[zango]

Originally Posted by Davesh_Manocha @ Wed Sep 29, 2004 6:03 pm

Hi.
Have joined PG just today.Was going thru the cows problems & the proposed solution.

My solution is :
40 cows graze in 40 days.
Each cow graze in 1 day = 1/(40*40)
30 cows will graze 30/1600 in 1 day.

Since 30 cows have been able to complete the job in 60 days then
60*30/1600=x =
where x is the grass that is to be grazed.

Now there are 20 cows
so 20*n/1600=x
where n is the no: of days required by 20 cows
=> n=90 Days.

Please let me know what everyone thinks.

Davesh - welcome to PG (PagalGuy.com)
Here, you are fixing your reference point as the 40 cows / 40 days data - and proceeding from there. From this - you have arrived at a reference for the amount of grass present for the 30 cows (which is 1.125 times the grass available for the 40 cows).
Next, you take this same reference point (1.125 times the grass available for 40 cows (which also is the same as the total amount of grass available for 30 cows in 60 days)) and used that info to calculate grass consumption for 20 cows. I don't think it works that way

Well, the question was a little lacking - in that - (i guess) it didn't supply all the info there is... I saw the same question a little later - where it was stated that the grass is also regrowing... so the amount of grass that will be present in 40 cows (40 days) may be less than amount of grass present for 30 cows (60 days) may be less than amount of grass present for 20 cows. we have to figure out the rate at which the cows are eating the grass - and the rate at which the grass is growing back - here is a solution posted by someone on a Yahoo Group called urpercentile (I believe that person is also a member here) - am posting it here for your perusal.

Pls solve this prob..
40 cows graze a patch in 40 days.30 cows in 60 days. If grass is
growing at a some constant rate find in how many days will 20
cows finish tht field?

ANS: Let the original qty. of grass be G and let it grow at L units
per day.
Then, 40 cows need G+40L and 30 cows need G+60L
Obviously, (G+40L)/(40x40) = (G+60L)/(30x60) (assuming all
cows eat the same
qty. of grass per day)
Or, 9(G+40L) = 8(G+60L)
Or, G=120L
So, each cow needs 1/10 x L grass per day

Let 20 cows finish in D days
Then, 120L + DL = 20xDxL/10
Or, DL = 120L and D=120

[silajeet]

Originally Posted by koolfriend4u

1.An escalator is descending at constant speed. A walks down and takes 50 steps to reach the bottom. B runs down and takes 90 steps in the same time as A takes 10 steps.
How many steps are visible when the escalator is not operating?

let speed of A = 1 step/min
speed of esc = n steps/min
while A moves 50 steps, esc moves 50n steps.
Again, Atakes 10 steps in 10 mins. So B takes 90 steps in 10 mins, while esc moves 10n in the same time.
so, 50 + 50n = 90 + 10n
so, n=1
no. of stairs=100