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Each of the numbers a, b, c, d is equal to 1, 0, or 1. What is the [#permalink]
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12 Nov 2014, 09:08
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Re: Each of the numbers a, b, c, d is equal to 1, 0, or 1. What is the [#permalink]
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12 Nov 2014, 19:28
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Bunuel wrote: 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.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.



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Re: Each of the numbers a, b, c, d is equal to 1, 0, or 1. What is the [#permalink]
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13 Nov 2014, 08:16
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Bunuel wrote: 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. 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.
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Each of the numbers a, b, c, d is equal to 1, 0, or 1. What is the [#permalink]
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13 Nov 2014, 08:58
Bunuel wrote: Bunuel wrote: Tough and Tricky questions: Algebra. 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. How did you conclude A=1 so quickly? Why not 2 or 3? edit: haha oops, the answer to my question is given...
Last edited by DangerPenguin on 13 Nov 2014, 09:11, edited 1 time in total.



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Re: Each of the numbers a, b, c, d is equal to 1, 0, or 1. What is the [#permalink]
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13 Nov 2014, 09:00
DangerPenguin wrote: Bunuel wrote: Bunuel wrote: Tough and Tricky questions: Algebra. 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. How did you conclude A=1 so quickly? Why not 2 or 3? We are given that each of the numbers a, b, c, d is equal to 1, 0, or 1.
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