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

$\begin{array}{1 1} (A)\;(7,8),(7,3) \\(B)\;(3,5),(3,1) \\(C)\;(5,7),(7,9) \\(D)\;(2,3),(4,5) \end{array}$

## 1 Answer

Toolbox:
• Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality.
• Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality.
• If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e $'>'$ sign changes to $'<'$ and $'<'$ changes $'>'$ .
Let the two consecutive odd positive integer be $x$ and $x+2$.
Both number are smaller than 10 Therefore
$x+2 <10$
Adding $-2$ to both sides,
$=> x < 10-2$
$=> x < 8$
Also sum of the two integers is more than 11.
So, $x + x+2 > 11$
$=> 2x +2 > 11$
adding $-2$ to both sides,
$=> 2x > 11-2$
$=> 2x > 9$
Adding -2 to both sides,
$=> 2x > 11-2$
$=> 2x > 9$
Diving by 2 on both sides,
$=> x > 9/2$
$=> x > 4.5$
Step 2 :
Since x is an odd integer number greater than $4.5$ and less than 8 (from 0) x can take values 5 and 7.
Thus the required pairs are $(5,7)$ and $(7,9)$
Hence C is the correct answer.
answered Jul 28, 2014 by

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