Help! - pls solve these puzzles - esp. escalator oriented probs how 2 handle?

[zango]

Originally Posted by quest4cat

hi,
was working with a friend on the paintings wala sum........he is sure that the answer is 25..........cud u plz explain the funda again.

ok 5/2 people work for 5/2 days to make 5/2 paintings.
Therefore, 1 person needs to work for 5/2*5/2 days to make 5/2 paintings
therefore, for 1 person to make 5/2*10 (=25) paintings, he will have to work for 5/2*5/2*10 = 250/4 days.
Now, we have 25 people working for x days to do same work as this 1 person
So they will take 1/25th of the time as 1 person takes.
so, 250/4/25 = 250/100 = 5/2
(PS: I am ignoring the 5/2 canvases part - i am not sure about its significance)

Originally Posted by quest4cat

also for the 6561 balls wala sum........i got the answer as 13........here's how i got it
6561/2=3280+3281
similarly 3281=1640+1641.........likewise go on.........the total no of weighings comes out as 13.........which is lesser than 24. do u agree? if u find it wrong plz let me know where i'm going wrong
cheers

Depends - if you have weights to weigh the balls, then yes, this is possible. If you dont have weights, then my theory was - you can weigh the (groups of) balls against one another.

Cheers!

[zango]

Originally Posted by zango

Depends - if you have weights to weigh the balls, then yes, this is possible. If you dont have weights, then my theory was - you can weigh the (groups of) balls against one another.

Cheers!

after thinking on this - it can be done in the following way:
Divide the balls into 3 groups each = 6561= 2187 * 3
Now, take one set - and weigh against each of the other two. One(or both) of these weights will be unequal - choose the heaviest group and proceed.
In this fashion, you will need fewer weighings = 8 * 2 = 16 (still not 13) but fewer than 24

[whocarez]

Originally Posted by zango

Divide the balls into 3 groups - each of size 6561/3 = 2187 - weigh them individually = 3 weighings.
Now take the heaviest - and sub-divide into 3 more groups 2187/3 = 729 - weigh them individually = 3 more weighings.
Take the heaviest - divide into 3 - 729/3 = 243 - weigh them individually = 3 more weighings.
Take the heaviest - divide into 3 - 243/3 = 81 - weigh them individually = 3 more weighings.
Take the heaviest - divide into 3 - 81/3 = 27 - weigh them individually = 3 more weighings.
Take the heaviest - divide into 3 - 27/3 = 9 - weigh them individually = 3 more weighings.
Take the heaviest - divide into 3 - 9/3 = 3 - weigh them individually = 3 more weighings.
Take the heaviest - divide into 3 - 3/3 = 1 - weigh them individually = 3 more weighings.
The heaviest now is your required ball. The total number of weighings required was: 3+3+3+3+3+3+3+3 = 24

actually you need only 1-1 for each of the 3-3 cases you have mentioned thus it should be 8. Infact as a rule of thumb 3^k should be greater than equal to the number of balls given then k is the answer as 6561=3^8 you answer automatically becomes 8

[zango]

Originally Posted by whocarez

actually you need only 1-1 for each of the 3-3 cases you have mentioned thus it should be 8. Infact as a rule of thumb 3^k should be greater than equal to the number of balls given then k is the answer as 6561=3^8 you answer automatically becomes 8

Damn, you're right - and i'm slow :P - but i never thought - damn damn damn
I guess were I given the options, I would have figured out the answer to be 8
Kool friend bhai - please post the answers once we post the solution - will help us decide whether our answers are correct or wrong.

ciao

[koolfriend4u]

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?

soln. here

Let D = number of steps

Let S = constant speed of escalator (that is, it goes down at the rate of S steps/second)
Let sa = speed of A (sa steps/second) when escalator is stationary.
Let sb = speed of B (sb steps/second) when escalator is stationary.
If the escalator is moving at speed S, and A is walking at speed sa, then the effective speed of A is S+sa
If the escalator is moving at speed S, and B is running at speed sb, then the effective speed of B is S+sb

Let T = time required by escalator to cover D steps.
Let ta = time required by A to cover D steps when escalator is stationary.
Let ta50 = time required by A to cover 50 steps when escalator is stationary.
Let taeff = effective time required by A to cover 50 steps when escalator is moving.

Let tb = time required by B to cover D steps when escalator is stationary.
Let tb90 = time required by B to cover 90 steps when escalator is stationary.
Let tbeff = effective time required by B to cover 90 steps when escalator is moving.

Therefore,
D = S*T
D = sa*ta
D = sb*tb

50 = sa*ta50 (for A walking on stationary escalator) --- 1
90 = sb*tb90 (for B running on stationary escalator) --- 2

D = (S+sa)*taeff (for A walking on moving escalator) --- 3
D = (S+sb)*tbeff (for B running on moving escalator) --- 4

For A, the time required to reach the bottom while escalator is moving is same as the time required to walk 50 steps.
Thus, ta50 == taeff.
or 50/sa = D/(S+sa) --- [from 1 and 3]
or D = ((S+sa)*50)/sa --- 5


For B, the time required to reach the bottom while escalator is moving is same as the time required to run 90 steps.
Thus, tb90 == tbeff.
90/sa = D/(S+sb) --- [from 2 and 4]
or D = ((S+sb)*90)/sb --- 6

Also, B coveres all 90 steps when A has covered only 10 steps == 1/5th the total steps for A. Thus, effective speed of B is five times effective speed of A

(S+sb) = 5*(S+sa)
Replacing S+sb with 5*(S+sa) in 6 gives
D = 5*(S+sb)*90/sb --- 7

Combine 5 and 7
(S+sa)*50/sa == 450(S+sa)/sb
thus sb/sa = 9

B goes 9 times faster then A

Solving for eq 5 and 6 using sb=9*sa gives S = sa.
Using S = sa in 5 gives D = 100.

Answer: 100 steps are visible when the escalator is not operating.
A walks at the same speed as the escalator and thus reaches in half the normal time = A covers 50 steps + escalator covers 50 steps.
B runs at the 9 times the speed of escalator and thus reaches in 1/10th the normal time = B covers 90 stpes + escalator covers 10 steps.

[koolfriend4u]

[quote="zango"]

Originally Posted by koolfriend4u

thanks a lot.

You are welcome

Originally Posted by koolfriend4u

some more puzzles here.

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?

Some information is missing here? If not - then I don't know how to answer this one

[quote="koolfriend4u"]

another shortcut method my friend mailed me (after i solved via conventional method) but am not clear wit his shortcuts.
cud anyone explain.

Here's the shortcut method.

Suppose A takes T time to reach bottom (or 50 steps) when escalater is moving

For B to take 90 steps it will require T/5 time (<- am not understanding how T/5 )


total no of steps in escalater= (90T-50T/5)/T-T/5
=80T/4T/5=100 STEPS

[koolfriend4u]

if any other escalator probs. pls post here friends.

[Davesh_Manocha]

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_Manocha]

How ur frined got T/5 is fairly simple.

Hes calculating the time T with respect to speed of A.
Suppose A takes T time to reach bottom (or 50 steps) when escalater is moving.
Now a wud take T/5 time to cover 10 steps.
As B covers 90 steps while a cover 10 steps therfore B takes T/5 time to cover 90 Steps.

But can u please explain whats the logic used in the formuale thereafter.

Davesh

[D e e p t i]

say x is the number of steps.
t1 - time taken by A
t2 - time taken by B

x = At1 + 50
x = At2 + 90

Eliminating A,

x/t2 = x/t1 - 50/t1 + 90/t2

x(1/t2-1/t1) = 90/t2 - 50/t1

x ( t1 - t2 / t1 t2) =( 90t1 - 50 t2 ) / t1 t2

Now subsitute t2 = t1/5 and solve for x