Official Quant thread for CAT 2013

MBA CAT 2024
20446
@mailtoankit said:
78^379 = 8^379 = 2^1337 = 2^7 * (2^10)113 = 8*24^113 = 8*24 = 92?second waala explain karna ....
78^379 = 8^379

ye kaise ho gaya

what are you dividing by ?

plz explain karrna
20447
@joyjitpal said:
78^379 = 8^379ye kaise ho gayawhat are you dividing by ?plz explain karrna
bhai divide kuch nahi kiya...bas 8 ki power consider kiya hai....
20448
@amresh_maverick 1> 92
2> 0
20449

time , speed n distance

Q. P , Q and R start frm same place X at a km/hr , (a+b)km/hr and (a+2b) km/hr . If Q starts p hrs aftr P , hw many hrs aftr Q shld R start so tht both Q and R overtake P at same time ?
(a>0 , b>0)
Ans.-> pa/(a+2b)
need soln. plz
20450
@mailtoankit said:
78^379 = 8^379 = 2^1137 = 2^7 * (2^10)^113 = 8*24^113 = 8*24 = 92?second waala explain karna ....
abcdef is div by 11 if def -abc is div by 11 (here 3 digits)

123456789 - Take 9 digits at a time , sum will result in zero, hence div by 11
20451
@joyjitpal said:
78^379 = 8^379ye kaise ho gayawhat are you dividing by ?plz explain karrna
I have used :

78^379 = 2^379 * 39^379
2^379 ends in 88
39^379 ends in 59
so, 78^379 ends in 88*59 ..... 92

20452

can anyone plz help me wid the soln of tht ques

20453
@Abir1103 said:
time , speed n distance
Q. P , Q and R start frm same place X at a km/hr , (a+b)km/hr and (a+2b) km/hr . If Q starts p hrs aftr P , hw many hrs aftr Q shld R start so tht both Q and R overtake P at same time ?
(a>0 , b>0)
Ans.-> pa/(a+2b)
need soln. plz
in p hours P travels ap Kms
now Q will reach P in ap/b hours
distance travelled by Q=ap(a+b)/b point of overtake
and R will reach distance this distance in ap(a+b)/b(a+2b) hours
time difference between r and q=ap/b (1-(a+b)/(a+2b))
=ap/b(b/(a+2b))
=ap/a+2b
20454
Rem when 123123123.......300 digits is divided by 504 ?
20455
@amresh_maverick said:
Rem when 123123123.......300 digits is divided by 504 ?
504=7*72=7*8*9
123123.... mod 7=0
123123.... mod 9=
6*100 mod 9=6
123 mod 8=3
7a=8b+3
b=4
56d+35=9c+6
56d+29=9c
d=8
504n+483
483?
20456
@amresh_maverick said:
OA :1> 922> 0
for 2nd que , the difference between the odd and even placed numbers is zero ..so it is divisble by 11, hence remainder is zero
20457
@Subhashdec2 said:
504=7*72=7*8*9123123.... mod 7=0123123.... mod 9=6*100 mod 9=6123 mod 8=37a=8b+3b=456d+35=9c+656d+29=9cd=8504n+483483?
bhai yaha pe B equal to 4 kaise

explain karna thora
20458
@joyjitpal said:
bhai yaha pe B equal to 4 kaiseexplain karna thora
7a=8b+3

when b=4 ; 32+3 =35 is div by 7 , isliye b=4


20459
@amresh_maverick said:
7a=8b+3when b=4 ; 32+3 =35 is div by 7 , isliye b=4
then we got to take the lcm
of 7 and 8 i.e 56 isi liye


56d+35=9c+6

thanx
20460
@amresh_maverick i donot know the answer
20461
@sagarcat donot know the answer....explain ...how?
20462

In a right angled triangle,side opposite to 90 degree angle is hypotenuse,but How to identify which side is base and which one is perpendicular ? ( i want to know so that i can apply sin,cos formulae to know the unkown side ).

20463
@nole said:
In a right angled triangle,side opposite to 90 degree angle is hypotenuse,but How to identify which side is base and which one is perpendicular ? ( i want to know so that i can apply sin,cos formulae to know the unkown side ).
you can find sin cos tan irrespective of that, those are just names.
for eg sin@=side opposite/hypotenuse. Here you have to know @ if hyp is given. Do you have any specific doubt other than this? I am not able to completely gauge your doubt.
20464
@emanresu I mean a triangle is given where it is lying on its hypotenuse( generally it is lying on its base ). now i want to find length of a side (i don't know what should i consider that side ? ) should i take that side as base or perpendicular ? for sine u said sin= angle opposite/hypotenuse.i guess that is the approach rite?
20465

@nole said:
@emanresu I mean a triangle is given where it is lying on its hypotenuse( generally it is lying on its base ). now i want to find length of a side (i don't know what should i consider that side ? ) should i take that side as base or perpendicular ? for sine u said sin= angle opposite/hypotenuse.i guess that is the approach rite?
see, forget about base and perpendicular,these names, as there is not much use in a right angled triangle. to explain you simply
consider rt ang triangle ABC right angled at B
then SinC=AB/AC CosC=BC/AC
similarly SinA=BC/AC CosA=AB /AC

Draw a diagram & then try to understand.