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by Rahul@gurome » Wed May 12, 2010 11:13 pm
2^(-3) * n = (2^3)^(-3) +(2^2 * 3)^(-5)
2^(-3) * n = 2^(-9) + 2^(-10) * 3^(-5)
2^(-3) * n = 2^(-3) {2^(-6)+ 2^(-7) * 3^(-5)}
n = 2^(-6) + 2^(-7) * 3^(-5)
n = 2^(-6) {1 + 1/(2*243)}
n = 1/64 {1 + 1/486}
So, 32n = 32/64 {487/486} = ½ {487/486}

Solving, we get 32n = 487/972
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by rajeshsources » Wed May 12, 2010 11:28 pm
ern5231 ----

we can write the given statement as
n * 2^-3 = (2*2*2)^-3 + (3*2*2)^-5

==> n * 2^-3 = 2^-3 [ ((2*2)^-3) + (3^-5 * 2^-5 * 2^-2)]
Cancel 2^-3 on both the sides, Then,

==> n = [ (2^-6) + (3^-5 * 2^-5 * 2^-2)]
As the question asked for 32n, we will take LCM as 32(or 2^5), So,
==> n = 2^-5 [(2^-1) + (3^-5 * 2^-2)]
we can write 2^-5 == 1/32. Then,
==> n = 1/32[(2^-1) + (3^-5 * 2^-2)]

In such a way, we will get 32n = [ 1/2 + 1/(3^5*2^2)],
==> 32n = 1459/2916. we can round it to 0.5

HTH, GOOD LUCK,

Thanks,
Rajesh,
Loves GMAT....!!!!
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