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# A person borrowed a certain sum at the rate of 20% per annum

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Joined: 01 Dec 2012
Posts: 39
Location: India
WE: Project Management (Retail)
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The median of the series sin[m]1^{ [#permalink]  31 Jan 2013, 06:27
Q1. The median of the series sin1^{\circ}, sin2^{\circ}, sin3^{\circ},...............,sin179^{\circ} is

a. 1
b. 1/\sqrt{2}
c. sin46^{\circ}
d. sin89^{\circ}
e. sin98^{\circ}

Q2. If the Standard deviation of 1,2,3,4,5,6,7,8,9,10,11 is M, then the standard deviation of 101,102,103,...................,111 is
a. M
b. 100+M
c. 100-M
d. M-100
e. 100M

1.
[Reveal] Spoiler:
b

2.
[Reveal] Spoiler:
a
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Re: The median of the series sin[m]1^{ [#permalink]  31 Jan 2013, 06:42
Q1. The median of the series sin1^{\circ}, sin2^{\circ}, sin3^{\circ},...............,sin179^{\circ} is

a. 1
b. 1/\sqrt{2}
c. sin46^{\circ}
d. sin89^{\circ}
e. sin98^{\circ}

Is this a GMAT question ?

Ans: B

sin function starts from sin0^{\circ} = 0 increases till sin90^{\circ} = 1 , then decreases till sin180^{\circ} = 0
and sinx^{\circ} = sin(180 - x)^{\circ}

So the mid value occurs at sin45^{\circ} or sin135^{\circ} = 1/\sqrt{2}

Q2. If the Standard deviation of 1,2,3,4,5,6,7,8,9,10,11 is M, then the standard deviation of 101,102,103,...................,111 is
a. M
b. 100+M
c. 100-M
d. M-100
e. 100M

Ans: A

Theory: The standard deviation remains unchanged if we add a constant to every term.
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Re: The median of the series sin[m]1^{   [#permalink] 31 Jan 2013, 06:42
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