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Computer Problem(Percent decrease)

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Computer Problem(Percent decrease)

by rkiran9 » Fri Jun 07, 2013 2:42 am
At the begining of the year, a certain computer problem requires 3.6 X 10^11 calculations to solve, And a computer used to solve it is capable of 1,000,000 calculations per second.At the end of the year, a new algorithm comes out which reduces the number of calculations by the factor of 10, and a new computer is buit capable of perfoming twice as many calculations per second.What percent decrease in computing time(in hrs) has been achived?

Options:60%,75%,85%,90%,95%

Am not getting this part " new algorithm comes out which reduces the number of calculations by the factor of 10" does it mean 3.6 X 10^11/10!?

Could u help me out in knowing the correct approach?[/u]
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by Atekihcan » Fri Jun 07, 2013 3:07 am
rkiran9 wrote:" new algorithm comes out which reduces the number of calculations by the factor of 10" does it mean 3.6 X 10^11/10!?
Factor is different from factorial.
Reduces the calculations by the factor of 10 means number of calculations is 1/10 of what was earlier, i.e. new number of calculations is (3.6 x 10^11)/10

Now, number of calculations has been reduced by a factor of 10 and computer speed has been increased by a factor of 2. So, time taken to complete computation has been reduced by a factor of 10*2 = 20

So, if previous time was 100 hours, new time is 100/20 = 5 hours.
So, percentage decrease = (100 - 5)/100 = 95/100 = 95%

Answer : E
Last edited by Atekihcan on Fri Jun 07, 2013 3:14 am, edited 1 time in total.
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by mkdureja » Fri Jun 07, 2013 3:11 am
rkiran9 wrote:
Am not getting this part " new algorithm comes out which reduces the number of calculations by the factor of 10" does it mean 3.6 X 10^11/10!?

Could u help me out in knowing the correct approach?[/u]
It doesn't mean 10!, it just means divided by 10, i.e. 3.6 X 10^11/10
1000 reduced by factor of 10 becomes 100. Understood?

Now, Computing Time, T = Calculations Required/Speed of computer(calculations/sec) = C/S
Given:
C is reduced by a factor of 10, i.e. new C, C' = C/10
Speed is doubled, S' = 2S

New computing time, T' = (C/10)/2S = C/S * 1/20 = T/20.

New computing time is 20th part of old computing time.
So, decrease in computing time in percent = 95% (5% is 20th part of 100%)
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