Official Quant thread for CAT 2013

MBA CAT 2024
30458
@hexagon said:
Vicky and Nicky run back and forth between the town hall and the county station at respective speeds of 12 kmph and 18 kmph. They start simultaneously - Vicky from the town hall and Nicky from the county station. If they cross each other for the first time 14 minutes from the start, at what distance from the county station do they cross each other for the fifth time?ans 4.2 km

my take will be 7-4.2=2.8 from county station and 4.2 from town hall
30459
@hexagon said:
Vicky and Nicky run back and forth between the town hall and the county station at respective speeds of 12 kmph and 18 kmph. They start simultaneously - Vicky from the town hall and Nicky from the county station. If they cross each other for the first time 14 minutes from the start, at what distance from the county station do they cross each other for the fifth time?ans 4.2 km
TH____________7KM____________CS
|.............2.8KM.......|..........4.2KM........|

in first meet V will cover = 12*14/60 = 2.8km
N will cover = 18*14/60 = 4.2km
total dist. b/w TH and CS = 7KM
total distance covered by them after the fifth meet = 7 + 4*2*7 = 63
now distance covered by N till the fifth meet = 18/30*63 = 37.8km
therefore N will be 2.8 km away from TH after the fifth meet....so distance b/w fifth meet and CS = 7 - 2.8 = 4.2 km
30460
@AsihekAdhvaryu said:
my take will be 7-4.2=2.8 from county station and 4.2 from town hall
my take will be 7-4.2=2.8 from town hall and 4.2 from county station

30461
@hexagon said:
Vicky and Nicky run back and forth between the town hall and the county station at respective speeds of 12 kmph and 18 kmph. They start simultaneously - Vicky from the town hall and Nicky from the county station. If they cross each other for the first time 14 minutes from the start, at what distance from the county station do they cross each other for the fifth time?ans 4.2 km
they meet for the 1st time at 2800m from Town hall, 4200m from county station.
their speeds are in ratio 2:3

when vicky travels 2x , nick travels 3x.
2x = 2800 => x=1400
3x = 4200
=> 2nd meeting pt at 1400m from CS , 3rd at town hall, 4th at 1400m from CS and 5th at 4200m from county station


30462

please solve it

30463
@hexagon said:
Vicky and Nicky run back and forth between the town hall and the county station at respective speeds of 12 kmph and 18 kmph. They start simultaneously - Vicky from the town hall and Nicky from the county station. If they cross each other for the first time 14 minutes from the start, at what distance from the county station do they cross each other for the fifth time?ans 4.2 km
First meeting point is same as the fifth meeting point. I think, it'll be easier to visualise now.
Answer will be 18(5/18)(14)(60)m = 4200m = 4.2 km :)

Team BV - Kamal Lohia
30464

What is the sum f this infinite series..


1 + 1/3 + 1/6 + 1/10...

Please explain how to approach aswell..
30465
@bvdhananjay said:
What is the sum f this infinite series..1 + 1/3 + 1/6 + 1/10...Please explain how to approach aswell..
1/1 + 1/3 + 1/6 + 1/10 + .... = 2/1*2 + 2/2*3 + 2/3*4 + 2/4*5 + ... = 2(1 - 1/2) + 2(1/2 - 1/3) + 2(1/3 - 1/4) + 2(1/4 - 1/5) + .... = 2(1) = 2
30466
@bvdhananjay said:
What is the sum f this infinite series..1 + 1/3 + 1/6 + 1/10...Please explain how to approach aswell..
2??
30467
@meenu05 said:
please solve it


Use the property that medians divide the triangle into 2 equal halves
So A(ACE)=2.40=80
Again BE is median of ACE
So A(BCE)=40
Now we need A(BCG)
Now BG || AD
So A(BCG) = 1/4(ACD = 40/4=10
Hence BEG = BCE - BCG = 40-10=30
Now we need to find area of FDE
We have A(FDE)/A(BGE) = (2/3)^2 [ Becoz D is midpoint ]
So A(FDE) = 40/3
Hence Shaded area = 30 - 40/3 = 50/3

Refer below links :
http://www.quantexpert.co.in/qenotes/quantnotes/geometry/54-intercept-and-midpoint-theorem.html
http://www.quantexpert.co.in/qenotes/quantnotes/geometry/53-triangles.html
30468
@catahead no dude asnwer is 50/3
30469

Can anyone please provide the link to " key of Wren and Martin". Sorry guys, I know I am posting it in wrong thread but I am in urgent need of it.

Thanks in advance.
30470
@meenu05 said:
@catahead no dude asnwer is 50/3
Sorry did not see the shade properly. Edited my post
30471
@heylady 2 is right.. please help me understand how to solve it.
30472
an = 2^n + 1 then (a1 + a2 + - - - - + a20) €“ a21 is?
30473
@mani0303 said:
an = 2^n + 1 then (a1 + a2 + - - - - + a20) €“ a21 is?
20??
((2^21 - 1 ) + 20 )- 2^21 +1
30474
@mani0303 said:
an = 2^n + 1 then (a1 + a2 + - - - - + a20) €“ a21 is?
18?

regards
scrabbler
30475
@manoj_msr said:
20??((2^21 - 1 ) + 20 )- 2^21 +1
((2^21 - 1 ) + 20 ) - (2^21 +1) hoga na?

regards
scrabbler
30476
@manoj_msr said:
20??((2^21 - 1 ) + 20 )- 2^21 +1
@scrabbler said:
18?regardsscrabbler
It's 18

P.S: Guys,It's not fair to solve one question/25 mins - please chip in guys:)
30477
@hexagon said:
@catahead hi how it is (2/3)^2 as FD=0.5 of BG
No, it is not.
Say DG=1
So CD=DE=2
So DE/GE = 2/3