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Question

If a, b, x, y are positive natural numbers such that 1x+1y=1 and axx+byy is-

A
ab
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B
ab
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C
=ab
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D
Can't be found out
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Solution

The correct option is A ab
Consider the opposite numbers
ax,ax,..... ky times and by,by,.....kx times
AM={ax+ax+.....kytimes}+{by+by+.....kxtimes}kx+ky=kyax+kxbyk(x+y)=yax+xby(x+y)
GM={(ax.ax....kytimes)(by.by....kxtimes)}1k(x+y)
(ax(ky).by(kx))1k(x+y)=(ab)kxyk(x+y)=(ab)xy(x+y).....(i)
As 1x+1y=1,x+yxy=1,i.e,x+y=xy
(i)becomesyzx+xbyxyab or axx+byyab
Hence, option 'B' is correct.

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