Question

Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.​

Answer 1

Answer:

(5, 7) and (7, 9)

Step-by-step explanation:

Let us assume x be the smaller of the two consecutive odd positive integers

∴ Other integer is = x + 2

It is also given in the question that, both the integers are smaller than 10

∴ x + 2 < 10

x < 8 … (i)

Also, it is given in the question that sum off two integers is more than 11

∴ x + (x + 2) > 11

2x + 2 > 11

x > 9/2

x > 4.5 … (ii)

Thus, from (i) and (ii) we have x is an odd integer and it can take values 5 and 7

Hence, possible pairs are (5, 7) and (7, 9)

HOPE IT IS HELPFUL

Answer 2

Step-by-step explanation:

Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is x+2. Since both the integers are smaller than 10.

x+2<10

⇒x<10−2

⇒x<8 ..... (i)

Also, the sum of the two integers is more than 11.

∴x+(x+2)>11

⇒2x+2>11

⇒2x>11−2

⇒2x>9

⇒x> 9/2

⇒x>4.5 ...... (ii)

From (i) and (ii), we obtain

Since x is an odd number, x can take values, 5 and 7.

Thus, the required possible pairs are (5,7) and (7,9).