonnegative integers and x+y=25 what is x? [#permalink]

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03 Dec 2016, 06:00

Bunuel wrote:

If x and y are nonnegative integers and x+y=25 what is x

Given: x+y=25 --> y=25-x. Question: x=?

(1) 20x+10y<300 --> 2x+y<30 --> 2x+(25-x)<30 --> x<5. Not sufficient. (2) 20x+10y>280 --> 2x+y>28 --> 2x+(25-x)>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

If x and y are nonnegative integers and x+y=25 what is x

Given: x+y=25 --> y=25-x. Question: x=?

(1) 20x+10y<300 --> 2x+y<30 --> 2x+(25-x)<30 --> x<5. Not sufficient. (2) 20x+10y>280 --> 2x+y>28 --> 2x+(25-x)>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

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.
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