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 \(N\)? [This Question was asked as TITA]

• A. 4
• B. 7
• C. 5
• D. 6

Answer 1

The Correct Option is D

Given, \(2<x<10\) and \(14<y<23\) \(⇒17 <(x+y)<32\) i.e. \(17<N<32\)

But \(N>25\) 

Hence \(25<N<32\)

\(N\) can take \(6\) distinct values.

Answer 2

Any value from the set {3, 4, 5, 6, 7, 8, 9} can be used for \(2 < x < 10 x\). \(14 < y < 23 \)

Any value from the set {15, 16, 17, 18, 19, 20, 21, 22} can be used for y. 

N, or x+y, can have a maximum value of 9+22 = 31. (at y = 22 and x = 9) 
30 is available at x = 9; y = 21
29 is available at x = 9; y = 20
28 is available at x = 9; y = 19
27 is available at x = 9; y = 18
26 is available at x = 9; y = 17
25 is available at x = 9; y = 16 
However, the intended sum is not x+y=25, which is why x+y have different values {31,30,29,28,27,26}.

So, x+y, and hence, N, can have six different values.