Official Quant thread for CAT 2013

MBA CAT 2024
9801
Find the number of divisors of 1080 excluding the throughout divisors, which are perfect squares?

Explain...

9802
@tmohan02 said:
Find the number of divisors of 1080 excluding the throughout divisors, which are perfect squares?Explain...
1080 = (2^3)(3^3)(5)

We know that perfect square factor will be of form {2^(2a)}{3^(2b)}{5^(2c)}

a can be 0 or 1, b can be 0 or 1 and c can only be 0 (as highest exponent of 2 and 3 is 3 and that of 5 is 1)

So, 2*2*1 = 4 perfect square factors
9803
@tmohan02 said:
Find the number of divisors of 1080 excluding the throughout divisors, which are perfect squares?Explain...
1080 = 3^3 * 2^3 * 5

Perfect square divisors = (select 3^0 or 3^2)*(Select 2^0 or 2^2)*(Select 5^0)
= 2*2*1 = 4

Total divisors = 4*4*2 = 32

Required number = 32 - 4 = 28 ?
9804
The perfect square includes 1^2, 2^2, 3^2 and (2*3)^2.....Am i right???
9805
@tmohan02 yes it includes 1, 4, 9 and 36

How many two degree polynomials are there with integral coefficients and with integral roots, such that f(0) = 210 ?
9806

In quadrilateral ABCD, sides AB and BC each have length sqrt(2), while side CD has length 2. What is the area of quadrilateral ABCD?

(1) The length of side AD is 2.
(2) The angle between side AB and side BC is 90.

1) using only 1
2) using 2
3) using both
4)using either
5)cannot be answered


9807
@bullseyes said:
In quadrilateral ABCD, sides AB and BC each have length sqrt(2), while side CD has length 2. What is the area of quadrilateral ABCD?(1) The length of side AD is 2.(2) The angle between side AB and side BC is 90.1) using only 12) using 23) using both4)using either5)cannot be answered
We can not say anything using statement (1) alone.

Using statement 2 also we can not find the area as we know only three sides.

But using both, we can find the area, as there will be one rt angle triangle and one equilateral triangle (I hope we have to consider only convex quadrilateral, else we can not answer)
9808
@chillfactor said:
I think its not correctIn case 2:- I I B _ _ _ _ _ _ _ _ _ _ _, you have excluded I I B B B, but you didn't remove I I B B I B B and I I B I B B B casesIn case 3:- I B I _ _ _ _ _ _ _ _ _ _ _, you have excluded I B I B B but not I B I B I B B, I B I I B B BSo, removing these 4 cases you will also get 637 ways which is exactly the same I got using Catalan numbersC(14, 4) - C(14, 3) = 637
But this seems to assume that all the goals of Brazil are different, and similarly all goals of India are different? Am not convinced...

regards
scrabbler
9809
@scrabbler said:
But this seems to assume that all the goals of Brazil are different, and similarly all goals of India are different? Am not convinced...regardsscrabbler
No, here order of the goals matter thats why in case of IIIB______, its C(11, 3) - 1
9810
@Harmeet89 said:
I am unable to understand the difference between these two cases:-In how many ways can 12 books be distributed1) into 3 parcels ?2) among 3 boys ?Please clarify the difference.
Parcel when made are all same
Boys when they have book are different entities
9811
@Harmeet89 said:
I am unable to understand the difference between these two cases:-In how many ways can 12 books be distributed1) into 3 parcels ?2) among 3 boys ?Please clarify the difference.
Boys are distinct. Parcels are identical.

Suppose you have 3 veg and 2 non-veg sandwiches. Case 1: You give them to 2 friends. It is important who gets which sandwiches as a friend might be pure veggie (or pure non-veggie). However Case 2: you are parceling the sandwiches in 2 takeaway packets. Will it matter which parcel gets the veg and which gets the non-veg?

regards
scrabbler
9812
@chillfactor said:
No, here order of the goals matter thats why in case of IIIB______, its C(11, 3) - 1
OK...I am not sure I get it, but will think on it. I made some wrong assumption while reading. Thanks!

Edit: ...wow, that's 10000 posts on this forum in just 5 weeks. Impressed...

regards
scrabbler
9813
@Harmeet89 said:
@AIM_IIM_2013@scrabbler But the problem here is that nothing is mentioned about the books. they could be same or different. That is what is confusing me. The books have to be divided, not the parcels or boys.
When we say that 4 objects have to divided in two groups of two OR 4 objects have to distributed among 2 persons such that both gets 2 then these are different cases

for the first one (AB, CD), (AC, BD) and (AD, BC) are the three possibilities

but for 2nd case, there will be 6 possibilities:-
for (AB, CD) we have two cases AB goes to person 1 and CD goes to person 2 (and similarly for other two cases)

I hope it helps
9814
@Harmeet89 Look it this way 12 books divd as abcd efgh ijkl
Now if first case "parcel" is considered then this group of 3 can be distributed in any order among parcel
but in case of boys change the order that is instead of abcd being given to boy1 it is given to boy2 then this is a separate case.
9815
@Harmeet89 said:
@AIM_IIM_2013@scrabbler But the problem here is that nothing is mentioned about the books. they could be same or different. That is what is confusing me. The books have to be divided, not the parcels or boys.
That is yet another story. You asked what is the difference when boys are changed to parcels 😃 My answer was for that. Each of those cases will have a further 2 options of the question depending on whether the books are distinct or identical. for example boys would give:

12 Distinct books, 3 (Distinct) boys - 3^12
12 Identical books, 3 (Distinct) groups - 14C2 (partitioning logic)

If the objects as well as groups are identical, then it is a good deal more painful. Listing is probably the best option. If distinct objects and identical groups.....well, good luck 😃 Well beyond CAT level, I'd suspect....at max you would look at 5-6 objects in 3-4 groups, beyond that it is just too painful. For the sake of my sanity I would assume the books identical here ;)

regards
scrabbler
9816
@chillfactor said:
@tmohan02 yes it includes 1, 4, 9 and 36How many two degree polynomials are there with integral coefficients and with integral roots, such that f(0) = 210 ?
f(x) = a*(x-p)*(x-q) , where a,p,q are integers

Now, f(0) = a*p*q = 210

210 = 2*3*5*7

Distribution of 4 distinct objects to 3 boxes (2 identical(p,q) and 1 different (a) from them), where a box can remain empty too [empty box means that variable is 1]

1) (4,0,0) -> 2 (ways to distribute) = 2
2) (3,1,0) -> (4!/3!1!0!) * (3!/2!) = 12
3) (2,1,1) -> (4!/2!1!1!2!) * (3!/2!) = 18
4) (2,2,0) -> (4!/2!2!2!) *(3!/2!) = 9

41 ways

Now, the integers can be negative too, in that case,

1) (4,0,0) -> 2 + 3C2 = 5 ways [because the distribution of negatives is over 210,1,1]
2) (3,1,0) -> 12*3C2 = 36
3) (2,1,1) -> 18*3C2 = 54
4) (2,2,0) -> 9*3C2 = 27

122 ways

Total = 41 + 122 = 163 ?

PS: Very very doubtful 😞 :splat:
9817
@chillfactor said:
@tmohan02 yes it includes 1, 4, 9 and 36How many two degree polynomials are there with integral coefficients and with integral roots, such that f(0) = 210 ?
Another method:

After a*p*q = 210 = 2*3*5*7

If I consider p,q to be different then, total ways = 3^4 = 81

Also, 3x + 6y = 81 , x = distribution containing two equal terms, y = distinct terms
x + 2y = 27
Now, x = 1 ( case of 210,1,1)

So, y = 13

Now, For positive integers, distribution = y*3!/2 + 2 (case of 210,1,1) = 41
And for negative integers, distribution = y*(3!/2)*3C2 + 5 (case of 210,1,1) = 122

Total = 163
9818
@tmohan02 said:
Find the number of divisors of 1080 excluding the throughout divisors, which are perfect squares?Explain...
1080=3^3*2^3*5
3^2(3*2^3*5)=2*4*2=16
2^2(3^3*2*5)=2*4*2=16
3^2*2^2(3*2*5)=2*2*2=8

16+16-8=28
9819
@chillfactor said:
@tmohan02 yes it includes 1, 4, 9 and 36How many two degree polynomials are there with integral coefficients and with integral roots, such that f(0) = 210 ?
a*x^2+b*x+c=0

a*x^2+b*x+210=0

sum of two roots =-b/a

Product of two roots=210/a

a is number of factors of 210
210=2*3*5*7

Number of factors =2*2*2*2=16

a can be negative
Total number is 32
9820
@gnehagarg I think you are missing some cases, answer is 163 I think