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Thelegend2631
Posts: 371
24 Jun 2021, 06:08
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
29% (02:20) correct
71%
(02:31)
wrong
based on 82
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How many whole number solutions are possible for the equation \(x+y+z = 30\) such \(x < y < z\)
A. 75
B. 150
C. 90
D. 298
E. 65
Posted from my mobile device
A. 75
B. 150
C. 90
D. 298
E. 65
Posted from my mobile device
24 Jun 2021, 11:24
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hD13
This can be done through partition in combination problems, but this method is not tested, and you may never see such question on GMAT.
Let us try doing this by method that may be useful from perspective of GMAT questions.
Whole numbers mean that 0 is included along with positive integers or we are looking at non-negative integers as per GMAT language.
\(x+y+z=30\)
Case I: x=0, then z+y=30.....Average is 15, so y can take values from 1 to 14....14 ways
Why?: x is 0, y can take any value less than the average as y<z. For each value of y taken in this range 1 to 14, there will be a corresponding value for z in range 29 to 16
Case II: x=1, then z+y=29.....Average is 14.5, so y can take values from 2 to 14....13 ways
Case III: x=2, then z+y=28.....Average is 14, so y can take values from 3 to 13....11 ways
Case IV: x=3, then z+y=27.....Average is 13.5, so y can take values from 4 to 13....10 ways
Case V: x=4, then z+y=26.....Average is 13, so y can take values from 5 to 12....8 ways
Case VI: x=5, then z+y=25.....Average is 12.5, so y can take values from 6 to 12....7 ways
So, we get a pattern, and the sum is \((14+13)+(11+10)+(8+7)+(5+4)+(2+1)=27+21+15+9+3=75\)
Updated on: 24 Jun 2021, 23:55
Originally posted by GMATGuruNY on 24 Jun 2021, 12:36.
Last edited by GMATGuruNY on 24 Jun 2021, 23:55, edited 1 time in total.
Last edited by GMATGuruNY on 24 Jun 2021, 23:55, edited 1 time in total.
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For a description of the SEPARATOR METHOD, check my post here:
https://gmatclub.com/forum/how-many-pos ... 10356.html
The task above is equivalent to distributing 30 identical chocolates among 3 children x, y and z.
Using the separator method, we get:
Number of ways to distribute 30 identical chocolates among 3 children = 32C2 \(= \frac{32*31}{2*1} = 496\)
Disallowed Case 1: All 3 values are the same
10-10-10 --> 1 option
Disallowed Case 2: 2 values are the same, the other value is different
Let R = the repeated integer and N = the non-repeated integer --> R+R+N = 30
Number of options for R = 15 (Of the 16 integers 0 through 15, any but 10, already considered in Case 1 above)
Number of options for N = 1 (Each option for R+R has only one mate N that will yield a sum of 30)
To combine these options, we multiply:
15*1 = 15
Since each option for R+R+N can be arranged 3 ways, we multiply by 3:
15*3 = 45 options
Subtracting the disallowed options, we get:
496 - 1 - 45 = 450
Each of the 450 options above is composed of 3 distinct nonnegative integers.
Number of ways to arrange 3 distinct integers = 3! = 6.
Of the 6 possible arrangements for any set of 3 distinct integers, ONLY ONE will list the 3 values in ascending order.
Thus, 1/6 of the 450 options above will satisfy the condition that x<y<z:
1/6 * 450 = 75
https://gmatclub.com/forum/how-many-pos ... 10356.html
hD13
The task above is equivalent to distributing 30 identical chocolates among 3 children x, y and z.
Using the separator method, we get:
Number of ways to distribute 30 identical chocolates among 3 children = 32C2 \(= \frac{32*31}{2*1} = 496\)
Disallowed Case 1: All 3 values are the same
10-10-10 --> 1 option
Disallowed Case 2: 2 values are the same, the other value is different
Let R = the repeated integer and N = the non-repeated integer --> R+R+N = 30
Number of options for R = 15 (Of the 16 integers 0 through 15, any but 10, already considered in Case 1 above)
Number of options for N = 1 (Each option for R+R has only one mate N that will yield a sum of 30)
To combine these options, we multiply:
15*1 = 15
Since each option for R+R+N can be arranged 3 ways, we multiply by 3:
15*3 = 45 options
Subtracting the disallowed options, we get:
496 - 1 - 45 = 450
Each of the 450 options above is composed of 3 distinct nonnegative integers.
Number of ways to arrange 3 distinct integers = 3! = 6.
Of the 6 possible arrangements for any set of 3 distinct integers, ONLY ONE will list the 3 values in ascending order.
Thus, 1/6 of the 450 options above will satisfy the condition that x<y<z:
1/6 * 450 = 75
