Official Quant thread for CAT 2013

MBA CAT 2024
11401
@vbhvgupta said:
Q2
to make all three playing least, students who are playing exactly two games have to be maximum and students who are playing exactly one game have to be least.

so, taking students who are playing exactly one game =0,
b+c=100
2b+3c=210

=> b=90, c=10

so, least percentage=10%
11402
@vbhvgupta said:
Q2
10%?
11403

the external angles in degrees,of a convex polygon are all integers which are in arithmetic progression..how many values are possible for n

16
14
10
8
11404
@pavimai 16? n=3 to 18
11405
In ΔABC, AB = 9. If BC : AC = 40 : 41 what is the largest possible area of ΔABC?
OPTIONS

1) 764 sq. units
2) 820 sq. units
3) 268 sq. units
4) 1508 sq. units
5) None of these
11406
@sujamait said:
In ΔABC, AB = 9. If BC : AC = 40 : 41 what is the largest possible area of ΔABC?OPTIONS1) 764 sq. units 2) 820 sq. units 3) 268 sq. units 4) 1508 sq. units 5) None of these
9,40x, 41x

Area = 820x^2 sinC

cosC = (3281x^2 - 81)/3280x^2 => sinC = (9/4*820x^2) * root( (81-x^2)*(81x^2 - 1) )

A = (9/4) * root[(81 - x^2)*(81x^2 - 1)]

dA/dx = 0 => x = 6.36 => A = 819.99 = 820 😐 ..
11407
@sujamait said:
In ΔABC, AB = 9. If BC : AC = 40 : 41 what is the largest possible area of ΔABC?OPTIONS1) 764 sq. units 2) 820 sq. units 3) 268 sq. units 4) 1508 sq. units 5) None of these
s=(9+40x+41x)/2=(9+81x)/2

Area=root[(9+81x)/2*((9+81x)/2-9)*((9+81x)/2-40x)*((9+81x)/2-41x)]
=root[(9+81x)*(81x-9)*(9+x)*(9-x)]/4
=root[(81x^2-1)*(81-x^2)]*9/4

d(Area)/dx=0 ==> x=6.3644

Area=820
11408
@pavimai said:
the external angles in degrees,of a convex polygon are all integers which are in arithmetic progression..how many values are possible for n1614108
Sum of exterior angle =360.
360 = 2^3*3^2*5

When n is odd, total number of possibility = 4 {n=3,9,5,15}
When n is even, total number of possibility = (3) {n=6,10,18} + (2) {n=4,12} +(1) {n=8} =6.

So, total should be 10?
11409
@YouMadFellow said:
9,40x, 41xArea = 820x^2 sinCcosC = (3281x^2 - 81)/3280x^2 => sinC = (9/4*820x^2) * root( (81-x^2)*(81x^2 - 1) )A = (9/4) * root[(81 - x^2)*(81x^2 - 1)]dA/dx = 0 => x = 6.36 => A = 819.99 = 820 ..
@krum said:
s=(9+40x+41x)/2=(9+81x)/2Area=root[(9+81x)/2*((9+81x)/2-9)*((9+81x)/2-40x)*((9+81x)/2-41x)]=root[(9+81x)*(81x-9)*(9+x)*(9-x)]/4=root[(81x^2-1)*(81-x^2)]*9/4d(Area)/dx=0 ==> x=6.3644Area=820
.
11410
@joyjitpal said:
A cube of side 1 has another cube of side 1/2 stuck to the top of it, on top of that cube we have another cube of side 1/4 stuck on... On top of that is another cube of side 1/8 and so on to infinity with each cube having a side length 1/2 the one it is on top of. What is the total surface area of the resulting solid?
nt sure if we need to take area of base of 1st sq too..here i'm not taking it.
I -> (5-1)+ (1-1/4) = 19/4
II -> (1) + (1/4 - 1/16 ) = 19/16
III -> 1/4 + (1/16 - 1/64) = 1/4 + 3/64 = 19/64
so it is a GP with r = 1/4
so, sum = 19/4*4/3 = 19/3 ?
11411
@krum said:
s=(9+40x+41x)/2=(9+81x)/2Area=root[(9+81x)/2*((9+81x)/2-9)*((9+81x)/2-40x)*((9+81x)/2-41x)]=root[(9+81x)*(81x-9)*(9+x)*(9-x)]/4=root[(81x^2-1)*(81-x^2)]*9/4d(Area)/dx=0 ==> x=6.3644Area=820
@YouMadFellow said:
9,40x, 41xArea = 820x^2 sinCcosC = (3281x^2 - 81)/3280x^2 => sinC = (9/4*820x^2) * root( (81-x^2)*(81x^2 - 1) )A = (9/4) * root[(81 - x^2)*(81x^2 - 1)]dA/dx = 0 => x = 6.36 => A = 819.99 = 820 ..

Mr. Meter was going from Kashmir to Kanyakumari by his car. When he started his journey the odometer
of his car showed a reading of 00000 km. On the way Mr. Meter realized that the odometer of his car was
not functioning properly and displayed 'δ' in between 4 and 5 (i.e. after 34, first it showed 3δ and then 35).
Odometer of the car at the end of journey showed a reading of 009δ5 km. Find the distance traveled by Mr.
Meter during the entire journey.
A. 1149 km
B. 594 km
C. 1171 km
D. 1271 km
E. 716 km

A hacker is trying to steal highly confidential information from the Department of Defense system that uses a
nine – step security process. Security at each level of the system can be bypassed by entering a numeric
sequence whose sum is equal to the number of that security level. A traitor has informed the hacker that the
desired numeric sequence only consisted of digits 1 or 2 and security levels are numbered 1 to 9. The
numeric sequence to each level is unique. What is the maximum possible number of attempts that will take
for the hacker to break the security of all the levels?
A. 55
B. 89
C. 512
D. 256
E. 142


11412
Find the largest prime factor of
3(3(3(3(3(3(3(3(3(3(3 + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1
a) 5
b) 7
c) 13
d) 73
e) None
11413
@sujamait said:
Find the largest prime factor of3(3(3(3(3(3(3(3(3(3(3 + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1a) 5b) 7c) 13d) 73e) None
3(3(3(3(3(3(3(3(3(3(3 + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1
=265720
= 2^3*5*7*13*73

So, 73.

Did it manually :banghead:
There should be some shorter method :-/
11414
@vijay_chandola said:
3(3(3(3(3(3(3(3(3(3(3 + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1) + 1=265720= 2^3*5*7*13*73So, 73.Did it manually There should be some shorter method :-/
Peculiarities of €˜Chalata- Firata €™ tribe are: neither an outsider can enter nor a resident can leave the tribe and
no one dies unless he/she is killed by someone else in the tribe. There are exactly seven beggars in the tribe.
Three of them are notorious criminals and killers; they rob and kill those who give alms to at least one of
them on the same day. Rubbagubba, a resident of the tribe, gives alms to exactly three of the seven beggars
on 23rd of a month. What is the probability that Rubbagubba would not be alive on 24th of the month if there
is no other killer in the tribe except the three?

11415
@sujamait
P(e)=1-(4C3/7C3)
11416
@sujamait said:
Peculiarities of €˜Chalata- Firata €™ tribe are: neither an outsider can enter nor a resident can leave the tribe andno one dies unless he/she is killed by someone else in the tribe. There are exactly seven beggars in the tribe.Three of them are notorious criminals and killers; they rob and kill those who give alms to at least one ofthem on the same day. Rubbagubba, a resident of the tribe, gives alms to exactly three of the seven beggarson 23rd of a month. What is the probability that Rubbagubba would not be alive on 24th of the month if thereis no other killer in the tribe except the three?
this one is easy. :)
1- C(4, 3)/C(7, 3)=31/35 :mg:

P.S. I'm also in your league now. (certified) :mg:
11417
@sujamait said:
cker is trying to steal highly confidential information from the Department of Defense system that uses anine €“ step security process. Security at each level of the system can be bypassed by entering a numericsequence whose sum is equal to the number of that security level. A traitor has informed the hacker that thedesired numeric sequence only consisted of digits 1 or 2 and security levels are numbered 1 to 9. Thenumeric sequence to each level is unique. What is the maximum possible number of attempts that will takefor the hacker to break the security of all the levels?A. 55B. 89C. 512D. 256E. 142

1 ---> 1 way
2 ---> 1 way
3 ---> 3 ways
4 ---> 6 ways
5 ---> 8 ways
6 ---> 13 ways
7 ---> 21 ways
8 ---> 34 ways
9 ---> 55 ways

Total = 142 ways ?

11418
@sujamait said:
Peculiarities of €˜Chalata- Firata €™ tribe are: neither an outsider can enter nor a resident can leave the tribe andno one dies unless he/she is killed by someone else in the tribe. There are exactly seven beggars in the tribe.
P ( @#%* ) = 1 - 4C3/7C3 = 31/35

11419
@ScareCrow28 said:
1 ---> 1 way 2 ---> 1 way 3 ---> 3 ways 4 ---> 6 ways 5 ---> 8 ways 6 ---> 13 ways 7 ---> 21 ways 8 ---> 34 ways 9 ---> 55 waysTotal = 142 ways ?
@ScareCrow28 said:
P ( @#%* ) = 1 - 4C3/7C3 = 31/35
@vijay_chandola said:
this one is easy. 1- C(4, 3)/C(7, 3)=31/35 P.S. I'm also in your league now. (certified)
@sumeet1489 said:
@sujamaitP(e)=1-(4C3/7C3)
11420
@sujamait said:
Mr. Meter was going from Kashmir to Kanyakumari by his car. When he started his journey the odometerof his car showed a reading of 00000 km. On the way Mr. Meter realized that the odometer of his car wasnot functioning properly and displayed 'δ' in between 4 and 5 (i.e. after 34, first it showed 3δ and then 35).Odometer of the car at the end of journey showed a reading of 009δ5 km. Find the distance traveled by Mr.Meter during the entire journey.A. 1149 kmB. 594 kmC. 1171 kmD. 1271 kmE. 716 km
10*121+5*11+6=1271km