Official Quant thread for CAT 2013

MBA CAT 2024
25388

Try a similar question on basis of Pick's Theorem:

How many points in the region enclosed by x>=0,y
14,15,13,25

OA: 25

25389
@toshalimitra said:
Q)find the remainder when (21!)^2 is divided by 43.(don't have the OA,so with approach please)

41! mod 43 = 1 (Wilson's Theorem)

(21!)*(22*23*...*41) mod 43 = 1

21! * (-21*-20*...*-2) mod 43 = 1

21!^2 mod 43 = 1
25390
@Aizen said:
Try a similar question on basis of Pick's Theorem:How many points in the region enclosed by x>=0,y14,15,13,25OA: 25
A = 17.5
Take Ceiling of this , So Area = 18
applying the theorem =
Area of region= I + B/2 -1
B= 13
18 = I + 13/2 -1
I = 12.5 ..... ye kya ho gaya mere se ?


how to use this ?
25391
@amresh_maverick said:
A = 17.5 Take Ceiling of this , So Area = 18applying the theorem =Area of region= I + B/2 -1B= 13 18 = I + 13/2 -1I = 12.5 ..... ye kya ho gaya mere se ?how to use this ?

@amresh_maverick bhai : Ceil function subtract karne ke baad lena he ... mene typo kar diya tha upar.

So, In this case :

Area of region= I + B/2 -1
B= Boundary points
I=Integral points

here we have a triangle of area 1/2*5*7=35/2
Boundary points are = (0,0) to (7,0)=8 points
(0,-1) to (0,-5)=5 points
Boundary points=13
So,
35/2 = I + 13/2 - 1
I=12

But since question has so 12+13 = 25

25392
@amresh_maverick said:
A = 17.5 Take Ceiling of this , So Area = 18applying the theorem =Area of region= I + B/2 -1B= 13 18 = I + 13/2 -1I = 12.5 ..... ye kya ho gaya mere se ?how to use this ?
i think here you must consider the equality so the points on the borders should be added
Hence I=[12.5](floor function)~12
I=12+13
25393
@Aizen said:
Try a similar question on basis of Pick's Theorem:How many points in the region enclosed by x>=0,y14,15,13,25OA: 25
area of region 1/2 * 5* 7 = 17.5 units
B.p = 6+7=13
i = Area - B/2 +1
17.5- 6.5 +1 = 12 points _/\_

25394
@Aizen said:
Try a similar question on basis of Pick's Theorem:How many points in the region enclosed by x>=0,y14,15,13,25OA: 25
Consider the rectangle formed by x = 0, y = 0, x = 7 and y = -5, there are 8*6 = 48 points having integral coordinates lies on or inside this rectangle

Now, two points having integral coordinates lie on the line 5x - 7y ≤ 0 and x ≥ 0 and y ≤ 0

So, total (48 - 2)/2 + 2 = 25 points will satisfy the given condition
25395
@Aizen said:
@amresh_maverick bhai : Ceil function subtract karne ke baad lena he ... mene typo kar diya tha upar.So, In this case :Area of region= I + B/2 -1B= Boundary pointsI=Integral pointshere we have a triangle of area 1/2*5*7=35/2Boundary points are = (0,0) to (7,0)=8 points(0,-1) to (0,-5)=5 pointsBoundary points=13So,35/2 = I + 13/2 - 1I=12But since question has so 12+13 = 25
mere khyal se answer 27 hoga bhai...

35/2 = i + (8+5+4)/2 - 1
35/2 = i + 17/2 - 1
35/2 = i + 15/2
i = 35/2 - 15/2 = 10

10+17 = 27 then??

mere 4 extra boundary points isliye hai coz i have considered the integer boundary points even on the slope of the triangle, why won't we do that??
25396
@Logrhythm said:
mere khyal se answer 27 hoga bhai...35/2 = i + (8+5+4)/2 - 135/2 = i + 17/2 - 135/2 = i + 15/2i = 35/2 - 15/2 = 10 10+17 = 27 then?? mere 4 extra boundary points isliye hai coz i have considered the integer boundary points even on the slope of the triangle, why won't we do that??
Bhai, there is no integral point on slope of triangle.

Please check the attached pic.

Vertical is X - axis ; Horizontal is Y axis.

p.s: @chillfactor saar \\___o//
25397

In how many ways 5 rings can be placed in 4 fingers ??

25398
@CrookDinu said:
In how many ways 5 rings can be placed in 4 fingers ??
if rings or different, each ring can go in to possible 4 fingers. hence 4*4*4*4*4 = 1024 ways
25399
@CrookDinu said:
In how many ways 5 rings can be placed in 4 fingers ??
1> If rings are different :

n= diff rings
r = fingers
No of ways = (n+r-1)! /(r-1)!

2> If rings are identical
No of ways = n+r-1Cr-1

so here , I am taking it as same
= 8C3= 56 ways
25400

Last three digits of 3^3^3^3 or (% 1000)?

25401
@CrookDinu said:
In how many ways 5 rings can be placed in 4 fingers ??
8!/3!
25402
Three horses : Kanishka, Silver Streak and Arabian Knight are the only horses competing in a race and only one of these three can win the race. If Kanishka is twice as likely to win as Silver Streak and Sliver Streak is twice as likely to win as Arabian Knight, then what is the probability of Arabian Knight losing this race?
25403
@CrookDinu said:
In how many ways 5 rings can be placed in 4 fingers ??
Fingers -> a + b + c + d
Rings-> r1, r2, r3, r4, r5

a+ b + c + d = 5

Total 8C3*5! ways = 8!/3!.. ? ( 5! because r1r2 is diff from r2r1) ..

25404
@amresh_maverick said:
Three horses : Kanishka, Silver Streak and Arabian Knight are the only horses competing in a race and only one of these three can win the race. If Kanishka is twice as likely to win as Silver Streak and Sliver Streak is twice as likely to win as Arabian Knight, then what is the probability of Arabian Knight losing this race?
1 - 1/7 = 6/7 ?
25405
@amresh_maverick said:
Three horses : Kanishka, Silver Streak and Arabian Knight are the only horses competing in a race and only one of these three can win the race. If Kanishka is twice as likely to win as Silver Streak and Sliver Streak is twice as likely to win as Arabian Knight, then what is the probability of Arabian Knight losing this race?
is that 1/7?
25406
@amresh_maverick said:
Three horses : Kanishka, Silver Streak and Arabian Knight are the only horses competing in a race and only one of these three can win the race. If Kanishka is twice as likely to win as Silver Streak and Sliver Streak is twice as likely to win as Arabian Knight, then what is the probability of Arabian Knight losing this race?
6/7
25407
@amresh_maverick said:
Three horses : Kanishka, Silver Streak and Arabian Knight are the only horses competing in a race and only one of these three can win the race. If Kanishka is twice as likely to win as Silver Streak and Sliver Streak is twice as likely to win as Arabian Knight, then what is the probability of Arabian Knight losing this race?
4x,2x,x be the probabilties of winning of respective horses. Now since Only three of them are there, 7x = 1,
x=1/7;
Hence probability of losing is 6/7