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Fractions

by JeetGulia » Tue Jul 20, 2010 12:37 pm
If x=a/2+b/2^3+c/2^4, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT

(A) 1/16
(B) 3/16
(C) 5/16
(D) 10/16
(E) 11/16

We can do it by trying all the combinations of a,b, and C. However is there any smarter approach?
Last edited by JeetGulia on Tue Jul 20, 2010 12:50 pm, edited 1 time in total.
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by aloneontheedge » Tue Jul 20, 2010 12:44 pm
JeetGulia wrote:If x=a/2+b/2^3+c^4, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT

(A) 1/16
(B) 3/16
(C) 5/16
(D) 10/16
(E) 11/16

We can do it by trying all the combinations of a,b, and C. However is there any smarter approach?
If we convert the denominator we get
a/2+b/8+c^4
a and b cannot be zero because the denominator will not be 16 if any one of these is zero.
so a and b are equal to 1.C can be either 1 or 0(C=0,None fo the options holds true for C=1)
Hence C
Last edited by aloneontheedge on Tue Jul 20, 2010 12:57 pm, edited 1 time in total.
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by JeetGulia » Tue Jul 20, 2010 12:49 pm
aloneontheedge wrote:
JeetGulia wrote:If x=a/2+b/2^3+c/2^4, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT

(A) 1/16
(B) 3/16
(C) 5/16
(D) 10/16
(E) 11/16

We can do it by trying all the combinations of a,b, and C. However is there any smarter approach?
If we convert the denominator we get
a/2+b/8+c^4
a and b cannot be zero because the denominator will not be 16 if any one of these is zero.
so a and b are equal to 1.C can be either 1 or 0
Hence D
Nope it is C , you can get 5/8. If a=1, b=1 and c=0
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by aloneontheedge » Tue Jul 20, 2010 12:55 pm
JeetGulia wrote:
aloneontheedge wrote:
JeetGulia wrote:If x=a/2+b/2^3+c/2^4, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT

(A) 1/16
(B) 3/16
(C) 5/16
(D) 10/16
(E) 11/16

We can do it by trying all the combinations of a,b, and C. However is there any smarter approach?
If we convert the denominator we get
a/2+b/8+c^4
a and b cannot be zero because the denominator will not be 16 if any one of these is zero.
so a and b are equal to 1.C can be either 1 or 0
Hence D
Nope it is C , you can get 5/8. If a=1, b=1 and c=0
Thanks for the correction.Will edit it. Hope the concept was clear
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by Stuart@KaplanGMAT » Tue Jul 20, 2010 1:06 pm
JeetGulia wrote:If x=a/2+b/2^3+c/2^4, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT

(A) 1/16
(B) 3/16
(C) 5/16
(D) 10/16
(E) 11/16

We can do it by trying all the combinations of a,b, and C. However is there any smarter approach?
Hi,

I'd start by writing out the values of the 3 terms when a, b and c equal 1.

If a=1, a/2 = 1/2
If b=1, b/2^3 = 1/8
If c=1, c/2^4 = 1/16

Next I'd convert to a common denominator of 16 (since that's the denominator in every answer choice):

if a=1, a/2 = 8/16
if b=1, b/2^3 = 2/16
if c=1, c/2^4 = 1/16

So, each term can equal 0 or the values above.

Finally, I'd evaluate each choice, starting with (E) and working my way up (Kaplan research has shown that, when problem solving questions include the phrase "which of the following" or "each of the following except", the answer is D or E more than 40% of the time; i.e. the answers are biased toward the bottom of the list).

(E) 11/16... 8 + 2+ 1 = 11... POSSIBLE
(D) 10/16... 8 + 2 + 0 = 10... POSSIBLE
(C) 5/16... no way to get a sum of 5... IMPOSSIBLE, therefore CORRECT!

No need to check (B) or (A).
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by JeetGulia » Tue Jul 20, 2010 1:10 pm
Stuart Kovinsky wrote:
JeetGulia wrote:If x=a/2+b/2^3+c/2^4, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT

(A) 1/16
(B) 3/16
(C) 5/16
(D) 10/16
(E) 11/16

We can do it by trying all the combinations of a,b, and C. However is there any smarter approach?
Hi,

I'd start by writing out the values of the 3 terms when a, b and c equal 1.

If a=1, a/2 = 1/2
If b=1, b/2^3 = 1/8
If c=1, c/2^4 = 1/16

Next I'd convert to a common denominator of 16 (since that's the denominator in every answer choice):

if a=1, a/2 = 8/16
if b=1, b/2^3 = 2/16
if c=1, c/2^4 = 1/16

So, each term can equal 0 or the values above.

Finally, I'd evaluate each choice, starting with (E) and working my way up (Kaplan research has shown that, when problem solving questions include the phrase "which of the following" or "each of the following except", the answer is D or E more than 40% of the time; i.e. the answers are biased toward the bottom of the list).

(E) 11/16... 8 + 2+ 1 = 11... POSSIBLE
(D) 10/16... 8 + 2 + 0 = 10... POSSIBLE
(C) 5/16... no way to get a sum of 5... IMPOSSIBLE, therefore CORRECT!

No need to check (B) or (A).
.....You are the dude!!
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