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# If X = a/2^2 + b/2^3 + c/2^5, does X = 5/16 ?

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If X = a/2^2 + b/2^3 + c/2^5, does X = 5/16 ?  [#permalink]

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18 Oct 2019, 03:14
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Difficulty:

85% (hard)

Question Stats:

31% (02:14) correct 69% (01:57) wrong based on 54 sessions

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If $$X = \frac{a}{2^2} + \frac{b}{2^3} + \frac{c}{2^5}$$, does X = 5/16 ?

(1) a, b, and c are each either 0 or 1.

(2) The greatest possible value of X is 13/32.

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If X = a/2^2 + b/2^3 + c/2^5, does X = 5/16 ?  [#permalink]

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Updated on: 20 Oct 2019, 06:27
Bunuel wrote:
If $$X = \frac{a}{2^2} + \frac{b}{2^3} + \frac{c}{2^5}$$, does X = 5/16 ?

(1) a, b, and c are each either 0 or 1.

(2) The greatest possible value of X is 13/32.

Asked: If $$X = \frac{a}{2^2} + \frac{b}{2^3} + \frac{c}{2^5}$$, does X = 5/16 ?

(1) a, b, and c are each either 0 or 1.
X = (8a + 4b + c)/32 = 10/32
It is not possible to arrive at (8a + 4b + c) = 10 if a, b & c are either 0 or 1.
SUFFICIENT

(2) The greatest possible value of X is 13/32.
X = a/4 + b/8 + c/32 = (8a + 4b + c)/32
max(8a+4b+c) = 13
Since values of 8a+4b+c is unknown
NOT SUFFICIENT

IMO A
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Originally posted by Kinshook on 18 Oct 2019, 04:26.
Last edited by Kinshook on 20 Oct 2019, 06:27, edited 2 times in total.
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Re: If X = a/2^2 + b/2^3 + c/2^5, does X = 5/16 ?  [#permalink]

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18 Oct 2019, 19:35
Bunuel wrote:
If $$X = \frac{a}{2^2} + \frac{b}{2^3} + \frac{c}{2^5}$$, does X = 5/16 ?

(1) a, b, and c are each either 0 or 1.

(2) The greatest possible value of X is 13/32.

Is x = 5/16 or 10/32

$$X = \frac{a}{2^2} + \frac{b}{2^3} + \frac{c}{2^5}$$
—> $$X = \frac{(8a + 4b + c)}{32}$$

(1) a, b, and c are each either 0 or 1.

When a, b, c takes either 0 or 1, we can see the numerator can never take the value 10

Possible values of
(a, b, c) = (0, 0, 0) —> X = 0/32
(a, b, c) = (0, 0, 1) —> X = 1/32
(a, b, c) = (0, 1, 0) —> X = 4/32
(a, b, c) = (1, 0, 0) —> X = 8/32
(a, b, c) = (0, 1, 1) —> X = 5/32
(a, b, c) = (1, 1, 0) —> X = 12/32
(a, b, c) = (1, 0, 1) —> X = 9/32
(a, b, c) = (1, 1, 1) —> X = 13/32

—> A definite no. X can never be 10/32
—> Sufficient

(2) The greatest possible value of X is 13/32
—> 8a + 4b + c = 13

We can adjust values of a, b & c to get the greatest value as 13 and one of the values as 10

Case 1: As per statement (1) maximum value is 13/32 and X is never equal to 10/32

Case 2: Let the Maximum values of a, b and c can be 1
—> Maximum value of X = (8+4+1)/32 = 13/32
We can also assume one possible set of values, a = 1, b = 1/2, c = 0
—> X = (8*1 + 4*1/2 + 0)/32 = 10/32

So, we get 2 outcomes —> Insufficient

IMO Option A

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Re: If X = a/2^2 + b/2^3 + c/2^5, does X = 5/16 ?  [#permalink]

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18 Oct 2019, 20:04
Bunuel wrote:
If $$X = \frac{a}{2^2} + \frac{b}{2^3} + \frac{c}{2^5}$$, does X = 5/16 ?

(1) a, b, and c are each either 0 or 1.

(2) The greatest possible value of X is 13/32.

$$X = \frac{a}{2^2} + \frac{b}{2^3} + \frac{c}{2^5} = \frac{a}{4} + \frac{b}{8} + \frac{c}{32}=\frac{8a+4b+c}{32}$$

So we have to find if $$\frac{8a+4b+c}{32}=\frac{5}{16}=\frac{10}{32}$$ or is $$8a+4b+c=10$$

(1) a, b, and c are each either 0 or 1.
Is $$8a+4b+c=10$$? c has to be even for that as all other terms 8a, 4b and 10 are even, so ONLY possible value is 0.
So, is 8a+4b=10... if you fill a and b as 0 or 1, the values of 8a+4b you will get is 0, 4, 8, 12.
Suff

(2) The greatest possible value of X is 13/32.
So, $$8a+4b+c=13$$..
When a=b=c=1, $$8a+4b+c=13$$
When a=0, b=3, and a=1, $$8a+4b+c=10$$
But, we can also have a=b=1, c=0, then 12/32, so there can be various possibilities from 0/32 to 13/32
Insuff

A
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Re: If X = a/2^2 + b/2^3 + c/2^5, does X = 5/16 ?   [#permalink] 18 Oct 2019, 20:04
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