Official Quant thread for CAT 2013

MBA CAT 2024
6381
@grkkrg said:
If A=(p,q,r,s,t) and B=(x,y,z), how many onto functions A->B exist?
If A and B are two sets having m and n elements respectively such that 1≤n≤m then number of onto function from A to B is
∑ (-1)^n-r (nCr) r^m where, r varies from 1 to n

put the values and hence the ans: 150
6382
@grkkrg said:
If A=(p,q,r,s,t) and B=(x,y,z), how many onto functions A->B exist?
@Brooklyn said:
2,2,1-> 5c2*3c2*23,1,1-> 5c3*2c1*2total = 100 ??
OA: 150


@Zedai said:
If A and B are two sets having m and n elements respectively such that 1≤n≤m then number of onto function from A to B is∑ (-1)^n-r (nCr) r^m where, r varies from 1 to nput the values and hence the ans: 150
6383
A raindrop consists of 75% water and the rest is dust. However, by the time it reaches the surface of Earth, it is left with 70% water as 2 ml water evaporates on the way. Find the original volume of the raindrop.
6384
@grkkrg said:
A raindrop consists of 75% water and the rest is dust. However, by the time it reaches the surface of Earth, it is left with 70% water as 2 ml water evaporates on the way. Find the original volume of the raindrop.
12 kiya hua hai
.75x-2=.70(x-2)
x=12
6385
@grkkrg said:
A raindrop consists of 75% water and the rest is dust. However, by the time it reaches the surface of Earth, it is left with 70% water as 2 ml water evaporates on the way. Find the original volume of the raindrop.
12ml??
6386
@grkkrg said:
A raindrop consists of 75% water and the rest is dust. However, by the time it reaches the surface of Earth, it is left with 70% water as 2 ml water evaporates on the way. Find the original volume of the raindrop.
12 ml
Let the initial volume be x ml
Water is 0.75x
After evaporation
(0.75x-2)/x-2=70/100
Solve to get x=12
6387
@grkkrg said:
A raindrop consists of 75% water and the rest is dust. However, by the time it reaches the surface of Earth, it is left with 70% water as 2 ml water evaporates on the way. Find the original volume of the raindrop.
original volume - x
dust = x/4
water=3x/4
water content after evaporation=x/4*10/3*7/10=7x/12

3x/4-7x/12=2
=>2x/12=2
=>x=12

12ml
6388
@grkkrg said:
A raindrop consists of 75% water and the rest is dust. However, by the time it reaches the surface of Earth, it is left with 70% water as 2 ml water evaporates on the way. Find the original volume of the raindrop.

V is the volume of drop when it reaches surface
=> water = 0.7V and dust = 0.3V

=>(0.7V+2)/(0.7V+0.3V+2) = 0.75
=>(0.7V+2) = 0.75(V+2)
=>0.5 = 0.05V
=>V=10 ml.

=> original volume = 10+2 =12.

ATDH.

6389
@grkkrg said:
If A=(p,q,r,s,t) and B=(x,y,z), how many onto functions A->B exist?
150
3C2*[5!/2!2!+5!/3!]
6390
How many numbers between 1 and 1000 are there such that n^2+3n+5 is divisible by 121 ?
6391
For how many integers 'm' is m^3-8m^2+20m-13 a prime number ?

6392
@krum said:
How many numbers between 1 and 1000 are there such that n^2+3n+5 is divisible by 121 ?
none??
6393
@krum said:
How many numbers between 1 and 1000 are there such that n^2+3n+5 is divisible by 121 ?
Zheero
6394
@krum said:
How many numbers between 1 and 1000 are there such that n^2+3n+5 is divisible by 121 ?
Is it 0?
6395
@krum said:
For how many integers 'm' is m^3-8m^2+20m-13 a prime number ?
3??



m^3-8m^2+20m-13=(m-1)(m^2-7m+13)

for it to be prime:
case1:
m-1=1 => m=2


case 2:
m^2-7m+13=1 =>m=3,4

therefore m=2,3,4
6396
@Zedai said:
none??
@deedeedudu said:
Zheero
@soumitrabengeri said:
Is it 0?
@Zedai said:
3??values: 2,3,4.
yup
6397
@krum said:
How many numbers between 1 and 1000 are there such that n^2+3n+5 is divisible by 121 ?
9 >= 4.1.(5-121k)
9 >= 20 - 484k
484k >= 11
K=1/11 only will satisfy
but,
n^2+3n+5 = 11
n^2+3n = 6
no integer satisfy :|
6398
@krum said:
How many numbers between 1 and 1000 are there such that n^2+3n+5 is divisible by 121 ?
non
6399

@krum said:
How many numbers between 1 and 1000 are there such that n^2+3n+5 is divisible by 121 ?
say (n^2+3n+5) = 121k
=> n^2 +7n -4n - 28 +33 = 121K
=> (n+7)(n-4) + 33 = 121K
=> (n+7)(n-4) = 11*(11k-3)

put n+7 = 11t => n-4 = 11t -11

Hence the eqn becomes

11t*(11t-11) = 11*(11k - 3)

=> 11*t(t-1) = 11k - 3

since LHS is multiple of 11 and RHS is not a multiple hence this is not possible.

So we can generalize that for no value of n, n^2 + 3n + 5 is divisible by 121.

ATDH.


6400
@sujamait said:
9 >= 4.1.(5-121k)9 >= 20 - 484k484k >= 11K=1/11 only will satisfybut,n^2+3n+5 = 11n^2+3n = 6no integer satisfy
how did you get the inequality.... I could not really understand your solution can you please elaborate a bit?

ATDH.