0 votes

The $a, b, c$ and $d$ satisfy the equations$$\begin{matrix} a & + & 7b & + & 3c & + & 5d & = &16 \\ 8a & + & 4b & + & 6c & + & 2d & = &-16 \\ 2a & + & 6b & + & 4c & + & 8d & = &16 \\ 5a & + & 7b & + & 3c & + & 5d & = &-16 \end{matrix}$$Then $(a+d)(b+c)$ equals

- $-4$
- $0$
- $16$
- $-16$

1 vote

**Let the given equations be 1,2,3,4 in the order they are given.**

**Adding equation 2 and 3 - **

10a + 10b + 10c + 10d = 0; **Let this be equation 5**

**Adding equation 1 and 4 - **

6a + 10b + 10c + 6d = 0; ** Let this be equation 6**

**Equation 5 minus 6 :**

we get : 4a + 4d =0

so (a+d) =0

if (a+d) =0 then (a+d)(b+c)=0

**therefore answer is B**