[.mba]

Originally Posted by jigar_p_civil

I have checked u r method and it is correct, so may be ans also correct

Because this puzzle from magazine , more trust on PG's ans compare to magazine

Hi

I tried the following method ...and getting a different ans...
Can anybody tell me wat mistake I am commiting...

the tens place can be 1,4,9
So , case 1: wen tens place is 1 :
100(1+2+3+...+9) +10+ (1+2+3+.... +9)
case 2:wen tens place is 4 :
100(1+2+3+...+9) +40+ (1+2+3+.... +9)
case 3: wen tens place is 9 :
100(1+2+3+...+9) +90+ (1+2+3+.... +9)
Adding these three, I am getting my answer as 13775 .... which is no where close to any of the options.

Waiting for the reply guys....
Thanks in advance...

[shanks4mba]

Originally Posted by bhavin422

Am not too sure abt my ans ...is it 1/(1 + root2)?

Approach:
Radius of outer circle = 1, let inner radius = r
Hence, line joining origin to centre of smaller circle is 1-r ..

This line is also diagonal of square of side r ...i.e r root 2...

Equating, 1-r = r root 2
Hence, r = 1/ (1 + root 2)

Pls correct fr errors..

ur absolutly rt...one thing can u suggest me a place(book or site or watever) for geometry practice

[rohitmodimodi]

There are 8 equally spaced points lying on a circle. How many different quadrilaterals can you make with these points as vertices?

my answer is c(8,4)=70 which happens to be the wrong answer,cud someone plz explain why this is wrong ,along with the rite answer.

thanks in advance

[Ankurb]

<wrong post>

[No_Talent]

Hey rohit,

The ans of your question is given in your question. As it is mentioned Different and also he points on the cirlce are equally spaced so there must be repetition of figures among your 8C4 quadrialaterals. Hence you need to subract them.

[rohitmodimodi]

@no_talent

why wud there be any repetition in my answer when i am using C instead of P,evry combination of 4 points wud give a different quad each time....

please correct me if i am wrong
regards

[No_Talent]

Originally Posted by aryan007

How many three-digit numbers are there such that no two adjacent digits of the number are consecutive?

Is it 783? I have hardly reached an accurate ans in this short of counting problems but can tell you how to approach such problems.

first you find out how many three digit numbers are there? it is 900.
now find out how many numbers among them have two or three consecutive digits. then subtract this value from 900. you will get your ans.
My approach to find out how many numbers are there with consecutive digits -
say the first position is occupied by 1 then the second postion by 2 and third position could be occupied in 7ways.
so taotal 9*7 = 63 ways (for ordered pairs like 1,2; 2,3;3,4; etc in 1st and2ndposition).
now say second position is occupied by 1 and third position position by 2
first position can be filled up in 6 ways. such a way if you count you get total 9*6 = 54 ways.
hence total 63+54 = 117 such numbers are there.
so ans = 900 - 117 = 783

not sure how accurate I am

[pt_ct]

Originally Posted by rohitmodimodi

There are 8 equally spaced points lying on a circle. How many different quadrilaterals can you make with these points as vertices?

my answer is c(8,4)=70 which happens to be the wrong answer,cud someone plz explain why this is wrong ,along with the rite answer.

thanks in advance

hi rohit..
70 is the total no of quads u can make..of these many will be of equal area..if u want 'DIFFERENT' quads..then this is how its done..
since i cant draw a circle and show, im representing the points 1 to 8 in a straight line
| . . . . . . . . |
remember point 1 and 8 are adjacent
case 1:
select all 4 points adj t each other..
only 1 type
case 2:
select 3 adj(say points 1,2,3)..( | . . . | . . . . . )
the other vertex may be points 5 or 6..so 2 possibilities
case 3:
select 2 adj ( | . . | . . . . . . )
the next 2 vertices can be adj (points 4 and 5) or ( 5 and 6 )...
or seperated..( points 4 and 6) or points (4 and 7)..
4 possibilities
case 4:
all 4 are seperated:
only 1 possibility..

therefore..the answer is 1+2+4+1=8
hope its clear

[No_Talent]

Originally Posted by rohitmodimodi

@no_talent

why wud there be any repetition in my answer when i am using C instead of P,evry combination of 4 points wud give a different quad each time....

please correct me if i am wrong
regards

arre no no what I mean is completely different. don't get so mathematical here.

I mean say you have chosen ponit 1,2,3,4 and have got a quadrilateral (probably a trapezium). again when you choose points 5,6,7,8 you will have another trapezium. These two may be identical in phyisical appearance so they should be counted as duplicate and count should be 1.

now coming to the concept of C an P.
C is selection. you would choose 4 points from 8 and would connect using lines to form a geometrical figure.
P is arrangement after selection. Now you first choose 4 points and then instead of drawing the figure you are going to arrange those points? how?

Please reply if it is OK? (Becos I am a person who has lot of doubt about his own permutation/combination funda)

[aaptuster]

Originally Posted by aryan007

How many three-digit numbers are there such that no two adjacent digits of the number are consecutive?

9^3 =729

First digit 1-9
2nd digit 0-9 minus 1 which is adjacent number to previous digit
3rd digit 0-9 minus 1 which is adjacent number to previous digit

thus, 9^3