[badshahkhan]

dont post questions one by one. give a set of 5-10 qns at one go. otherwise the no. of pages wud become very large n it wud take students a lot of time to get thru the qns

[thakarsagar]

Originally Posted by clsuresh

sagar,

Is it 229. confirm.........

closer, actual answer is 242....But it may be wrong too...(as it was solved by one of my friends who's the "god" of mathematics)....let see...
-sagar

[clsuresh]

Hi friends,

Originally Posted by eagles_dare13
without multiplying actually, is there a way to tell which is greater ..
say 61*94 and 92 * 65.
This comes in handy in DI.
Also , please tell me exactly how remebering reciprocals upto 30 will fhelp.
becasue i can calculate percentages by adjusting numerator and denominator until the denom is 100.

One of the sol posted is :

61* 94 and 92 * 65 ..
y dont u compare ..61/65 and 92/94
61/65 is 1-(4/65) and the other is 1- (2/94)
clearly... 92/94 is bigger => 92 * 65 is greater..
cos if a/b > c/d then ad > bc...
just a thot...
andy.


Are there any other ways please.........

[thakarsagar]

Originally Posted by clsuresh

Hi friends,

Here comes our next question.
Find the total number of ways in which two sqares (1x1) can be selected from a chess board such that they will have a side in common.

is the answer 2*56 = 112??
solution:

In any row, we have n-1 pairs which are adjacent to each other....(where n is the no of columns)
so in all we have n*(n-1) pairs in n rows....Multiply by 2 because it's a square...

-sagar

[clsuresh]

Sagar,

Your answer is right .Let me give explanation more lucidly.

Let me say that the squares are named as

A1 A2 A3 A4 A5 A6 A7 A8
B1 B2 B3 B4 B5 B6 B7 B8
C1 C2 C3 C4 C5 C6 C7 C8
D1 D2 D3 D4 D5 D6 D7 D8
E1 E2 E3 E4 E5 E6 E7 E8
F1 F2 F3 F4 F5 F6 F7 F8
G1 G2 G3 G4 G5 G6 G7 G8
H1 H2 H3 H4 H5 H6 H7 H8

Now as you said I can select two squares that share a common side from first row (i.e row A) in 7 ways i.e(A1,A2),(A2,A3), (A3,A4),(A4,A5),(A5,A6),(A6,A7),(A7,A.
Similarly since there are 8 rows total ways are 8 x 7 = 56.

Now thinking in the same way I can select two squares that share a coomon side from the first column in 7 ways and since there are 8 columns I get another 56 ways.

So total no. of ways is 112.

BUT I HAVE MY OWN DOUBT ABOUT THIS APPROACH IF THE SQUARES TO BE SELECTED ARE SQUARES OF DIMENSION 2 X 2, 3 X 3, .........

SAGAR CAN U PROCEED FURTHER NOW AND CAN U GET ME A GENERALISATION FOR ANY M X N DIMENSIONAL BOARD

[thakarsagar]

Originally Posted by clsuresh

Sagar,

Your answer is right .Let me give explanation more lucidly.

Let me say that the squares are named as

A1 A2 A3 A4 A5 A6 A7 A8
B1 B2 B3 B4 B5 B6 B7 B8
C1 C2 C3 C4 C5 C6 C7 C8
D1 D2 D3 D4 D5 D6 D7 D8
E1 E2 E3 E4 E5 E6 E7 E8
F1 F2 F3 F4 F5 F6 F7 F8
G1 G2 G3 G4 G5 G6 G7 G8
H1 H2 H3 H4 H5 H6 H7 H8

Now as you said I can select two squares that share a common side from first row (i.e row A) in 7 ways i.e(A1,A2),(A2,A3), (A3,A4),(A4,A5),(A5,A6),(A6,A7),(A7,A.
Similarly since there are 8 rows total ways are 8 x 7 = 56.

Now thinking in the same way I can select two squares that share a coomon side from the first column in 7 ways and since there are 8 columns I get another 56 ways.

So total no. of ways is 112.

BUT I HAVE MY OWN DOUBT ABOUT THIS APPROACH IF THE SQUARES TO BE SELECTED ARE SQUARES OF DIMENSION 2 X 2, 3 X 3, .........

SAGAR CAN U PROCEED FURTHER NOW AND CAN U GET ME A GENERALISATION FOR ANY M X N DIMENSIONAL BOARD

Suresh,
In general ,
1) For a m * n n & required of size p=1 -> [m-(p-1)] * [n-(p-1)] + [n -(p-1]*[m-(p-1)].
-sagar

[clsuresh]

SAGAR,

I think this will not work out for the squares of dimension 2 x 2 on a chess board.

[thakarsagar]

Originally Posted by clsuresh

SAGAR,

I think this will not work out for the squares of dimension 2 x 2 on a chess board.

suresh, i gave the solution for p = 1....

For a general m * n rectangle and the no of p squares are to be found which are adjacent, its

ceil(m/p) * [ceil(n/p) - 1] + ceil(n/p) * [ceil(m/p) - 1]....

here, ceil refers to ceiling...ceil(2.5) = 3...smallest no greater than the given no...
regards,
-sagar

[Neeraj.Gupta]

keep it up buddy

giv some more interestingrules and mthds....

Neeraj


Cheers!!!!!!!

[clsuresh]

Quote:
Originally Posted by clsuresh
SAGAR,

I think this will not work out for the squares of dimension 2 x 2 on a chess board.

Sagar,
Applying this to chess board (8 x and for sqares of side 2 x 2
ceil(m/p) * [ceil(n/p) - 1] + ceil(n/p) * [ceil(m/p) - 1]....
we get it as
4*(4-1) + 4*(4-1) = 24
But the answer is > 24.
regards,
Suresh.