The number of water lilies on a certain lake doubles every

Notice that we are told that "the number of water lilies on a certain lake doubles every two days", thus if initially there were two water lilies instead of one, we can consider that two days are over and therefore only 58 days are left.

Answer: D.

Similar question to practice: it-takes-30-days-to-fill-a-laboratory-dish-with-bacteria-140269.html

Hope it helps.
_________________

Starting from 1 Water Lilly it takes 60 days.
If there are already two present, it can be taken as the first day is over.
It will take 59 more days.

I understand the logic, but am not able to solve it algebraically.

since the series is in the geometric progression with the common ration (r) = 2, initial condition can be rewritten as:

a(n) = 1.2^60-1 {a(n = a.r^n-1)} === which gives us total number of lillies in the pool ==>2^59.....no this is equated when the the pool starts with 2 lillies...==> 2^59 = 2.2^n-1 ===>n=59..

Where am I going wrong?

We are told that "the number of water lilies on a certain lake doubles every TWO days".

If there are two lilies, then in order to cover the lake they would need to double one time less than in case with 1 lily. Since lilies double every two days, then 60-2=58 days are needed.

_________________

Bunuel,
would you please illustrate this question using the population formula rule used in the MGMAT.

Thank you

Sorry not familiar with that one. The following threads might help:
a-certain-bacteria-colony-doubles-in-size-every-day-for-144013.html
it-takes-30-days-to-fill-a-laboratory-dish-with-bacteria-140269.html
a-certain-culture-of-bacteria-quadruples-every-hour-if-a-52258.html
_________________

Bunuel or anyone,

Please confirm if my approach is correct.

Sum of lillies for 30 days using Geo Series:
a= 1+2+2^2+2^3..2^30 --(1)
2a = 2+2^2...2^31 -- (2)
Subtract 1 from 2
a=2^31 - 1 (Total lillies in the pond)

Now let x be number of times, both lillies expanded at once
lilly 1 -> a=1+2+2^2...2^x
sum of lilly 1 using Geo series described above = 2^x+1 - 1
lilly 2 -> a=1+2+2^2....2^x
sum of lilly 2 using Geo series described above = 2^x+1 - 1
--> sum of lilly 1 + sum of lilly 2 = 2^31 -1
so 2(2^x+1 -1) = 2^31 - 1
2^x+2 - 2 = 2^31 -1
approximately 2^x+2 = 2^31
x+2 = 31, x= 29 times ....so 58 days as lillies doubles evry 2 days

No, that's not correct. Neat algebraic manipulations though...

Notice that the total number of lilies is not 1+2+2^2+2^3..2^30, it's 2^30.

Initially = 1;
After 2 days = 2, not 1+2;
After 4 days = 2^2 = 4, not 1+2+4.
...
After 60 days = 2^30, not 1+2+2^2+2^3+...+2^30.

Similarly, if initially there are 2 lilies, then the total number would be 2*2^x.

So, we'd have that 2^30 = 2*2^x --> x = 29.

Similar questions to practice:
a-certain-bacteria-colony-doubles-in-size-every-day-for-144013.html
it-takes-30-days-to-fill-a-laboratory-dish-with-bacteria-140269.html
a-certain-culture-of-bacteria-quadruples-every-hour-if-a-52258.html
the-population-of-locusts-in-a-certain-swarm-doubles-every-90353.html
the-population-of-a-bacteria-culture-doubles-every-2-minutes-167378.html

Hope it helps.
_________________

Thank you Bunuel. My interpretation of question is incorrect.

That is pretty easy one.
Full = 60 days, knowing that the number of lilies doubles each 2 days we can deduce that the half of the lake was full at 58 days.
Taking initial information that we have 2 lilies at day 1 we can just simply multiply 2 lilies by 1/2 of the lake which means that the lake will be full at 58 days.

I thought this way :
30 days will take to complete the pond with lillies count as 2^30 (since it takes 2 days to double hence will take 30 days of 60)
on first day - 2^0 =1 lilly
on day 2 - 2^1 = 2
on day 3 - 2^2 = 4 so on ...
now since the there are two lillies already it will take 2^30/ 2^1 = 2^29 ...this will take complete 2* 29 days i.e 58 days

I hope it is clear .....

I am going by this formula : y(t) = y(0) x K^t
where
y(t) = desired value after t period
y(0) = initial value
k = multiplier (or the factor by which the value increases every t period)
t = time period

Given - # of lilies doubles every two days
==> t= 2 days
k^t = k^2 = 2
==> k = sqrt(2)
Now,
it takes 60 days for a lake to be fully covered with water lilies starting from 1 lily
so, y(0) = 1
t = 60 days
y(t) i.e no. of lilies after 60 days
y(t) = 1 x sqrt(2)^60

now, we have the final count of lilies after 60 days if we start from 1 lily.
we can calculate the time period if we start from 2 lilies ( the # of lilies after 60 days will not change as the multiplier is constant)

1 x sqrt(2)^60 = 2 x sqrt(2)^t

Solving this will give t= 58 days.

I hope it helps.

The number of water lilies on a certain lake doubles every two days. If there is exactly one water lily on the lake, it takes 60 days for the lake to be fully covered with water lilies. In how many days will the lake be fully covered with lilies, if initially there were two water lilies on it?

(A) 15
(B) 28
(C) 30
(D) 58
(E) 59

hey Bunuel,

i have a doubt in the first part of the problem it is given that if there is one lily it will take 60 days and number of water lillies double every 2 days.

so, it is in GP and the terms will be a, ar, ar^2, ar^3 etc. here it is 1,2,4,8....

we need to find the 30th term which will be ar^n-1 gives us ar^29 that leads to 1(2^29) but you got it as 2^30

what is wrong with what i did?

If you take first term as 1, then you'd have 31 terms: 1st day plus 30 divisions.
_________________

since it doubles every 2 days..
on day 2 - we have 2
on day 4 - we have 4 or 2^2
on day 6 - we have 8, or 2^3
as we see, the power is nr of days/2
so in 60 days, we'll have 2^30 lilies.

now, we start with 2...
so on day2 - we have 4
on day 4 - we have 8...
day 6 -> 2^4
day 8 -> 2^5
day 10 -> 2^6
day 12 -> 2^7

we can notice a pattern, that when # of days is divisible by 2, the power is +3 than for the last nr of days divisible by 6.
so: day 54 -> 4+3+3+3+3+3+3+3 -> so 28th power.
on day 56 - we'll have 2^29
on day 58 - we'll have 2^30 - the number we need.
so 58 days.

Or just think about it logically - you start with
1 lily on Day 1 beginning,
get 2 by Day 2 end (or Day 3 beginning ),
get 4 by Day 4 end
get 8 by Day 6 end and so on
till you get pond full of lilies by Day 60 end.

If there are already 2 water lilies, you are just starting with Day 3 beginning and skipping the first 2 days. So to cover the pond you will need 2 days less i.e. 60 - 2 = 58 days.
_________________

I found this formula to be easy to apply and it consistently gives me the right answer on questions like this.

Final population = S * P ^ (t/l)
S = starting population
P = progression (doubles = 2, triples = 3 etc.)
t/l = total amount of iterations
t = time
I = intervals

Let X be final number of lilies that covered the lake after 60 days, which means that x = 1 lily * 2 ^ ( 60 days /2 days)
From here => x = 2 ^30 = that is the final number of lilies needed to cover the lake
now if we start with 2 lilies => 2 ^ 30( which is the total number of lilies needed to cover the lake ) = 2 * 2 ^ t/ 2

=> 30 = t/2 +1 => 29 = t/2 => t = 58 - is the time needed to cover the lake when starting with 2 lilies

Every 2 days it doubles
start with 1 means full in 60 days
so start with 2 lilies means 60-2 days
answer should be 58. D

Could someone evaluate whether I did it right?
2^60 =\(\frac{2^n}{2}\)
60=n-2
n=58 option D