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x and y are integers. x + y < 11 , and x > 6. What is the smallest pos

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x and y are integers. x + y < 11 , and x > 6. What is the smallest pos [#permalink]

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New post 13 May 2016, 05:00
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B
C
D
E

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Question Stats:

70% (01:14) correct 30% (01:18) wrong based on 123 sessions

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Kudos [?]: 132894 [0], given: 12391

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x and y are integers. x + y < 11 , and x > 6. What is the smallest pos [#permalink]

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New post 13 May 2016, 13:50
Constraints: X>6, x,y are integers (no fractions) and x+y<11 (can be positive or negative)

Test Case 1: X and Y are positive:
X(>6) Y(positive integer) X+Y<11 X-Y
7 3 10 4
8 2 10 6
9 1 10 8
10 0 0 10

Least difference in the above is 4

Test Case 2 : X is positive Y is negative:

X(>6) Y(positive integer) X+Y<11 X-Y
7 -8 -1 15
8 -7 1 15
9 -6 3 15
10 -4 6 14

Least Difference X-Y as you can is far higher when Y is negative. Hence test case 2 is rules out , going by test case 1 and other constraints defined in the problem only answer is Least difference in the above is Choice C- 4

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Re: x and y are integers. x + y < 11 , and x > 6. What is the smallest pos [#permalink]

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New post 06 Jun 2016, 21:46
Bunuel wrote:
x and y are integers. x + y < 11 , and x > 6. What is the smallest possible value of x - y ?

A. 1
B. 2
C. 4
D. -2
E. -4



Because x & y are integers, the two equations can be simply written as :
x+y<= 10 --- eq. 1
x>= 7 ---- eq. 2

For smallest value of x-y, x has to be the smallest whereas y has to be the largest.
From the above eq. 2, smallest value of x = 7
Plugging it in eq. 1: 7+y<=10, or y<=3
Therefore, largest value of y=3 & smallest values of x=7
x-y = 7 - 3 = 4
Answer is C

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Re: x and y are integers. x + y < 11 , and x > 6. What is the smallest pos [#permalink]

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New post 06 Jun 2016, 23:03
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Bunuel wrote:
x and y are integers. x + y < 11 , and x > 6. What is the smallest possible value of x - y ?

A. 1
B. 2
C. 4
D. -2
E. -4


Focus on the transition points and plug in the values.
x > 6 so x could be 7, 8, 9, 10, 11 etc

Look at x = 7
7 + y < 11
y < 4
y could be 3, 2, 1, 0, -1 etc
When y is 3, we get x - y = 4

This must be the smallest value because as x increases, y reduces so x - y increases.
Take another example. x = 11 so y < 0.
x - y will give a value greater than 11.

Answer (C).
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Kudos [?]: 17830 [1], given: 235

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Re: x and y are integers. x + y < 11 , and x > 6. What is the smallest pos [#permalink]

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New post 06 Jun 2016, 23:43
Bunuel wrote:
x and y are integers. x + y < 11 , and x > 6. What is the smallest possible value of x - y ?

A. 1
B. 2
C. 4
D. -2
E. -4


Smallest value of x-y will be obtained when x and y are nearest to one another.

x > 6. This means that for x-y to be negative value y has to be more than 7. in such a situation x+y cannot be less than 11.

So D and E are out.

Now if negative value is not there, next best option is zero. x+y< 11. Say x=y=5, then x+y = 10 which is less than 11.

But x has to more than 6.

So: values of x = 7, 8, 9 , 10. Then values of y 3, 2, 1 and 0

Smallest value of x-y will be obtained when x and y are nearest to one another.

smallest difference will be 4. (any effort to reduce this difference below 4 will not be possible and will negate x+y <11.

C is the answer.

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Re: x and y are integers. x + y < 11 , and x > 6. What is the smallest pos [#permalink]

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New post 15 Oct 2017, 09:37
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Kudos [?]: 283 [0], given: 0

Re: x and y are integers. x + y < 11 , and x > 6. What is the smallest pos   [#permalink] 15 Oct 2017, 09:37
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