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12 Nov 2014, 10:08
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
47% (02:32) correct
53%
(02:35)
wrong
based on 148
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Tough and Tricky questions: Algebra.
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) \(\frac{a}{4} + \frac{b}{16} + \frac{c}{64} + \frac{d}{256} = \frac{11}{64}\)
Kudos for a correct solution.
13 Nov 2014, 09:16
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Bunuel
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.
Answer: B.
General Discussion
12 Nov 2014, 20:28
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Bunuel
1) a/2+b/4+c/8+d/16=1/8
=> 4a+2b+c+d/2=1
so the solution of this equation can be-
a=b=d=0 and c=1 Sum=1
or, a=1, b=c=-1 and d=0 Sum=-1
Not Sufficient
2) Solving similiarly, 16a+4b+c+d/4=11
Only one solution is possible
a=1, b=c=-1,and d=0
So, B is the answer.
Correct me if I am wrong.
