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If a, b and c have the values shown, which of the following shows

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If a, b and c have the values shown, which of the following shows  [#permalink]

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New post 01 Dec 2019, 06:04
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a = \(\sqrt{2} + \sqrt{6}\)
b= \(\sqrt{3} + \sqrt{5}\)
c= 4


If a, b and c have the values shown, which of the following shows their order from least to greatest?

A) a, b, c

B) b, c, a

C) b, a, c

D) c, a, b

E) c, b, a

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Re: If a, b and c have the values shown, which of the following shows  [#permalink]

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New post 01 Dec 2019, 10:06
Let us take the square of a , b , and c and compare the squares
\(a^2 = (\sqrt{2} + \sqrt{6})^2)= 2+6+2*(\sqrt{12})\)
\(b^2 = (\sqrt{3} + \sqrt{5})^2) = 3+5+2*(\sqrt{15})\)
\(c^2\) = 16

Now it is clear that \( 8+2*(\sqrt{12})<8+2*(\sqrt{15})<16\)
Hence \(a^2<b^2<c^2\)
or, \(a<b<c\)

Answer = A

CAMANISHPARMAR wrote:
a = \(\sqrt{2} + \sqrt{6}\)
b= \(\sqrt{3} + \sqrt{5}\)
c= 4


If a, b and c have the values shown, which of the following shows their order from least to greatest?

A) a, b, c

B) b, c, a

C) b, a, c

D) c, a, b

E) c, b, a

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Re: If a, b and c have the values shown, which of the following shows  [#permalink]

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New post 01 Dec 2019, 10:22
a=\(\sqrt{2} + \sqrt{6}\)

\(a^2\)= \(2+6+2\sqrt{6*2}\)= \(8+2\sqrt{12}\)

b= \(\sqrt{3} + \sqrt{5}\)

\(b^2\)= \(3+5+2\sqrt{5*3}\)= \(8+2\sqrt{15}\)

c= 2+2= \(\sqrt{4} + \sqrt{4}\)

\(c^2\)= \(4+4+2\sqrt{4*4}\)= \(8+2\sqrt{16}\)

\(\sqrt{12}<\sqrt{15}<\sqrt{16}\)

a<b<c

CAMANISHPARMAR wrote:
a = \(\sqrt{2} + \sqrt{6}\)
b= \(\sqrt{3} + \sqrt{5}\)
c= 4


If a, b and c have the values shown, which of the following shows their order from least to greatest?

A) a, b, c

B) b, c, a

C) b, a, c

D) c, a, b

E) c, b, a
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Re: If a, b and c have the values shown, which of the following shows   [#permalink] 01 Dec 2019, 10:22
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If a, b and c have the values shown, which of the following shows

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