@gs4890 said:u have some lr sets ? plz post here
right now i am doing ques from test papers, no soft copy..so i cnt type that whole set..sorry! :)
@invincible2910 said:the radius of the front tyre of a gypsy is 5/pie m long and that of the rear tyre is 7.5/pie m long. what is the distance travelled by the gypsy after the first tyre has undergone 6 more revolutions than the rear tyre?150 m ; 100 m ; 144m ; 180 m
D/(2*pi*5/pi) - D/(2*pi*7.5/pi)=6
=>D/10-D/15=6
=>5D=6*150
=>D=180
=>D/10-D/15=6
=>5D=6*150
=>D=180
@krum said:D/(2*pi*5/pi) - D/(2*pi*7.5/pi)=6=>D/10-D/15=6=>5D=6*150=>D=180
thanks..i was doing a silly mistake 
@invincible2910 there was a mistake when i typed it should be 90 only 1960 cm2 for 360 deg and 490 cm2 for 90degree. But im getting 490cm2 and not 419 as there in the question.. 144 was for that tyre question, but i edited my answer after the calculation. its 180..
C1=15,C2=10
15n=10(n+6)
n=12
So distance 15*12=180
Find the ratio of a to b .
Please explain ur approach.
its not a spam i have attached a qn to make up for my sin 😃 i will try to become a regular here.
hi everyone ,ďťżi just cant express how grateful i am towards the quant thread i have lived the thread for about 1500 pages and i can say for sure it is the best teaching i ever got, i never made any comments and nothing but just studied it like a book. thanks you guys and i mean it all of you. its the strangest place where people teach their competitors in this rat race. i am overflowing with emotions probably best not said here. will come with that on my blog.
@pussu01 said:its not a spam i have attached a qn to make up for my sin i will try to become a regular here.hi everyone ,i just cant express how grateful i am towards the quant thread i have lived the thread for about 1500 pages and i can say for sure it is the best teaching i ever got, i never made any comments and nothing but just studied it like a book. thanks you guys and i mean it all of you. its the strangest place where people teach their competitors in this rat race. i am overflowing with emotions probably best not said here. will come with that on my blog.
a+b=-m; ab=1
y+z=-n; yz=1
=>y^2+z^2=n^2-2
(a-y)(b-y)(a+z)(b+z)
=>(ab-(a+b)y+y^2)(ab-(a+b)z+z^2)
=>(1+my+y^2)(1+mz+z^2)
=>(1+(mz+my)+m^2yz+(myz^2+y^2mz)+y^2+z^2+y^2z^2)
=>(1-mn+m^2-mn+1+n^2-2)
=>m^2+n^2-2mn
=>(m-n)^2
y+z=-n; yz=1
=>y^2+z^2=n^2-2
(a-y)(b-y)(a+z)(b+z)
=>(ab-(a+b)y+y^2)(ab-(a+b)z+z^2)
=>(1+my+y^2)(1+mz+z^2)
=>(1+(mz+my)+m^2yz+(myz^2+y^2mz)+y^2+z^2+y^2z^2)
=>(1-mn+m^2-mn+1+n^2-2)
=>m^2+n^2-2mn
=>(m-n)^2
@pankaj1988 said:I is a set of natural numbers1) is a set of 3 digit numbers leaves remainder of 2 when divided by 4.2) is divisible by 11 but not by 13.Number of elements in I ?
based on 1. number should be of 2*(2n+1)
based on 2. it is divisible by 11 hence 2n+1 must be of form 11(2k+1)
=>given number must be of form 22*(2k+1)
also (2k+1) must not be a multiple of 13
Now k=0,1,2,3......12,...................22 (21 values -> rejecting k=6 and k=19)
But we reject 0,1 as number is a 3 digit number. Hence 19 values of K.
ATDH.
@anytomdickandhary said:based on 1. number should be of 2*(2n+1)based on 2. it is divisible by 11 hence 2n+1 must be of form 11(2k+1)=>given number must be of form 22*(2k+1)also (2k+1) must not be a multiple of 13Now k=0,1,2,3......12,...................22 (22 values)But we reject 0,1 as number is a 3 digit number. Hence 20 values of K.ATDH.
sir, 0-22 23 values, so removing 0,1 leaves 21
out of these, k=6 and k=19 are divisible by 13
so 21-2=19 values
out of these, k=6 and k=19 are divisible by 13
so 21-2=19 values
@pussu01 said:its not a spam i have attached a qn to make up for my sin i will try to become a regular here.hi everyone ,i just cant express how grateful i am towards the quant thread i have lived the thread for about 1500 pages and i can say for sure it is the best teaching i ever got, i never made any comments and nothing but just studied it like a book. thanks you guys and i mean it all of you. its the strangest place where people teach their competitors in this rat race. i am overflowing with emotions probably best not said here. will come with that on my blog.
n^2-m^2..option 1
@krum said:sir, 0-22 23 values, so removing 0,1 leaves 21out of these, k=6 and k=19 are divisible by 13so 21-2=19 values
i missed on that 19 wala value....thanks for correction. I had thought only of 6 as the possible number that needed to be excluded hence I had written 22 values. (i.e excluding k=6).
Thanks for correction again.
These are classic ways by which CAT paper setters steal marks out of us.
ATDH.
Triplets consisting of three different numbers are formed from number 1 to 10.
1. How many of these triplets are such that sum of numbers is divisible by 3.
37,42,36,38
2.How many of triplets formed are such that sum of numbers is divisible by 9 and do not have 9 in them? 7,6,9,10
What are the no. of factors of 1000000 which are less then 1000 but do not divide 1000 ?
@mbajamesbond - aajao sir letz start nw
Consider the set S = {1, 2 , 3 €Ś €Ś.. 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?
(1) 3
(1) 3
(2) 4
(3) 6
(4) 7
@pankaj1988 said:Triplets consisting of three different numbers are formed from number 1 to 10.1. How many of these triplets are such that sum of numbers is divisible by 3.37,42,36,382.How many of triplets formed are such that sum of numbers is divisible by 9 and do not have 9 in them? 7,6,9,10
2. 7 ?
@soumitrabengeri said:Consider the set S = {1, 2 , 3 €Ś €Ś.. 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?(1) 3 (2) 4 (3) 6 (4) 7
7??