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# Power saving of a bulb is directly proportional to square root

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Manager
Joined: 15 Apr 2016
Posts: 71
Power saving of a bulb is directly proportional to square root  [#permalink]

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04 Sep 2016, 09:41
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Difficulty:

45% (medium)

Question Stats:

64% (02:00) correct 36% (02:14) wrong based on 85 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%

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

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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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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

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Status: Stepping into my 10 years long dream
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Power saving of a bulb is directly proportional to square root  [#permalink]

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13 Nov 2017, 10:03
1
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]

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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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- Stne

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

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20 Jan 2019, 01:26
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Re: Power saving of a bulb is directly proportional to square root   [#permalink] 20 Jan 2019, 01:26
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