Question : If $a,b,c$, and $d$ satisfy the equations:
$a+7b+3c+5d=0$, $8a+4b+6c+2d=–4$, $2a+6b+4c+8d=4$, $5a+3b+7c+d=–4$, then $\frac{a+d}{b+c}$?
Option 1: 0
Option 2: 1
Option 3: –1
Option 4: – 4
Answer (1)
16th Jan, 2024
Correct Answer: –1
Solution :
Given: $8a+4b+6c+2d=–4$ (equation 1)
$2a+6b+4c+8d=4$ (equation 2)
Adding equations 1 and 2, we get,
$8a+4b+6c+2d+2a+6b+4c+8d=–4+4$
⇒ $10(a+b+c+d)=0$
⇒ $(a+d)=–(b+c)$
⇒ $\frac{a+d}{b+c}$ = –1
Hence, the correct answer is –1.
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