official quant thread for cat08

[IIM maniac]

Originally Posted by stalwart_itsme

thanks bro... i missed out... its a squre with side 4 units ... so area 16...you got it...

Guys i want some explaination, even though my doubt is very pitty. if we take co-ordinates as (2,0),(0,2),(-2,0),(0,-2) this also satisfies the equation, also co-ordinates (2,1) satisfies and even (2,2) satisfies, so what made us to decide on the co-ordinates to be (2,2)?

[parmindersaluja]

hi

hope this image will help u, purple ones are the co-ordinates given by me and red ones by stalwart.

Originally Posted by IIM maniac

Guys i want some explaination, even though my doubt is very pitty. if we take co-ordinates as (2,0),(0,2),(-2,0),(0,-2) this also satisfies the equation, also co-ordinates (2,1) satisfies and even (2,2) satisfies, so what made us to decide on the co-ordinates to be (2,2)?

[IIM maniac]

Originally Posted by parmindersaluja

hi

hope this image will help u, purple ones are the co-ordinates given by me and red ones by stalwart.

I am able to come to these figures but one thing i am not able to understand is even though three sets of co-ordinates satisfies the equation, so how could (2,2) be the unique co-ordinates?? Sorry if my query if very silly but still it is.

[stalwart_itsme]

Originally Posted by IIM maniac

Guys i want some explaination, even though my doubt is very pitty. if we take co-ordinates as (2,0),(0,2),(-2,0),(0,-2) this also satisfies the equation, also co-ordinates (2,1) satisfies and even (2,2) satisfies, so what made us to decide on the co-ordinates to be (2,2)?

dude ... the coordinates (2,0),(0,2),(-2,0),(0,-2) does not cover the full area... full area occupied by the eqation |x+y|+|x-y|=4 is covered only when the points (2,2)(2,1),(2,-2),(-2,2),(2,-1),(-2,1) are all considered...

Please correct me if i am wrong...

[parmindersaluja]

Originally Posted by IIM maniac

I am able to come to these figures but one thing i am not able to understand is even though three sets of co-ordinates satisfies the equation, so how could (2,2) be the unique co-ordinates?? Sorry if my query if very silly but still it is.

because |x+y| + |x-y| = 4, can take values such as (4,0), (0,4), (3,1), (1,3), (2,2)...and their negative counterparts...
now, solving for (x,y) for the extreme cases (4,0), (0,4), (-4,0) and (0,-4) will give u values of x and y which will form a square encompassing all other values.

P.S. if u still have any doubt, dont hesitate to ask.

[IIM maniac]

Originally Posted by parmindersaluja

because |x+y| + |x-y| = 4, can take values such as (4,0), (0,4), (3,1), (1,3), (2,2)...and their negative counterparts...
now, solving for (x,y) for the extreme cases (4,0), (0,4), (-4,0) and (0,-4) will give u values of x and y which will form a square encompassing all other values.

P.S. if u still have any doubt, dont hesitate to ask.

Thanks for the reply, so you mean to say we should consider the extreme case after which this equation does not get satisfied? is it? or in proper words , we should consider the case or co-ordinates that covers the full range which is in case of (4,0), (0,4), (-4,0) and (0,-4). thats why area is 16, is it?

[80199]

Originally Posted by stalwart_itsme

The numbers are..

1001
1003
1005
1007
1009
1011
1013
1015
1017
1019
1021
...
...
1191
1193
1195
1197
1199
(since only odd numbers...)
100
so divisible by 3 are 1005,1011,1017...goes on till 1197
so a=1005 d=6 last number = 1197
tn=a+(n-1)*d
1197=1005+(n-1)*6
192=6n-6
n=33...
something went wrong i suppose...... pls help me out...

i am not very much sure but question is saying that every digit in n is odd so i eliminated all the numbers containing 10.. earlier also i was getting the same result but ans does not matched. its a question of CAT 05. have a discussion and come to any conclusion

[parmindersaluja]

Originally Posted by 80199

i am not very much sure but question is saying that every digit in n is odd so i eliminated all the numbers containing 10.. earlier also i was getting the same result but ans does not matched. its a question of CAT 05. have a discussion and come to any conclusion

hey i posted this on page 10, anyways please confirm the answer
for all the digits to be odd, the thousands and the hundreds digit needs to be 1 and 1
so 1 + 1 = 2
the other two digits can be 1,3,5,7,9
but the sum should be divisble by 3 so
2 + 4 = 6---possible 2 numbers
2 + 7 = 9 --not possible
2 + 10 = 12---possible 5 numbers
2 + 13 = 15 --not possible
2 + 16 = 18 --possible 2 numbers
so 9 numbers possible

[IIM maniac]

Originally Posted by 80199

i am not very much sure but question is saying that every digit in n is odd so i eliminated all the numbers containing 10.. earlier also i was getting the same result but ans does not matched. its a question of CAT 05. have a discussion and come to any conclusion

Kindly check earlier posts as PUYs have solved this question, still FYI we have 9 values for it, 1113, 1119,1131,1137,1155,1173,1179,1191,1197.

[IIM maniac]

Originally Posted by IIM maniac

Thanks for the reply, so you mean to say we should consider the extreme case after which this equation does not get satisfied? is it? or in proper words , we should consider the case or co-ordinates that covers the full range which is in case of (4,0), (0,4), (-4,0) and (0,-4). thats why area is 16, is it?

@parmindersaluja,Can you please comment on this?