Official Quant thread for CAT 2013

MBA CAT 2024
24548
@viewpt said:
84504
approach ???
24549
@pm89 said:
N = 121212121212..................300 digits. find the remainder when N is divided by 99 ?1.) 18 2.) 1 3.) 98 4.) none of these.Regards,pm89
12*150 mod 99=1800 mod 99=200 mod 11 = 2

2*9=18
24550
@pm89 said:
N = 121212121212..................300 digits. find the remainder when N is divided by 99 ?1.) 18 2.) 1 3.) 98 4.) none of these.Regards,pm89
18 hai bhai..
24551
@bodhi_vriksha said:

Ext. Qs. What is the number of distinct triangles with integral valued sides and perimeter 2014?Team BV - Kamal Lohia
n^2/48 = 2014*2014/48 = 1007*1007/12 =84504
24552
@pm89 said:
N = 121212121212..................300 digits. find the remainder when N is divided by 99 ?1.) 18 2.) 1 3.) 98 4.) none of these.Regards,pm89
99 = 9*11

1*150 + 2*150 % 9 = 450 % 9 = 0

so we can straight away eliminate 1 and 98, but since we have NOT in the options we need to find the remainder with 11

2*150 - 1*150 % 11 = 150 % 11 = 7

so 11k+7...hence 18 is the remainder...
24553
@pm89 said:
N = 121212121212..................300 digits. find the remainder when N is divided by 99 ?1.) 18 2.) 1 3.) 98 4.) none of these.Regards,pm89
18

1212( 10^296+ 10^292+..............)
1212[( 10^2)^148+..............)]
1212[ 1+......+.. 75 times]
1212*75 mod 99= 18

24554
@viewpt said:
None of these .. I am getting 931212( 10^296+ 10^292+..............)1212[( 10^2)^148+..............)]1212[ 1+......+.. 74 times]1212*74 mod 99= 93??
1212*75 h bhai wo
10^0 tak jaega
fir 18 aayega
24555
@bodhi_vriksha said:

Ext. Qs. What is the number of distinct triangles with integral valued sides and perimeter 2014?Team BV - Kamal Lohia
[n^2/48] when n (perimeter) is even
[(n+3)^2/48] when n (perimeter) is odd

2014^2/48 = 84504..??
24556
@viewpt said:
None of these .. I am getting 931212( 10^296+ 10^292+..............)1212[( 10^2)^148+..............)]1212[ 1+......+.. 74 times]1212*74 mod 99= 93??plz chk kahan gadbad hai..
[1 +.... 75 times]....
24557
@pm89 said:
N = 121212121212..................300 digits. find the remainder when N is divided by 99 ?1.) 18 2.) 1 3.) 98 4.) none of these.Regards,pm89
18 Hoga
24558
@jain4444 said:
9720000 = 2^6 * 5^4 * 7^3 x^3 = 2^3a * 5^3b * 7^3c y^2 = 2^2p * 5^2q * 7^2r z^4 = 2^4x * 5^4y * 7^4z 3a + 2p + 4x = 6 => 3 solutions 3b + 2q + 4y = 4 => 2 solutions 3c + 2r + 4z = 3 => 1 solution and we could have 2 values for "y" and "z" so , total solutions = 3*2*1*2*2 = 24
Sir can you please explain the last part again.....
Thanks...
24559
@saurav205 said:
Sir can you please explain the last part again.....Thanks...
y can be positive and negative as well , same with z

that's why we have 2*2
24560
@amresh_maverick said:
@bodhi_vriksha@scrabbler how to find non existing terms . total terms will be 76 as pointed out above(1 + x + x^5)^15 - no of terms ?
For this one we can use Chicken McNuggets theorem also

Here coefficients are 0, 1 and 5. So if we start from 0, 1, 2, 3, ... we can see that all can be formed (as 5*1 - 5 - 1

But if we start from the end then it will be 75 - (5a + 4b + 0c)
Here coefficients are 0, 4, 5. So, (5 - 1)(4 - 1)/2 = 6 numbers can not be formed

Hence out of 76 exponents we can not get 6 exponents.

=> 76 - 6 = 70 terms weill be there in the expansion
24561
@chillfactor said:
For this one we can use Chicken McNuggets theorem alsoHere coefficients are 0, 1 and 5. So if we start from 0, 1, 2, 3, ... we can see that all can be formed (as 5*1 - 5 - 1 But if we start from the end then it will be 75 - (5a + 4b + 0c)Here coefficients are 0, 4, 5. So, (5 - 1)(4 - 1)/2 = 6 numbers can not be formedHence out of 76 exponents we can not get 6 exponents.=> 76 - 6 = 70 terms weill be there in the expansion


regards
scrabbler
24562
@bodhi_vriksha said:
Hint: General term = 15C(a,b,c)*1^a*x^b*x^5c
AND a+b+c=15.

You need to find the possible values of b + 5c

Vineet, Team BV
I was in a cyber cafe while typing this one earlier today and made a typo in hurry. Thankfully, now the net is functioning at my residence :)
Basically we need to find distinct values of b+5c. The possible unique values are completely exhausted if we take b=0 to 4 and c= 0 to 15-b. Hence, total terms=12+13+14+15+16=70
Vineet, Team BV
24563
@jain4444 said:
9720000 = 2^6 * 5^4 * 7^3

x^3 = 2^3a * 5^3b * 7^3c
y^2 = 2^2p * 5^2q * 7^2r
z^4 = 2^4x * 5^4y * 7^4z

3a + 2p + 4x = 6
=> 3 solutions

3b + 2q + 4y = 4
=> 2 solutions

3c + 2r + 4z = 3
=> 1 solution

and we could have 2 values for "y" and "z"

so , total solutions = 3*2*1*2*2 = 24
There is a small calculation mistake. 9720000 = 2^6 * 5^4 * 3^5.
Your third equation would become 3c+2r+4z=5
Vineet, Team BV
24564
@iLoveTorres said:
i)27ii)27iii)34
Gaurav..please post your approach as well.

As per me, 2 of the 3 answers are correct.

Team BV - Kamal Lohia
24565
@bodhi_vriksha said:
Gaurav..please post your approach as well.As per me, 2 of the 3 answers are correct.Team BV - Kamal Lohia
My approach:

as per the question 93 subscribed THE DON 79 JAN NAYAK AND 85 LOK NETA
Now to maximise only JAN NAYAK OR LOK NETA basically there should be maximum overlap between the other two and hence out of the remaining (120-93)=27 can only be the maximum who would have subscribed to only JAN NAYAK OR LOK NETA

for the third question maximum who subscribed to JAN NAYAK & LOK NETA but NOT DON would be when there is maximum overlap between the two but this should not include the don hence again (120-93)=27 will be the maximum.

sorry last time did a blunder while considering the overlap

Hence the answer would be
1)27
2)27
3)27
24566

Evening Teaser: (Easy) LCM and HCF of two numbers is 11! and 7! respectively. How many such pairs of two numbers are possible?

Team BV - Kamal Lohia

24567
@bodhi_vriksha said:
Evening Teaser: (Easy) LCM and HCF of two numbers is 11! and 7! respectively. How many such pairs of two numbers are possible?Team BV - Kamal Lohia
8?

The difference is 2^4 * 3^2 * 5^1 * 11^1 which can be split into co-prime pairs in 2^(4-1) = 8 ways...

regards
scrabbler