[msksent]

Originally Posted by ®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.

The answer for the problem is 250 gm for each orange.

[Apple]

Originally Posted by msksent

The answer for the problem is 250 gm for each orange.

pls post ur solution as i m getting 200 gms.

[msksent]

Originally Posted by Apple

pls post ur solution as i m getting 200 gms.

The situation is ike this. The arms of the balance are not equal which is the reason for unequal number of oranges getting balanced by the 1kg weight. Here the 1 kg balanced should be equal to (8*2)^0.5 = 4. Hence 4 oranges weigh 1 kg. 1 orange will weigh 250 gms. A similar problem is available in Shakumthaladevi.

[®ajat]

Originally Posted by msksent

The situation is ike this. The arms of the balance are not equal which is the reason for unequal number of oranges getting balanced by the 1kg weight. Here the 1 kg balanced should be equal to (8*2)^0.5 = 4. Hence 4 oranges weigh 1 kg. 1 orange will weigh 250 gms. A similar problem is available in Shakumthaladevi.

umm are u sure that this is the correct answer! i didnt get why the geometric mean of the two quantities is taken here. Anyway I saw a similar problem, actually the exact one in this thread --- 3rd question >>>HERE<<< (Infy Puzzles)
And the person has written the answer as 200 gms.

[sapanjain]

Plz post the complete solutions to following problems :

1.There were 2 systems A n B.14 degrees in A is equivalent to 36 in system B.and 133 in A is equivalent to 87 in B.now what is the temperature where they both r equal?

2.There was a community in which there were 1000 couples.In that 2/3rd of men who r taller r also heavier n 3/4th of the men who r heavier r also taller n there were 120 women who were both heavier n taller than men.So how many men r both taller n heavier than men?

3.There are 6561 balls out of them 1 is heavy.Find the min. no. of
times the balls have to be weighed for finding out the haevy ball.

4.In a railway station, there are two trains going. One in the
harbour line and one in the main line, each having a frequency of 10
minutes. The main line service starts at 5 o'clock and the harbour line starts
at 5.02A.M. A man goes to the station every day to catch the first
train that comes. What is the probability of the man catching the first
train?

[msksent]

Originally Posted by sapanjain

Plz post the complete solutions to following problems :

1.There were 2 systems A n B.14 degrees in A is equivalent to 36 in system B.and 133 in A is equivalent to 87 in B.now what is the temperature where they both r equal?

2.There was a community in which there were 1000 couples.In that 2/3rd of men who r taller r also heavier n 3/4th of the men who r heavier r also taller n there were 120 women who were both heavier n taller than men.So how many men r both taller n heavier than men?

3.There are 6561 balls out of them 1 is heavy.Find the min. no. of
times the balls have to be weighed for finding out the haevy ball.

4.In a railway station, there are two trains going. One in the
harbour line and one in the main line, each having a frequency of 10
minutes. The main line service starts at 5 o'clock and the harbour line starts
at 5.02A.M. A man goes to the station every day to catch the first
train that comes. What is the probability of the man catching the first
train?

For your previous post some answers were given. U have not confirmed at all. whether the y are right.
Coming to this post for problem1, the value is 52.5
Problem 3 8 balls
Problem 4 mail line will be 0.2 and harbour line will be 0.8

[sapanjain]

Originally Posted by msksent

For your previous post some answers were given. U have not confirmed at all. whether the y are right.
Coming to this post for problem1, the value is 52.5
Problem 3 8 balls
Problem 4 mail line will be 0.2 and harbour line will be 0.8

thnx for the reply......can u plz post the complete solution.

about the previous post answers are 300 chickens and 10:59(actual) and 11:54(claimed).

[msksent]

Originally Posted by sapanjain

thnx for the reply......can u plz post the complete solution.

about the previous post answers are 300 chickens and 10:59(actual) and 11:54(claimed).

For the first problem, we can write
(A -- 14)/(133 -- 14) = (B--36)/(87-36)
(A -- 14)/119 = (B--36)/51
we will get 7B -- 3A = 210 which is the equation in terms of A and B. If A = B, then we get 4B = 210 and B = 52.5

[NoopS]

Originally Posted by sapanjain

Plz post the complete solutions to following problems :

3.There are 6561 balls out of them 1 is heavy.Find the min. no. of
times the balls have to be weighed for finding out the haevy ball.

well, the problem 2 doesnt specify whether to use a spring balance or a 2-pan balance, since answer would differ for both of them.

considering 2-pan balance, answer would be 8 times.

solution:

assume the no. of balls to be 3. it is obvious that u ll use the balance once. weigh two balls. if they weigh different, u get the heavier one. if they weigh same, 3rd ball is heavy.

take different cases of 4,5,...,8,9 balls. u will find in all the cases, u ll have to weigh two times, making groups of balls and weighing them together. this can be generalised like:

no. of balls ------------- no. of weighing

1 to 3 i.e. 3^1 -----------------1
4 to 9 i.e. 3^2----------------- 2
10 to 27 i.e. 3^3 --------------3
28 to 81 i.e. 3^4 --------------4
82 to 243 i.e. 3^5 -------------5
244 to 729 i.e. 3^6 ------------6
730 to 2187 i.e. 3^7----------- 7
2188 to 6561 i.e. 3^8 ----------8

hence for 6561 balls, u need to weigh 8 times.

in the case of a spring balance, take the powers of 2 instead of 3. answer would be 13 ( as 2^12=4096 n 2^13=8192 and 4096<6561<8192)

hope this helps.:smile:

[NoopS]

Originally Posted by sapanjain

Plz post the complete solutions to following problems :

4.In a railway station, there are two trains going. One in the
harbour line and one in the main line, each having a frequency of 10
minutes. The main line service starts at 5 o'clock and the harbour line starts
at 5.02A.M. A man goes to the station every day to catch the first
train that comes. What is the probability of the man catching the first
train?

for each 10 min interval, if man comes in first 2 min, he'll catch the 1st train, if he comes in next 8 min, he'll catch the 2nd train.

hence for harbor line - (2/10) 0.2 and for main line 0.8.

solution for problem 1 has already been given by mksent, me still working on problem 2....