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17 Nov 2018, 10:01
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C
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Difficulty:
95%
(hard)
Question Stats:
26% (02:14) correct
74%
(02:20)
wrong
based on 96
sessions
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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
a - (1/a) = 1 or (1/a) - a = 1, since the question says the numbers differ by 1.
which gives, a^2 - 1 = a or a^2 - a - 1 = 0 OR 1 - a^2 = a or a^2 + a - 1 = 0
Solutions to the 1st quadratic equation are: [1 + sqrt(5)]/2 and [1 - sqrt(5)]/2
Solutions to the 2nd quadratic equation are: [-1 + sqrt(5)]/2 and [-1 - sqrt(5)]/2
Question says that a and b are positive integers, so out of the above 4 solutions, the positive ones are [1 + sqrt(5)]/2 and [-1 + sqrt(5)]/2, which must be a and b.
Hence a + b = 0.5 + sqrt(5)/2 - 0.5 + sqrt(5)/2 = 2*sqrt(5)/2 = sqrt(5)
Option C!
which gives, a^2 - 1 = a or a^2 - a - 1 = 0 OR 1 - a^2 = a or a^2 + a - 1 = 0
Solutions to the 1st quadratic equation are: [1 + sqrt(5)]/2 and [1 - sqrt(5)]/2
Solutions to the 2nd quadratic equation are: [-1 + sqrt(5)]/2 and [-1 - sqrt(5)]/2
Question says that a and b are positive integers, so out of the above 4 solutions, the positive ones are [1 + sqrt(5)]/2 and [-1 + sqrt(5)]/2, which must be a and b.
Hence a + b = 0.5 + sqrt(5)/2 - 0.5 + sqrt(5)/2 = 2*sqrt(5)/2 = sqrt(5)
Option C!
17 Nov 2018, 10:22
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Each number a and b differs from its reciprocal by 1.
this implies,
a-(1/a)=1 ------ eq 1
b-(1/b)=1 -------eq 2
on solving eq 1
(a^2)-1=a
a=(1+\sqrt{5})/2 or (1+\sqrt{5})/2
similarly b=(1+\sqrt{5})/2 or (1-\sqrt{5})/2
when trying all 4 possibilities for a+b
(i.e) a+b=(1+\sqrt{5})/2+(1+\sqrt{5})/2 = 1+\sqrt{5}
or a+b= (1+\sqrt{5})/2+(1-\sqrt{5})/2=\sqrt{5}
or a+b=(1-\sqrt{5})/2+(1-\sqrt{5})/2=1-\sqrt{5}
or a+b=(1-\sqrt{5})/2+(1+\sqrt{5})/2=\sqrt{5}
The only option which satisfies one of the possibilities is C
IMO: OPTION C
NOTE: EVEN IF EQUATION ONE IS TAKEN AS (1/a)-a=1 AND EQUATION TWO AS (1/b)-b=1, the answer remains the same
this implies,
a-(1/a)=1 ------ eq 1
b-(1/b)=1 -------eq 2
on solving eq 1
(a^2)-1=a
a=(1+\sqrt{5})/2 or (1+\sqrt{5})/2
similarly b=(1+\sqrt{5})/2 or (1-\sqrt{5})/2
when trying all 4 possibilities for a+b
(i.e) a+b=(1+\sqrt{5})/2+(1+\sqrt{5})/2 = 1+\sqrt{5}
or a+b= (1+\sqrt{5})/2+(1-\sqrt{5})/2=\sqrt{5}
or a+b=(1-\sqrt{5})/2+(1-\sqrt{5})/2=1-\sqrt{5}
or a+b=(1-\sqrt{5})/2+(1+\sqrt{5})/2=\sqrt{5}
The only option which satisfies one of the possibilities is C
IMO: OPTION C
NOTE: EVEN IF EQUATION ONE IS TAKEN AS (1/a)-a=1 AND EQUATION TWO AS (1/b)-b=1, the answer remains the same
