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Quant Challenge for CAT 2008 Aspirants
by pradeep pandey - Thursday, 21 February 2008, 09:58 PM
  cat 2010 cat 2009 mba 2009 xat 2010 quant challengeThe following quant challenge has been prepared for all CAT 2008 aspirants by Mr. Pradeep Pandey, an experienced quant trainer of MBA aspirants and author of the book "Quantitative Aptitude for CAT." Mr. P. Pandey has been creator of many interesting mathematics and data interpretation problems and he has graciously agreed to share many of them with students on TG to help them in their CAT preparation. So try these problems and post your solutions in this thread itself. Mr. P. Pandey will post the solutions to them later- Total Gadha

 

Do you really want to crack CAT? Try these out first smile

quant challenge for cat 2008

 

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Re: Quant Challenge for CAT 2008 Aspirants
by lets try - Friday, 22 February 2008, 10:02 AM
  Hi,

I am not good at P&C...so trying all others:

1)......P&C
2) (2)60
3) (3)20
4) (5)21
5) (5)None of these...thoughtful
speed of A can be 36 or 64 if both the buses are moving.
if one reaches N and stops then also the speeds can be 34 or 69. so no unique answer. hence none of these thoughtful
6) my answer is coming out as 51.9%...unable to figure out my mistake
7) (3)52
8) (2)1 ....if we consider both numbers are equal (i.e. 13) otherwise 0.
9) (4)80
10) thoughtful
11) (4)24
12) (1)0
13) (2)68
15) (4)3......4th point is on the x-axis.

Thanks..
Re: Quant Challenge for CAT 2008 Aspirants
by Raghavan Prabhu - Friday, 22 February 2008, 12:17 PM
 

I tried the first sum and got the answer as 126 . In fact, we can eliminate the options 2,3 & 4 due to the following reasons.

1.) options 2 & 4 are the same and 1296 is the max possible solution one can get when 4 identical dice are thrown together. But the question asks for " "How many different ways",so it just can't be the answer as it would involve many repetitions.

2.) 4> 1296 , which leaves behind with options 1 & 5.

Of course you have to work out from here on to find the solution, but eliminating three options would make you more confident.

Correct me if my answer is wrong TG smile

 

Re: Quant Challenge for CAT 2008 Aspirants
by Little Star - Friday, 22 February 2008, 03:04 PM
  2th ans=60, angle is same in same arc
4th ans=21
13th ans=           (x-y)(x+y)=17807*2 so no any value possible for X & Y

Little Star
Re: Quant Challenge for CAT 2008 Aspirants
by Little Star - Friday, 22 February 2008, 03:32 PM
  1st ans 6^4 
Re: Quant Challenge for CAT 2008 Aspirants
by Little Star - Friday, 22 February 2008, 04:01 PM
  15 th ans x^4 + x^3 + 5x^2 +x-4=2x^4+x^3+x

so x^4-5x^2+4=0
put x^2=t

t^2-5t+4=0 so t=4 ans 1    so x=+/- 1 and +/- 2

Little Star
Re: Quant Challenge for CAT 2008 Aspirants
by Little Star - Friday, 22 February 2008, 05:06 PM
  7th ans

if in school n student then
n(n-1)-(n-2)(n-3)=102

so n=27         so total belt=702
if n=26          so ans=702-650=52

Little Star
Re: Quant Challenge for CAT 2008 Aspirants
by Little Star - Friday, 22 February 2008, 05:27 PM
  7th ans

if in school n student then
n(n-1)-(n-2)(n-3)=102

so n=27         so total belt=702
if n=26          so ans=702-650=52

8th ans =0

Little Star
Re: Quant Challenge for CAT 2008 Aspirants
by Prateek Agarwal - Friday, 22 February 2008, 08:44 PM
 

15th. 3 points

There r 4 points in all but only 3 above x asis

(1,4),(2,42),(-1,0) and (-2,22)

 

 

Re: Quant Challenge for CAT 2008 Aspirants
by Small Wonder - Friday, 22 February 2008, 08:58 PM
 

A sincere request to all friends,

Please post the approach as well.

Small

Re: Quant Challenge for CAT 2008 Aspirants
by PRIYANKA KESHRI - Friday, 22 February 2008, 09:14 PM
 

1) 4^6

2) 60

3)20

4)21

5)none

7)52

12)13

13)0

15) more than 3

 

Re: Quant Challenge for CAT 2008 Aspirants
by sMILING GadhUUU - Friday, 22 February 2008, 10:45 PM
 

1)6^4
2)60
3)20
4)21
5)64 km/hr
6)26.66 %
7)52 
8)3
9)80
10)31
11)30
12)17
13)1
14)68
15)2


SOLUTIONS:
1)6*6*6*6

2)Consider the center of the two concentric circles as 0.
Now  Triagle OBC would be an equilateral triangle.(2*30)
Now Consider the triangle OAD, With Triangle OAB and Triangle OCD
you would get Angle OAD=120

Therefore Angle AQD=120/2= 60

3)Number of Minutes in 4 hours = 240
All alarms are set with the integral muntiples of minutes.

We are basicallly looking for the number of factors of 240.

240=2^4 *  3 * 5

Number of Factors = 5 * 2 * 2 =20

Therefore Maximum num of Clocks in the room= 20

4)Any number such that its last two digits are 0 would be divisible by 25(25*4).
Therefore we have to take care of the last two digits only to get the remainder.

The Muntiplication of 21,we would always get the Last to digits as
21 41 61 81 01 and then it would start repeating.

This has cyclicity of 5

Consider 21^21
This has cyclicity 21mod5=1

Therefore 21.

So everytime we raise a power of 21 we would get a remainder 21
21*21=(rem)21
If we repeat this 21 times the final remainder would be 21.


5)Total travelling Time of  Bus B = 6:30 am to 9:30 pm = 15 hours
Total distance it had reduced ,between Bus A and B = 5 *15( 5 kms every hour)
=75 km
At 9:30 pm the distance between A and B.=21.

So at Morning from 5:00 am to 6:30 am (1 and 1/2 hours).
The distance Bus A had travelled = 75+21 = 96 kms
Distance Travelled by A in 1 hour = 64 km/hr

6) Let  P,A and G be the respective volumes of Poison,Alcohol and Glycerine in V1,V2,V3
Total  Poison in V2=
2P/9[  2/3(P/3)]     +   P/81[1/3(1/3(1/3(P/3)))]
=19/81

Total Alcohol in V2=
2A/3 + A/27[ 1/3(1/3(A/3))]
19/27

Therefore Ratio of Poison and Alcohol in V2

=
{ (19/81)P/(19/27)A)

But  P:A = 4:5
Therefore
(27*4) / (81*5) =   26.66 %


7)When 2 Students, A and B were absent No. of Bands left = 102
If there are x + 2 students in the class, 

These 102 bands include
x bands which student A would tie to the Class
x bands which the class would tie to A.

x bands which Student B would tie to the Class
x bands which the class would tie to B.

1 band which A ties to B.
1 band which B ties to A.
 
=4x +2 bands =102

Therefore   x= 25,
Total students in the class =27.

No one bands left when only 1 student A is absent(B turns up) =
26 bands which B ties to rest of the class
26 bands which rest of the class ties to A.

=52

8)1.    37,1,1,
   2.    33,3,3,
   3.   21,15,3 

9)To divide by 999 means to divide into groups such that each group has 999,Whatever is left is the remainder.
Let us Suppose that We Divide this Number 123456789123456789341 by  1000.

Now this would leave a remainder 341.Now number of  Small parts of 1000 Formed is 123456789123456789
Now if we want the Groups to be of 999,Each of  123456789123456789  groups of 1000 will give us 1 member each.
Along with this we have 341.

Now we again divide What we have received 123456789123456789 members into group of 1000.
123456789123456 groups of 1000 Formed at the same time 789 members are left.

If we again want this to be divided into groups of 999 instead of 1000 We will get 123456789123456 .Along with this we will get the 789 Who could not be divided.

If we follow this process in recursion( Beleive me if we make a program using recursive function to find a solution to this ,it  is a cakewalk).
The final group of 1000 members that we would get will be 123.
So we again take 1 member out from these 123 groups of 1000 members each.

Finally we have been able to sort out the whole lot that we have into groups of 999..
Now let us also sort out into GROUPS of 999, the members that remained in each of the Rounds.


   341
+ 789
+ 456
+ 123
+ 789
+ 456
+ 123
_____

3077
____


Now if we divide these 3077 into groups of 999, We could form 3 groups of 999, But still 80 members will be left.

Which is the remainder.

10)Is very easy to understand if you draw the diagram, Break Everything into Right Angled Triangle and Equilateral Triangles

                                       E    
                                         *
                                      * 60 *
                                   *            *
                                 *                 *
                              *                       *
                           *                             *        A
                        *                           2       *
                     * 2    Z          5               * 60  *      
             D  ******************************* * *******  *    X
                    *60     *90              *     *  Y
                      *      *                 *  * 
                         *   *                 * *  4
                           * *   5            *   
                             * **************30 
                         C                          B



 Consider  the equilateral triangle formed by the ED X 


Now,the lengh of one of the arms of this equilateral triangle EDX is DZ + ZY + YX   = 2 + 5 + 2 = 9

Therefore ED = EX = 9.
But  EX = EA   +  AX
 
SO EA=    9  -  2   =  7

Now we have EA  = 7
                      ED  =  9
                      AB  =  4
                      BC  =  5
                      CD  =  6

Therefore perimeter of ABCDE = 31

 

11)
Let the number be abcd.

Now a+b+c+d=18
b+c = a+d
b+d  = 5 ( a + c )

a+c + 5(a+c) =6(a+c) = 18
Therefore a+c =3
b+d =15
Possible vaules of b And d

b      d
__  ___  
9      6
8      7
7      8
6      9

Therefore sum of all the possible values is 30.


12) D is one of the vertices of the triangle and it is fixed.
Now we want to have two lines from D on BA and AC  such that they should be of minimum length.
The minimum length can be achieved if the two sides of the triangle  FDE coincide with XD AND DY. So the three sides of the Triangle DFE would be of lengths 2.5,6 and The diagonal  XY of  the rectangle XDYA
13 is eleminated as the perimeter has to be more than 6 + 6+ 2.5 =14.5
Therefore Option 2) 17 cms.
So the


13) 1
(17805,17803)

14)To find the Heaviest box the weight of the other boxes should be as much close to each other such that the  difference between the Heaviest and the one less than the Heaviest is the maximum.

Consider the Sum of weights of the 2 lightest Boxes = 122 ,So the lightest box cannot be 61 kg,It has to be less than that.
If the lightest box is 60,the second box would be 62,Similarly considering the Sum of weights the next boxes would be 64 and 65.. Now comparing all the Sum of boxes ,
The heaviest box would be of weight =68 Kg.
Therfore 60+62 =122     60+64 =124   62+64 =126  60+68=128 60+65 = 125 62+65=127 65+64 =129 62+68 =130 64+68=132 65+68 =133
All Comply.

So Heaviest box =68 kg.


15)Solving the two equations we will get x^2 - 5x + 4 = 0

No of positive values of X. ,
 x=1 and x=2

 

 

Re: Quant Challenge for CAT 2008 Aspirants
by Neo Sinha - Friday, 22 February 2008, 10:55 PM
 

Sol. 5)


The Time taken by A T(A)=16.5 hrs (From 5:00 AM To 9:30 PM)

The Time taken by B T(B)=15 hrs (From 6:30 AM To 9:30 PM)

By question, Velocity of B, V(B)=5+Velocity of A =5+V(A) km/h

  S(B)-S(A)=21

 [V(A)+5]*15-V(A)*16.5=21 {Since velocity(speed)=Displacement(Distance)/Time

  On solving, we get V(A)=34 km/hr which is choice(1)

Neo Sinha

 

 

Re: Quant Challenge for CAT 2008 Aspirants
by Neo Sinha - Friday, 22 February 2008, 11:42 PM
  some of ur answers are wreong viz. q5) & q15. resolve it.
Re: Quant Challenge for CAT 2008 Aspirants
by rishu batra - Saturday, 23 February 2008, 02:44 AM
 

1) (1) 126

2) (2) 60

3) (3) 20

4) (5) 21

5)

6)

7) (3) 52

8) 4 possibilities-(18,3),(19,1),(7,4),(13,13)

9) (4)  80

10) (1) 35

11) (4) 24

12)

13) (1) 0

14) (2) 68

15) 3 above x-axis

 

Re: Quant Challenge for CAT 2008 Aspirants
by jai prakash - Saturday, 23 February 2008, 02:39 PM
  Explain Q8.
Re: Quant Challenge for CAT 2008 Aspirants
by Raghavan Prabhu - Saturday, 23 February 2008, 09:01 PM
  Question No 5 :

No information has been given as to whether the buses are still on the move or one of the bus has reached. So its better and safe to assume that the buses are still on the move. Also, no information has been given as to which bus leads the other. So we get two different equations considering that either bus A leads or B leads and obviously two different answers 64km/hr and 36km/hr respectively. So as " lets try " has pointed out in the very first post , there is no unique answer.
It has to be option 5.) none of these


Re: Quant Challenge for CAT 2008 Aspirants
by shweta jain - Sunday, 24 February 2008, 05:23 PM
  hi
i m weak in quants
plz help me out from my doubts.

please explain me the q no 4,8,9.
in 8th ques sum should be 39??
but smiling gadha has given the options whose sum is not 39.plz explain this.

IN QUES 10 AND 11, I M GETTING ANS 35
WHICH IS THE RIGHT ANS.

how to find the factors of 35614 in ques 13.

thanks!!
Re: Quant Challenge for CAT 2008 Aspirants
by Nishant Verma - Wednesday, 27 February 2008, 12:09 AM
 

HI

Here are my answers..

1) 6*6*6*6

2)

3) 20

4) 21

5) None of these

6) 51%

7) 52

8)

9)

10)

11) 24

12)

13) 2

14) 67

15) 3 points

 

Re: Quant Challenge for CAT 2008 Aspirants
by rohit Kaushal - Thursday, 28 February 2008, 05:44 PM
 

1)6^4
2)60
3)20
4)21
5)64 km/hr
6)26.66 %
7)52 
8)3
9)80
10)31
11)30
12)17
13)1
14)68
15)2

Re: Quant Challenge for CAT 2008 Aspirants
by jahir alam - Thursday, 28 February 2008, 09:56 PM
 

1.126

2.60

3.20

7.52

8.3

13.1

15.3

Hi TG the quant challenge is really very useful forCAT 08 aspirants like me.

Expecting similar Quant Challenges in future before CAT.Thanks TG and thanks Pradeep Pandey.

Re: Quant Challenge for CAT 2008 Aspirants
by pradeep pandey - Tuesday, 4 March 2008, 08:58 PM
  Hi All,

Please download the answers to the above questions.

P. Pandey
Re: Quant Challenge for CAT 2008 Aspirants
by asad khatau - Monday, 10 March 2008, 12:00 PM
 

Hello frnds ,m relatively new to TG.i have just strtd my prep fr CAT08.I have a doubt in Number SYS .Actually i have multiple methods fr doing dis type of problem ,bt i seem 2 b getin diff ans with them.Can ne 1 plz help me out to solve it in a simple way?

Q.Find the units digit of the following

1.(12^55)/(3^11) + (8^48)/(16^18)

Re: Quant Challenge for CAT 2008 Aspirants
by bimal mohan - Monday, 10 March 2008, 10:04 PM
 

hi  asad,

   here  is   my  approach  :

     12^55/3^11 +8^48/16^18

  =3^55*4^55/3^11    +    2^144/ 2^72

  = 3^44*4^55 +2^72

  ---> 1*4 +6 --> 0

Re: Quant Challenge for CAT 2008 Aspirants
by Mrinal Kumar - Thursday, 20 March 2008, 09:42 AM
 

 "D is one of the vertices of the triangle and it is fixed.
Now we want to have two lines from D on BA and AC  such that they should be of minimum length.
The minimum length can be achieved if the two sides of the triangle  FDE coincide with XD AND DY. So the three sides of the Triangle DFE would be of lengths 2.5,6 and The diagonal  XY of  the rectangle XDYA
13 is eleminated as the perimeter has to be more than 6 + 6+ 2.5 =14.5"

 

May I know what concept made you to infer this.

Re: Quant Challenge for CAT 2008 Aspirants
by swapnil srivastava - Friday, 30 May 2008, 08:01 PM
  great article...
can you give me the proper solution of this quest.
Q.which is greater: 570 or 750 ?

thanx...
Re: Quant Challenge for CAT 2008 Aspirants
by siva nagarajan - Friday, 6 June 2008, 07:13 PM
 

hi sir,

for q1 can u please explain the solution as i feel for case 2 in place of 6C1,it  should be 6C3. Please clarify.

for q6. I did not understand why have u assumed v1,v2,v3 are (3^4*4),(3^4*5),(3^4*6). please clarify.

 

Re: Quant Challenge for CAT 2008 Aspirants
by Neo Sinha - Saturday, 7 June 2008, 02:53 AM
 

570=(57)10 =(625*125)10             

750= (75)10=(343*49)10

          so 570       >    750

 

 

Neo

 

Re: Quant Challenge for CAT 2008 Aspirants
by swapnil srivastava - Saturday, 7 June 2008, 06:29 PM
  thanx for ur solution.
Re: Quant Challenge for CAT 2008 Aspirants
by Ankit agarwal - Wednesday, 11 June 2008, 03:29 AM
 

solution for q.7.

102 bands were nt utilised at 10.a.m. it means dere must be 50 students in d class because enery studenr will fasten the band 2 other.50 students will tie to both the student means 50*2=100 are used n those who r absent will fasten to each other.remaning two will b used there.at 11.a.m a student comes means 50 students will tie band 2 him.102-50=52 will  remain

Re: Quant Challenge for CAT 2008 Aspirants
by Anshu Airan - Tuesday, 21 October 2008, 12:03 AM
 

Hello Sir,

In question 8, there is one more result if we take H = 1 i.e. 7 & 4

7 * 4 = 28

7 + 4 + 28 = 39

 

Please clarify

Re: Quant Challenge for CAT 2008 Aspirants
by aritreyee chaudhuri - Saturday, 14 November 2009, 11:23 AM
 

hello sir

shouldnt there be 4 possible pairs for Q.8??

i guess these should be the pairs:

1, 19 -> lcm=19

3, 18 -> lcm=18

13,13 -> lcm=13

4, 7 -> lcm=28

please clarify this.

Re: Quant Challenge for CAT 2008 Aspirants
by Ajay Parsana - Wednesday, 13 October 2010, 03:29 PM
 

Hi,

Answer to question 5 is None of this because in question it is mentioed that speed of bus B is more than that of bus A and none of the option from 1 to 4 have speed less than 5 so only choice left is None of this.

Ajay Parsana