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Power saving of a bulb is directly proportional to square root

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Shrivathsan
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Power saving of a bulb is directly proportional to square root [#permalink] New post   04 Sep 2016, 10:41
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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%
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E
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abhimahna
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Re: Power saving of a bulb is directly proportional to square root [#permalink] New post   04 Sep 2016, 11:11
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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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stne
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Power saving of a bulb is directly proportional to square root [#permalink] New post   13 Nov 2017, 10:53
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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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Power saving of a bulb is directly proportional to square root [#permalink] New post   13 Nov 2017, 11:03
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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?
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Re: Power saving of a bulb is directly proportional to square root [#permalink] New post   13 Nov 2017, 11:15
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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Udniez
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Re: Power saving of a bulb is directly proportional to square root [#permalink] New post   20 Feb 2024, 04:29
You can also multiply by 100 both powers, now you have p1=100*sqrt0.8 and P2=100*sqrt 0.2, P1 is 89 and something while P2 is 45 and something, so the answer is 100%
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Re: Power saving of a bulb is directly proportional to square root [#permalink] New post   27 Mar 2024, 00:19
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you can simply put it in the following way:

square root of 0.8 --> square root of (2 * 2 * 0.2)

--> 2 square root of 0.2

compared to the second bulb, it's x 2

hence 100% greater
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Re: Power saving of a bulb is directly proportional to square root [#permalink]
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