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

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

Answer 2

x>4.5 ------


Now wat to do

Answer 3

4.5<x<8

Answer 4

Then..hw to find the pairs!!

Answer 5

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

Answer 6

find the integers in this interval in such a way that if one is a then other is a + 2

Answer 7

(5,7) , (6,8).......

Answer 8

see here
http://yougems.reflectionsinfos.com/queries/viewquery/3114

Answer 9

gud

Answer 10

is that correct!

Answer 11

yes

Answer 12

so...... wat abt (7,9)

Answer 13

but in my book there are only two answer (5,7) (7,9)............

Answer 14

because 6 and 8 are not odd

Answer 15

No need for x and y here. Just enumerate
1,3
3,5
5,7
7,9
Only the last two meet the second condition.

Answer 16

Gud @telliott99 but it seems that this question is based on inequality

Answer 17

@Yahoo!
Ur book is correct

Answer 18

See, u get x can be 5,6,7 because it lies between 4.5 and 8
thus, x+2 can be 7,8,9
Now, 6 and 8 are even

Thus only answer is (5,7) , (7,9)

Answer 19

aha..........thxx

Answer 20

is mine is correct

Answer 21

Yes, @best.shakir but u did what he had already done...... :)