  TestFunda will be down for maintenance on Tuesday, 04th April 2017 from 08:30 AM to 12:00 PM.  4Likes 
Available on PM
Quant Thread For CAT 2011
Hi People,
Please use this thread for Quant discussions (for CAT 2011 and other MBA exams of the coming season)
Some very simple rules
1. Please do not spam.
2. Try to post a solution and not just the key.
3. Post any and every doubt you have, it helps everyone learn.
4. Be polite and civil
Good Luck!

The Following 3 Users Say Thank You to sumit2goody For This Useful Post:
ajay_h (28Mar11), rashi_goyal (16May11), tusharsem (01Apr11)

Can any1 please tell me hw to go about this problem? what will be the remainder when 2^133 is divided by 133? 

Virtuoso
Originally Posted by shon.n Can any1 please tell me hw to go about this problem? what will be the remainder when 2^133 is divided by 133? You must be using Euler's theorm for solving these kind of questions.
I guess you can find related info in one of TF's thread.


Time Speed distance
Here`s an interesting question 
A man leaves office daily at 7pm, when a driver comes to pick him up from his home. One day he gets free at 5.30pm and instead of waiting for the driver, he starts walking towards his home. On the way, he meets the car and returns home. He reaches home 20min earlier than usual. Had he become free at 6.00pm, how much early would he have reached ?
a. 11 mins
b. 13 mins
c. 40/3 min
d. 40/7 mins
e. None of these.
Kindly post your solution alongwith the method. Assume constant speed for both the man and the car

The Following User Says Thank You to mohasan02 For This Useful Post:

TSD question
C`mon guys.... looking forward to some innovative solutions ! It would be great to compare approaches !!...


Originally Posted by mohasan02 C`mon guys.... looking forward to some innovative solutions ! It would be great to compare approaches !!... Answer : 40/3 minutes
I just took few variables here!
a is the speed of the man , b is the speed of the car
t = time taken by the car
T = in next case to total time taken by man and car ( when he starts at 6:00 pm in evening )
a ' is the next time speed of the man and b' is the next time speed of the car
for first case we got this equation :
a/v + b/x = t 20
t = l/x l = total length of the road.
if you solve above equation we get this the ratio of v and x
i.e v/x = 11/9 ( given a=90v)
Now , in next case :
T = a'/v+b'/x ( a'=60v and b'=l60v)
if we solve this eqaution we get this :
T = l/x  40/3
which is why man will take 40/3 minutes lesser time to reach in next case.


i did it like this..
133=2^7 + 2^2 + 2^0
now... 2^133 / (2^7+2^2+2^0)...
now 2^133= (2^7)^19
4m denominator we take 2^7 as common..and we get deno in form
2^7(1+2^5+2^7)..
now..remainder when 2^133 is divided by 2^7 is zero..so
the overall remainder will be zero..


Originally Posted by Mandreak2k3 Answer : 40/3 minutes
I just took few variables here!
a is the speed of the man , b is the speed of the car
t = time taken by the car
T = in next case to total time taken by man and car ( when he starts at 6:00 pm in evening )
a ' is the next time speed of the man and b' is the next time speed of the car
for first case we got this equation :
a/v + b/x = t 20
t = l/x l = total length of the road.
if you solve above equation we get this the ratio of v and x
i.e v/x = 11/9 ( given a=90v)
Now , in next case :
T = a'/v+b'/x ( a'=60v and b'=l60v)
if we solve this eqaution we get this :
T = l/x  40/3
which is why man will take 40/3 minutes lesser time to reach in next case. i appreciate your response. The answer is correct. However the concept is a bit wrong !!.... To help you understand the concept , try another question on tha same lines.
Q. Every Day Rupa's Husband meets he at the bus stop at 6:30 p.m sharp and drives her back home. Today Rupa left Office early, caught an earlier bus and reach the bus stop at 6:00 pm. Not seeing her husband, she began walking back home. Her husband met her on the way, and they reach the home 10 minutes earlier than usual.
1> How Long had Rupa been Walking?
2.>What is the Speed Of Rupa And her husband.(can we determine this or not?)
I will post the solution for both the question and the explanation tomorrow. Looking forward to some good responses !


Explanation to Rupa`s question
Q. Every Day Rupa's Husband meets he at the bus stop at 6:30 p.m sharp and drives her back home. Today Rupa left Office early, caught an earlier bus and reach the bus stop at 6:00 pm. Not seeing her husband, she began walking back home. Her husband met her on the way, and they reach the home 10 minutes earlier than usual.
1> How Long had Rupa been Walking?
2.>What is the Speed Of Rupa And her husband.(can we determine this or not?)
Explanation :
Let bus stand be at point A and home be at point B.
A .................................................................................................B
The sequence of events are  the car reaches A at 6.30pm and takes, say t mins, to reach B.
If Rupa leaves early i.e. at 6.00pm, which means the car meets Rupa while its on its way to the bus stand i.e. to say, car meets Rupa before 6.30pm. Now the question is  how much minutes before 6.30pm, does the car meet Rupa ???? We have two ways to calculate this  Method 1
Let,
speed of car = x
speed of Rupa = v
& let car meet Rupa t` mins before 6.30pm.
Therefore, the time saved as compared to previous days = 10 mins.
Corresponsding distance saved = 10x (since the time taken for the journey is evaluated from 6.30pm onwards. Any distance covered from 6.00pm to 6.30pm is a bonus !)
Distance saved = distance travelled by Rupa in (30t`)min + distance travelled by car (carrying Rupa) in t` mins.
10x = (30t`)v + xt` .................... (i)
Refering to the diagram below 
A........................X........................................................................B
a = AX= distance travelled by Rupa
b = BX= distance travelled by car after picking up Rupa at point X.
Thus 'a' was travelled by Rupa in 30t` mins, whereas 'a' would have been covered by car in t` mins. Thus for constant distance case 
(x/v) = (30t`)/t` ....................... (ii)
Solving (i) & (ii) 
we get t` = 5 mins
Thus Rupa travelled for 25 mins. Method 2
Now lets look at the shorter method !
Total time saved = 10mins
As seen from above  the car first reaches the bus stand and then starts home bound with Rupa. Having intercepted Rupa on the way itself, it saved some portion of to & fro travelling. As speed is constant, thus, refering to the above diagram... car was saved from covering distance 'a' twice due to the reduced to & fro travelling. This translates into a saving of 5 min in each direction !As the question deals with Rupa`s travel to home (one way travel), hence the car catches Rupa at 6.25pm !!... i.e. time travelled by Rupa = 25mins !!!


Hedonist
Originally Posted by mohasan02 Q. Every Day Rupa's Husband meets he at the bus stop at 6:30 p.m sharp and drives her back home. Today Rupa left Office early, caught an earlier bus and reach the bus stop at 6:00 pm. Not seeing her husband, she began walking back home. Her husband met her on the way, and they reach the home 10 minutes earlier than usual.
1> How Long had Rupa been Walking?
2.>What is the Speed Of Rupa And her husband.(can we determine this or not?)
I will post the solution for both the question and the explanation tomorrow. Looking forward to some good responses !
well.... this is a simple prob... simply see she starts at 6... while husband wld reach at 6 30.. but he saves ten min in all... so 5 in each direction.... so in effect they meet at 6 25.. so she has travelled for 25 min
as for speeds... u wld only know the ratio of their speeds and not the value in itself... here... speed of car to rupa is 5:1


Hedonist
Originally Posted by sameerkishan i did it like this..
133=2^7 + 2^2 + 2^0
now... 2^133 / (2^7+2^2+2^0)...
now 2^133= (2^7)^19
4m denominator we take 2^7 as common..and we get deno in form
2^7(1+2^5+2^7)..
now..remainder when 2^133 is divided by 2^7 is zero..so
the overall remainder will be zero.. O.O how can overall remainder be zero here..... ???? since when did an odd number divide an even number.... well i think u made some mistake in between ...check again....... this is more about euler theory... n^p1 form were p is a prime number....


Sequence & series
Heres another question from my side 
If S_{1} = (1/1x); S_{2} = (1/1S_{1}); S_{3} = (1/1S_{2}); & S_{n} = (1/1S_{n}_{}_{1}) [for n more than equal to 2; x not equal to 0]
Then S_{2}_{0}_{4} is ?
a. 1/(1  x)
b. 1/x
c. x
d. (x  1)/x


Originally Posted by mohasan02 Heres another question from my side 
If S_{1} = (1/1x); S_{2} = (1/1S_{1}); S_{3} = (1/1S_{2}); & S_{n} = (1/1S_{n}_{}_{1}) [for n more than equal to 2; x not equal to 0]
Then S_{2}_{0}_{4} is ?
a. 1/(1  x)
b. 1/x
c. x
d. (x  1)/x Op c.x
Period of 3, and as 204 is a mult of 3, S_{204}= S_{3}=x
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 
The Following User Says Thank You to shaunak_87 For This Useful Post:

Sequence & series
Originally Posted by shaunak_87 Op c.x
Period of 3, and as 204 is a mult of 3, S_{204}= S_{3}=x Bravo !!... Correct 

Hedonist
Originally Posted by mohasan02 Heres another question from my side 
If S_{1} = (1/1x); S_{2} = (1/1S_{1}); S_{3} = (1/1S_{2}); & S_{n} = (1/1S_{n}_{}_{1}) [for n more than equal to 2; x not equal to 0]
Then S_{2}_{0}_{4} is ?
a. 1/(1  x)
b. 1/x
c. x
d. (x  1)/x well well.....
periodicity 3, so S204 = S3
so option c 
The Following User Says Thank You to grimreaper For This Useful Post:

Hedonist
Q3
Lets number the Question ok.... so that way its easier to keep track..... Q3) N is the sum of squares of 3 consecutive odd numbers such that all digits of N are same. If N is a four digit no. then N is 
a)9999
b)7777
c)2222
d)5555
also what are the three consecutive numbers ?
p.s when someone posts a good solution feel free to show yo appreciation by clicking on the thanks button...... 
The Following 5 Users Say Thank You to grimreaper For This Useful Post:
anujcat (26Jul11), Chinmay010 (22Jul11), manasvr (11Aug12), pragyashubham (30Oct11), rgrohitgrover (01Aug11)

Originally Posted by grimreaper Lets number the Question ok.... so that way its easier to keep track..... Q3) N is the sum of squares of 3 consecutive odd numbers such that all digits of N are same. If N is a four digit no. then N is 
a)9999
b)7777
c)2222
d)5555
also what are the three consecutive numbers ?
p.s when someone posts a good solution feel free to show yo appreciation by clicking on the thanks button...... Op a)9999
The nos are 3331,3333,3335
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Hedonist
Originally Posted by shaunak_87 Op a)9999
The nos are 3331,3333,3335 DUde.... sum of "squares " of 3 consecutive odd numbers
hehehe...try again


Originally Posted by grimreaper Lets number the Question ok.... so that way its easier to keep track..... Q3) N is the sum of squares of 3 consecutive odd numbers such that all digits of N are same. If N is a four digit no. then N is 
a)9999
b)7777
c)2222
d)5555
also what are the three consecutive numbers ?
p.s when someone posts a good solution feel free to show yo appreciation by clicking on the thanks button......
Option d : 5555
The numbers are 41,43,45.


Hedonist
Originally Posted by madhu89 Option d : 5555
The numbers are 41,43,45. correct... post your method.... 
Similar Threads 
By baburaoapate in forum Quantitative Ability Forum
Replies: 633
Last Post: 25Sep14, 11:57 AM 
By tusharsem in forum Quantitative Ability Forum
Replies: 359
Last Post: 10Jan11, 9:00 PM 
By sumit2goody in forum Quantitative Ability Forum
Replies: 241
Last Post: 06Oct09, 11:09 PM 
By StreetHawk in forum Quantitative Ability Forum
Replies: 133
Last Post: 03Oct09, 9:02 PM 
By implex in forum Quantitative Ability Forum
Replies: 182
Last Post: 01Nov08, 12:50 PM Tags for this Thread Posting Permissions  You may not post new threads
 You may not post replies
 You may not post attachments
 You may not edit your posts
Forum Rules     
 
Bookmarks