$\begin{array}{1 1} (A)\;(7,8),(7,3)\;and\;(2,3) \\(B)\;(6,8),(8,10)\;and\; (10,12) \\(C)\;(5,7),(7,9)\;and \;(2,6) \\(D)\;(2,3),(4,5)\;and\;(3,1) \end{array} $

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- 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 $'>'$ .

Step 1:

Let x and x+2 be two consecutive even positive integers.

Since both the integer are larger then 5. $ x > 5$ -----(1)

Also sum of two is less than 23

$x+x+2 < 23$

$=> 2x+2 <23$

Adding -2 to both sides

$2x < 23-2$

$2x < 21$

Diving by 2 on both sides,

$\large\frac{2x}{2}$$ < 21$

$x < \large\frac{21}{2}$

$x < 10.5$-----(2)

Step 2:

Since x is an even positive integer greater than 5 and less then 10.5 x can take value $6,8,10$ .

Thus the required pair of numbers is $ ( 6,8), (8,10)$ and $ (10,12)$

Hence B is the correct answer.

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