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Power saving of a bulb is directly proportional to square root

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Joined: 15 Apr 2016
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Power saving of a bulb is directly proportional to square root [#permalink]

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New post 04 Sep 2016, 09:41
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Difficulty:

  55% (hard)

Question Stats:

55% (01:12) correct 45% (01:57) wrong based on 55 sessions

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Power saving (in %) of a bulb is directly proportional to square root of its efficiency. By how much % is power saving of bulb-1(efficiency 0.8) greater than power saving of bulb-2 (efficiency 0.2)?

A. 60%
B. 70%
C. 80%
D. 90%
E. 100%
[Reveal] Spoiler: OA

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Kudos [?]: 22 [0], given: 66

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Re: Power saving of a bulb is directly proportional to square root [#permalink]

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New post 04 Sep 2016, 10:11
Shrivathsan wrote:
Power saving (in %) of a bulb is directly proportional to square root of its efficiency. By how much % is power saving of bulb-1(efficiency 0.8) greater than power saving of bulb-2 (efficiency 0.2)?

A. 60%
B. 70%
C. 80%
D. 90%
E. 100%


Power, \(P1 = k * sqrt(0.8)\)

Power, \(P2 = k * sqrt(0.2)\)

Difference = \(P1-P2 = k * sqrt(0.8) - k * sqrt(0.2)\)

% greater =\(k * sqrt(0.8) - k * sqrt(0.2)/k * sqrt(0.2) * 100\)

= 100%.

Hence, E
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Power saving of a bulb is directly proportional to square root [#permalink]

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New post 13 Nov 2017, 09:53
abhimahna wrote:
Shrivathsan wrote:
Power saving (in %) of a bulb is directly proportional to square root of its efficiency. By how much % is power saving of bulb-1(efficiency 0.8) greater than power saving of bulb-2 (efficiency 0.2)?

A. 60%
B. 70%
C. 80%
D. 90%
E. 100%


Power, \(P1 = k * sqrt(0.8)\)

Power, \(P2 = k * sqrt(0.2)\)

Difference = \(P1-P2 = k * sqrt(0.8) - k * sqrt(0.2)\)

% greater =\(k * sqrt(0.8) - k * sqrt(0.2)/k * sqrt(0.2) * 100\)

= 100%.

Hence, E



Can you please elaborate how to simply this.

Quote:
% greater =\(k * sqrt(0.8) - k * sqrt(0.2)/k * sqrt(0.2) * 100\)

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- Stne

Kudos [?]: 90 [0], given: 487

1 KUDOS received
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User avatar
D
Status: Aiming MBA
Joined: 18 Jul 2015
Posts: 2865

Kudos [?]: 972 [1], given: 71

Location: India
Concentration: Healthcare, Technology
GPA: 3.65
WE: Information Technology (Health Care)
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Power saving of a bulb is directly proportional to square root [#permalink]

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New post 13 Nov 2017, 10:03
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stne wrote:
Can you please elaborate how to simply this.

Quote:
% greater =\(k * sqrt(0.8) - k * sqrt(0.2)/k * sqrt(0.2) * 100\)


Here I go:

\(sqrt(0.8)\) = sqrt(8)/sqrt(10) = [2*sqrt(2)]/sqrt(10)

\(sqrt(0.2)\) = sqrt(2)/sqrt(10)

You know we can cancel 'k' from denominator and numerator.

Similarly, you have sqrt(10) both at numerator and denominator, so you can cancel that out.

Therefore, you are left with [2*sqrt(2) - sqrt(2) ] / sqrt(2) ] *100 = [sqrt(2) ] / sqrt(2)] *100 = 100%

Does that make sense?
_________________

How I improved from V21 to V40! ?

Kudos [?]: 972 [1], given: 71

Senior Manager
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Joined: 27 May 2012
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Re: Power saving of a bulb is directly proportional to square root [#permalink]

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New post 13 Nov 2017, 10:15
abhimahna wrote:
stne wrote:
Can you please elaborate how to simply this.

Quote:
% greater =\(k * sqrt(0.8) - k * sqrt(0.2)/k * sqrt(0.2) * 100\)


Here I go:

\(sqrt(0.8)\) = sqrt(8)/sqrt(10) = [2*sqrt(2)]/sqrt(10)

\(sqrt(0.2)\) = sqrt(2)/sqrt(10)

You know we can cancel 'k' from denominator and numerator.

Similarly, you have sqrt(10) both at numerator and denominator, so you can cancel that out.

Therefore, you are left with [2*sqrt(2) - sqrt(2) ] / sqrt(2) ] *100 = [sqrt(2) ] / sqrt(2)] *100 = 100%

Does that make sense?


Took a while to realize this but finally got it. Yes it makes perfect sense.
Thanks a ton +1
_________________

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Kudos [?]: 90 [0], given: 487

Re: Power saving of a bulb is directly proportional to square root   [#permalink] 13 Nov 2017, 10:15
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Power saving of a bulb is directly proportional to square root

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