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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Find all pair of consecutive even positive integers, both of the which are larger than 5 such that their sum is less than 23.

$\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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1 Answer

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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 $'>'$ .
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.
answered Jul 28, 2014 by meena.p
 

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