Official Quant thread for CAT 2013

MBA CAT 2024
24648
@bodhi_vriksha said:
Very good scrabble. This is the construction i was looking for.Team BV - Vineet
I think I can see an even better one. Let me check and revert....(and draw :banghead: bahut time lagta hai usmein!!)

And btw it's scrabbler.

regards
scrabbler
24649
@bodhi_vriksha said:
Nice try Aizen. Can anyone try a pure geometrical approach here? It is not more than 4-5 liner for sure.Team BV - Vineet
ABC is the trinagle with G is as centroid and D, E, F as mid points of BC, CA and Ab respectively.

Extend GD to P such GD = DP, then PC = BG = 2BE/3, PG = AG = 2AD/3 and GC = 2CF/3

area of triangle PCG = 2*area of triangle GDC = 2*(1/6)*area of of triangle ABC
= (1/3)*area of triangle ABC

Since dimensions of triangle PCG are 2/3 of the traingle having dimensions 3, 4, 5, we can say that ar(PCG) = (4/9)*area of triangle having dimensions 3, 4, 5

=> (4/9)*area of triangle having dimensions 3, 4, 5 = (1/3)*area of triangle ABC

=> area of triangle ABC = (4/3)*area of triangle having dimensions 3, 4, 5
24650

10000! = (100!)^k * p .,p and k are integers .What can be the maximum value of 'k'?

24651

From four positive real nos a, b, c and d, 4 distinct combination of sum of three numbers are formed S1, S2, S3 and S4. IF abcd = 5. FInd the minvalue of S1S2S3S4

24652
@aditi88 said:
10000! = (100!)^k * p .,p and k are integers .What can be the maximum value of 'k'?
We would look for power of 97, since it is the largest prime in 100!
So ,power of 97 in 10000! = 103 + 1 = 104
So, max "k" = 104..
24653
@aditi88 said:
10000! = (100!)^k * p .,p and k are integers .What can be the maximum value of 'k'?
is it 104?
24654
@techgeek2050
27*(5^3/4)
@aditi88
103
24655
@ChirpiBird said:
is it 104?
nop 103
24656
@bodhi_vriksha said:
Nice try Aizen. Can anyone try a pure geometrical approach here? It is not more than 4-5 liner for sure.Team BV - Vineet
Another proof. Longer to type out in Word 😞 but quicker to use.

regards
scrabbler
24657
@anantn said:
@techgeek205027*(5^3/4)@aditi88103
it's 405
24658
@anantn said:
@techgeek205027*(5^3/4)@aditi88103
soln??
24659

@techgeek2050
ah yes i only took S1*S2*s3 and ate up the s4 :). itll be 5^4*81=405.
Attention lapses causing silly mistakes at midnight..
24660
@anantn need a lil more than that :)
24661
@aditi88 said:
nop 103
can u plz post the solution

my approach :
number of zeroes in 10000! is 2499.. .. [this i found ...10000!/(5^n)]
number of zeroes in 100! is 24

so i can write
S*10^2499 = L*10^24*P
24x = 2499
x = 104
24662
@aditi88
since we want the highest coefficient of 100!, we would ideally want to take the number which acts as the highest constraint for a power of 100! to be complete, which would be the highest prime number under
24663
@anantn said:
@aditi88since we want the highest coefficient of 100!, we would ideally want to take the number which acts as the highest constraint for a power of 100! to be complete, which would be the highest prime number under
Power of 97 is 104 in 100!
24664
PLZ help..
From 6 couples who attended a party, four people are selected at random. Find the probability that exactly one couple is selected.
24665
@techgeek2050
As you say......expand (a+b+c)(a+b+d)(b+c+d)(a+c+d).
and then use the am>= gm logic to find that a=b=c=d for that expansion to be minimum....so each number =5^1/4 hence answer is 5^4/4*81
24666
@anantn said:
@aditi88since we want the highest coefficient of 100!, we would ideally want to take the number which acts as the highest constraint for a power of 100! to be complete, which would be the highest prime number under
but the highest power of 97 in 10000! will be 103 + 1 = 104, isnt it ?
24667
ah yes its 104.....my bad
half asleep, pardon my silly mistakes..