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‘a’ and ‘b’ are two integers such that (4a – 17) (a + 1) = 0 and

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‘a’ and ‘b’ are two integers such that (4a – 17) (a + 1) = 0 and  [#permalink]

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New post 15 Sep 2018, 21:23
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‘a’ and ‘b’ are two integers such that \((4a – 17) (a + 1) = 0\) and \((4b – 1) (b – 1) = 0\). Which one among the following is a possible value of \(\frac{a}{b}\) ?


A) \(-17\)

B) \(-1\)

C) \(\frac{−1}{4}\)

D) \(\frac{17}{4}\)

E) \(17\)

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Re: ‘a’ and ‘b’ are two integers such that (4a – 17) (a + 1) = 0 and  [#permalink]

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New post 15 Sep 2018, 21:46
workout wrote:
‘a’ and ‘b’ are two integers such that \((4a – 17) (a + 1) = 0\) and \((4b – 1) (b – 1) = 0\). Which one among the following is a possible value of \(\frac{a}{b}\) ?


A) \(-17\)

B) \(-1\)

C) \(\frac{−1}{4}\)

D) \(\frac{17}{4}\)

E) \(17\)


As a & b are integers, a = -1, b = 1.
\(a/b = -1.\)
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Re: ‘a’ and ‘b’ are two integers such that (4a – 17) (a + 1) = 0 and  [#permalink]

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New post 15 Sep 2018, 21:56
(4a-17)(a+1)=0
So a can be 17/4 or -1 but a is an integer so -1 can be considered
Similarly, (4b-1)(b-1)=0
So b can be 1 or 1/4 but b is an integer so 1 can be considered
When a=-1 and b=1 we get 1/ab=-1
So B is the correct choice.
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Re: ‘a’ and ‘b’ are two integers such that (4a – 17) (a + 1) = 0 and  [#permalink]

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New post 16 Sep 2018, 13:47
\((4a-17)(a+1) = 0\) \((4b-1)(b-1)=0\)
\(a=17/4\) \(a=-1\) \(b=1/4\) \(b=1\)

Question say a & b are integers so we can remove the fractions

\(a=-1\) \(b=1\)

and so

\(a/b=-1\)

Answer:B
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Re: ‘a’ and ‘b’ are two integers such that (4a – 17) (a + 1) = 0 and   [#permalink] 16 Sep 2018, 13:47

‘a’ and ‘b’ are two integers such that (4a – 17) (a + 1) = 0 and

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