Official Quant thread for CAT 2013

MBA CAT 2022
1581
@maddy2807 said:
@rachit_28 u need to select 5 coins out of the 7 distinct coins. so i guess ans will be 21*5!*25correct me if wrng?
mujhe khud clear nhi hai ! 😃
@krum bhai aapka answer sahi hai, mere explanation ko edit/ append kar do to samajh mein aa jaeyga :)
1582
How many scalene triangles can be drawn with distinct sides a,b and c which take integral values less than 8?
1583
@Sixty said:
How many scalene triangles can be drawn with distinct sides a,b and c which take integral values less than 8?
options
1584
@rachit_28 said:
mujhe khud clear nhi hai ! @krum bhai aapka answer sahi hai, mere explanation ko edit/ append kar do to samajh mein aa jaeyga
coins distinct hain isliye unme selection hoga 7c5, baki sahi hai
1585
@rachit_28 said:
There are 7 distinct coins which r 2 b distributed randomly into 5 distinct pots in a row. Any pot can receive any no. of coins. All coins need to b distributed.In how many ways can these coins b distributed so that no pot is empty?I approached like this:First I distributed 1 coin in each of the pots as no pot can remain empty which can be done in 5! ways, now the remaining 2 coins can be distributed in 5*5(25) ways, so total ways = 25*5! ways. Please point out the mistake.@soumitrabengeri@sujamait@krum
you have not selected the coins here first..7C5 !!! also I am not sure whether your approach will be matching mine..

my approach..
2,2,1,1,1
7C3*6 * 5!

3,1,1,1,1
7C4*5!

add them..
8400
1586
@Sixty said:
How many scalene triangles can be drawn with distinct sides a,b and c which take integral values less than 8?
ans = 6?
1587
@rachit_28 said:
There are 7 distinct coins which r 2 b distributed randomly into 5 distinct pots in a row. Any pot can receive any no. of coins. All coins need to b distributed.In how many ways can these coins b distributed so that no pot is empty?I approached like this:First I distributed 1 coin in each of the pots as no pot can remain empty which can be done in 5! ways, now the remaining 2 coins can be distributed in 5*5(25) ways, so total ways = 25*5! ways. Please point out the mistake.@soumitrabengeri@sujamait@krum
Whats the OA ??
iss thread ki speed tends to 0.. mann hee nhn karta aane ka ab toh..IIFT type Qs bhi nhn hai kisi ke pass..
1588
Let P(x) be a polynomial with integer coefficients such that P(17) = 10 and P(24) = 17.

If P(n) = n + 3 has two distinct integer solutions n1 and n2, then find the sum n1 + n2.
OPTIONS

1) 41
2) 27
3) 30
4) 34
5) 53
1589
@sujamait said:

iss thread ki speed tends to 0.. mann hee nhn karta aane ka ab toh..IIFT type Qs bhi nhn hai kisi ke pass..
chill man.....abhi to season hardly shuru hua h....the thread will pick up speed gradually...
1590
H.C.F. of three positive integers €“ A, B and C €“ is 6. If the sum of the three integers is 48, how many triplets (A, B and C) are possible?
1591
@sujamait said:
Let P(x) be a polynomial with integer coefficients such that P(17) = 10 and P(24) = 17.If P(n) = n + 3 has two distinct integer solutions n1 and n2, then find the sum n1 + n2.OPTIONS1) 41 2) 27 3) 30 4) 34 5) 53
p(x)=ax^2+bx+c
=>10=289a+17b+c
=>17=576a+24b+c

==>287a+7b=7
==>41a+b=1
==>(a,b)= (-1,42)/(1,-40)

p(n)=an^2+bn+c
==>an^2+bn+c=n+3
==>an^2+(b-1)n+c-3=0

sum of roots = (1-b)/a = (1-42)/-1 or (1+40)/1 = 41
1592
@raakesh88 said:
H.C.F. of three positive integers €“ A, B and C €“ is 6. If the sum of the three integers is 48, how many triplets (A, B and C) are possible?
6(a+b+c)=48
=>a+b+c=8

5+3-1c2-1
=>7c2
=>21

this includes cases which will make hcf > 6

such cases, 2,2,4 - 3

so total = 21-3=18

edited
1593
@sujamait Ye lo bhai..IIFT 2006 question

If a is not equal to n*(pi) and tan(a) is the GM of sina and cosa, then find the square of..
2-4sin^2(a)+3sin^4(a)-sin^6(a)

1) 1
2) 4
3) 1/4
4) NOT
1594
Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?

I am getting 25. Please confirm if it's right.
1595
@raakesh88 said:
H.C.F. of three positive integers €“ A, B and C €“ is 6. If the sum of the three integers is 48, how many triplets (A, B and C) are possible?
A= 2*3*p1 (where p1,p2,p3 are products of unique primes themselves)
B=2*3*P2
C=2*3*p3

A+B+C = 6(P1 + P2 +P3 ) =48
=> P1+P2+P3 = 8
=>(P1,P2,P3)= ( 2,3,3)
therefore nos are 12,18,18.
only 1 triplet????????

1596
@rachit_28 said:
Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?I am getting 25. Please confirm if it's right.
me getting 5
1597
@soumitrabengeri said:
@sujamait Ye lo bhai..IIFT 2006 questionIf a is not equal to n*(pi) and tan(a) is the GM of sina and cosa, then find the square of..2-4sin^2(a)+3sin^4(a)-sin^6(a)1) 12) 43) 1/44) NOT
bhai 2) 4 hai kya ?
1598
@krum said:
p(x)=ax^2+bx+c=>10=289a+17b+c=>17=576a+24b+c==>287a+7b=7==>41a+b=1==>(a,b)= (-1,42)/(1,-40)p(n)=an^2+bn+c==>an^2+bn+c=n+3==>an^2+(b-1)n+c-3=0sum of roots = (1-b)/a = (1-42)/-1 or (1+40)/1 = 41
@sujamait said:
Let P(x) be a polynomial with integer coefficients such that P(17) = 10 and P(24) = 17.If P(n) = n + 3 has two distinct integer solutions n1 and n2, then find the sum n1 + n2.OPTIONS1) 41 2) 27 3) 30 4) 34 5) 53


wats the relevance of P(n) = n+3???? and @krum : y did u assume quadratic? its only said polynomial
1599
@rkshtsurana said:
me getting 5
even i am getting the answer as 5..seems wrong..but can't figure out where
1600
@rachit_28 said:
bhai 2) 4 hai kya ?
Nahi bhai..please try again