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22 Dec 2025, 11:24
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Given , bacteria population at 10AM is 8000. At 7 PM , the population is 64000. Let r be the percentage rate growth per hour.
Then , (1+r/100)^9 *8000 = 64000
(1+r/100)^3 = 2
Now we need to find the time in which the population becomes 256000
So , (1+r/100)^t *8000 = 256000
(1+r/100)^t = 32 = 2^5
Now earlier we know that (1+r/100)^3 = 2
So , (1+r/100)^15 =2^5 , which denotes that t=15 hours.
So time will be 1 AM of the next day .
E is the answer.
Then , (1+r/100)^9 *8000 = 64000
(1+r/100)^3 = 2
Now we need to find the time in which the population becomes 256000
So , (1+r/100)^t *8000 = 256000
(1+r/100)^t = 32 = 2^5
Now earlier we know that (1+r/100)^3 = 2
So , (1+r/100)^15 =2^5 , which denotes that t=15 hours.
So time will be 1 AM of the next day .
E is the answer.
Bunuel
22 Dec 2025, 11:45
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consider 8000 as the basic variable
then at 7:00 pm pop. will be 64000 it will be 8 * base value i.e 2 ^3 in 9 hours time diff b/w 10:00 am and 7:00 pm
it means doubling in every three hours 9;/3 = 3hours
and 256000 is basically 32* base value :
2 ^ 5 and 2 ^ 3 has 2^ 2 difference so total hours would be 2 * 3 = 6 hours from 7: 00 pm that is 1:00Am next day
then at 7:00 pm pop. will be 64000 it will be 8 * base value i.e 2 ^3 in 9 hours time diff b/w 10:00 am and 7:00 pm
it means doubling in every three hours 9;/3 = 3hours
and 256000 is basically 32* base value :
2 ^ 5 and 2 ^ 3 has 2^ 2 difference so total hours would be 2 * 3 = 6 hours from 7: 00 pm that is 1:00Am next day
22 Dec 2025, 11:45
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given hourly fixed percentage rate-> let r be the rate -> 8000(1 + r/100)^9 = 64000 => (1 + r/100) = 8^1/9
time (x) for reaching 256000 -> 64000(1 + r/100)^x = 256000 => (1 + r/100)^x = 4
2^(3x/9)=2^2 => x = 6 hours
hence 7 PM + 6 hours -> 1 AM next day
time (x) for reaching 256000 -> 64000(1 + r/100)^x = 256000 => (1 + r/100)^x = 4
2^(3x/9)=2^2 => x = 6 hours
hence 7 PM + 6 hours -> 1 AM next day
