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Question

Let (x1,y1,z1) and (x2,y2,z2) be 2 sets of solution satisfying the following equations:
log10(2xy)=4+(log10x1)(log10y2)
log10(2yz)=4+(log10y2)(log10z1)
log10(zx)=2+(log10z1)(log10x1)
such that (x1>x2),
then match the elements of List - I with the correct answer in List -II.

List -IList -II(I)y1x1(P)2(II)z1x2(Q)100(III)z1x2z2(R)1000(IV)y2+z1x2(S)150

Which of the following is the only 'INCORRECT' combination?

A
(I)(P)
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B
(II)(Q)
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C
(III)(S)
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D
(IV)(S)
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Solution

The correct option is C (III)(S)
Let log10x1=A
log10y2=B
log10z1=C
So (A1)(B1)=log102
(B1)(C1)=log102
(A1)(C1)=1
(A1)(B1)(C1)=±log102
A1=±1C1=±1B1=±log102
(A,B,C)=(2,1+log102,2) or (0,1log102,0)
(x,y,z)=(1000x1,2000y1,1000z1),(10x2,500y2,10z2)
Now match correct combinations

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