Let N, x and y be positive integers such that N = x + y, 2

Question

Let N, x and y be positive integers such that N = x + y, 2 < x < 10 and 14 < y < 23. If N > 25, then how many distinct values are possible for N?

A. 4
B. 5
C. 6
D. 7
E. 8

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ManifestDreamMBA · Answer

x = {3,9} and y = {15,22}
N>25

Given the constraints, x cannot be 3
If x = 4, y = 22 , N = 26
If x = 5, y = 21 or 22 , N = 26, 27
If x = 6, y = 20, 21 or 22 , N = 26, 27, 28
If x = 7, y = 19, 20, 21 or 22 , N = 26, 27, 28, 29
If x = 8, y = 18, 19, 20, 21 or 22 , N = 26, 27, 28, 29, 30
If x = 9, y = 17, 18, 19, 20, 21 or 22 , N = 26, 27, 28, 29, 30, 31

We have 6 distinct values of N

GMATinsight · Answer

N_{max} = 9+23 = 31

for N > 25
Possible values of N are {also notice that  y_{min} = 15 }

N_{min} = 26  (possible)
N= 27, 
N= 28, 
N= 29, 
N= 30, 
 N_{max}= 9+22 = 31,

Hence, 6 values

Answer: Option C

N_{min} =

Dereno · Answer

Given N =x+y

2 < x < 10 and 14 < y < 23

Question : If N >25, possible value of N ?

x can take values ( 3,4,5,6,7,8,9)

similarly y can take values ( 15,16,17,18,19,20,21,22)

N= x+y =  minimum  value can be  18  (3+15).

maximum  value can be (9+22) =  31.

The distinct values greater than 25 is : 26,27,28,29,30,31. Thus, we have   6 values  .

Option C

Ayeka · Answer

N= x+y ; 2<x<10 ; 14<y<23 
x can be (3,4,5,6,7,8,9) and y can be (15,16,17,18,19,20,21,22) 
We will get Minimum value of N when x=3 and y=15 
And maximum value when x=9 and y=22 
N minimum = 3+15=18 
N maximum = 9+22=31 
We need values of N which are more than 25. So, N can be 26,27,28,29,30,31. It can take 6 values.

Answer: C

Let N, x and y be positive integers such that N = x + y, 2 < x < 10

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