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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 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: Technology, General Management
GMAT 1: 650 Q48 V31
GMAT 2: 770 Q50 V47
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Followers: 17

Kudos [?]: 242 [0], given: 142

Re: The median of the series sin[m]1^{ [#permalink]  31 Jan 2013, 06:42
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, 06:42

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