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For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?

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For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?  [#permalink]

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New post 06 Oct 2017, 04:30
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Project DS Butler: Day 28: Data Sufficiency (DS56)


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For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?

(1) a/b = 3/5

(2) a and b have no common factors greater than 1

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Re: For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?  [#permalink]

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New post 06 Oct 2017, 05:43
Bunuel wrote:
For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?

(1) a/b = 3/5

(2) a and b have no common factors greater than 1


Given \(\frac{a}{(b-c)}=1 => b-c=a\)

to find \(\frac{(b-c)}{a}=\frac{a}{b}\)

Statement 1: Directly provides the answer. Hence Sufficient

Statement 2: when you divide \(a\) by \(b\) common factors get cancelled out but the non common factors / primes remain whose information is not provided. So \(\frac{a}{b}\) cannot be calculated. Insufficient

Option A
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Re: For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?  [#permalink]

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New post 06 Oct 2017, 06:07
a,b and c are integers
a/(b-c) = 1
means a= (b-c)
(b-c)/b = ?

stmt 1
a/b = 3/5
substitute a= (b-c)
(b-c)/b = 3/5
Hence Sufficient.

stmt 2
a and b have no common factors greater than 1
means a and b are prime, that does not give us any idea about a/b ratio.
Hence Insufficient.

Answer is Option A.

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Re: For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?  [#permalink]

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New post 06 Oct 2017, 07:41
Bunuel wrote:
For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?

(1) a/b = 3/5

(2) a and b have no common factors greater than 1


A = B-C

What is (b-c)/b , i.e. what is a/b??

from 1

suff

from 2

a, b are co prime ...no mention of values or ratio ( depends on the unknown prime factors of each) .. insuff

A
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Re: For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?  [#permalink]

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New post 06 Oct 2017, 15:20
Bunuel wrote:
For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?

(1) a/b = 3/5

(2) a and b have no common factors greater than 1


The answer should be A as follows.

Given \(\frac{a}{(b-c)}=1\)

1. Condition A gives \frac{a}{b}=\(3/5\) == \(a=\frac{3}{5}*b\)
Putting this value of "a" in the given condition we get \(\frac{a}{(b-c)}=1\)==>\(((3/5)*b)/(b-c)=1\)==>\(\frac{b}{(b-c)}\)=\(1*\frac{5}{3}\)==>\(\frac{b}{(b-c)}\)=\(\frac{5}{3}\)==>\(\frac{(b-c)}{b}\)=\(\frac{3}{5}\)
SUFFICIENT

2. a and b have no common factors greater than 1 ==> This neither provide any clue about the value of a and b nor provide any relation between a and b that can be substituted in the given equation to solve.
INSUFFICIENT

Hence answer is A
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Re: For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?  [#permalink]

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New post 29 Oct 2018, 06:25
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Bunuel wrote:
For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b?

(1) a/b = 3/5

(2) a and b have no common factors greater than 1


Given: a/(b - c) = 1

Target question: What is the value of (b-c)/b ?
This is a good candidate for rephrasing the target question.
Aside: The video below has tips on rephrasing the target question
If a/(b - c) = 1, then we know that a = b - c
The target question asks "What is the value of (b-c)/b ?"
Since a = b - c, we can replace b - c with a to get:
REPHRASED target question: What is the value of a/b ?

Statement 1: a/b = 3/5
Perfect!
The answer to the REPHRASED target question is a/b = 3/5
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: a and b have no common factors greater than 1
There are several values of a and b that satisfy statement 2. Here are two:
Case a: a = 1 and b = 2. In this case, the answer to the REPHRASED target question is a/b = 1/2
Case b: a = 2 and b = 3. In this case, the answer to the REPHRASED target question is a/b = 2/3
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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Re: For integers a, b, and c, a/(b-c) = 1. What is the value of (b-c)/b? &nbs [#permalink] 29 Oct 2018, 06:25
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