NBME 4- SARIM n math experts, plz help - iqbalian01

[ArchivalUser]

hi (freaking #s .i hate'm)

10 raise to the power of 9. just put nine "0" in front of "1"

1,000,000,000 ( # of bacterias)

every time we knock off a "0" we are left with the 10% (90% bacterias killed)= takes 10 minutes

in 80 minutes we knock off eight"0" and we are left with 10 bacterias.

so to knock off nine "0" it will take us 90 minutes and we'll be left with 1 bacteria and since 90 is not the choice and 100 minutes is the shortest amount of time listed to bring the # of bacterias below 10, that would be the answer.

good luck

[ArchivalUser]

thanks Sarim! of course, you make numbers 'fun', ;-)

[ArchivalUser]

thanx a lot iqbalian01, redstorm, hellfireomega, sarim, and all other friends

i think i got your point...
they are not asking exact time needed to reduce number of bacteria... they are interested with out basic knowledge of this concept... concept is important here not numbers....
and i agree calculations in this question should be easy... b/c we would have only about 72 seconds per question in real exam...

[ArchivalUser]

I suggest you to use https://www.easyunitconverter.com/ It helps me a lot while solving math problem.

[ArchivalUser]

Hello,
Yeah, your right. Most of the math I learned, I learned more or less on my own from books. To be sure, I was a math major an "advanced" one, perhaps. I took a bunch of math classes by using the cool eSingapore Math program.

[ArchivalUser]

all of the above answers are wrong. a constant fraction of bacteria are killed every 10 min. bit like 1st order reaction. the answer is 10m. no matter what the initial bacteria count , 90% will be killed in 10m.

[ArchivalUser]

to add to the above...

90% of 10^9 = 9 * 10^8 you are now left with 10^9 “ (9 * 10^8) = 10^8 (10min)
90% of 10^8 = 9 * 10^7 you are now left with 10^8 “ (9 * 10^7) = 10^7 (20min)
90% of 10^7 = 9 * 10^6 now left with 10^7 “ (9 * 10^6) = 10^6 (30min)

10^5 (40min)
10^4 (50min)
10^3 (60min)
10^2 (70min)
10^1 (80min)
10^0 (90min)

i wonder if there is an easier way?