[ashish_123]

hi i am a newbie !..i wanna join infy and i am currently trying to solve shakuntala devi puzzles...i hope i get help from this forum and i ll try to give my best as well....
good luck..

[ashish_123]

hey guys.
.i need help on the first problem of the book....prob named "tall men next door"...

has anyone posted it before?

[sapanjain]

plz give the complete solution :

1.There r some chickens and some chicken field. If we sell 75 chickens Then the feed will last 20 days longer.If we buy 100 chickens more, then the feed will get over before 15 days.What is the present number of chickens?

2.Sometime after 10:00 PM a murder took place. A witness claimed that
the clock must have stopped at the time of the shooting. It was later
found that the postion of both the hands were the same but their
positions had interchanged. Tell the time of the shooting (both actual and
claimed).

[msksent]

Originally Posted by sapanjain

plz give the complete solution :

1.There r some chickens and some chicken field. If we sell 75 chickens Then the feed will last 20 days longer.If we buy 100 chickens more, then the feed will get over before 15 days.What is the present number of chickens?

2.Sometime after 10:00 PM a murder took place. A witness claimed that
the clock must have stopped at the time of the shooting. It was later
found that the postion of both the hands were the same but their
positions had interchanged. Tell the time of the shooting (both actual and
claimed).

Is the answer for the first problem 300. and the second one 10.59 pm

[ankursanghi]

Originally Posted by sapanjain

plz give the complete solution :

1.There r some chickens and some chicken field. If we sell 75 chickens Then the feed will last 20 days longer.If we buy 100 chickens more, then the feed will get over before 15 days.What is the present number of chickens?

2.Sometime after 10:00 PM a murder took place. A witness claimed that
the clock must have stopped at the time of the shooting. It was later
found that the postion of both the hands were the same but their
positions had interchanged. Tell the time of the shooting (both actual and
claimed).

Normal days d, and normally chicken are r, assume each chicken eats q qty of feed / day. So
total feed = rqd.

rqd = (r-75)*(d+20)q and rqd = (r+100) (d-15)q

So, we get

rd = rd-75d+20r-1500 => 20r-75d = 1500--- (1)
and
rd = rd+100d-15r-1500 => 100d-15r =1500----(2)

From 1 & 2
20r-75d = 100d-15r
175d = -35r

d/r = -7/17 <<-- seems like the given conditions are not possible.

For the second question, the question, as I percieve, asks the time when both hands are together between 10 & 11. Because the murder could happen even at 12 noon and it will satisfy all conditions here.

If my understanding of question is correct then the answer would be like:

The minute hand has to catch up with the hour hand and there is an initial difference of 300 degrees at 10.00. The relative speed of minute hand wrt to hour hand is 11/2 degrees per minute. So, the answer will be 600/11 minutes past 10.00.

[msksent]

[QUOTE=ankursanghi]Normal days d, and normally chicken are r, assume each chicken eats q qty of feed / day. So
total feed = rqd.

rqd = (r-75)*(d+20)q and rqd = (r+100) (d-15)q

So, we get

rd = rd-75d+20r-1500 => 20r-75d = 1500--- (1)
and
rd = rd+100d-15r-1500 => 100d-15r =1500----(2)

From 1 & 2
20r-75d = 100d-15r
Here is the mistake. It should be 35 r = 175 d
Hence r = 5d. Backsubstitution will give u 25 d = 1500 and d = 60. Hence r = 300. Hope u got it.

[msksent]

there is a unbalanced common balance ie., one side is lengthier than the other. If a 1kg weight is placed in left side it is equal to 8 oranges on the otherside. If we weigh 1 kg on the right side it is equal to 2 oranges on the left side.find the weight of each orange.

[®ajat]

Originally Posted by msksent

there is a unbalanced common balance ie., one side is lengthier than the other. If a 1kg weight is placed in left side it is equal to 8 oranges on the otherside. If we weigh 1 kg on the right side it is equal to 2 oranges on the left side.find the weight of each orange.

I am getting the answer as 1/5 Kg. Is it correct ?

[Apple]

Originally Posted by msksent

there is a unbalanced common balance ie., one side is lengthier than the other. If a 1kg weight is placed in left side it is equal to 8 oranges on the otherside. If we weigh 1 kg on the right side it is equal to 2 oranges on the left side.find the weight of each orange.

I think the answer shud be 200gm.
Solution:
Let the left side be x kg heavier than the right side.(that means if we put x kg weight on right side it will be balanced).
first case:
left:1kg
right:8oranges =8y kg (ykg be weight of each orange)
=> 1+x=8y
second case:
left:2oranges=2y
right: 1kg
=>2y+x=1
Solving these for y we get
y=200 gm.
Cheers
Apple

[®ajat]

I used this approach.

Let the left be of greater length.
and let the weight of each orange be x kg.

So the actual weight of 8 oranges = 8x Kg
But the balance is showing it as 1 Kg.
So the error in measurement on left side = 1-8x kg

Similarly error in measurement on right side = 2x - 1 Kg.

The errors on both sides should be same.
So, 1-8x = 2x-1
Solving for x we get x = 1/5 kg.