Q58 / 89 · KVPY 2017 (5 Nov SB/SX) · Mathematics
Let $a, b, c, d, e$, be real numbers such that $a+b < c+d, b+c < d+e, c+d < e+a$ $\mathrm{d}+\mathrm{e} < \mathrm{a}+\mathrm{b}$. Then
- Correct answerAThe largest is a and the smallest is $\mathrm{b}$
- BThe largest is a and the smallest is $c$
- CThe largest is $c$ and the smallest is e
- DThe largest is $\mathrm{c}$ and the smallest is $\mathrm{b}$
Solution
(i) $a+b < c+d$
(ii) $b+c < d+e$
(iii) $\mathrm{c}+\mathrm{d} < \mathrm{e}+\mathrm{a}$
(iv) $\mathrm{d}+\mathrm{e} < \mathrm{a}+\mathrm{b}$
from (i) \& (iii)
$a+b < e+a$
$\Rightarrow \mathrm{b} < \mathrm{e}$
from (ii) \& (iv)
$b+c < a+b$
$\Rightarrow \mathrm{c} < \mathrm{a}$
(i) - (ii)
$a-c < c-e$
$\Rightarrow c>e$
(i) $-(i v)$
$(a-e)+(b-d) < (c-a)+(d-b)$
from thus $d>b$
(i) $+($ iii) $-($ ii)
$\quad c>d$
Hence, the correct answer is option: 'A'.