# If 200 increases by 2% each day, in how many days will it reach 2000?

Relevance

Let n = # of days:

200(1.02)^n = 2000

1.02^n = 2000/200

1.02^n = 10

log(1.02^n) = log(10)

n*log(1.02) = 1

n = 1 / log(1.02)

n = 116.27

It will take 117 days.

Note: The people who say the answer is 450 days are wrong because they are not taking into account the fact that it's not 2% of 200 each day, rather it's 2% of 200 the first day, 2% of 204 the second, etc... This is basically a compound interest problem where the interest is compounded daily.

• if 200 increases by 2%, it will increase by .02(200) or 4 per day. At this rate, it will take 450 days for 200 to increse to 2000.

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• if n is the number of days

then 2000= 200(1+0.02)^n

10=(1.02)^n

log10=n(log 1.02)

n = 1/log(1.02)=116.27

after 117 days the money will be more than 2000

• use formula

2000 = 200 (1+2/100)^n where n= no. of days

or 10=(1+0.02)^n

or 1.02^n=10

or nlog(base 10) 1.02=log10=1

or n= 1/log1.02

find log 1.02 and proceed

you will get n= 117 approx

• Anonymous

in 10 dayz

• 200*(1.02)^x = 2000

(1.02)^x = 10

x=ln(10) / ln(1.02)