Appurv and Vikram play a game in which they roll a 6-faced die alternately starting with Appurv.Each of them keeps on adding the numbers rolled by him and the first one to get to a sum of at least 3 wins the game. What is the probability of Vikram winning the game?
(a)2/9

(b) 293/6^4

(c) 299/6^4

(d) None of these

total 10-digit numbers in base 10 => 2*9 = 512

we will find out numbers in which there are no 2 consecutive 1s and subtract them from 512

1st digit will always be 1, so 2nd will be 0
now out of remaining 8 digits=>
if there is no 1 => 1 way
if there is 1 1 => 8c1 ways
if there are 2 1s => 7c2 ways
if there are 3 1s => 6c3 ways
if there are 4 1s => 5c4 ways
if there are 5 1s => not possible

so it will be 512 - (1+8+21+20+5) = 457 numbers

Culdip bhai, kuch mistakes lagi yahan... Please check and pardon me if I am wrong and tried questioning your solution.

Edited... :)

For case if there are two 1s, your solution will also include, 1001100000
as two 1s can take any of the remaining 7 positions.

For case of four 1s, there will be only two possibilities according to me,
1010101010
1001010101

Similarly, I am not sure for case of three 1s.

___/\___

Unmukt bhai kal karna bhul gaya...aj zarur kar denge ghar pohochke bro...
Find the number of 10 digit numbers in base 2 which have atleast 2 consecutive 1's?

Total no. =512

No. of 10 digit numbers which does not have 2 consecutive 1's=fibonacci10 with f1=2,f2=3
So f10=144

So asnwer=512-14=368

total 10-digit numbers in base 10 => 2*9 = 512

we will find out numbers in which there are no 2 consecutive 1s and subtract them from 512

1st digit will always be 1, so 2nd will be 0
now out of remaining 8 digits=>
if there is no 1 => 1 way
if there is 1 1 => 8c1 ways
if there are 2 1s => 7c2 ways
if there are 3 1s => 6c3 ways
if there are 4 1s => 5c4 ways
if there are 5 1s => not possible

so it will be 512 - (1+8+21+20+5) = 457 numbers

there can be two consecutive 1's among 6c3 also or 7c2,5c4

saurabh3091 Says

in a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, ii and iii. 50% of those asked favoured proposal i, 30% favoured proposal ii and 20% favoured proposal iii. If 5% of those asked favoured all three of the proposals, what percentage of those asked favoured more than one of the three proposals?

17 ... ? .

saurabh3091 Says

In a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, II and III. 50% of those asked favoured proposal I, 30% favoured proposal II and 20% favoured proposal III. If 5% of those asked favoured all three of the proposals, what percentage of those asked favoured more than one of the three proposals?

let there be 100 people
22 favoured none
78 favoured at least one

50 fp1
30 fp2
20 fp3
5 all

now a=fp1,fp2
b=fp2,fp3
c=fp3,fp1
d=fp1 only
e=fp2 only
f=fp3 only
g=all=5
a+b+c+d+e+f=78-5=73
a+c+d=50-5=45
a+b+e=30-5=25
b+c+f=20-5=15
then use 2 3 4
2(a+b+c)+d+e+f=85
sub 1 from it
a+b+c=12
we have to find a+b+c+5
so ans. 12+5 = 17% since total 100 people

saurabh3091 Says

In a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, II and III. 50% of those asked favoured proposal I, 30% favoured proposal II and 20% favoured proposal III. If 5% of those asked favoured all three of the proposals, what percentage of those asked favoured more than one of the three proposals?

a+b+c+2d+2e+2f+3g=100---1

a+b+c+d+e+f+g=78---2

we want d+e+f+g

using 2 in 1 we get

d+e+f+2g=22

g=5
hence d+e+f+g=17

hence should be 17 percent

saurabh3091 Says

In a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, II and III. 50% of those asked favoured proposal I, 30% favoured proposal II and 20% favoured proposal III. If 5% of those asked favoured all three of the proposals, what percentage of those asked favoured more than one of the three proposals?

my take -17%

I + II + III = 78
I +2 II +3 III = 100

=>II+ 2 III = 22
SO II = 12%
SO II+III = 12+5 = 17%

NOTE - I = no. of people favouring single proposal
II = no. of people favouring 2 proposals

III = no. of people favouring all the 3 proposals

Unmukt bhai kal karna bhul gaya...aj zarur kar denge ghar pohochke bro...
Find the number of 10 digit numbers in base 2 which have atleast 2 consecutive 1's?

991 aa raha hai mera... zarur kuch galti hogi. But aas paas bhi hai iske to accha hai. Batao Sam bhai.

saurabh3091 Says

In a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, II and III. 50% of those asked favoured proposal I, 30% favoured proposal II and 20% favoured proposal III. If 5% of those asked favoured all three of the proposals, what percentage of those asked favoured more than one of the three proposals?

Edited:17% hoga???

Sabhi ko pranaam,,, __/\__ aur chill sir ko 5000 posts hone ki khushi mein,,double namashkar,, ___//\\___

Unmukt bhai kal karna bhul gaya...aj zarur kar denge ghar pohochke bro...
Find the number of 10 digit numbers in base 2 which have atleast 2 consecutive 1's?

total 10-digit numbers in base 10 => 2*9 = 512

we will find out numbers in which there are no 2 consecutive 1s and subtract them from 512

1st digit will always be 1, so 2nd will be 0
now out of remaining 8 digits=>
if there is no 1 => 1 way
if there is 1 1 => 8c1 ways
if there are 2 1s => 7c2 ways
if there are 3 1s => 6c3 ways
if there are 4 1s => 5c4 ways
if there are 5 1s => not possible

so it will be 512 - (1+8+21+20+5) = 457 numbers

Unmukt bhai kal karna bhul gaya...aj zarur kar denge ghar pohochke bro...
Find the number of 10 digit numbers in base 2 which have atleast 2 consecutive 1's?

1 000 000 000
total posibble no.s in base 2 10digit =2^9
now in these with no consecutive one's are
10-0-0-0-0-
in these five places we can fill either 0 or 1
so we have to select any 4 places
5C4*2^4=80
so ans. 512-80=432

In a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, II and III. 50% of those asked favoured proposal I, 30% favoured proposal II and 20% favoured proposal III. If 5% of those asked favoured all three of the proposals, what percentage of those asked favoured more than one of the three proposals?

Bahut bahut dhanyawad varun.tyagi & sam05....!
One request- can u pls PM me Number system material that covers such techniques in Number System? I am following Arihant & the stuff is very limited there.!
Thanks!!

Unmukt bhai kal karna bhul gaya...aj zarur kar denge ghar pohochke bro...
Find the number of 10 digit numbers in base 2 which have atleast 2 consecutive 1's?

Bahut bahut dhanyawad varun.tyagi & sam05....!
One request- can u pls PM me Number system material that covers such techniques in Number System? I am following Arihant & the stuff is very limited there.!
Thanks!!

Pls give the detailed solution...
Also, Why dividing by 9?

Thanks...

If a number N = abcdef. It can be written as

N = 10^5*a + 10^4*b + 10^3*c + 10^2*d + 10*e + f

Take remainder by 9 on both sides

N %9 = (10^5*a + 10^4*b + 10^3*c + 10^2*d + 10*e + f ) %9

=> N%9 = 1*a + 1*b + 1*c + 1*d + 1*e + f


=> N%9 = a + b + c + d + e + f = sum of the digits of the number = digit sum

Therefore remainder by 9 is nothing but the digit sum of the number

---------------------------------------------------------------------------------------------------------
For this particular question:

Q.) 5^(1! + 2! + 3! + ....)

E(9) = 6

Since 5 and 9 are co-prime. 5^6k will give a remainder of 1 with 9.
So, we need to find the remainder of 1!+2!+3!+..... with 6

All the numbers from 3! are multiples of 6. So, we need to find the remainder by 1! + 2! = 3

Therefore 5^(1! + 2! + 3! + ....) = 5^(6k + 3)

5^(6k + 3) will give a remainder of 5^3 = 125

Therefore digit sum = 1+2+5 = 8

Pls give the detailed solution...
Also, Why dividing by 9?

Thanks...

Consider the number say 12345

The digit sum is 15
then we add 5+1 to get 6 as the digit sum...It is important to note that digit sum is a one digit number...Now when divided by 9 the rem would be

1+2+3+4+5mod9=6..which is nothing but the digit sum found in last step
We check divisibility by 9 by finding the sum of digits and then dividing by 9...

Hence the remainder by 9 would essentially give the digit sum which is needed over here...

For digit sum simply find the remainder by 9

E(9)=6

(1!+2!)mod 6 is 3
hence 5^3mod9=8

Pls give the detailed solution...
Also, Why dividing by 9?

Thanks...

solve:

5^(1!+2!+............+13!) find the sum of digits down to single digit for above no?
options:

6, 7, 8, 9

sum of digit of 5 repeat in order 5 7 8 4 2 8..
so we have to check remainder of (1!+2!+............+13!) by 6.. i.e 3
so ans is 8

60^40 + 4(60^30)(11^10) + 6(60^20)(11^20) + 4(60^10)(11^30) + 11^40

I got the solution but what is the Pascal's triangle method. please PM me or tell it here itself.
Thanx in advance..

It tells u the powers to which the expression is raised.And we can see that powers are decreasing and increasing by same amount i.e sum of powers is always 40.

1
11
121
1331 -------------->3
14641 -------------->4
15101051 --------->5


So we can directly write the expression as (60^10+11^10)^4

By seeing the coefficients as 1 4 6 4 1 -->Remember Pascals Triangle.

So The term will reduce to (60^10+11^10)^4.

So remainder should be 0 .

I got the solution but what is the Pascal's triangle method. please PM me or tell it here itself.
Thanx in advance..

What will be the remainder when 60^40 + 4(60^30)(11^10) + 6(60^20)(11^20) + 4(60^10)(11^30) + 11^40 is divided by 3721 ?

PS: Thanks Varun, Faruq and Koshti

The expression is the binomial expansion of (60^10 + 11^10)^4, also the denominator 3721=3600+121=60^2 + 11^2
Now consider 60^2=a and 11^2=b

we get the expression: (a^5 + b^5)^4 divided by a+b, we know the property, (a^n + b^n) is always divisible by (a+b) if n is odd. n=5 in this case. Therefore the answer should be 0.

Hi puys, if somebody wishes to buy the career launcher test series including proc and unproc with test gym at discounted price then PM me.

What will be the remainder when 60^40 + 4(60^30)(11^10) + 6(60^20)(11^20) + 4(60^10)(11^30) + 11^40 is divided by 3721 ?

PS: Thanks Varun, Faruq and Koshti

By seeing the coefficients as 1 4 6 4 1 -->Remember Pascals Triangle.

So The term will reduce to (60^10+11^10)^4.

So remainder should be 0 .

What will be the remainder when 60^40 + 4(60^30)(11^10) + 6(60^20)(11^20) + 4(60^10)(11^30) + 11^40 is divided by 3721 ?

PS: Thanks Varun, Faruq and Koshti

Pranam bhai log...
Heavy Lunch

Is it 0???
Yeh binomial expansion he of (60^10+11^10)^4

What will be the remainder when 60^40 + 4(60^30)(11^10) + 6(60^20)(11^20) + 4(60^10)(11^30) + 11^40 is divided by 3721 ?

PS: Thanks Varun, Faruq and Koshti

Pranam bhai log...
Heavy Lunch

Is it 0???
Yeh binomial expansion he of (60^10+11^10)^4

What will be the remainder when 60^40 + 4(60^30)(11^10) + 6(60^20)(11^20) + 4(60^10)(11^30) + 11^40 is divided by 3721 ?

PS: Thanks Varun, Faruq and Koshti

0 hona chahiye

This will reduce to, (60^10 + 11^10)^4 / (60^2 + 11^2)
Let, 60^2 = m and 11^2 = n

So, (m^5 + n^5)^2 / (m+n)
m^5 + n^5 = (m+n)(m^4 + something + n^4)

So divisible.

What will be the remainder when 60^40 + 4(60^30)(11^10) + 6(60^20)(11^20) + 4(60^10)(11^30) + 11^40 is divided by 3721 ?

PS: Thanks Varun, Faruq and Koshti

Congratulation CHILL Sir . 5000 post complete

Let m=10^10^10^10 if m ends with n zeros and n ends with p zeros how many digits are there in p?


don't have OA

I think its 11.

As M will end in 10^10^10 zeros, which is N

Now N will end in 10^10 zeros, Which is P

So Number of digits in P = [ 10* log 10] + 1 = 11



Ps : Pranaam Bhailog _______/\______

PS : Congratulation Chill Sir on 5000 th post

Congratulation CHILL Sir . 5000 post complete

Let m=10^10^10^10 if m ends with n zeros and n ends with p zeros how many digits are there in p?


don't have OA

m ends with n zeroes = 10^10^10 zeroes

n ends with p zeroes = 10^10 zeroes

There are 11 digits in p.

Itna simple hai ya maine koi locha kiya hai isme??


Hemant bhai 5000 post complete kar li.....mubarak ho :clap:

Congratulation CHILL Sir . 5000 post complete

Let m=10^10^10^10 if m ends with n zeros and n ends with p zeros how many digits are there in p?


don't have OA

Puys Can u tell me why this method of finding Rank doesn't work here:See the link:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/55747-cat-2010-concepts-fundas-tips-2.html

as there we were using all the letters but here we are not using all the letters

letters starting from K - 1+ C(4, 1) + C(4, 2)*2! + C(4, 3)*3! + 1*4! = 65 such letters
Similarly starting from N will be 65
Same for R, 65
Total = 195, but we need to find rank of RUN

Lets find letters that will appear after RUN and starts from R
RUN _ , we can fill the blanks in 2 ways
RUN _ _ , we can fill the blanks in 2 ways
similarly for RUT, we have 5 words

So, 195 - 9 = 186 th word

Goodmrng bhaiyo __/\__

1)The sides PQ,QR and RS of a traingle PQR have 4,5 and 6 points(not the end points) respectively on them...The number of triangles that can be constructed using these points as vertices is
a)455 b)34 c)425 d)421

2)There are 8 different books and 2 identical copies of each in a library...the number of ways in which one or more books can be selected is
a)2^8 b)3^8-1 c)2^8-1 d)3^8

3)The number of four digit telephone numbers that have at least one of their digits repeated is?
a)9000 b)4464 c)4000 d)3986

1) 15C3 - (4C3+5C3+6C3)
455-(4+10+20) -> 455-34 -> 421

2)in this case we have three choice for every book
first -> original
Second-> copy
third-> no any
So total cases we have 3^8. we have to select one or more so 3^8-1

3)Total No. of cases- No. digit is repeated
9*10*10*10 -> 9000

No digit is repeated -> 9*9*8*7 -> 4536
-> 9000-4536 -> 4464

Puys Can u tell me why this method of finding Rank doesn't work here:See the link:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/55747-cat-2010-concepts-fundas-tips-2.html

It is because in the given question, the word can be of 1 alphabet, 2 alphabets, 3 alphabets or 4 alphabets, whereas in the link provided, all the alphabets must be used simultaneously to form a word.

correct ans is 186. :clap::clap::clap::clap:

a string of three letters is formed using the vowels a, e, i and consonants b, c, d, f, g such that

i) the first letter is not a vowel
ii) the second letter is a vowel
iii) if the second letter is a, then the third letter is a consonant different from the first letter
iv) if the second letter is e, then the third is either c or d
v) if the second letter is i, then the third letter is the same as the first.

how many of the strings end with d?

4 + 5 + 1 =10

edited.......

K_ _ _ _ = C(4,0) + C(4,1) + C(4,2)*2! + C(4,3)*3! + C(4,4)*4! = 1 + 4 + 12 + 24 + 24 = 65

N_ _ _ _ = 65

R = 1 way
R _ = 4 ways
RK_ = 3 ways
RN_ = 3 ways
RT_ = 3 ways
RUK = 1
RUN = 1

Sum = 65*2 + 1 + 15 = 146th rank

saurav4489 Says

149 aa raha hai....

Fix K _ _ _ _

No of ways = (1+4p1+4p2+4p3+4p4) = 65

Same for N _ _ _ _ ---> 65 ways

Now R K _ _ _ ---> (1+3p1+3p2+3p3)=(1+3+6+6) =16

Same cases for :
R N _ _ _
R T _ _ _

So total 48 cases here.

R --> 1case

R U --->1case

R U K_ _ --->5 case

RUN at last

So I am getting:

65*2+48+7+1=186 rank

when word starts with K
K _ _ _ _ than possible combination
4c0+4c1 + 4c2*2! +4c3*3! +4c4*4! =65

similarly when it starts with N and R=65
total 65*3 = 195

now when the word is R U _ _ _
possible combination
3c0+3c1 +3c2*2! +3c3*3! = 16

possible combination till R T _ _ _ is 195-16 =179

RU ---> 1 way
RUK _ _ ---> 2c0 +2c1 +2c2*2! = 5
RUN

179+1+5+1 =186

Correct ans is 186. :clap::clap::clap::clap:

A string of three letters is formed using the vowels a, e, i and consonants b, c, d, f, g such that

i) the first letter is not a vowel
ii) the second letter is a vowel
iii) if the second letter is a, then the third letter is a consonant different from the first letter
iv) if the second letter is e, then the third is either c or d
v) if the second letter is i, then the third letter is the same as the first.

How many of the strings end with d?

All possible words from the letters of the word "TRUNK", used without repetition, are written down in alphabetical order. What is the rank of the word RUN?


Ans 186

Puys Can u tell me why this method of finding Rank doesn't work here:See the link:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/55747-cat-2010-concepts-fundas-tips-2.html

Goodmrng bhaiyo __/\__

1)The sides PQ,QR and RS of a traingle PQR have 4,5 and 6 points(not the end points) respectively on them...The number of triangles that can be constructed using these points as vertices is
a)455 b)34 c)425 d)421

2)There are 8 different books and 2 identical copies of each in a library...the number of ways in which one or more books can be selected is
a)2^8 b)3^8-1 c)2^8-1 d)3^8

3)The number of four digit telephone numbers that have atleast one of their digits repeated is?
a)9000 b)4464 c)4000 d)3986

Chinese Remainder Theorem is very powerful when the divisor is large...
In that case using just Euler might be difficult.
So Chinese Remainder Theorem can be used to get the Divisor down to smaller value and then apply Euler...

Well then how will we apply BOTH to:
Find rem when 3^1001 is divided by 1001......?

All possible words from the letters of the word "TRUNK", used without repetition, are written down in alphabetical order. What is the rank of the word RUN?


Ans 186

when word starts with K
K _ _ _ _ than possible combination
4c0+4c1 + 4c2*2! +4c3*3! +4c4*4! =65

similarly when it starts with N and R=65
total 65*3 = 195

now when the word is R U _ _ _
possible combination
3c0+3c1 +3c2*2! +3c3*3! = 16

possible combination till R T _ _ _ is 195-16 =179

RU ---> 1 way
RUK _ _ ---> 2c0 +2c1 +2c2*2! = 5
RUN

179+1+5+1 =186

All possible words from the letters of the word "TRUNK", used without repetition, are written down in alphabetical order. What is the rank of the word RUN?


Ans 186

Fix K _ _ _ _

No of ways = (1+4p1+4p2+4p3+4p4) = 65

Same for N _ _ _ _ ---> 65 ways

Now R K _ _ _ ---> (1+3p1+3p2+3p3)=(1+3+6+6) =16

Same cases for :
R N _ _ _
R T _ _ _

So total 48 cases here.

R --> 1case

R U --->1case

R U K_ _ --->5 case

RUN at last

So I am getting:

65*2+48+7+1=186 rank

All possible words from the letters of the word "TRUNK", used without repetition, are written down in alphabetical order. What is the rank of the word RUN?


Ans 186

149 aa raha hai....

My take : 10;)

As hadn't got any standard approach,here are all the possible outcomes

(a1,a2,a3);(a1,a4,a5);(a1,a6,a7);(a1,a8,a9);
(a2,a4,a6);(a2,a5,a7);(a2,a8,a9);
(a3,a4,a7);(a3,a5,a6);(a3,a8,a10)

Mani bhai, here two cases are repeating...

All possible words from the letters of the word "TRUNK", used without repetition, are written down in alphabetical order. What is the rank of the word RUN?


Ans 186

K_ _ _ _ = C(4,0) + C(4,1) + C(4,2)*2! + C(4,3)*3! + C(4,4)*4! = 1 + 4 + 12 + 24 + 24 = 65

N_ _ _ _ = 65

R = 1 way
R _ = 4 ways
RK_ = 3 ways
RN_ = 3 ways
RT_ = 3 ways
RUK = 1
RUN = 1
Edit: Oh yeah Mistake :)
R = 1
RK _ _ _ = 1 + C(3,1) + C(3,2)*2! + C(3,3)*3! = 1 + 3 + 6 + 6 = 16
RN_ _ _ = 16
RT_ _ _ = 16
RU = 5
RUK = 1
RUN = 1


Sum = 65*2 + 1 + 16*3 + 6 + 1 = 186th rank

a1 not with a10 ; did it mean that we can recluse some cases

even in that case how can we be assure that how many outcomes ?

we can go for only (a1,a2,a3);(a1,a4,a5);(a1,a6,a7) :o)

if a1 paired w/ a10,then we may have to select the one which is already paired w/ a1,and yes,as far as i know,a1 can go for 4 days;)

All possible words from the letters of the word "TRUNK", used without repetition, are written down in alphabetical order. What is the rank of the word RUN?


Ans 186

edited.......

Faruq Says

All possible words from the letters of the word "TRUNK", used without repetition, are written down in alphabetical order. What is the rank of the word RUN?

My take: 69

2*4! + 3*3! + 1*2! + 1 = 69

mani0303 Says

My take : 10;)

As hadn't got any standard approach,here are all the possible outcomes
(a1,a2,a3);(a1,a4,a5);(a1,a6,a7);(a1,a8,a9);
(a2,a4,a6);(a2,a5,a7);(a2,a8,a9);
(a3,a4,a7);(a3,a5,a6);(a3,a8,a10)

a1 not with a10 ; did it mean that we can recluse some cases
even in that case how can we be assure that how many outcomes ?
we can go for only (a1,a2,a3);(a1,a4,a5);(a1,a6,a7) :o)

All possible words from the letters of the word "TRUNK", used without repetition, are written down in alphabetical order. What is the rank of the word RUN?


Ans 186

10 students have joined an institute for summer intern..
On each day 3 of them are supposed to clean the class after class hours..
It was found that every pair of students has been on duty exactly once..
Then what is the duration of the summer intern??

My take : 10;)

As hadn't got any standard approach,here are all the possible outcomes

(a1,a2,a3);(a1,a4,a5);(a1,a6,a7);(a1,a8,a9);
(a2,a4,a6);(a2,a5,a7);(a2,a8,a9);
(a3,a4,a7);(a3,a5,a6);(a3,a8,a10)

Unmukt Says

Puys I am confused as to when to apply Chinese Remainder theorem? Euler works pretty good- does it obviate the need for Chinese in all cases or does Chinese address a different domain as well?

Chinese Remainder Theorem is very powerful when the divisor is large...
In that case using just Euler might be difficult.
So Chinese Remainder Theorem can be used to get the Divisor down to smaller value and then apply Euler...

@ Neetujain1987
Exactly!
Try it with a smaller space or any other instance, it seems that this is not possible! at least one of the pairs is missed out.....

PS: Any ans for my earlier query, anyone?

Pranam Jain bhai ,Aizen bhai ___/\___
10 students have joined an institute for summer intern..
On each day 3 of them are supposed to clean the class after class hours..
It was found that every pair of students has been on duty exactly once..
Then what is the duration of the summer intern??

yeh 15 students kaha se aya??i mean why 15C2??
10C2 hoga na???

I have one more doubt, here it is 3 person to choose. But only 1 time a pair is allowed.

Lets say :

A B C D E F G H I J
if we chose AB => ABC then ABD is not allowed as AB will apear again.

So it can be ABC, ADE, AFG, AHI and then AJ remains. But the problem is with AJ , now B,C,D,E,F,G,H,I can not go so no three people for last one.

every pair of students has been on duty exactly once.. : Here pair means 2 people only.
Am I interpreting the question wrong, as always?

Let the course ran for x days.
Let S denote the no of interns who worked together for all x days.
and As every day 3 students worked.

So, S=3x --- (A)

Also given every pair of students has been on duty exactly once.

So C(15,2)*1 = 105 ---(B)
This is nothing but S.

So x=S/3 =35

p.s: srry for posting late manager aa gya tha...

yeh 15 students kaha se aya??i mean why 15C2??
10C2 hoga na???

10 students have joined an institute for summer intern..
On each day 3 of them are supposed to clean the class after class hours..
It was found that every pair of students has been on duty exactly once..
Then what is the duration of the summer intern??

Let the course ran for x days.
Let S denote the no of interns who worked together for all x days.
and As every day 3 students worked.

So, S=3x --- (A)

Also given every pair of students has been on duty exactly once.

So C(10,2)*1 = 45 ---(B)
This is nothing but S.

So x=S/3 =15

p.s: srry for posting late manager aa gya tha...

Puys I am confused as to when to apply Chinese Remainder theorem? Euler works pretty good- does it obviate the need for Chinese in all cases or does Chinese address a different domain as well?

Did like this :

2/M+3/W = 1/6
3/M =1/x
9/W = 1/x+5

X =4
M =12
W =81

Not getting 6 times :(

Bhai apke method main M aur W woking days hain....
9/W represents work done by 9 men in one day I don't think we can add 3/M and 5
W/9= M/3+5 ho sakta hain...

It takes 6 days for 3 women and 2 men working together to complete the work.Three men would do the same work 5 days sooner than the 9 women.How many times does the output of the man exceeds that of a woman..???
A) 3 times B) 4 times C) 5 times D) 6 times E) 7 times

Did like this :

2/M+3/W = 1/6
3/M =1/x
9/W = 1/x+5

X =4
M =12
W =81

Not getting 6 times

35 hai kya???
p.s:Sam bhai __/\__

donno the OA bro...but can you tell the method??
I am getting 15..
basically 10C2=45 pairs..
now 3 pairs will make one day...so 45/3=15
but i think i am making some wrong assumption