If N, x, and y are three positive integers such that N = x + y, 2

Question

If N, x, and y are three positive integers such that N = x + y, 2 < x < 10, and 14 < y < 23. If N > 25, then find the number of distinct values of integer N.

A. 2
B. 4
C. 6
D. 8
E. 11

chetan2u · Answer

The minimum value of x+y is (minimum x+minimum y)=3+15=28
The maximum value of x+y is (maximum x+maximum y)=9+22=31
As N>25, so N can be 26, 27, 28, 29, 30 or 31. => 6 values

C

kalyanchilaka · Answer

Every number starting from x=4 will produce one unique number greater than 25.
eg:
when x=4 and y=22 N=26
when X=5 we have N=26,27
when x=6 we have N=26,27,28
when x=7...
 at every possibility of X we have one distinct value for N.
Since X can take 4,5,6,7,8,9 there are 6 possibilities

gmathack93 · Answer

Hi  chetan2u

Thx for your answer.

I tried to solve this using addtion of inequalities, however that method seems to give a wider range than yours. I believe I am missing somthing in my solution. Could you please help understand why below approach is wrong ?

2 < x < 10
14 < y < 23
-----------------
16 < x+y < 33

If I add the two inequalities, my range comes to be 17 to 32

However, when I use min(x+y) or (3+15) = 18 and max(x+y) or (9+22) = 31.

I am not able to understand why there is a difference between the two. Could you please help.

Thx

btsaami · Answer

x can take values 3 to 9, y from 15 to 22.
N>25.
N should start with 26. maximum value 22+9= 31
N can take the values from 26 to 31. 27= 18+9, 29= 19+9 and so on
N has 6 distinct values.

If N, x, and y are three positive integers such that N = x + y, 2 < x

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