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# S95-13

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Math Expert
Joined: 02 Sep 2009
Posts: 58370

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16 Sep 2014, 01:49
00:00

Difficulty:

95% (hard)

Question Stats:

33% (01:49) correct 67% (02:02) wrong based on 82 sessions

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Each of the numbers $$a$$, $$b$$, $$c$$, $$d$$ is equal to -1, 0, or 1. What is the value of $$a + b + c + d$$?

(1) $$\frac{a}{2} + \frac{b}{4} + \frac{c}{8} + \frac{d}{16} = \frac{1}{8}$$

(2) $$\frac{a}{4} + \frac{b}{16} + \frac{c}{64} + \frac{d}{256} = \frac{11}{64}$$

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16 Sep 2014, 01:49
Official Solution:

We must determine the value of $$a + b + c + d$$. Since each of these numbers can be -1, 0 or 1, it is necessary to determine which value each letter has.

Statement 1 says that $$\frac{a}{2} + \frac{b}{4} + \frac{c}{8} + \frac{d}{16} = \frac{1}{8}$$. Note that all of the denominators in this equation are factors of 16, which makes finding a common denominator straightforward: $$\frac{8a}{16} + \frac{4b}{16} + \frac{2c}{16} + \frac{d}{16} = \frac{2}{16}$$. Multiplying the whole equation by 16 gives us $$8a + 4b + 2c + d = 2$$. Note that each of the first three terms on the left hand side is even, since an even number (e.g. 8, 4, 2) multiplied by an odd OR even number gives an even result. The sum of three even numbers is also even. Furthermore, the right hand side, 2, is even. The only way for the equation to be true, then, is for $$d$$ to be 0; if $$d$$ were -1 or 1, the left hand side would be odd, since the sum of an odd and an even number is odd.

Since $$d = 0$$, the equation becomes $$8a + 4b + 2c = 2$$. This equation is true if $$a = 1$$, $$b = -1$$, and $$c = -1$$. However, it is also true if $$a = 0$$, $$b = 0$$, and $$c = 1$$. Since these sets of values give different results for the sum $$a + b + c + d$$, we do not have enough information to answer the question. Statement 1 is NOT sufficient. Eliminate answer choices A and D. The correct answer choice must be B, C, or E.

Statement 2 says that $$\frac{a}{4} + \frac{b}{16} + \frac{c}{64} + \frac{d}{256} = \frac{11}{64}$$. Note that all denominators are factors of 256. We can rewrite the equation: $$\frac{64a}{256} + \frac{16b}{256} + \frac{4c}{256} + \frac{d}{256} = \frac{44}{256}$$. Multiplying both sides by 256 gives us $$64a + 16b + 4c + d = 44$$.

We can see that $$a$$ must equal 1; $$a$$ is not 0 because there is no way for $$16b + 4c + d$$ to add up to 44, and $$a$$ is not -1 because there is no way for $$-64 +16b + 4c + d$$ to add up to 44. By the same reasoning we used above, we know that $$d = 0$$. Substitute to get: $$64 + 16b + 4c = 44$$. Simplify: $$16b + 4c = -20$$. Thus, $$b = -1$$ and $$c = -1$$.

We have solved for the value of each variable, so we can find the only possible value for the sum $$a + b + c + d$$. Statement 2 is sufficient to answer the question.

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18 Jun 2016, 07:37
It is a 700 level question, isn't it ?
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19 Jun 2016, 10:23
Alex75PAris wrote:
It is a 700 level question, isn't it ?

Yes, check the stats here: each-of-the-numbers-a-b-c-d-is-equal-to-1-0-or-1-what-is-the-188466.html
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24 Dec 2016, 12:54
I am not understanding how 64a+8b+4c+d=44.When a=1,and b & c=-1, I am getting 52. Please explain.
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25 Dec 2016, 02:09
pdub18 wrote:
I am not understanding how 64a+8b+4c+d=44.When a=1,and b & c=-1, I am getting 52. Please explain.

It's 64+16b+4c=44.
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01 Aug 2017, 09:44
Bunuel wrote:
Alex75PAris wrote:
It is a 700 level question, isn't it ?

Yes, check the stats here: http://gmatclub.com/forum/each-of-the-n ... 88466.html

Its a great DS question; how much time shall I take to solve this kind of question?
Can you post some more similar questions in which value plugging is required to reach the final answer.....
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01 Aug 2017, 09:47
buan15 wrote:
Bunuel wrote:
Alex75PAris wrote:
It is a 700 level question, isn't it ?

Yes, check the stats here: http://gmatclub.com/forum/each-of-the-n ... 88466.html

Its a great DS question; how much time shall I take to solve this kind of question?
Can you post some more similar questions in which value plugging is required to reach the final answer.....

The average time for a correct answer is 2:46 minutes. So, I'd say anything less than that is good.
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10 Aug 2017, 04:49
St 1:
Making denominator as 8,
Equation becomes
4a+2b+c+0.5d=1
Now, LHS=RHS for 2 sets of values
a=b=d=0, c=+1and
a=+1, b=c=-1, d=0
Hence, Insufficient.

St 2:
Making denominator as 64,
Equation becomes
16a+4b+c+0.25d=11
LHS=RHS for only one set of value
a=+1, b=c=-1, d=0
Hence, sufficient

Ans B
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29 Mar 2018, 14:33
Bunuel wrote:
Each of the numbers $$a$$, $$b$$, $$c$$, $$d$$ is equal to -1, 0, or 1. What is the value of $$a + b + c + d$$?

(1) $$\frac{a}{2} + \frac{b}{4} + \frac{c}{8} + \frac{d}{16} = \frac{1}{8}$$

(2) $$\frac{a}{4} + \frac{b}{16} + \frac{c}{64} + \frac{d}{256} = \frac{11}{64}$$

MathRevolution how do you use a variable approach on this?
S95-13   [#permalink] 29 Mar 2018, 14:33
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# S95-13

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