Quant by Arun Sharma

MBA CAT 2024
3454
every year number of matches played

1919 = a= 891/34.27 = 26 (approx)
1920 = b = 949/28.75 = 33
1921 = c = 1329/42.87= 31
1922 =d 1101/36.70 = 29

number of matches played in 4 years= 26+33+31+29 = 119

so 4 year avg = (891+949+1329+1101) / 119 = 37 approx

this is fine...but i am finding division part little tedious...how to do that fast...:(
3455

inally Posted by setu_sin:
Qn no. 6 : LOD 2..
There are three circular garbage
cans each of diameter 2m. the
three are touching each other
extarnally at one point only.find
the circumferance of the rope
encompassing the three cans.
Ans : 1. 2pi + 6
2. 3pi +4
3. 4pi + 6
4. 6pi + 6

3456
inally Posted by setu_sin:
Qn no. 6 : LOD 2..
There are three circular garbage
cans each of diameter 2m. the
three are touching each other
extarnally at one point only.find
the circumferance of the rope
encompassing the three cans.
Ans : 1. 2pi + 6
2. 3pi +4
3. 4pi + 6
4. 6pi + 6



My take on this one:
Answer: 2pi + 6
Explanation:
The angle made by twocontact points for a can by rope is 120 degrees. We have 2 such arcs.
So, total length of these arc = pi*d .... d=diameter
Plus we have three straight lengths of rope each one eqaul to d.
So, totally we have 3d + pi*d = 6+2pi
3457
A does a job in 10 days, B does in 15 days, C does in 20 days. They all start working on a job together but A leaves after 2 days. B leaves a day before completion. The job was completed in:-

a.)7 3/7 days b.)6 3/7 days c.)5 3/7 days. d.)7 2/7 days. e)none of these.




My approach:-

LCM of 10,15,20=60
units of work done per day by A=6,B=4,C=3.
implies, in 2 days 2*(6+4+3) =26 units of work is done.
implies, 60-26=34 units of work remaining.
Now if B and C continue the work they would be done in 34/7 days but B left a day before implies B and C worked for 34/7-1 days=27/7 days.
so the units of work done during this time 27/7 *4 +27/7*3=27
so the units of work remaining =34-27=7
C does this work alone implies time taken=7/3
so the total time taken to complete the work
2+27/7+7/3=8 4/21 days.
hence my answer was e.) none of these
but the correct answer seems to be a.)

can anyone help me to see the error in my approach??
3458
A does a job in 10 days, B does in 15 days, C does in 20 days. They all start working on a job together but A leaves after 2 days. B leaves a day before completion. The job was completed in:-

a.)7 3/7 days b.)6 3/7 days c.)5 3/7 days. d.)7 2/7 days. e)none of these.




My approach:-

LCM of 10,15,20=60
units of work done per day by A=6,B=4,C=3.
implies, in 2 days 2*(6+4+3) =26 units of work is done.
implies, 60-26=34 units of work remaining.
Now if B and C continue the work they would be done in 34/7 days but B left a day before implies B and C worked for 34/7-1 days=27/7 days.
so the units of work done during this time 27/7 *4 +27/7*3=27
so the units of work remaining =34-27=7
C does this work alone implies time taken=7/3
so the total time taken to complete the work
2+27/7+7/3=8 4/21 days.
hence my answer was e.) none of these
but the correct answer seems to be a.)

can anyone help me to see the error in my approach??

Answer is a..
My approach..
Now for first two days all of them have worked..
and let after two days B worked for x days and c work for x+1 days..
Hence..we get..
2/10+2/15+2/20+x/15+(x+1)/20=1
solving we get x=31/7=4(3/7)
Now we need to add this to 3 days to get total number of days..
so total=3+31/7..
7 3/7
hence a...
3459

The number of ways in which four particular persons A,B,C,D and six more persons can stand in a queue so that A always stand before B,B always before C and C always before D is

a)10!/4!
b)10(P)4
c)10(C)4
d)(6!)*(4!)

3460

My approach:-

Now if B and C continue the work they would be done in 34/7 days but B left a day before implies B and C worked for 34/7-1 days=27/7 days.
so the units of work done during this time 27/7 *4 +27/7*3=27
so the units of work remaining =34-27=7

can anyone help me to see the error in my approach??


Here is the error in bold.....see by these calculations u get 7 units of work remaining which C takes more than 2 days to complete.....which means B has left more than 2 days before the work is completed......and that is contradicting the question....
3461
A does a job in 10 days, B does in 15 days, C does in 20 days. They all start working on a job together but A leaves after 2 days. B leaves a day before completion. The job was completed in:-

a.)7 3/7 days b.)6 3/7 days c.)5 3/7 days. d.)7 2/7 days. e)none of these.




My approach:-

LCM of 10,15,20=60
units of work done per day by A=6,B=4,C=3.
implies, in 2 days 2*(6+4+3) =26 units of work is done.
implies, 60-26=34 units of work remaining.
Now if B and C continue the work they would be done in 34/7 days but B left a day before implies B and C worked for 34/7-1 days=27/7 days.
so the units of work done during this time 27/7 *4 +27/7*3=27
so the units of work remaining =34-27=7
C does this work alone implies time taken=7/3
so the total time taken to complete the work
2+27/7+7/3=8 4/21 days.
hence my answer was e.) none of these
but the correct answer seems to be a.)

can anyone help me to see the error in my approach??

Error in your approach..
now after 2 days we get 34 unit of work left out of which C works alone for one day..
hence total work remain..
34-3=31..
remaining work is done by b n c==4+3..7 unit/day..
hence time to complete 31 unit==31/7..
add 3 days..
hence we get a
3462
A does a job in 10 days, B does in 15 days, C does in 20 days. They all start working on a job together but A leaves after 2 days. B leaves a day before completion. The job was completed in:-
a.)7 3/7 days b.)6 3/7 days c.)5 3/7 days. d.)7 2/7 days. e)none of these.


Answer (a) 7 3/7 days
My approach:
LCM 60, so assuming 60 units of work.
A, B and C do 6, 4 and 3 units per day respectively.

A does 12 units and leaves.
So, we jsut need to find out time taken by B and C for 48 units.
Now, B leaves 1 days early. So, B and C together work for 45 units.
And C takes one more day for 3 units.

So, time taken= 45/7 + 1 => Choice (a)
3463
The number of ways in which four particular persons A,B,C,D and six more persons can stand in a queue so that A always stand before B,B always before C and C always before D is

a)10!/4!
b)10(P)4
c)10(C)4
d)(6!)*(4!)



Total ways of arranging 10 people = 10!
Now, per 4! possible arrangements of A, B, C and D only 1 is allowed.

So, answer = 10! / 4!
3464

Hi puys..i need arun sharma for quants.I am staying in Chennai. Where will I get these books?
Will it be available in landmark or odyssey type of book shops?

3465

Hi Puys,
Please solve and explain the following question

the ratio of Jims salary for October to his salry for November is 1.5:1.333 and the ratio of salary for November ato December is 2:2.6666. the worker got 40 rs. More for December than for October and received a bonus constituting 40% of salary for 3 months. Find the bonus.( assume that the no. of workdays is same in every month)
a) 368.888 b) 152.5555 c) 265.6 d) 222.22 e)none

3466

You just need to simplify the ratios

1.5=3/2
1.33=4/3
2.66= 8/3

So finally u'll get Oct:Nov = 9:8
and Nov:Dec = 3:4

So Oct:Nov:Dec= 27:24:32

So total for three months = 83 k (say)

Also Dec-Oct= 5k= Rs. 40
So k=8

Bonus= 40% of 83k

0.4*83*8= 265.6 option c

3467

thanks buddy,
even i was expecting the same answer bt answer in arun sharma is given as 222.22. I guess thats the wrong answer.

3468

i was trying to figure this problem out using wilson's theorem ie:
if P is a prime number then Remainder=0 but wasn't able to comprehend it.

Q. Find the remainder when 39! is divided by 41.

would really appreciate a good explanation

3469
i was trying to figure this problem out using wilson's theorem ie:
if P is a prime number then Remainder=0 but wasn't able to comprehend it.

Q. Find the remainder when 39! is divided by 41.

would really appreciate a good explanation

As stated by u already, for prime 'p', (p-1)! = -1modp

40! = -1mod41
40*39! = 40mod41
39! = 1mod41

Remainder = 1
3470

100 students appeared for 2 exams. 60 passed the 1st , 50 passed the 2nd and 30 passed both. Find the probability that a student selected in random has failed in both ?

a)1/5
b)1/7
c)5/7
d)5/6

3471
100 students appeared for 2 exams. 60 passed the 1st , 50 passed the 2nd and 30 passed both. Find the probability that a student selected in random has failed in both ?

a)1/5
b)1/7
c)5/7
d)5/6


60 passed in first, let A
50 passed in second, let B
30 passed in both, i.e., A (intersection) B
Now,
A u B = 60 + 50 - 30 = 80
So, 100- 80 = 20 are the students that failed in both.
As there are 100 students in total,
So, required answer is 20 / 100 = 1/5.

What's the OA ?
3472

Cant this be solved in any other way , this wilson's theorem is a bit confusing ...

3473
ggl26ban Says
Cant this be solved in any other way , this wilson's theorem is a bit confusing ...


which part do u find confusing?