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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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15 Sep 2018, 21:23
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35% (medium)

Question Stats:

73% (01:23) correct 28% (01:58) wrong based on 52 sessions

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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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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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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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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

$$a/b=-1$$

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