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If x, y and z are non-negative integers such that x < y < z,

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If x, y and z are non-negative integers such that x < y < z,  [#permalink]

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New post 05 Nov 2018, 10:28
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15
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

45% (02:27) correct 55% (02:10) wrong based on 122 sessions

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If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78

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If x, y and z are non-negative integers such that x < y < z,  [#permalink]

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New post Updated on: 05 Nov 2018, 22:11
Misinterpreted non-negative for positive.

Originally posted by philipssonicare on 05 Nov 2018, 21:23.
Last edited by philipssonicare on 05 Nov 2018, 22:11, edited 1 time in total.
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Re: If x, y and z are non-negative integers such that x < y < z,  [#permalink]

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New post 05 Nov 2018, 22:04
3
X, Y, Z Is non negative integer, and x+y+z=11, and x<y<z
We can fix value of x and calculate distinct solutions
Case-1
X=0
Y can have 1,2,3,4,5 (Total 5 values) to reach x+y+z =11.
For each value of y z will have a fix value. So total 5 distinct solutions.
Case-2
x=1
Y can be 2,3,4, for each value, z will have a fix value ( 6,7, and 8)
Total 3 distinct solutions in this case.
Case -3
X=2
Y can be 3,4 for each value, z will have a fix value ( 5,6,)
Total 2 distinct solutions in this case.
Case -4
X=3
No value of y can satisfy x+y+z=11, and x<y<z.

Total
5+3+2=10
B.

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If x, y and z are non-negative integers such that x < y < z,  [#permalink]

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New post 07 Nov 2018, 07:43
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GMATPrepNow wrote:
If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78


I created this question to show that there can be times when the best (i.e., fastest) way to solve a counting question is by listing and counting

How do we know when it's not a bad idea to use listing and counting?
The answer choices will tell us (ALWAYS scan the answer choices before beginning any answer choices)
Here, the answer choices are reasonably small, so listing and counting shouldn't take long.
ASIDE: Yes, 78 (answer choice E) is pretty big. However, if you start listing possible outcomes and you eventually list more than 22 outcomes (answer choice D), you can stop because the answer must be E.

Okay, let's list possible outcomes in a systematic way.
We'll list outcomes as follows: x, y, z

Since x is the smallest value.
Let's list the outcomes in which x = 0. We get:
0, 1, 10
0, 2, 9
0, 3, 8
0, 4, 7
0, 5, 6


Now, the outcomes in which x = 1. We get:
1, 2, 8
1, 3, 7
1, 4, 6


Now, the outcomes in which x = 2. We get:
2, 3, 6
2, 4, 5

Now, the outcomes in which x = 3. We get:
3, 4...hmmm, this won't work.

So, we're done listing!
Count the outcomes to get a total of 10

Answer: B

Cheers,
Brent
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Re: If x, y and z are non-negative integers such that x < y < z,  [#permalink]

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New post 11 Oct 2019, 07:48
1
GMATPrepNow wrote:
If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78


0≤x<y<z
x+y+z=11
x=0, y=1, z=11-1=10: average (10+1)=11/2=5.5 so y={1 to 5}={1,2,3,4,5}=5
x=1, y=2, z=11-3=8: average (8+2)=10/2=5 so y={2 to 4}={2,3,4}=3
x=2, y=3, z=11-5=6: average (6+3)=9/2=4.5 so y={3 to 4}={3,4}=2
x=3, y=4, z=11-7=4: invalid

total solutions: 5+3+2=10

Answer (B)
GMAT Club Bot
Re: If x, y and z are non-negative integers such that x < y < z,   [#permalink] 11 Oct 2019, 07:48
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