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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)
Followers: 1

Kudos [?]: 35 [0], given: 10

The median of the series sin[m]1^{ [#permalink]

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New post 31 Jan 2013, 07:27
Q1. The median of the series sin\(1^{\circ}\), sin\(2^{\circ}\), sin\(3^{\circ}\),...............,sin\(179^{\circ}\) is

a. 1
b. \(1/\sqrt{2}\)
c. sin\(46^{\circ}\)
d. sin\(89^{\circ}\)
e. sin\(98^{\circ}\)

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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Concentration: Finance, Technology
GMAT 1: 650 Q48 V31
GMAT 2: 770 Q50 V47
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Kudos [?]: 353 [0], given: 142

Re: The median of the series sin[m]1^{ [#permalink]

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New post 31 Jan 2013, 07:42
Q1. The median of the series sin\(1^{\circ}\), sin\(2^{\circ}\), sin\(3^{\circ}\),...............,sin\(179^{\circ}\) is

a. 1
b. \(1/\sqrt{2}\)
c. sin\(46^{\circ}\)
d. sin\(89^{\circ}\)
e. sin\(98^{\circ}\)

Is this a GMAT question ?

Ans: B

sin function starts from sin\(0^{\circ}\) = 0 increases till sin\(90^{\circ}\) = 1 , then decreases till sin\(180^{\circ}\) = 0
and sin\(x^{\circ}\) = sin\((180 - x)^{\circ}\)

So the mid value occurs at sin\(45^{\circ}\) or sin\(135^{\circ}\) = \(1/\sqrt{2}\)

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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― Voltaire


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Re: The median of the series sin[m]1^{   [#permalink] 31 Jan 2013, 07:42

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

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