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Question

If [x+32y+xz14az]=[0732a], then (x+y+z+a) is:

A
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B
0
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C
1
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D
8
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Solution

The correct option is C 1
If A=B, then all the elements of A and B should be same.

x+3=0x=3

2y+x=7..................(i)

z1=3z=4

z4a=2az=2a

Putting value of x in eq.(i)

2y=37y=42=2

y=2, z=4 & a=2

So, x+y+z+a=32+4+2 =1

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