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Quant Challenge for CAT 2008
by pradeep pandey - Monday, 28 April 2008, 04:16 PM
 

ca 2008 cat 2009 xat 2008 cat questions cat coaching cat paper cat delhi mba 2008If you are a serious CAT 2008 student, then you cannot escape the charm of TG.com. evil If you are a serious CAT trainer teaching at any coaching institute, then also you cannot avoid peeking at TG for good questions to cover in your classes. tongueout Whatever may be the case, you would be spreading education around you in your own way. The best thing about education in India is that it is touches everyone. And once education touches someone’s soul it permeates through his entire life and touches everyone around him in some way or the other. No matter which part of the world you are, we hope that TG.com helps you become a better person. In this regard, we continue with our Quant Challenge Series, prepared by Mr. Pradeep Pandey, an experienced CAT trainer in the industry. The pains that Mr. Pandey takes to create his questions can be easily gauged from their quality. While you gear up to attack the CAT paper in November, solving these questions will help you boost your confidence. Share these problems with your fellow CAT aspirants over a cup of coffee and enjoy some good problem-solving sessions.- Total Gadha


cat 2008 quant challenge

Mr. Pradeep Pandey is 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.
Re: Quant Challenge for CAT 2008
by TG Team - Monday, 28 April 2008, 06:18 PM
 

Q1 - 4

Q3 - 1

Q6 - 4

Q10 - 1

Re: Quant Challenge for CAT 2008
by Little Star - Monday, 28 April 2008, 06:27 PM
 

 

Q-10

ans is 25
value for a/b/c/d/e=7/5/9/3/1


Little Star

Re: Quant Challenge for CAT 2008
by TG Team - Monday, 28 April 2008, 06:54 PM
  Q9 - 2
Re: Quant Challenge for CAT 2008
by Kabuli Waala - Monday, 28 April 2008, 07:00 PM
  Q1 - 4
Q7 - 1
Q8 - 5
Q10 - 1
Q14 - 2
Re: Quant Challenge for CAT 2008
by Varun Hans - Monday, 28 April 2008, 09:52 PM
  Hello Sir,

Can u let me know frm where can i get your book on quants?

im stayin in chennai.
Re: Quant Challenge for CAT 2008
by Gagandeep Sharma - Tuesday, 29 April 2008, 12:28 AM
  Q1-4
Q3-3
Q8-5
Q10-1

This is what i got in my first look..
Re: Quant Challenge for CAT 2008
by Lord Aragorn - Tuesday, 29 April 2008, 12:41 AM
  some of the answers are:

1. 3

2. 1

3. 5 ( answer is 60)

4. 5

7. 1

8. 5

9. 2

10. 1

rest tomorrow... smile







Re: Quant Challenge for CAT 2008
by Lord Aragorn - Tuesday, 29 April 2008, 11:12 AM
 

q 11 -> 5 a travels for 170 hrs befoe meeting B at strtng point

q 12 -> 3.  1:3 similarity of trngles

 

Re: Quant Challenge for CAT 2008
by Jim Braddock - Tuesday, 29 April 2008, 11:20 AM
  1. 4
3. 60 (NOT)
9. 2
10. 1
13. 4
Re: Quant Challenge for CAT 2008
by pradeep krishnan - Tuesday, 29 April 2008, 12:15 PM
 
3.  5 
7. 1
8. 5
9. 4
10. 1
11. 4
12. 3
13. 4
14. 2
15. I am getting it as 23.32 mixed!
can sumone giv some clues on how to solve 5,6 thoughtful??
 
Re: Quant Challenge for CAT 2008
by Dipanjan Biswas - Tuesday, 29 April 2008, 12:54 PM
  I am wdout pencil nd paper rit nw....So try to solve sm of dm verbally...

Q1. Z=5X^4+2Y^4, so by narrowin down we get 6 isimpossible one..
Option (4) is correct....

Q3.As (1/2 x 10) x h is d nswer.So last digit shud b either 5 or zero....
Option (4) is correct....

Q6. Largest no. 9876543210
So,210=2(mod 8)
Option (1) is correct.....

Q7.Easy one...Formed d equations nd get Option (1) as correct...

Q8. If we just keep b=1 fixed nd keepon changin d value of a=9,18,27,36(multiple of 9) dn nswer shud be an integer...
Option (5) is correct.....

Q10. (6-a)(6-b)(6-c)(6-d)(6-e)=45
So a=5,b=3,c=9,d=7,e=1
a+b+c+d+e=25....
So option (1) is correct.....

TG Rockzzzz
Re: Quant Challenge for CAT 2008
by nirmesh sinha - Wednesday, 30 April 2008, 11:46 AM
 

 4.5

11.4

14.2

 

Re: Quant Challenge for CAT 2008
by pradeep pandey - Wednesday, 30 April 2008, 11:49 AM
 

Hi Varun,

I hope u will enjoying the recent  quant challenge.

The quantitative Aptitude for CAT by K.Kundan & P.Pandey available in several book store of T.Nagar,Chennai i.e Jayanti Book Depot, PCM Book depot etc.

keep solving the quant challenges.

P.Pandey.

Re: Quant Challenge for CAT 2008
by amol malani - Wednesday, 30 April 2008, 03:43 PM
 

Q1  4

Q3 all answer options are wrong or printing mistake in answer option 5 it has to be 60.

Q7   1

Q10  1

Re: Quant Challenge for CAT 2008
by pradeep pandey - Thursday, 1 May 2008, 01:53 PM
 

Hi Amol'

Yes u are absolutely right.There are two printing mistakes in this quant challenge.

Question (3) Option(5) 60

Question (8) in the question part 14a/9b may be read as 14b/9a.

thanks.

P. Pandey

Re: Quant Challenge for CAT 2008
by jupiter singh - Thursday, 1 May 2008, 05:58 PM
 

sir,

ijust wanted to know is the awnser of question 15 u posted 15 questions is 2?(ie 26.6 im getting 27)

jupiter singh

Re: Quant Challenge for CAT 2008
by jupiter singh - Thursday, 1 May 2008, 06:02 PM
 

hey dude how u getting for 15--> 23.32...

 

Re: Quant Challenge for CAT 2008
by Pirates Are Back In Town - Thursday, 1 May 2008, 09:30 PM
  1>4
3>(60)
4>5
7>1
8>5
9>4
10>1
11>4
12>3
Re: Quant Challenge for CAT 2008
by Nishant Verma - Friday, 2 May 2008, 05:05 PM
 

Here are my answers:

1) 6

2)

3)60

4)4.15

5)

6)

7)750

8)2

9)12

10)None of these

11)330

12)3:1

13)64

14)3

15)23cm

 

Re: Quant Challenge for CAT 2008
by S A R SHAH - Friday, 2 May 2008, 07:16 PM
 

Q1-4

Q3-5

Q9-5

Q10-1

Re: Quant Challenge for CAT 2008
by sMILING GadhUUU - Friday, 2 May 2008, 10:50 PM
 

Answers:

1.   4) 6

2.    1)2002

3.     5)60

4.     5)2.22 

5.     4)1692

6.     1)2

7.     1)750

8.      1)1

9.      1)6

10.     1)25

11.      4)330

12.      3)3:1

13.      3)64

14.      2)3

15.      2)26.6

Attempt:

 

1) Z = 5x^4  + 2y^4

When we divide Z by 10

We get

Z  /  10   =    (5X^4)  / 10   +    (2Y^4)  /  10

We would get   ( x^ 4) /  2   +   (y^4)  / 5

Remainder of first term can be either 1 or 0.

Remainder for second term is either 1 or 0.

The last digit of fourth power of any integer
is 5 , if the integer ends in 5.
is 10, if the number ends in 0
otherwise its  1 for an odd number
and 6 for an even number

So the final remainders possible for the 2 parts are

5 * 0   or 5* 1        And  2  * 0  or 2 * 1

The combinations possible are   0 + 0,5 + 0 , 0 + 2 , 5 + 2
Therfore 6 is not possible

Ans:-   4) 6
********

2)  There are 9 people and 6 floors , People are identical to the operator
  Consider  9 balls to be put into 6 boxes( floors in this case)

I dont mind any boxes or a number of boxes be empty.(There can be floors where no person gets off)

0 0 0 0 0 0 0 0 0
I add 5  BALLS from my side as i want to divide the balls in 6 groups.

0 o o o o o O 0 0 0 0 0 0 0 0

Now ,All the possible ways i can select these 5 balls out of these 14 balls would give me the possible combinations

for 9 persons to be dropped at 6 floors.
14 C  5... As we are not concerned only with combinations.  

Ans : - 1)2002

*************
3) Given the Perimeter  = 36
One side = 10
Now we know by herons Formula to calculate the area of a triangle

Area =   Root (  S ( S-A ) ( S-B ) ( S-C) )

Where S = perimeter / 2,  A,B,C are the three sides.
Out of this let A =10.
We wamt the value of this Area to be maximum.
For this value of S ( S-A ) ( S-B ) ( S-C)  has to be maximum
But  S and S-A are already fixed.

So value of  (S-B)(S -C) should be maximum.
As A = 10,  B +  C = 26.

So, S-B + S -C  = 2S -(B + C )  = 36 - 26  = 10  (Constant )

Applying the rule of inequality,
IF a + b is constant value of  a * b is maximum  when  a = b

Therefore take  S-A   =  S - B
So  A = B = 13.

Using this values in   Area = root   (S ( S-A ) ( S-B ) ( S-C))

Max Area = 60 cm  square.

Ans :- 5) 60

*******************

 

Consider a coss section of Half the cone


*
*    * 
*        *
*             *
****************
*               *     *  
*  R             *         * 
*               *              *  
*               *                  *   
*    R/2      *                      *
********************************** 


Consider side of the cube as r.

Now the two trinangles formed ae similar triangle

Therefore    ( 5 - r ) / 5  = r / 4

r= 20 / 9  = 2.22 cm

Ans :- 5)2.22 cm

********* 

5) The 10 digit number is divisible by 11111

Also as all the digits are distinct the digits in this 10 digit number would always be  0,1,2,3,4,5,6,7,8,9

The sum of these digits is 45 or 9.
So If the   10 digit number    N = 11111 x   A
As 11111 is not a muntiple of
Then A must be a muntiple of 9.
In other words the number N is divisible by 11111 x 9  = 99999
We can write this number as 100000 - 1

So our 5 digit number should be such that  abcde00000  -  abcde should give us distinct digits.
or to be more precise
a b  c d  e  (9-a) (9-b)  (9-c)  (9-d)  and (10 -e) should be distinct..

Total combinations that would have sum up as 9
0,9
1,8
2,7
3,6
4,5   ( The selection can be made from any of these 10,But once a single number of a pair is selected the other number cannot be selected as it would aleady be covered  in 9 - (number).
Taking e as 1 - 9
possibilities for e =9,
Possibilities for  a = 7 ( We cannot have zero otherwise the final number would be a 9 digit number)
Posiibility for b = 6
Possibility for c = 4
= 1512  Combinations.
 Another 180 combinations should come from the possibility of e = 0 .

I am doubtful could not get these 180 case, am getting 384

The closest answer seems to be 1692.


******************
6) Greatest Such number  =  9876543210
When we divide it by 8.
Remainder = 2 .


8) Let the marked price of the book be 600.
30 % discount = 180 Rs
With  420 Rs  Profit is 20 % ,120 % 

100 % = 350 Rs Now  if there had been loss of 20 %
Then price would have been 280 .

The difference   320 is given to be 400..
If for 320 the price should be 600
For 400 it should be   ( 400 x 600  )  / 320
=  30 x 25

So in case the discount of 400 leads to a loss of 20 %
The   Marked price should be 750.

Ans   1) 750
*************

Could think of only one pair 2,3 but hope there is a better solution.

( 9a^2  +  14 b ^2 ) /  9 ab

The Numerator   9a^2  + 14 b^2 must be divisible by 9 .So both of them should be divisible by 9.

14* b^2 is divisible by 9 . So b*2 must be divisible by 9.
But then we can check the different  values of b  where  b = 3 * Any prime number

Huh..
**************
9) Surface area = 24

area of one surface = 24/6  = 4
one side = 2
Diameter of inner circle = 2
Diagonal of cube inside the circle =2
side of inner cube = 1 ,  ( 2a^2 =2   Therefore a = 1 )

Area of one surface = 1*1 =1
Total suface area = 1* 6 = 6

Ans: -  1) 6

*****************888
factors of 45  = 3 x 3 x 5 x 1
To make 5 distinct factors

3 x -3 x -1 x 1 x 5
(6-3)(6-9)(6-5)(6-7)(6-1)

a+b+c+d+e  =25

Ans :- 1)25
****************
11) Consider P Q as the two end points of the diameter of a circle .
The lengh of two semi circular arcs be 55  km.

Now   When either A OR B go from P --------> Q   We consider the upper semi circular arc   PQ.
When either A or B go from  Q--------> P we consider the lower semi circular arc QP.

For understanding we can also Consider P Q as the two end points of the diameter of a circle .
The lengh of two semi circular arcs be 55  km.

Now   When either A OR B go from P --------> Q   We consider the upper semi circular arc   PQ.
When either A or B go from  Q--------> P we consider the lower semi circular arc QP.

With Relative Speed Q has to cover  55 kms with a speed of -0.5 kms/sec (Or 5 kms in opposite direction)

So the time it would take to reach A (Which we consider stable at P) = 55 / 0.5   = 110 hrs.

So both A and B would travel for 110 hours each ,Before meeting at P.

Distance travelled by A in 110 hours is 110 x 3 = 330 KMS.

Ans:-  4) 330 KMS
*********************

                              *

                        ***********

                   *********************              9 equal   area  regions on this side

               *****************************

     F    ***************************************    G   

   *    *     *    *    *     *    *    *    *    *    *    *       (diagonal)1 equal area    

   D     ****************************************    E              region    FGDE

                *****************************

                      ******************         9 equal   area  regions on this side

                      A   *********   B  

                               *    C     smallest triangle

Form  parallel lines 9 on both sides of a diagonal of a Paralellogram.

Now the Smallest triangle ABC formed by the last parallel line would be a similar traingle as compared to the largest Triangle  DCE formed by the Parallel line Closest to the diagonal on that side.

The Ratio of areas of the smallest triangle considered : to that of  largest triangle =    1 : 9
(9 regions of equal  area on one side of the diagonal ,Of which the small triangle is 1 such regiion)

Now the ratio of sides of similar triangles ^ 2 = ratio of areas of the two similar triangles.

Both the smallest and the largest parallel lines considered are propotional sides of the two triangles.

Ratio of area = 1: 9
So ratio of sides   = ROOT ( 1 : 9)
= 1: 3

Ans :- 3) 3: 1

*******************

13)
   The numbers are in the series    21 + 19( n - 1)
If we want it not to have 1 as HCF with 240( 3x16x5 )

The number should not contain 2 3 and 5.

Total Numbers in the series  = 240
As 19 is added to every next number   

a,Total number of factors of 2 = 120.

b,Total number of factors of 3 = 80

c,Total number of factors of 5  = 48

So total factors out of 240 = 120 + 80 + 48  = 248

But factors of  6  ,15 and 10 have been counted twice so we have  to remove them
So  factors of 6 = 40
factors of 15    = 16
factors of 10   = 24

248 - 80 = 168
Adding factors of 30 Which have been substracted Twice  = 8

So  168 + 8  = 176 out of  would have HCF more than one with 240
So those numbers Having HCF 1 with 240   = 240 -176  = 64


Ans      3)64

******************

Number is of the Form 5a  +  3   , 7b  +  2  ,  8c +  3
So the number is (a muntiple of 5 + 3)
Its last digit is either 3 or 8.

But  FOR 8C + 3 to be 8  , 8c has to be odd which is not possible
So last digit of the number is 3.
Also number lies between 600 and 700..
We check such numbers of the form 8c + 3 that end in 3
603, 653 ,693,

Out of this 653 satisfies 5a +  3 and 7b + 2

If we divide 653 into rows of 14 chairs.
3 chairs will be left

Ans  : -    2) 3

***********************

Consider Triangle PSQ its a triangle in semi circle so a Right angled triangle.

Triangle PQR and Triangle PSQ are similar triangles.
Taking ratio of sides   PS /PQ  = PQ /PR
PR.PS  = 400   -----(1)

Now applying the rule of secants     PS . PR   =   QR ^2
But QR  =   QT + TR  = QT +  6  ------(2)

From (1) and (2)

400  =   QT^2  +  12 QT  + 36
Solving this we get     QT^2  + 26 QT   - 14 QT  - 364  =0

QT  =  14

QR = 14 + 6 = 20

Solving for   pr ^2    =  pq^2  +   qr ^2
                               = 400 + 400
                      pr    =   root (800)
PR approx = 28
Or closest to 26.6
Ans   :- 2)26.6       

Re: Quant Challenge for CAT 2008
by Varun Hans - Saturday, 3 May 2008, 05:23 PM
  Hi Sir,

Since yesterday, i have visited around 10 bookshops in t nagar chennai including jayanthi and PCM. But i have still not found the book i was looking for. Can u help me find it??
Re: Quant Challenge for CAT 2008
by avijit mohapatra - Sunday, 4 May 2008, 12:48 PM
  Hi
    Pradeep sir thanx for this wonderful set of questions
     It would be great if could post the answers
My answers:-
 Same as smiling gadha;gr8 work smile
except for :--
2 - 6^9---none of the above
5 -b
9 -d
 
question number --8   I am not sure of the method,anybody who could help me out with this.............
Re: Quant Challenge for CAT 2008
by pradeep krishnan - Monday, 5 May 2008, 02:26 AM
 
Hi Smiling Gadha,

First, In Q.15,shudn't the application of secant rule lead to RS*RP = QR^2 ?! 
i.e.(external part*whole part)= tangent^2..thn I guess u wud get 23.32 as me n nishant did!
or i went this way..PQR ~ PSQ ~ QSR  so..RS*RP = QR^2. I din't use(actually forgot) the formula(read shortcutwink).

Second, I cudn't solve Q.5. I went thru' ur answer for Q.5. nice
But i think 'e' shud be replaced by 'e-1'(as it lends) or alternatively 'e' n '9-e' n thn we can proceed.

@avijit; For Q.8, I too did what Dipanjan did and got the choice as 4.   

For Q.9, for inner cube a*root(2)=2(guess u solved it on the comp(and cudnt find the sq.rt symbolapprove)!).
k, so edge of inner cube, A=root(2).and so s.a.= 6*(A^2) = 12.

Re: Quant Challenge for CAT 2008
by sMILING GadhUUU - Monday, 5 May 2008, 10:38 AM
 

Hi pradeep ,

I think you are using it correctly it should be (external part * whole part) nice way to remember, you and nishant are getting it correctly as 23.32

About question 5, There are 2 possibilities with e.

(1) e is not zero (2) e is zero

(1) e is non zero..

e - 1 should be considered(as you said e would lend) but the other number to be considered would be 10 -e  as e would be the last digit to be substracted.  eg if you substract 46 from 200, 6 would be substracted from 10 and 4 would be substracted from 9.

so the numbers to be considered for combinations  will be a,  b ,c ,d ,(e -1) , (9-a) ,(9-b) ,(9-c) ,(9-d) ,and (10 -e)  for 0,1,2,3,4,5,6,7,8,9 digits.

(2)  e is zero .. in this case it is difficult to find the combinations.

as in the substaction   abcd000000  - abcd0.

digits 0 and 9 would already be reserved . And we would have to find the combinations of  a ,b ,c ,(d-1)  {as d would lend this time}   , (9-a) ,(9-b) ,(9-c)  and (10 - d)..... it is getting difficult to find the combinations here..

Pradeep sir please enlighten..

Re: Quant Challenge for CAT 2008
by CATendra 2009 - Tuesday, 6 May 2008, 11:43 AM
  Hi Smiling Gadha,

Can you provide a revised solution to Q15.I cannot arrive at the answer as 23.32

Re: Quant Challenge for CAT 2008
by dipak mishra - Tuesday, 6 May 2008, 04:16 PM
  For 15 you wrote " Now applying the rule of secants     PS . PR   =   QR ^2"
I think it is wrong,because  according to the rule of secants RS.PR=QR^2.The answer hence is not correct.
Re: Quant Challenge for CAT 2008
by jupiter singh - Tuesday, 6 May 2008, 06:58 PM
 

hey pradeep ithink 26.6 is the awnser in getting as iposted earlier...how ur getting 23.32 the logic is same tangent secant extended to it...or rather..can u plz let me knw ur logic..isuppose shortcuts are to be used as last alternative when nuthin is strikin..

Re: Quant Challenge for CAT 2008
by pradeep krishnan - Wednesday, 7 May 2008, 01:29 PM
 
hey jupiter, For Q.15 this is wat i did ->Join QS. Thn triangle PQR~PSQ~QSR => <PQS = <SRQ.
Thn <QPS = <SQR = <QST.(Angles in alternate segment)
Triangle STR will be isosceles with ST=TR.
Also, SQT will be isosceles with QT=ST.
=> QT=TR=6.
So, we get PQ=20, QR=12. PQR= 90 degree.
QR= 4 root(34) = 23.32.
Re: Quant Challenge for CAT 2008
by Sandeep Bhandari - Thursday, 8 May 2008, 09:13 PM
  1-4
2-1
3-5
4-
5-1
6-1
7-
8-
9-2
10-1
11-4
12-3
13-4
14-1
15-
Re: Quant Challenge for CAT 2008
by dr naina singh - Saturday, 10 May 2008, 11:21 PM
  Hello Sir,

Can u let me know from where can i get your book on quants?

im staying in ludhiana . punjab
Re: Quant Challenge for CAT 2008
by sMILING GadhUUU - Sunday, 11 May 2008, 11:35 PM
 

U can try at Gulati Chowk, Model Town.(Its a huge place but with some Book stalls)

(2)  Best,You can go to Ghantaghar(I think it was previously called Chauda Bazaar),

There you can ask someone for Kitaab Bazaar.. There are a number of shops,

 that offer CAT books (I dont remember perfectly but i believe i tried at Chawla Book Depot)..

Re: Quant Challenge for CAT 2008
by jack sparrow - Monday, 12 May 2008, 02:07 PM
 

Dear Pandey ji,

 

Question number 8 has no solution..

 

Re: Quant Challenge for CAT 2008
by pradeep pandey - Monday, 12 May 2008, 05:02 PM
 

Dear Jack Sparrow,

       I am just giving u one pair of values of (a,b) where a and b are coprime no. and (a/b)+(14b/9a) is an integer.

(a,b) = (1,3)

jack keep trying. I will give u more pair of vaues of (a,b) at the end of discussions.

 

P.Pandey.

 

Re: Quant Challenge for CAT 2008
by Sanchit Srivastavav - Monday, 12 May 2008, 06:56 PM
 

Excellent material TG kudos!!!

My attempts are for now are:

1)4    3)5, 6)1, 7)1  10)1 11)1 13)3

Re: Quant Challenge for CAT 2008
by deepesh jhamar - Monday, 12 May 2008, 08:18 PM
   I am getting 3456 as a answer to the question no 5

That is option 2

Explanation:

As all the number are distinct, they will hold the values (0,1,2,3,.......9)

The sum of the all the digit is 45, means the number is also divisible by 9

As the number is also divisible by 11111, the number will be divisible by 99999

The numbers which are divisible by 99999  can be written in the form of

abcde(9-a)(9-b)(9-c)(9-d)(9-e)

{ As 99999 can be written as 10,0000 -1

let abcde is the number multiplied by 99999

(10,0000-1)abcde=abcde00000-abcde

which equals abcd(e-1)(9-a)(9-b)(9-c)(9-d)(10-e)

let e-1 be m then abcdm(9-a)(9-b)(9-c)(9-d)(9-m)

as m is just a variable we can replace it by e

so we get abcde(9-a)(9-b)(9-c)(9-d)(9-e)

}

Now a can have total 9 values(9,8,7,6,5,4,3,2,1)

so the 9-a will have corresponding have means the two digits are consumed

Suppose a has 9 so 9-a will be 0

Therefore b can have 8 values (8,7,6,5,4,3,2,1) and corresponding value for 9-b suppose b have 8 so 9-b will be 1

similarly c can have 6 values

d can have 4 values and e can have 2 values.

So the total numbers of possible combinations possible are

9*8*6*4*2=3456

 

Re: Quant Challenge for CAT 2008
by deepesh jhamar - Monday, 12 May 2008, 08:29 PM
 

For question number 6

The greatest number which sastify the contition mentioned in question number 5 is 987650121234  not 9876543210

which when divided by 8 leaves the reminder as 2

 

 

 

Re: Quant Challenge for CAT 2008
by Mission IMpossible - Wednesday, 14 May 2008, 02:54 PM
 

Hi Guys

For 15th ques I am also getting 23.32 as answer.

I followed this approach

Connect O and S.

Now <QPR + <QRP = 90 and <PSO + <RST = 90 (As <OST = 90)

As <PSO = <QPR ( Since OP = OS = Radius) --> <QRP = <RST --> TS = TR

Since TS = TQ(Tangents from a point to a circle are equidistant) --> TQ = TR = 6

Now QR = 12 & PQ = 20 --> PR = sqroot(12^2+20^2) = 23.32

MI

Re: Quant Challenge for CAT 2008
by pradeep pandey - Saturday, 17 May 2008, 12:15 PM
 

Hi Varun Hans,

    Now the Quant book is available in almost all book store in T.Nagar, Chennai. Thanks for ur feedback regarding availability of this book.

P.Pandey.

Re: Quant Challenge for CAT 2008
by nitesh agarwal - Sunday, 18 May 2008, 02:39 PM
 

for q 8) the nswer is more than 4.

with b=3 a=2,7,11,16,20,... satisfy the condition.

did it the hard way of substituiting b=3 and finding posible values of A.

Thus the answer ring any bell for the Genius's around

Re: Quant Challenge for CAT 2008
by himanshu sharma - Tuesday, 20 May 2008, 09:47 AM
  which books should be referred for quants in which most of the topics are covered?
Re: Quant Challenge for CAT 2008
by Abhishek Rajan - Friday, 23 May 2008, 08:28 PM
  Q1 is i guess 4 th option
for second one i guess the answer will be 81* 81*9
Re: Quant Challenge for CAT 2008
by pradeep pandey - Monday, 26 May 2008, 05:48 PM
  Hi All,

Please find the answers.

P. Pandey
Re: Quant Challenge for CAT 2008
by Total Gadha - Monday, 26 May 2008, 05:53 PM
  Hi Himanshu,

You can refer to the book by Mr. Pandey.

Total Gadha
Re: Quant Challenge for CAT 2008
by khushboo Sharma - Thursday, 5 June 2008, 05:04 PM
 

Hi pradeep,

Not able to download the PDF.

It says "file damaged".

Khushboo.

Re: Quant Challenge for CAT 2008
by sri jamalpur` - Sunday, 8 June 2008, 11:47 AM
  hi pradeep. i m leaving in hyderabad can u tell where can i get the book of Quant Challenge.

and i also not good in Quant's becoz i m B.com graduate. so how i will improve my maths...
Re: Quant Challenge for CAT 2008
by Sachin Gupta - Tuesday, 10 June 2008, 06:29 AM
  1-4
7-1
8-3
10-1
Re: Quant Challenge for CAT 2008
by Shambhu Thakur - Wednesday, 11 June 2008, 11:41 PM
  Ans for first question is 4th option but i didn't got the logic.Can someone pls explain the logic.
Re: Quant Challenge for CAT 2008
by Shambhu Thakur - Thursday, 12 June 2008, 05:48 PM
 

Hi Pradeep,

The solution given below for 4th question is totally illogical.

Can u please provide the solution since its not clear that which two traingles are similar and the ratio taken is different fron those traingles.

I have solved it and got the answer near to 1.81but not exactly.

So could u please provide right solution for it. 

In solution you have taken the half of cube side in the simmetry but it will be r/root2 in place of r/2. you havesolved the question in 2 Dimension but its in three dimension.in place if r/2 in your solution it should be r/root2.

Since the side of cube will not be as you have assumed.The sides will lie on a circle of radius r/root2.

So please look over it AS SOON AS POSSIBLE since you have provided the wrong solution. 

Re: Quant Challenge for CAT 2008
by ATOM ANT - Saturday, 14 June 2008, 09:19 PM
  Pradeep pandey sir,

Went to all the bookshops in Chennai including Jayanthi  and PCM  book house..
cud not get that book..
Re: Quant Challenge for CAT 2008
by bids p - Tuesday, 17 June 2008, 11:07 AM
 

Hi SG,

Would you please explain the answer to question no - 02. Why are u adding 5 more balls ?

Thanks

Re: Quant Challenge for CAT 2008
by kanika bhutani - Tuesday, 17 June 2008, 02:48 PM
  plz tell me hw to get answer to various questions on ur site..
Re: Quant Challenge for CAT 2008
by Total Gadha - Tuesday, 17 June 2008, 04:35 PM
  quant challenge

So I guess your answer is correct.

 

Total Gadha

Re: Quant Challenge for CAT 2008
by sumit kumar - Thursday, 19 June 2008, 11:10 AM
 

hi sir

plz tell me where can i get your book at bhopal (MP).

i inquired in  few book stores but they are not aware of the book .

kindly help

sumit

Re: Quant Challenge for CAT 2008
by meena chandran - Saturday, 21 June 2008, 07:08 AM
  I am looking for answers to data interpretation and puzzles and logical reasoning tests. Can someone plz tell me where i can find them.
Re: Quant Challenge for CAT 2008
by PRIYANKAR SINGHA - Monday, 23 June 2008, 11:21 AM
 

ans 1: option 4

i couldn't understand what u ment by stating prople book alike

ans 3: option 5

 

Re: Quant Challenge for CAT 2008
by saurav lohia - Monday, 23 June 2008, 10:51 PM
  the answer to question no.10 is 25.(6-1)(6-3)(6-5)(6-7)((6-9)=45and 1+3+5+7+9=25.
Re: Quant Challenge for CAT 2008
by arnab _quant-um - Thursday, 3 July 2008, 01:15 PM
  this book is npt avail in kolkata.....help.
Re: Quant Challenge for CAT 2008
by rishabh bhandari - Wednesday, 30 July 2008, 11:56 AM
  hello,
I wanted to buy book by Mr.Pandey but it is isn't available in mumbai
can i buy online !!
pls mail me at brainzunlimited17@yahoo.com
Re: Quant Challenge for CAT 2008
by Gaurav Talwar - Tuesday, 19 August 2008, 03:24 PM
  Hi,

Jst another way to solve q 11 .... solve using LCM ... total time taken by both in a round trip

165/3 and 165/2.5 ... take LCM of these two ... the answer is 330 ... enjoy .......
Re: Quant Challenge for CAT 2008
by rashi agarwal - Friday, 10 October 2008, 12:28 PM
  my answes are

1.4
2.1
3.5
4.5
5.
6.
7.1
8.3
9.1
10.1.
11.4
12.3
13.3
14.2
15.4

i have not got the solution of 5th and 6th question.please help sir.
Re: Quant Challenge for CAT 2008
by sumit jamwal - Sunday, 8 March 2009, 04:49 PM
  hi,
in 2nd question is that on reaching at 6th floor the lift is vacant or else..
i mean when lift leaves 5th floor every1 except operator has left the lift. ..

then in that case it wud be 13C4 .
confusion ..  sad

Regards,
sumit
 
Re: Quant Challenge for CAT 2008
by Venkata Narsi Reddy G - Wednesday, 17 June 2009, 05:58 PM
 

1)4

3) 5

13)3

10)1

9)2

7)1

 

 

Re: Quant Challenge for CAT 2008
by chandan sinha - Thursday, 11 March 2010, 12:01 AM
 

sir where can i get ur book for cat quants......  in kolkata or any website through which i can buy it online. plz do rply sir i desperately need your book.

 

thanking u sir.

Re: Quant Challenge for CAT 2008
by Nitin Kumar - Wednesday, 24 March 2010, 01:13 AM
 

Can any one please let me know, how to solve below problem?

Direction: In a game b/w two ppl A and B, a heap of pebbles is kept on the floor. A persone can pick 1,2 or 3 pebbles. the one who picks the last pebble loses. both A and B are very intelligent. If a person picks the pebbles once, then it is called one step.

Que1. If B begins the games and A has to win, then what is minimum number of pebbles greater than 5 that must have been lying on the floor?

Que2: If B is to begin the game and A has to win the number of pebbles is 13, then what is minimum number of steps that the game ends in?

Re: Quant Challenge for CAT 2008
by TG Team - Friday, 26 March 2010, 04:21 PM
 

Hi Nitin smile

Search regarding "Controlled Sum Logic" in the forums - I have it discussed somewhere.

1. 9 pebbles - Whatever is the first move of B(say he picks x pebbles), A will pick 4 - x pebbles (as 4 is the controlled sum here) leaving 5 pebbles in the heap. Again following same strategy A will make B picking the last pebble and ensure a win.

2. This must be clear from previous answer. As both are very intelligent, minimum nuber of steps required is 7. smile

Re: Quant Challenge for CAT 2008
by mohil mittal - Friday, 16 September 2011, 02:10 PM
  Sir,

In the solution sheet remainder theorem is applied for question no. 5

abcdefghij=10^5abcde+fghij which is a factor of 99999

so abcdefghij=99999 multiplied by M
=(10^5-1)M

M=abcde
M=-fghij
so
abcde+fghij=0??

please xplain anyone!!!!
Re: Quant Challenge for CAT 2008
by TG Team - Friday, 16 September 2011, 02:54 PM
 

Hi Mohit smile

Mr Pandey has explained it thoroughly in his solution. Let's look at the question again.

N = abcde fghij ≡ 0 mod11111 where a, b, c, d, e, f, g, h, i, j are all non-negative single digit integers from 0-9 not necessarily in order.

Also we know that N is divisible by 9 (Think why??)

That means N is divisible by 99999 (because HCF(9, 11111) = 1)

Now N = abcde fghij = abcde×105 + fghij ≡ abcde + fghij ≡ 0 mod99999.

So abcde + fghij should be multiple of 99999 and it can be easily seen that this sum cannot be greater than 99999.

Now we are having five pair of numbers each of which is adding upto 9 and we just need to arrange them to make two numbers abcdef and fghij.

Five pairs can be arranged in 5! = 120 ways and from each pair the numbers can be assigned to two numbers in 2 ways i.e. total ways should be 5!×25

Remember, this is not the final answer because these cases include the cases when 'a' was assigned '0'. So final valid cases will be (9/10)×5!×25 = 3456. smile

This is a question of Quant Challenge, so it is supposed to use multiple concepts.

Kamal Lohia   

Re: Quant Challenge for CAT 2008
by mohil mittal - Saturday, 17 September 2011, 11:57 AM
  SIR,

M ABSOLUTELY SATISFIED WITH YOUR LOGIC...

BUT CAN U PLEASE TELL ME WHERE DID I FALTER IN THE ABOVE XPLANATION? LOGICALLY I AM NOT ABLE TO GET WHERE M GNG WRONG?

PLEASE SEE MA LOGIC IN THE PREVIOUS POST AND CORRECT ME

REGARDS
MOHIL