@bhatkushal said:guys try doing it without the options and please do share the approach.......
0?
@bhatkushal said:guys try doing it without the options and please do share the approach.......
none of these ...
The HCF of 2472, 1284 & a third number "N" is 12. If their LCM is 2^3 x 3^2 x 5 x 103 x 107, then the number "N" is:
a) 2^2 x 3^2 x 7 b) 2^2 x 3^3 x 103 c) 2^2 x 3^2 x 5 d) None of these
Do share the approach
@anilapex said:The HCF of 2472, 1284 & a third number "N" is 12. If their LCM is 2^3 x 3^2 x 5 x 103 x 107, then the number "N" is:a) 2^2 x 3^2 x 7 b) 2^2 x 3^3 x 103 c) 2^2 x 3^2 x 5 d) None of theseDo share the approach
2472 = 2^3*3*103
1284=2^3*3*107
so N=2^2*3^2*5
1284=2^3*3*107
so N=2^2*3^2*5
@anilapex I think None of these.
Now the HCF of 1284 and 2472 is 12.
If the total HCF is 12 and the HCF of these two Numbers is 12,then the resulting number should have the combination of 2^2*3.
I am not sure. Waiting to get the right answer.
@krum said:2472 = 2^3*3*1031284=2^3*3*107so N=2^2*3^2*5
How are you inferring the last step?
I didn't get that altogether.
@anilapex knowing that the powers in LCM needs to be equal or more than the individual numbers, option a and b are ruled out.....option c and d possible......generally none of these something u can overllook when u have less time 😃
rest u always have the standard method..
@anilapex 2472 = 2 x 2 x 2 x 3 x 103
1284 = 2 x 2 x 3 x 107
Since the HCf of these 3 numbers is 12, so N must at least have 2 x 2 x 3.
Now, from the given LCM, we need 3 2s, 2 3s, one 5, one 103 and one 107. So from N, we need one 3 and a 5 only. Rest all are already there. So, N must 2x2x3x3x5 = 180.
Just subscribed to this thread... I thought i was early in starting my prep for CAT 2013.. and i find that this thread had 611 pages of posts.. 
@anilapex said:The HCF of 2472, 1284 & a third number "N" is 12. If their LCM is 2^3 x 3^2 x 5 x 103 x 107, then the number "N" is:a) 2^2 x 3^2 x 7 b) 2^2 x 3^3 x 103 c) 2^2 x 3^2 x 5 d) None of theseDo share the approach
for any no.s
HCF*LCM=n1 *n2* n3
PS: not sure its long tym since i made the last post
sabhi bhaiyo ko ________/\________
HCF*LCM=n1 *n2* n3
PS: not sure its long tym since i made the last post
sabhi bhaiyo ko ________/\________
@bhatkushal said:guys try doing it without the options and please do share the approach.......
0 as all have to be 1 in order to make 150
else taking any in 25 P or 50 P will make sum less than 150
else taking any in 25 P or 50 P will make sum less than 150
@piyushrohella12
@piyushrohella12 said:for any no.s
HCF*LCM=n1 *n2* n3
PS: not sure its long tym since i made the last post
sabhi bhaiyo ko ________/\________
If u do this -- u will not reach to any answer.
I assume that the formuale is: The product of two numbers=HCF*LCM
I am really bad in Quants -- would be starting from scratch after 5years with Quantum CAT from next Monday onwards.
@maneeshp said:bro,are you sure it will work for 3 numbers?
@SBRPhoenix said:@piyushrohella12 If u do this -- u will not reach to any answer.I assume that the formuale is: The product of two numbers=HCF*LCM
ya man thats why i added the "PS" thing
Its been ages my mistake
applies to only 2 no.s :)
Its been ages my mistake
applies to only 2 no.s :)
@piyushrohella12 said:ya man thats why i added the "PS" thing
Its been ages my mistake
applies to only 2 no.s
Are sorry boss... i completely overlooked that
@anilapex said:The HCF of 2472, 1284 & a third number "N" is 12. If their LCM is 2^3 x 3^2 x 5 x 103 x 107, then the number "N" is:a) 2^2 x 3^2 x 7 b) 2^2 x 3^3 x 103 c) 2^2 x 3^2 x 5 d) None of theseDo share the approach
2472=12*2*103
1284=12*107
d lcm of dis 3 nobs consist of three 2's, two 3's, one 5, one 103 n one 107
now from above two nob we get three 2's, one 3, 103 n 107
so for lcm to b 2^3 x 3^2 x 5 x 103 x 107 we require one additional 3 n one 5
as option a and b dont consist of 5 dey can be eliminated and also the third nob has tob multiple of 12( 2^2*3) so ans must be 2^2*3*3*5 i e c
1284=12*107
d lcm of dis 3 nobs consist of three 2's, two 3's, one 5, one 103 n one 107
now from above two nob we get three 2's, one 3, 103 n 107
so for lcm to b 2^3 x 3^2 x 5 x 103 x 107 we require one additional 3 n one 5
as option a and b dont consist of 5 dey can be eliminated and also the third nob has tob multiple of 12( 2^2*3) so ans must be 2^2*3*3*5 i e c
@anilapex said:The HCF of 2472, 1284 & a third number "N" is 12. If their LCM is 2^3 x 3^2 x 5 x 103 x 107, then the number "N" is:a) 2^2 x 3^2 x 7 b) 2^2 x 3^3 x 103 c) 2^2 x 3^2 x 5 d) None of theseDo share the approach
@SBRPhoenix said:Are sorry boss... i completely overlooked that
@maneeshp said:bro,are you sure it will work for 3 numbers?
i rekon we can not find the ans to the question as
the nos are
n1= 2^3 * 3 * 103
n2=2^2*3*107
N=??
HCF= 2^2*3
so N atleast is 2^2*3*...........
LCM=2^3 x 3^2 x 5 x 103 x 107
hence N= 2^2*3^2*5*......=180*....
here N=180/180*103/180*107/180*103*107
so NOT may b
the nos are
n1= 2^3 * 3 * 103
n2=2^2*3*107
N=??
HCF= 2^2*3
so N atleast is 2^2*3*...........
LCM=2^3 x 3^2 x 5 x 103 x 107
hence N= 2^2*3^2*5*......=180*....
here N=180/180*103/180*107/180*103*107
so NOT may b
@ShriR said:Just subscribed to this thread... I thought i was early in starting my prep for CAT 2013.. and i find that this thread had 611 pages of posts..
same hear bro..jst subscribed..
@bhatkushal said:Guys soln on this one .........please tell the approach as well.........
before joining of new students..
let per day expenditure of each student be x, n total expenditure b y den
42x=y....(1)
after joining of 13 new students
(42+13)(x-3)=y+31..(2)
on solving above two equations,
y=(196*42)/13=633.23
let per day expenditure of each student be x, n total expenditure b y den
42x=y....(1)
after joining of 13 new students
(42+13)(x-3)=y+31..(2)
on solving above two equations,
y=(196*42)/13=633.23