Official Quant thread for CAT 2013

MBA CAT 2024
11601
@sujamait said:
There were 320 students in the premises of an ongoing college fest. A student can take part in any of thethree activities €” Dance, Debate and Drama. Ten students took part in all the three activities and201 students took part in Debate. Number of students who took part in Dance and Drama but not in Debateis one-third the number of students who took part only in Drama. Number of students who took part inDance and Debate but not in Drama is equal to the number of students who took part in Debate and Dramabut not in Dance and is 1 more than the number of students who took part only in Debate. Twenty-ninestudents just came to enjoy the performances. Number of students who took part in Dance is 27 more thanthe number of students who took part in Drama. Find the number of students who took part in Dance.A. 54B. 45C. 137D. 154E. 68
player=320-29=291
debate=x+x+x-1+10=201
=>x=64..
dance=drama+27
[dance+drama]=drama/3
k+[k+27]+k/3=291-201=90
4k/3+k+27=90
7k+81=270
7k=270-81=189
k=27
nly dance=27+27=54..
dance=54+64+9+10=137

11602
@sujamait said:
What is the remainder when 9 × 99 × 999 × … × 99… 9 is divided by 1000?The last number in the product series contains 999 number of 9s.OPTIONS1) 89 2) 109 3) 129 4) None of these
9*99*999*999*999*999-----------------------------------------999(999 th term)

9*99*(999)^997

9*99*(-1)^997

-891

109
11603
@happy3475 said:
player=320-29=291debate=x+x+x-1+10=201=>x=64..dance=drama+27[dance+drama]=drama/3k+[k+27]+k/3=291-201=904k/3+k+27=907k+81=2707k=270-81=189k=27dance=27+27=54..
C. 137
11604
@viewpt said:
R=? 61^67^71^101/103
e(103)=102

67^71^101 mod 102

e(102)=32

71^101 mod 32

e(32)=16

101mod16=5

71^5mod32=7

67^7mod102=67

61^67mod103

61^5mod103=9

==> 9^13*61^2mod103

9^3mod103=8

==> 8^4*9*61^2mod103
==> 79*9*61*61mod103
==> 76


now tell me the right way :splat:
11605
@sujamait said:
C. 137
sorry..that was nly dance...when question was for WHOLE of dance...
11606
@viewpt said:
R=? 61^67^71^101/103
67^71^101 mod 102 = 67 itself check with 2 and 51

61^67 mod 103 is left now...which looks like shady expression
11607
@krum said:
e(103)=10267^71^101 mod 102e(102)=3271^101 mod 32e(32)=16101mod16=571^5mod32=767^7mod102=6761^67mod10361^5mod103=9==> 9^13*61^2mod1039^3mod103=8==> 8^4*9*61^2mod103==> 79*9*61*61mod103==> 76now tell me the right way
it seems correct.. OA i don't have.
11608
The function f(x,y) satisfies: f(0,y) = y + 1,
f(x+1,0) = f(x,1),
f(x+1,y+1) = f(x,f(x+1,y)) for all non-negative integers x, y.
Find f(4, 1981).
11609
The function f(x,y) satisfies: f(0,y) = y + 1,
f(x+1,0) = f(x,1),
f(x+1,y+1) = f(x,f(x+1,y)) for all non-negative integers x, y.
Find f(4, 1981).
11610
The function f(x,y) satisfies: f(0,y) = y + 1,
f(x+1,0) = f(x,1),
f(x+1,y+1) = f(x,f(x+1,y)) for all non-negative integers x, y.
Find f(4, 1981).
11611
@sujamait said:
What is the remainder when 9 × 99 × 999 × … × 99… 9 is divided by 1000?The last number in the product series contains 999 number of 9s.OPTIONS1) 89 2) 109 3) 129 4) None of these
option 2 : 109...
11612
@sujamait said:
What is the remainder when 9 × 99 × 999 × … × 99… 9 is divided by 1000?The last number in the product series contains 999 number of 9s.OPTIONS1) 89 2) 109 3) 129 4) None of these
9*99*999*9999*......
= 9*99*(-1)^997
= 9*99*-1
= -891 = 109
11613
@sujamait said:
There were 320 students in the premises of an ongoing college fest.
C.137?
11614
@19rsb 496?
11615
@sujamait

I dont know what you are asking for.


Two identical marked dices are brought together and kept with one of their faces in full contact. How many different arrangements are possible?(1) 36 (2) 60 (3) 72 (4) 84 (5) none of these

My approach : Fix one cube and then paste it with each number of another cube so 6 ways for every number => total 36.

Now see the surface of 2 cube. it can be arranged in 2 ways. fix one cube and second cube can be rolled upside or downside so total 36*2 =72 ways.

please somebody confirm.
11616
1,5,13, 26
what is 100th number?
11617
@krum

Answer is D for series question. I didnt get how.
11618
@sparklingaubade said:
1,5,13, 26 what is 100th number?
176650 ?
@sparklingaubade said:
@krumAnswer is D for series question. I didnt get how.
will check again , gtg for breakfast for now
11619

1=0^2+1^2
5=1^2+2^2
13=2^2+3^2
25=3^2+4^2
Shouldnt it be 25 instead of 26?

100th number = 99^2+100^2
=9801+10000
=19801

What are the options?
OA?

11620

An instrument measures the temperature accurately but there is a problem in the instrument, such that it skips the digit 5 and moves directly from 4 to 6. What is the actual temperature it the instrument shows 3016?


a) 2201 (b) 2202 (c) 2600 (d) 2960

OA is (a). Please mention your approach.