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Integer constraints with inequalities question

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Integer constraints with inequalities question

by JGoode » Thu Jun 10, 2010 5:05 pm
This problem is from MGMAT's equations book. I'd be v.grateful if someone could explain Manhattan's working out that I've posted below in the spoiler section. I've posted my confusion in capitals also.

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

(1) 20x + 10y < 300
(2) 20x + 10y > 280



[spoiler]Ans. is= (C); both statements together are sufficient.

Manhattan explains this as: "first, we should note that since x and y must be positive integers, the smallest possible value for 20x + 10y is 250, when x=0 and y=25. If we combine the above two statements together we get:

so 280 < 20x + 10y < 300 (we can substitute (25-x) for y)
so 280 < 20x + 10(25-x) < 300
so 30 < 10x < 50
so 3 < x < 5

since x must be an integer, x must equal 4."

I CAN'T UNDERSTAND WHY MANHATTAN STATES THAT X AND Y MUST BE POSITIVE INTEGERS WHEN ZERO IS NOT A POSITIVE INTEGER. ALSO, IF WE SUBSTITUTE X AS 0 AND Y AS 25 INTO THE 20x + 10y > 280, WE GET 0 + 250 > 280, BUT HOW CAN 250 BE GREATER THAN 280?

Thanks


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by selango » Thu Jun 10, 2010 9:12 pm
Jgoode,

The approach is as below,

From both sttmts,we can rephrase as

280<20x+10y<300

Since x+y=25,y=25-x

Substitute this in the above ineqaulties.

280<20x+10(25-x)<300

280<10x+250<300

3<x<5

From this we can deduce x=4

There is no need for substituting x=0 and solving the equation.
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by ankurmit » Fri Jun 11, 2010 12:10 am
If x and y are nonnegative integers and x + y = 25, what is x?

(1) 20x + 10y < 300
(2) 20x + 10y > 280



from 1st : 2x+y<30

from 2nd : 2x+y>28

as given x and y are integers so from 1st and 2nd 2x+y=29..got it?

now we got 2 equations and 2 variables and we can get value of X...
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by Testluv » Fri Jun 11, 2010 3:04 am
I CAN'T UNDERSTAND WHY MANHATTAN STATES THAT X AND Y MUST BE POSITIVE INTEGERS WHEN ZERO IS NOT A POSITIVE INTEGER.
In the solution, where it mentions that it is a "positive integer", Manhattan must have meant "non-negative integer", which accords to the question itself. 0 is non-negative but also not positive.

____

The above poster's approach is the best way of solving but we could have also solved it through substitution if the deduction evaded us:

What is x?

From the question stem, we have: x+y = 25 or y = 25 - x

(1) 2x + y < 30

so 2x + (25 - x) < 30

or x < 5...insufficient

(2) 2x + y > 28

so 2x + (25 - x) > 28

or x > 3...insufficient.

Combined, you know that x is an integer that is greater than 3 but less than 5, so it must be 4. Choose C.
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by outreach » Fri Jun 11, 2010 3:29 am
Option 1

20x + 10y < 300
10x+10(x+y)<300
10x+250<300
10x<50
x<5

not suff

option 2
20x + 10y > 280
10x+10(x+y)>280
10x+250>280
10x>30
x>3
not suff

combining 1 and 2

3<x<5
hence x = 4



JGoode wrote:This problem is from MGMAT's equations book. I'd be v.grateful if someone could explain Manhattan's working out that I've posted below in the spoiler section. I've posted my confusion in capitals also.

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

(1) 20x + 10y < 300
(2) 20x + 10y > 280



[spoiler]Ans. is= (C); both statements together are sufficient.

Manhattan explains this as: "first, we should note that since x and y must be positive integers, the smallest possible value for 20x + 10y is 250, when x=0 and y=25. If we combine the above two statements together we get:

so 280 < 20x + 10y < 300 (we can substitute (25-x) for y)
so 280 < 20x + 10(25-x) < 300
so 30 < 10x < 50
so 3 < x < 5

since x must be an integer, x must equal 4."

I CAN'T UNDERSTAND WHY MANHATTAN STATES THAT X AND Y MUST BE POSITIVE INTEGERS WHEN ZERO IS NOT A POSITIVE INTEGER. ALSO, IF WE SUBSTITUTE X AS 0 AND Y AS 25 INTO THE 20x + 10y > 280, WE GET 0 + 250 > 280, BUT HOW CAN 250 BE GREATER THAN 280?

Thanks


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by JGoode » Fri Jun 11, 2010 10:43 am
Thank you very much everyone for your posts, the various explanations really help a lot and make things much easier!

Could I ask for some brief clarification regarding the first part of Manhattan's explanation: "first, we should note that since x and y must be positive integers, the smallest possible value for 20x + 10y is 250, when x=0 and y=25." What purpose is there for finding the smallest possible value of 20x + 10y as being 250? 250 can't be greater than 280 as statement 1 indicates.

Outreach, you indicate in your approach:

20x + 10y < 300
10x+10(x+y)<300

could you explain why you went from 20x to 10x, I can't seem to figure out the reason (it's probably really obvious too).

Sorry if the questions seem basic as it's been a long time since I studied algebra/equations/inequalities so I very much appreciate all the help.
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by Testluv » Fri Jun 11, 2010 11:01 am
20x + 10y < 300
10x+10(x+y)<300

could you explain why you went from 20x to 10x, I can't seem to figure out the reason (it's probably really obvious too).
20x + 10y < 300

or 10x + 10x + 10y < 300

or 10x + 10(x + y) < 300

But I think you are better off just dividing both of the inequalities by 10 as I did in my post above.
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by JGoode » Sun Jun 13, 2010 4:29 am
I see now, thanks a lot Testluv for the explanation! That method is pretty ingenious actually and great to know. Much clearer now.

I've still not managed to understand why in the first part of Manhattan's explanation, they state: "the smallest possible value for 20x + 10y is 250, when x=0 and y=25." What purpose is there for finding the smallest possible value of 20x + 10y as being 250? 250 can't be greater than 280 as statement 1 indicates. Surely all we need to know is that x+y=25 and substitute the terms where required?
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