Official Quant thread for CAT 2013

MBA CAT 2024
24608
@bodhi_vriksha said:
Guys, now try this: Medians of a triangle are of length 3,4 and 5 respectively. Find the area of the triangle.Team BV - Vineet
area = [2(x^2*y^2 + y^2*z^2 + z^2*x^2) - (x^4 + y^4 + z^4)]^(1/3) = [2(144+400+225) - (962)]^(1/3) = [1538 - 962]^(1/3) = [576]^(1/3) = 8.3x ??
24609
@bodhi_vriksha said:
Guys, now try this: Medians of a triangle are of length 3,4 and 5 respectively. Find the area of the triangle.Team BV - Vineet
8 something??
24610
@nramachandran said:
Apologies4/3 * sqrt (36) = 8??
Pl share the approach...
24611
@bodhi_vriksha said:
Guys, now try this: Medians of a triangle are of length 3,4 and 5 respectively. Find the area of the triangle.Team BV - Vineet
sm=1/2(m1+m2+m3)=6
A=4/3 rt(sm(sm-m1)(sm-m2)sm-m3))
=4/3 rt(6*3*2*1)=8 ?????
24612
@ankitmishraiim

Area = 4/3 (sqrt (m * (m-a) * (m-b) * (m-c))

Where a,b,c are the medians

m = a+b+c/2
24613
@bodhi_vriksha said:
Guys, now try this: Medians of a triangle are of length 3,4 and 5 respectively. Find the area of the triangle.Team BV - Vineet
Area of Triangle when medians are known:

Let length of medians be a,b,c.

So, Area= 4/3 * Heron's Formula for Triangle

= 4/3 * _/(S*(S-a)*(S-b)*(S-c))

where S= (a+b+c) / 2
24614
@Aizen :TaichO __/\__

Sorry guys 😃
24615
@bodhi_vriksha said:
One more: What is the largest n such that 2^n divides [10^30/(10^10 + 3)], where [] represents greatest integer function?Team BV - Kamal Lohia
10^30 = 10^30 + 3*10^20 -3*10^20 - 9*10^10 + 9*10^10 + 27 - 27
=>10^30 = 10^20*(10^10 + 3) -3*10^10*(10^10+3) + 9*(10^10 +3) - 27

hence the expression inside [] = [10^20 - 3*10^10 + 9 - 27/(10^10+3)]
=[(10^10)*(10^10 - 3) + (9 - e)] { where e = 27/(10^10+3)}

since (9-e) is an integer just less than 9 hence the value of above expression

= (10^10)*(10^10-3) + 8

clearly (10^10-3) is an odd number say we put it as (2k+1) to get the above expression as

10^10(2k+1) + 8

= 8*{ 125*10^7*(2k+1) + 1} [taking out 8 as common factor]

clearly number in the {} is odd hence 2 is not a factor of expression inside {}.

=>the largest power of 2 that divides is 3.

ATDH.
24616
@Aizen said:
Area of Triangle when medians are known:

Let length of medians be a,b,c.

So, Area= 4/3 * Heron's Formula for Triangle

= 4/3 * _/(S*(S-a)*(S-b)*(S-c))

where S= (a+b+c) / 2
That's correct.
But can you prove this formula?
Team BV - Vineet
24617
@iLoveTorres said:
is it 8?
I am afraid to say that this is incorrect.

It'd be better to post your approach so that it might be possible to help.

Team BV - Kamal Lohia
24618
@nramachandran said:
Apologies
4/3 * sqrt (36) = 8??k
Correct :). And how do you arrive at this formula?
Team BV -Vineet
24619
How many sets of three distinct factors of the number N = 2^6 — 3^4 — 5^2 can be made such that the factors in each set has a highest common factor of 1 with respect to every other factor in that set?

24620
@bodhi_vriksha said:
One more: What is the largest n such that 2^n divides [10^30/(10^10 + 3)], where [] represents greatest integer function?Team BV - Kamal Lohia
10^30 = (-3)^3 mod (10^10 + 3) = (10^10 - 24) mod (10^10 + 3) = 2^3(even - odd) mod (10^10 + 3) = 2^3(odd) mod (10^10 + 3).


Now 10^30 = Quotient*(10^10 + 3) + Remainder
10^30 = Quotient*Odd + 2^3(odd)

Clearly Quotient must be odd multiple of 2^3 only as LHS is even multiple of 2^3.

Recall that [10^30/(10^10 + 3)] is the Quotient obtained when 10^30 is divided by (10^10 + 3). :)

Team BV - Kamal Lohia
24621
@fireatwill said:
How many sets of three distinct factors of the number N = 2^6 — 3^4 — 5^2 can be made such that the factors in each set has a highest common factor of 1 with respect to every other factor in that set?
5 ?
24622
@fireatwill said:
How many sets of three distinct factors of the number N = 2^6 — 3^4 — 5^2 can be made such that the factors in each set has a highest common factor of 1 with respect to every other factor in that set?
Layman approach: make cases

(1, p1, p2) - (6*4 + 6*2 + 4*2) = 44 sets
(1, p1, p2*p3) - (3*6*4*2) = 144 sets
(p1, p2, p3) - (6*4*2) = 48 sets.

So total 44 + 144 + 48 = 236 sets :)

Team BV - Kamal Lohia
24623
@fireatwill said:
How many sets of three distinct factors of the number N = 2^6 — 3^4 — 5^2 can be made such that the factors in each set has a highest common factor of 1 with respect to every other factor in that set?
236 ??

n= p^l*q^m*r^n
no. of sets= lm+mn+ln+4lmn
24624
@bodhi_vriksha said:
I am afraid to say that this is incorrect.It'd be better to post your approach so that it might be possible to help.Team BV - Kamal Lohia
the would reach the starting point when A(3kmph) and B(5kmph) cover 3 and 5 complete to and fro motion with distance d=120(assumed)
So they together travel 240*3+240*5=1920
They woulld meet for the first time when they together cover a distance of d
the next meeting would happen when they cover a distance of 2d together.. hence there would be total 8 meetings. Goes by my analysis there would be one instance when they would meet travelling in the same direction.. hence they would exchange 8-2=6 flowers
24625
@bodhi_vriksha said:
Layman approach: make cases(1, p1, p2) - (6*4 + 6*2 + 4*2) = 44 sets (1, p1, p2*p3) - (3*6*4*2) = 144 sets(p1, p2, p3) - (6*4*2) = 48 sets.So total 44 + 144 + 48 = 236 sets Team BV - Kamal Lohia
Could you explain the bold part, I think I am misinterpreting the question ..
24626
@bodhi_vriksha said:
why don't you want to solve it further where I left it.Anyways..f(k) has to be greater than 19 and has sum of digits 19. So smallest value of f(k) is 199.Similarly k has to be greater than 199 but has sum of digits 199, so smallest value of k is 1 followed by 22 9's i.e. a 23 digit number. Team BV - Kamal Lohia
i understood the logic after reading ur post..thanks...
24627
@ScareCrow28 said:
Could you explain the bold part, I think I am misinterpreting the question
Vibhor:)

There is nothing bold in your above post.
Anyway, it says for a number N, how many sets of three co-prime factors (a, b, c) can be formed?

For example for 6 we have only one such set (1, 2, 3) but for 120, we have (1, 2, 3), (3, 4, 5) and many more.

I hope it is clear now. :)

Team BV - Kamal Lohia