I got this question off a online practice test, but it looks fishy...
1/(2exp12) + 2/(2exp13) + 3/(2exp14) + 4/(2exp15) = ?
a) 1/(2exp10)
b) 1/(2exp12)
c) 15/(2exp15)
d) 2/(2exp10)
e) 23/(2exp16)
** exp = ^ and the above are the fraction 1/2 , with denominator raised to different powers
I know how to do this question through simplifying, but the official solution started simplifying by what looks to me is either a typo or incorrect assumption.
Can someone give their opinion on how to solve?
OA to follow
Exponents - Does this make sense?
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No matter how I do it, I get 13/2^14.
We can reduce to:
1/2^12 + 1/2^12 + 3/2^14 + 1/2^13
Our common denominator is 2^14, so:
2^2/2^14 + 2^12/2^14 + 3/2^14 + 2^1/2^14
(4+4+3+2)/2^14 = 13/2^14
You could also use 2^15 as your common denominator. After conversion, you'd get 26/2^15 = 13/2^14.
We can reduce to:
1/2^12 + 1/2^12 + 3/2^14 + 1/2^13
Our common denominator is 2^14, so:
2^2/2^14 + 2^12/2^14 + 3/2^14 + 2^1/2^14
(4+4+3+2)/2^14 = 13/2^14
You could also use 2^15 as your common denominator. After conversion, you'd get 26/2^15 = 13/2^14.
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The OA is A.
It says: If we express the numerators as powers of 2, then we would get
1/(2exp12) + 2/(2exp13)+ 4/(2exp14) + 8/(2exp15)
however I keep getting
1/(2exp12) + (2/2)/(2exp12)+ (3/4)/(2exp12) + (4/8)/(2exp12)
but the OA goes on to say that their simplification reduces to
1/(2exp12) + 1/(2exp12)+ 1/(2exp12) + 1/(2exp12) = 4/(2exp12)
4/(2exp12) = 1/(2exp10)
The way I understand it, if youre simplifying your denominator your factor stays in the demonator which results in the fractions in the manner that I simplified.
Help anyone?
It says: If we express the numerators as powers of 2, then we would get
1/(2exp12) + 2/(2exp13)+ 4/(2exp14) + 8/(2exp15)
however I keep getting
1/(2exp12) + (2/2)/(2exp12)+ (3/4)/(2exp12) + (4/8)/(2exp12)
but the OA goes on to say that their simplification reduces to
1/(2exp12) + 1/(2exp12)+ 1/(2exp12) + 1/(2exp12) = 4/(2exp12)
4/(2exp12) = 1/(2exp10)
The way I understand it, if youre simplifying your denominator your factor stays in the demonator which results in the fractions in the manner that I simplified.
Help anyone?
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the question stem has a '3' in the numerator and that has to be accounted somewhere in the answer , when it is all 2s in the denominator.
OA is not convincing.
I guess some typo is there in the question.
OA is not convincing.
I guess some typo is there in the question.
- Stuart@KaplanGMAT
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For this to be true, the original question would have to be:evansbd wrote:The OA is A.
It says: If we express the numerators as powers of 2, then we would get
1/(2exp12) + 2/(2exp13)+ 4/(2exp14) + 8/(2exp15)
1/2^12 + 2^1/2^13 + 2^2/2^14 + 2^3/2^15
or
1/2^12 + 2/2^13 + 4/2^14 + 8/2^15
which would indeed reduce to:
4(1/2^12) = 2^2/2^12 = 1/2^10
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