[sapanjain]

Originally Posted by NoopS

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:

thnx yaar ! dat was very neatly put.

[sapanjain]

Originally Posted by NoopS

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....

Thnx yaar !

[NoopS]

now its ur turn to do ur bit by atleast confirming the answers...

[sapanjain]

Originally Posted by 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?

let the height of the first man is = x
then according to the problem
height of the second man is = x+2
height of the third man is = x+4 and
height of the fourth man is = x+10

given that their average height is 74 inches

therefore

{x+(x+2)+(x+4)+(x+10)}/4 = 74

solving the equation u get x=70

so the heights are 70,72,74,80.

[sapanjain]

Originally Posted by NoopS

now its ur turn to do ur bit by atleast confirming the answers...

sorry dude i dont have the answer !

[msksent]

Originally Posted by sapanjain

sorry dude i dont have the answer !

I think the second problem is wrong.

[sapanjain]

plz post the complete solution for the following problems :

1.If 5/2 artists make 5/2 paintings using 5/2 canvases in 5/2 days then how many artists r required to make 25 paintings using 25 canvases in 25 days?
Ans : 2500 artists

2.Ramesh sit around a round table with some other men. He has one rupee more than his right person and this person in turn has 1 rupee more than the person to his right and so on, Ramesh decided to give 1 rupee to his right & he in turn 2 rupees to his right and 3 rupees to his right & so on. This process went on till a person has 'no money' to give to his right. At this time he has 4 times the money to his right person. How many men are there along with Ramesh and what is the money with poorest fellow.

[sapanjain]

Originally Posted by msksent

I think the second problem is wrong.

dont know if the problem is wrong. i got it from one of infy's old papers.

[sapanjain]

how to solve this one ?

1.Two friends A and B are running up hill and then to get down. Length of road is 440 yards. A on his return journey met B going up at 20 yards from top. If A has finished the race 5 minutes earlier than B,then how much time A had taken to complete the race?.

[msksent]

Originally Posted by sapanjain

how to solve this one ?

1.Two friends A and B are running up hill and then to get down. Length of road is 440 yards. A on his return journey met B going up at 20 yards from top. If A has finished the race 5 minutes earlier than B,then how much time A had taken to complete the race?.

I think some statement is missing again