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Two different positive numbers a and b each differ from their reciproc

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Two different positive numbers a and b each differ from their reciproc  [#permalink]

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New post 18 Mar 2019, 04:00
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

33% (02:11) correct 67% (02:19) wrong based on 33 sessions

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Re: Two different positive numbers a and b each differ from their reciproc  [#permalink]

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New post 18 Mar 2019, 04:07
Bunuel wrote:
Two different positive numbers a and b each differ from their reciprocals by 1. What is a + b ?


(A) 1
(B) 2
(C) \(\sqrt{5}\)
(D) \(\sqrt{6}\)
(E) 3


I am not sure whether such quadratic properties are tested on gmat
this question will yield an eqn x^2+x-1 and x^2+x+1
we need to use formula b^2+/-√4ac/2a
IMO C would be correct...
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Re: Two different positive numbers a and b each differ from their reciproc  [#permalink]

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New post 18 Mar 2019, 07:45
1
Bunuel wrote:
Two different positive numbers a and b each differ from their reciprocals by 1. What is a + b ?


(A) 1
(B) 2
(C) \(\sqrt{5}\)
(D) \(\sqrt{6}\)
(E) 3


\(\frac{1}{a} - a = 1\) and \(\frac{1}{b} - b = -1\) (since both a and b are different)

on solving both the equations we get
\(a^2 + a - 1 = 0\) and \(b^2 - b - 1 = 0\)

Since a and b are positive
so \(a = \frac{-1}{2}+\frac {\sqrt{5}}{2}\)

and \(b = \frac{1}{2}+\frac {\sqrt{5}}{2}\)

\(a + b = \frac{-1}{2}+\frac {\sqrt{5}}{2} + \frac{1}{2}+\frac {\sqrt{5}}{2} = \sqrt{5}\)

hence the answer is C.
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Re: Two different positive numbers a and b each differ from their reciproc  [#permalink]

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New post 18 Mar 2019, 22:40
Could anyone tell me why 1/b -b is taken as -1 since both the numbers differ from their reciprocals by 1 only

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Re: Two different positive numbers a and b each differ from their reciproc  [#permalink]

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New post 19 Mar 2019, 07:35
vishumangal wrote:
Could anyone tell me why 1/b -b is taken as -1 since both the numbers differ from their reciprocals by 1 only

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The question says both the numbers differ from their reciprocals which means \(\frac{1}{a} - a = 1\) and \(\frac{1}{b} - b = 1\) but since both the numbers are different that's why one of the number will differ from its reciprocal by \(b - \frac{1}{b} = 1\) or we can rewrite as \(\frac{1}{b} - b = -1\).
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Re: Two different positive numbers a and b each differ from their reciproc  [#permalink]

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New post 19 Mar 2019, 19:08
Bunuel wrote:
Two different positive numbers a and b each differ from their reciprocals by 1. What is a + b ?


(A) 1
(B) 2
(C) \(\sqrt{5}\)
(D) \(\sqrt{6}\)
(E) 3


It must be true that one number is more than its reciprocal, and the other is less than its reciprocal. We can let a be the one that is more than its reciprocal and hence b is the one that is less than its reciprocal. So we have:

a - 1/a = 1 and 1/b - b = 1

Subtracting these two equations, we have:

a - 1/a - 1/b + b = 0

a + b = 1/a + 1/b

a + b = (b + a)/(ab)

ab(a + b) = a + b

Since both a and b are positive, a + b ≠ 0. So we can divide both sides of the equation by a + b and obtain:

ab = 1

Now, let’s multiply the original first equation by a. We have:

a^2 - 1 = a

Similarly, let’s multiply the original second equation by b. We have:

1 - b^2 = b

Adding the two new equations, we have:

a^2 - b^2 = a + b

(a + b)(a - b) = a + b

Again, dividing both sides of the equation by a + b, we have:

a - b = 1

Squaring the above equation, we have:

a^2 - 2ab + b^2 = 1

Since ab = 1, we have:

a^2 - 2(1) + b^2 = 1

a^2 + b^2 = 3

Now add 2ab = 2 to the above equation, we have:

a^2 + 2ab + b^2 = 5

(a + b)^2 = 5

Taking the square root of both sides, we have:

a + b = √5

Answer: C
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Re: Two different positive numbers a and b each differ from their reciproc  [#permalink]

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New post 20 Mar 2019, 04:07
Abhishek292kumar, hi can you please tell how did you come to the value of a as -1/2 + ((5)^(1/2))/2 ?
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Re: Two different positive numbers a and b each differ from their reciproc  [#permalink]

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New post 20 Mar 2019, 06:29
abhishek31 wrote:
hi can you please tell how did you come to the value of a as -1/2 + ((5)^(1/2))/2 ?

Hi,
By using the formula of quadratic equation:
\(a{x^2} + bx + c = 0\) then \(x =\frac { -b + \sqrt{{b^2}-4ac}}{2a}\) and \(x =\frac { -b - \sqrt{{b^2}-4ac}}{2a}\)
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Re: Two different positive numbers a and b each differ from their reciproc   [#permalink] 20 Mar 2019, 06:29
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