[Guest]

A colony of 60 beetles in search of food came near a plant X.

They can feed on the arbitary plant X for 40 days.

Assume here that the plant X grows at an even rate.

Also 40 beetles can feed on plant X for 80 days.

So how many days could 20 beetles feed on the plant X?

Edited July 20, 2011 by 14.swapnil.14

[Guest]

Indefinitely. The plant grows faster than the feeding rate of the 20 beetles.

Here's the proof:

Let Uo, U be the starting size and current size (units of food) of the plant. If it grows at a uniform rate, U = Uo + (a - bN)T where a is the growth rate and b is the eating rate of a beetle, N is the number of beetles and T is the time in days.

plugging in the givens above, we now have the 2 equations:

Uo + 40(a - 60b) = 0

Uo + 80(a - 40b) = 0

Which gives a = Uo/40; b = Uo/1200

So U = Uo[1 + (1/40 - N/1200)T]

putting N = 20 gives U = Uo(1 + T/120) which shows a positive growth rate despite the 20 beetles feeding on it

Edited July 20, 2011 by rparry

[Guest]

<deleted>

Edited July 20, 2011 by rparry

[k-man]

Assuming that the "even rate of growth" means a constant amount added every day regardless of the current size of the plant, 20 beetles can feed on this plant forever as the amount of daily growth will be equal to the daily consumption of 20 beetles.

Let's call the daily growth rate Y. In 40 days 60 beetles eat X + 40Y. In 80 days 40 beetles consume X + 80Y. From this we can figure out the daily consumption of a single beetle.

(X + 40Y) / (40 * 60) = (X + 80Y) / (40 * 80)

80X + 3200Y = 60X + 4800Y

20X = 1600Y

X = 80Y

After the substitution we get the daily consumption of a single beetle is (80Y + 80Y) / (40 * 80) = Y/20

So a single beetle eats 1/20 of the daily growth per day, so 20 beetles eat exactly the amount of daily growth Y.

[Guest]

Assuming that the "even rate of growth" means a constant amount added every day regardless of the current size of the plant, 20 beetles can feed on this plant forever as the amount of daily growth will be equal to the daily consumption of 20 beetles.

Let's call the daily growth rate Y. In 40 days 60 beetles eat X + 40Y. In 80 days 40 beetles consume X + 80Y. From this we can figure out the daily consumption of a single beetle.

(X + 40Y) / (40 * 60) = (X + 80Y) / (40 * 80)

80X + 3200Y = 60X + 4800Y

20X = 1600Y

X = 80Y

After the substitution we get the daily consumption of a single beetle is (80Y + 80Y) / (40 * 80) = Y/20

So a single beetle eats 1/20 of the daily growth per day, so 20 beetles eat exactly the amount of daily growth Y.

This also assumes that the eaten parts of the plant keep growing at the same rate

Edited July 20, 2011 by rparry

[k-man]

Indefinitely. The plant grows faster than the feeding rate of the 20 beetles.

Here's the proof:

Let Uo, U be the starting size and current size (units of food) of the plant. If it grows at a uniform rate, U = Uo + (a - bN)T where a is the growth rate and b is the eating rate of a beetle, N is the number of beetles and T is the time in days.

plugging in the givens above, we now have the 2 equations:

Uo + 40(a - 60b) = 0

Uo + 80(a - 40b) = 0

Which gives a = Uo/40; b = Uo/1200

So U = Uo[1 + (1/40 - N/1200)T]

putting N = 20 gives U = Uo(1 + T/120) which shows a positive growth rate despite the 20 beetles feeding on it

Welcome to the Den, rparry. Please use spoilers when posting the answers.

Your equations are correct, buy you made a mistake in your calculations. Solving these equations results in a=Uo/80 and b=Uo/1600.

[k-man]

This also assumes that the eaten parts of the plant keep growing at the same rate

That's right. I stated that assumption in my answer. We both make the same assumption basically treating the plant as a renewing source of food with constant daily increment.

[dark_magician_92]

i gave it a try, but i am not sure about this. i am taking 'y' as the growth per day, which occurs once per day, not per second.

let 'a' be plant size, 'y' be the grwoth, then in 40 days, total plant length they ate is a+40y, by 60 beetles in 40 days, so plant length eaten/per beetle/ day= a+60y/2400,

similarly for 2nd case, a+80y/3200, hence 1 and 2 can be equated, a+40y/2400 = a+80y/3200, therefore a=80y.

now say, theuy take 'k' days for 3rd case, then a+ky/20k=a+40y/2400, whick gives k= 12800/0 i.e. infinite time, thinking about it, 1 beetle , 1 day = a+40y/2400 = 80y+40y/2400 =y/20, 20 beetles, 1 day= y, so eating = growth, so infinite time AND LUCKY BEETLES

[Guest]

Welcome to the Den, rparry. Please use spoilers when posting the answers.

Your equations are correct, buy you made a mistake in your calculations. Solving these equations results in a=Uo/80 and b=Uo/1600.

Sorry, you're right. I ran out of space on the post-it

[Peekay]

Without even going into equations, it is apparent from the problem statement that 20 beetles were just keeping up with the growth while the rest were devouring the original height (or length) of the plant. So, assuming that the growth rate of the plant is independent of the height of the plant, 20 beetles can happily feed on the plant for ever.

Why can't we get a source of food like this growing in our backyard?

[Guest]

A colony of 60 beetles in search of food came near a plant X.

They can feed on the arbitary plant X for 40 days.

Assume here that the plant X grows at an even rate.

Also 40 beetles can feed on plant X for 80 days.

So how many days could 20 beetles feed on the plant X?

My guess would be 160 days for 20 beetles.

[Guest]

Also the size of plant X would remain same