Official Quant thread for CAT 2013

MBA CAT 2024
7141
@IIM-A2013 said:
In a certain country the average monthly income is calculated on the basis of 14 months in a calendar year while the average monthly expenditure was to be calculated on the basis of 99 months per year. this leads people having an underestimation of their savings , since there would be an underestimation of the income and an overestimation of the expenditure per month.then,1. mr.jack comes back from ussr and convinces his community comprising 273 families to start calculating the average income on the basis of 12 months per year. now if it is know that the average estimated income in his community is (according to old system) 87 rs per month.then what will be the change in the average estimated savings for the country (assume that there are no other change)a. 251.60 rs b.282.75 rs c. 312.75 rs d . cannot be determined.2. miss rose comes back from usa and convinces his community comprising 546 families to start calculating the average income and average expenditure on the basis of 12 months per year. now if it is know that the average estimated income on the island is (according to old system) 87 rs per month.then what will be the change in the average estimated savings for the country (assume that there are no other change)a.251.60 rs b. 565.5 rs c. 625.5 rs d . cannot be determined.please tell.still waiting for answer of these so posting it again
1. CBD

2. CBD
7142
Can an irrational number raised to an irrational power yield a rational result?

A. Yes
B. Lol what?
C. No
D. CBD
7143
@mohnish_khiani said:
X and Y are playing a game. There are 11 coins on the table and each player must pick up at least 1 coin but not more than 5 coins. The person picking up the last coin loses. X starts. How many coins should he pick up to ensure a win no matter what strategy Y employs.A. 4B. 3C. 2D. 5What is correct approach to solve such problems???
It's A.4


7144
@Ibanez said:
Can an irrational number raised to an irrational power yield a rational result?A. YesB. Lol what?C. NoD. CBD
sqrt(10)^log (9)=10^log 3=3

So, yes.
7145
@chillfactor said:
Always try to find a pattern in these kind of question.If there is just 1 coin on the table, then the person whose turn it is will loose.In case of 2, 3, 4, 5, 6 coins, person whose turn it is can pick 1, 2, 3, 4, 5 coins and 5^c*7then win the game.In case of 7 coins no matter what a person does, he will always loose (as he will leave 2 or 3 or .. or 6 coins for the other person)Then for 8 or 9 or 10 or ... or 12, person whose turn it is will always win (as he can leave 7 for the other person)Now, here we have 11 coins on the table. So, if X can manage to leave 7 coins for Y, then he is certain of winning the gameHence, X will pick 4 coins to win the gameHow many numbers n less than 1000 are there such that number of even factors of n is 3 times the number of odd factors of n.
2^a*3^b*5^c*7^d
Number of even factors=3*number of odd factors
3*(a+1)=(b+1)*(c+1)*(d+1)
1->b+1=3 ,a=c=d=0 no solution
2->c+1=3,a=b=d=0 no solution
3->d+1=3,a=b=c=0 no solution

4->(b+1)(c+1)=3 ,a=d
b=0,c=2,a=d =1
b=2,c=0 ,a=d=1

5-> (b+1)(d+1)=3,a=c
b=2,d=0,a=c=1,2
b=0,d=2,a=c=1

6->(c+1)(d+1)=3,a=b
c=2,d=0,a=b=1,2
d=0,c=2 ,a=b=1

1->2*7*5^2
2->2*7*3^2
3->2*5*3^2
4->2^2*5^2*3^2
5->2*5*7^2
6->2*3*5^2
7->2^2*3^2*5^2
8->2*3*7^2

8?
7146
@krum said:
4 N=2^a*x^b*y^c ...=> a*(b+1)(c+1)... = 3*(b+1)*(c+1)...=> a=3max. possible number = 992 = 8*124of these only half satisfy the condition , so 62
r u able to find answer to that "area of field" question?? my answers are not matching the given options still
7147
@bullseyes said:
r u able to find answer to that "area of field" question?? my answers are not matching the given options still
nope :(
7148
Log(1/2+1/5+1/9+1/14+........)^2 = 2
base of log is (x)

find x
7149
@gnehagarg said:

3->2*5*3^24->2^2*5^2*3^25->2*5*7^26->2*3*5^27->2^2*3^2*5^28->2*3*7^28?
Answer is 62 (as solved by others)

You have considered only 3, 5, 7 as odd prime factors, but 11, 13, ... are also odd prime numbers

Solution:-
Suppose n = (2^p)*A, where A is an odd number
If A has k factors, then 'n' will have 'k' odd factors and kp even factors

=> kp = 3k
p = 3

So, n = 8A, where A is an odd number

Now the problem reduces to finding the number of odd multiples of 8
n = 8(2m + 1) = 16m + 8, where m can vary from 0 to 61

Hence 62 such numbers
7150
@bullseyes said:
Log(1/2+1/5+1/9+1/14+........)^2 = 2 base of log is (x) find x
(1/2 + 1/(2+3) + 1/(2+3+4) + 1/(2+3+4+5).....)

hence tn = 1/(2+3+4.......(n+1)) = 1/[(n+1)(n+2)/2 - 1] = 2/[n^2 +3n] = (2/3)*[1/n - 1/n+3]
as n -> infinity, sum = (2/3)*[1/1 + 1/2 + 1/3] = 11/9

hence log(base x)(11/9)^2 = 2

=>base = x = 11/9

ATDH.
7151
@bullseyes said:
Log(1/2+1/5+1/9+1/14+........)^2 = 2 base of log is (x) find x
1?
7152
@bullseyes said:
Log(1/2+1/5+1/9+1/14+........)^2 = 2 base of log is (x) find x
First lets find the general term of the series:-
2, 5, 9, 14, ....
S = 2 + 5 + 9 + 14 + ....... + t(n)
=> S = 2 + 5 + 9 + 14 + ......... + t(n)

Subtract them to get
0 = 2 + 3 + 4 + 5 + ....... + (n + 1) - t(n)
t(n) = (n + 1)(n + 2)/2 - 1 = n(n + 3)/2

So, general term of the series 1/2, 1/5, 1/9, 1/14, .... will be 2/{n(n + 3)}

2/{n(n + 3)} = (2/3){1/n - 1/(n + 3)}

=> 1/2 + 1/5 + 1/9 + 1/14 + ........
= (2/3)(1 - 1/4 + 1/2 - 1/5 + 1/3 - 1/6 + 1/4 - ............)
= (2/3)(1 + 1/2 + 1/3)
= 11/9

Log(1/2 + 1/5 + 1/9 + 1/14 + ........)^2 = 2
=> 2Log(1/2 + 1/5 + 1/9 + 1/14 + ........) = 2
=> Log(11/9) = 1, base is x
So, x = 11/9
7153
m and n are natural numbers such that hcf(m, n) + lcm(m, n) = m + n, then which of the following is definitely correct?

a) m = n
b) One of m or n is multiple of the other one
c) Both are correct
d) None of these

7154
@chillfactor said:
m and n are natural numbers such that hcf(m, n) + lcm(m, n) = m + n, then which of the following is definitely correct?a) m = nb) One of m or n is multiple of the other onec) Both are correctd) None of these
option c
7155
@chillfactor said:
First lets find the general term of the series:-2, 5, 9, 14, ....S = 2 + 5 + 9 + 14 + ....... + t(n)=> S = 2 + 5 + 9 + 14 + ......... + t(n)Subtract them to get0 = 2 + 3 + 4 + 5 + ....... + (n + 1) - t(n)t(n) = (n + 1)(n + 2)/2 - 1 = n(n + 3)/2So, general term of the series 1/2, 1/5, 1/9, 1/14, .... will be 2/{n(n + 3)}
u explained it so eloquently.. answer is 11/9

Four angles of trapezium are in AP. if one of the angles is 65, find the largest possible angle that trapezium can have ?


7156
@bullseyes said:
u explained it so eloquently.. answer is 11/9 Four angles of trapezium are in AP. if one of the angles is 65, find the largest possible angle that trapezium can have ?
165 ?

x 65 | 115 180 -x
90

gap of 50

x = 15
180-x = 165
7157
@rkshtsurana said:
165 ?x 65 | 115 180 -x------------90------------------gap of 50x = 15180-x = 165
yes correct.
7158

A welder can finish job A and job C in 10 days and 20 days respectively.A Blacksmith can finish job B and job c in 15 days and 10 days respectively.On the first day ,two welders begin work on job A and a day later three blacksmiths begin work on job B .what is the least time required to complete all the three jobs???

1.33/4
2.37/4
3.31/4
4.none of these
7159
@chillfactor said:
m and n are natural numbers such that hcf(m, n) + lcm(m, n) = m + n, then which of the following is definitely correct?a) m = nb) One of m or n is multiple of the other onec) Both are correctd) None of these
i would go with option B, as m=n (i.e option A) is a specific case of option B.

ATDH.
7160
@rkshtsurana said:
option c
Read the question carefully !

a, b, c are natural numbers such that a² + b² = c², what can be said about a*b*c

i) It will always be a multiple of 30 (but not always a multiple of 60)
ii) It will always be a multiple of 60 (but not always a multiple of 120)
iii) It will always be a multiple of 120
iv) None of the above

@anytomdickandhary Sir, its good to see that you are active here :)