Quant by Arun Sharma

MBA CAT 2024
3554
7^2008= (7^4)^502=(2401)^502
The units digit will be 1 and for the tens digit, just multiply the tens digit of the number (here, 0) with the last digit of the exponent (2), which will give you 0 => last 2 digits will be 01.



The method must be a general one

What if we take 7^2008= (7^2)1004=(49)^1004 (due to ease of multiplication);

Then 1's digit=1
10's digit =6 ??

a bit confused on this...
3555
Working steps needed-

1)3 yrs ago,average age of A,B & C was 27 and of B & C 5 yrs ago was 20 yrs.What is A's present age?

2)The average salary of all the workers is 95.Avg salary of 15 workers is 525 and the avg salary of rest of the workers is 85.Find the total no. of workers?

3)In a bag,there r 150 coins of Re.1,50p & 25p denominations.If the total value of coins is 150.Find how many rupees can be constituted by 50p coins.


1)A,B,C present av. age = 27+3=30



~ly, B and C av present age =25

Question becomes now, what no should be added to 2 no's having avg 25 so that avg becomes 30.

Visualize like this:

Prev:25 25 A

Now:30 30 30

The no 'A' should be 30 + 5 + 5

Hence, A = 40


B)The average salary of all the workers is 95.Avg salary of 15 workers is 525 and the avg salary of rest of the workers is 85.Find the total no. of workers?

Same as above,

Visualise,
525 525 525 ...... 525 (15 terms)
85 85 85 85.....n terms

95 95 95 95................................95 (n+15)

Here, total no of terms is not given,

but we know that a total of n+15 terms become 95 each when 15 terms of 525 each are added to n terms of 85 each

Now increase of 525-85 = 440, 15 times is responsible for an increase of 10, n+15 times.

hence , 440*15 = (n+15) *10
Or total workers (n+15)= 660

C) 0
3556

can any one decipher how to gear up probability stuffs..bcz many students knock down only in this aspects...

3557

can any1 plz try this ques of progressions

pg 75 quest 41 of arunsharma quant
It is based on progressions.
How many three digit numbers have the property that their digits taken frm left to right form an ap or gp?
15
20
36
42

3558

Hey !

I got the answer with the same approach as yours .

A B C LCM
No. of days 10 15 20 60 - Total work done
No.of units/day 6 4 3


2 days work of A : 12 units

Remaining work :48 units



Let the total number of days : x

=> B worked for x-1 days


so 4(x-1) + 3(x) = 48 => x= 7 3/7

3559
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??



Hey !

I got the answer with the same approach as yours .

A B C LCM

No. of days 10 15 20 60 - Total work done
No.of units/day 6 4 3


2 days work of A : 12 units

Remaining work :48 units



Let the total number of days : x

=> B worked for x-1 days


so 4(x-1) + 3(x) = 48 => x= 7 3/7
3560
can any1 plz try this ques of progressions

pg 75 quest 41 of arunsharma quant
It is based on progressions.
How many three digit numbers have the property that their digits taken frm left to right form an ap or gp?
15
20
36
42


I'm getting 36..

32 Ap's and 4 Gp's
3561
The method must be a general one

What if we take 7^2008= (7^2)1004=(49)^1004 (due to ease of multiplication);
Then 1's digit=1
10's digit =6 ??
a bit confused on this...

My approach is applicable only for the numbers ending with 1. So before applying that, you have to make sure that the units digit is 1.
3562

hi puys,

Just wanted to know if there are any solution books available for the Quant book by arun sharma
TIA

3563
Not sure if you are answering or asking.

But anyway here's how it goes

C(7,3)-First step

Next, out of these 3 any 2 can be on one side and one on the other, so C(3,2)

Next 4 on each side can be arranged among themselves in 4!, 4! ways each

so C(7,3)*C(3,2)*4! *4! = 60480


couldn't get the solution:(
Can you explain it elaborately please..
3564
if no is abc then abc will be in AP
and all nos of the form aaa
if u calculate total comes out to be 112..


How to calculate the total no.?

as in manually calculating would be too tedious..
3565

An A.P P consist of n terms .from the progressions three different progression P1 P2, P3are created such that P1 is obtained by 1st,4th ,7th ....terms of P ,P2 has the 2nd.5th ,8th ...terms of P and P3 has 3rd,6th ,9th ...terms of P.It is found that of P1, P2 and P3 two progressions have the property that their average is itself a term of the original progression P.what are the possible values of n?

( solution approach please)

3566

In an office where working in one dept is necessary 78% of the employees are in operations,69% are in finance and 87% are in HR. what are the maximum and minimum %ages of employees that could have been working in all 3 depts.

shouldn't the max answer be 69%.. ? considering all in finance are in other twos as well..?

please explain both min and max.

3567
An A.P P consist of n terms .from the progressions three different progression P1 P2, P3are created such that P1 is obtained by 1st,4th ,7th ....terms of P ,P2 has the 2nd.5th ,8th ...terms of P and P3 has 3rd,6th ,9th ...terms of P.It is found that of P1, P2 and P3 two progressions have the property that their average is itself a term of the original progression P.what are the possible values of n?

( solution approach please)


It is clear that P1, P2, P3 also are in progression.
Now, to have the property that their average is itself a term of the original progression P, there must be odd number of terms in that progression.

If P1 and P2 have the property that their average is itself a term of the original progression P, then they must have odd number of digits then P3 must have even number of digits.

=> number of terms in P must be 6k + 2, as then P1 and P2 both will have 2k + 1 terms and P3 will have 2k terms

If P2 and P3 have the property that their average is itself a term of the original progression P, then they must have odd number of digits then P1 must have even number of digits.

=> Number of terms in P must be of form 6k + 4, as then P1 will have 2k + 2 terms and P2 & P3 both will have 2k + 1 terms

=> For n = 6k + 2 or 6k + 4, given condition will satisfy.

How to calculate the total no.?

as in manually calculating would be too tedious..


Check post #3432 of this page

http://www.pagalguy.com/discussions/quant-by-arun-sharma-25023813
3568
In an office where working in one dept is necessary 78% of the employees are in operations,69% are in finance and 87% are in HR. what are the maximum and minimum %ages of employees that could have been working in all 3 depts.

shouldn't the max answer be 69%.. ? considering all in finance are in other twos as well..?

please explain both min and max.

let a b and c % work only in one of Ops, F or HR
and x y z .. in only Ops and F or F and HR or HR and Ops
and m all three
so we know a+x+z+m=78
b+x+y+m=69
c+y+z+m=87... add up the three a+b+c+2+3m= 234
and x+y+z+a+b+c+m=100
subtract from above x+y+z+2m= 134
we need to max m
m max is when x y z are 0
m= 67%
and min m=0
3569
In an office where working in one dept is necessary 78% of the employees are in operations,69% are in finance and 87% are in HR. what are the maximum and minimum %ages of employees that could have been working in all 3 depts.

shouldn't the max answer be 69%.. ? considering all in finance are in other twos as well..?

please explain both min and max.


Say I is the number of employees working in at least one department = 100%
II is the number of employees working in at least two department
III is the number of employees working in all three department

=> I II III
Exactly 1 = I - II
Exactly 2 = II - III
Exactly 3 = III

To find the maximum value of III, put II = III, i.e, exactly 2 = 0

Also, I + II + III = 69 + 78 + 87 = 234
=> 100 + III + III = 234
=> III(max) = 67

To get the least value of III, out I = II = 100
=> III(min) = 234 - 200 = 34
3571
Hey , i have a few doubts in P&C;,
1.If there are 3 variants of a test given to 12 students (so that each variant is used for 4 studnts) and there should be no identical variants side by side and that the student sitting one behind the other should have the same variant. Find the number of ways this can be done.
options- 6!^2, 6x6!x6!, 6!^3, 30x12C6x6!x6!, NOT
2.A dice is rolled six times. One, Two, 3,4,5,6 appear on consecutive throws of the dices. HOw many ways are possible of having 1 before 6.
options- 120,360,240,380,280


Is it 360 for 2nd one?
3573
phoenix0203 Says
yup..can u explain how u got that


As in ques we have to put 1 always before 6
1 _ _ _ _ 6 =4! Cases
_1 _ _ _ 6 = 4!
_ _1 _ _ 6=4!
_ _ _1 _ 6=4!
_ _ _ _1 6=4!

Similarly fix 1 at 1st place and move 6
By now v get 216 cases.

_1_ _ 6_ =4!
_ _1_ 6 _ =4!
_ _ _1 6_ =4!
_1 _ 6_ _ = 4!
_1 6 _ _ _ = 4!
_ _ 16_ _ =4!

Add all the cases you Will get answer

Hope iam clear not all clear;:):):)
3575
For the first sum , C) is the ans and i think i have got the same .... u check again ... i hav already solved it

for the second problem,as far as i know ..... the previous origin is (0,0) and new one is 3,3)

if i write

X=x+3
and put X=3 and x=0 it makes sense

but if i write

x=X+3
and put x=0 n X=3.....it doesnt work ..... 0!=6 right??
So i will still go with that ..... i m not bothered about options ..... ur concept shud be right

but still then ,u check if the new origin is (3,3) and not something else



Hey this discussion seems old but i Just happened to check this thread ... I think with the changed origin to (3,3) ; the equation is 2x+5y=4 ; that is when the answer will come out to p= -17
3576

Hi setu ,


With reference to the diagram given by cloudsonfire , Just draw the diagram once again ... Every tangent makes an angle of 90 degrees at the point of contact ...so if you consider a common tangent .. A rectangle will be formed with sides as

1) Line joining both the centres
2) common tangent
3) The radii of the resp circles to the point of tangent


If you see carefully the lines joining the centres of the three circles form an equilateral triangle => angle = 60degrees



So the angle of the arc = 360 - (90+90+60) = 120 degrees



Hope this was useful 😛