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05 Nov 2018, 10:28
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B
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Difficulty:
95%
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Question Stats:
44% (02:34) correct
56%
(02:10)
wrong
based on 326
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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
A) 5
B) 10
C) 11
D) 22
E) 78
05 Nov 2018, 22:04
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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.
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.
General Discussion
07 Nov 2018, 07:43
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GMATPrepNow
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
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
