Find all pairs of consecutive odd positive integers, both of which are larger than 10 such that their sum is less than 40. |
$4, 6; 6, 8; 8, 10; 10, 12; 12 , 14$ $5,7; 7, 9; 9,11; 11, 13;13, 15$ $11, 13; 13, 15; 15, 17; 17, 19$ $10, 12; 12, 14; 14, 16; 16, 18$ |
$11, 13; 13, 15; 15, 17; 17, 19$ |
The correct answer is option (3) : 11, 13; 13, 15;15, 17; 17, 19 Let x , x+2 be the 2 consecutive odd positive integer. As are larger than 10 and their sum is less than 40, we get $x> 10$ and $x+ (x+ 2) < 40$ $2x + 2< 40$ $10< x $ and $ x < 19$ $⇒10< x < 19$ Since x is odd positive integer and it lies between 10 and 19, the positive values of x are 11, 13, 15, 17, than the corresponding values of the other odd integer i.e x + 2 are 13, 15, 17, 19 Hence the required pairs are $11, 13; 13, 15; 15, 17; 17, 19$ |