Let a, b, c, d be real numbers such that |a-b|=2 , |b-c|=3,|c-d|=4 . Then, sum of all possible value

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4-2.Quadratic Equations and Inequations

hard

Let $a, b, c, d$ be real numbers such that $|a-b|=2$, $|b-c|=3,|c-d|=4$. Then, the sum of all possible values of $|a-d|$ is

A

$9$

B

$18$

C

$24$

D

$30$

(KVPY-2011)

Solution

(b)

Given, $|a-b|=2,|b-c|=3$ and $|c-d|=4$

$\therefore a-b=\pm 2, b-c=\pm 3$ and $c-d=\pm 4$

Possible value of $(a-d)$ are $\pm 9, \pm 5, \pm 3, \pm 1$.

$\therefore \quad|a-d|=9,5,3,1$

Sum of $|a-d|=9+5+3+1=18$

Standard 11

Mathematics

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