Official Quant thread for CAT 2013

MBA CAT 2022
2781
@rkshtsurana said:
Sum of AM = 13/2 * 2n/2sum of a term from last and start is same for an AP..so a+ b = 13/2then A1+ A2n = 13/2...so total 2n AM..taking two at a time so multiplied by n
a1+a2+..+a2n = 2n/2 ( a1+a2n)
or (2n+2)/2 * (a+b) = (a1+a2+...+a2n) - 13/2
bhai tmne jo kia wo smj nai aya
2782
@karl @rkshtsurana Ans given is option 1) 12 and i can't guarantee you the legitimacy of the answer..so
2783
There are 6 boxes numbered 1,2, €Ś 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
1. 5
2. 21
3. 33
4. 60
2784
A final year MBA student gets 50% in the exam and 80% in the assignments. If the exam should count for 70% of the final result and the assignment for 30%, what will be the final score of the student, if professors decide to use weighted harmonic mean to uneven performances?(1) 56.34%(2) 60.53%(3) 64.83%(4) 66.59%
OA is (1)......share ur approach........
2785
@ScareCrow28 said:
There are 6 boxes numbered 1,2, €Ś 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is 1. 5 2. 21 3. 33 4. 60
6 + 5 +4 + 3 + 2 + 1 = 21
2786
@ScareCrow28 said:
There are 6 boxes numbered 1,2, €Ś 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is 1. 5 2. 21 3. 33 4. 60
21
2787
@ScareCrow28 said:
There are 6 boxes numbered 1,2, €Ś 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is 1. 5 2. 21 3. 33 4. 60
1 green 5 red - 6 ways
2 green 4 red - 5 ways
3 green 3 red - 4
... 1
so,21 ways ?
2788
@ScareCrow28 said:
There are 6 boxes numbered 1,2, €Ś 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is 1. 5 2. 21 3. 33 4. 60
21 ?
2789
@19rsb said:
A final year MBA student gets 50% in the exam and 80% in the assignments. If the exam should count for 70% of the final result and the assignment for 30%, what will be the final score of the student, if professors decide to use weighted harmonic mean to uneven performances?(1) 56.34%(2) 60.53%(3) 64.83%(4) 66.59%OA is (1)......share ur approach........
1/sum = 0.7/e + 0.3/a
= 0.7/0.5 + 0.3/0.8
= 1.4 + 0.375 = 1.775

sum = 1/1.775 = 56.34%
2790
We have k identical mugs. In an n-storey building, we have to
determine the highest floor from which, when a mug is dropped, it still
does not break. The experiment we are allowed to do is to drop a mug
from a floor of our choice. How many experiments are necessary to
solve the problem in any case, for sure?
P.S. I just know the answer..so i won't give any options
2791
@ScareCrow28 said:
Between two numbers whose sum is 6 ˝ an even number of arithmetic means is inserted;the sum of these means exceeds their number by unity. How many means are there?(1) 12 (2) 6 (3) 24 (4) None
6 means ??
2792
@Estallar12 said:
6 means ??
Ans was 12..apne kaise kia?? please help
2793
@ScareCrow28 said:
We have k identical mugs. In an n-storey building, we have to determine the highest floor from which, when a mug is dropped, it still does not break. The experiment we are allowed to do is to drop a mug from a floor of our choice. How many experiments are necessary to solve the problem in any case, for sure? P.S. I just know the answer..so i won't give any options
smallest integer greater than or equal to log n (base 2) ?
2794
@grkkrg said:
smallest integer greater than or equal to log n (base 2) ?
I think you are correct!
going for lunch..so won't be available for a while..
2795
@fedbite said:
P1=(x-1)^3+313(x-1)^2-77(x-1)-8
1+2+3+......+20=210
P20=(3*210*210-313*210*2-77)=763
can you explain in detail every step?
2796
@Estallar12 said:
6 means ??
approach?
2797
@ScareCrow28 said:
We have k identical mugs. In an n-storey building, we have to determine the highest floor from which, when a mug is dropped, it still does not break. The experiment we are allowed to do is to drop a mug from a floor of our choice. How many experiments are necessary to solve the problem in any case, for sure? P.S. I just know the answer..so i won't give any options
Analogous to binary search for finding a correct floor
Worst case complexity be O~log2(n)
Average case complexity be O~log(n)
Best case complexity be O~n/2
logs are of base 2
Answer would be average case ~ log n base 2 and it will be less than that
2798
@sujamait 400
all no.s of the foprm 3n+1 and which are prime hav 2 b of the form 6k+1
2799
In a locality, the most read newspapers are " Hindu" , "TOI" and "IE" . 60% of the residents read at most one of these newspapers. 30% of the residents read exactly two of these newspapers. The number of residents who read all these papers is 5/6th of those who read none of the papers. 10% read "Hindu" and "TOI" but not "IE" . 40% read Hindu or "TOI" but not "IE" . 44% of residents read "Hindu".. 12% read only "TOI" and "IE". 2400 residents read only "Hindu" and "IE"

Q) How many residents read TOI
a) 2800 b)4200 c)3500 d) 4900

Q) How many residents don't read any of the 3 newspapers
a) 2400 b)3600 c) 3000 d) 4800

Please provide answers with detailed solutions :)
2800
@ScareCrow28

Given, x+y = 13/2...(i), Now, let d be the CD..Thus, AP will be of the form::
=> 1st term= x
=>2nd term/1st mean= x+d
=>3rd term/2nd mean= x+2d
.
.
.
=>2nd last term(2n+1)th term/last mean(2nth mean)= x +2nd
=>last term(2(n+1)th term)=x+ (2n+1)d...Now, x + (2n+1)d=y..(ii)

Thus, 2x + (2n+1)d = 13/2, or, n[2(x +nd) + d]=13/2...(iii)

Now, Sum of 2n mean=2n/2[2(x+d) + (2n-1)d]
=> n[2(x+ nd) +d]...(iv)

Given, n[2(x+ nd) + d] - 2n=1
=> n(13/2) - 2n =1 [Comparing eqn(iii) & eqn(iv)]
=> n(9/2)=1

Thus, n=2/9???