Q.) 2010 coins, each of value 1, 2 or 3 are arranged in a row such that between any two coins of value 1 there is at least one coin, between any two of value 2 there are at least two coins, and between any two of value 3 there are at least three coins. The maximum number of coins of value 3 that can be in the row?

a.)502............b.)670............c.)503.............d.)669..............e.)501

3 1 2 1 3 1 2 1.........3 1 2 1 3 1 2

cyclicity of four so , 2010 = 502*4 + 2

so , 502 times will 3 appear...doubt lag raha hai varun bhai

skashwin89 Says

please justify why 4-4-3 only?

coz we have 11 members and 3 cars and each car has maximum seating capacity of 4

read this line again again and think about why 4-4-3 only by this you will get it for life time

My take is 20160

Number of people in car can be (4, 4, 3) only.
Now, we have 4 males for 3 cars. So, it has to be (2, 1, 1).
Now, we get two cases: 2 men in a car of 3 and 2 men in a car of 4.

Case 1: Two men in a car of 3
These two men can be selected in 4C2 ways. Then, they can pick a car in 3 ways. Remaining two men can pick cars in 2*1 ways.
Now, we have 7 women and (1, 3, 3) psostions in car.
One lady can be chosen in 7 ways. Second car can have ladies in 6C3 ways while other car will have remaining ladies.

So, total = 4C2 * 3 * 2*1 * 7 * 6C3 ways = 5040 ways

Case 2: Two men in a car of 4
These two men can be selected in 4C2 ways and they can pick a car in 3 ways. Remaining men can pick cars in 2*1 ways.
Now, we have 7 women and (2, 3, 2) posistions in cars of capacity (4, 4, 3) resp.
So, first car can have women in 7C2 ways. Second car can have them in 5C3 ways and remaining two will go to third car.
One more point here that the cars which have 1 man each, have capacity of 4 and 3. So, the car with capacity of 4 can be chosen in 2 ways. Other will have capacity of 1.

So, total = 4C2 * 3 * 2*1 * 7C2 * 5C3 * 2*1 ways = 15120 ways

In all, we have (5040+15120) ways = 20160 ways

please justify why 4-4-3 only?

pratyasha1990 Says

2 values-11011 and 12221

I didnt get the answer .
How did you find that 2 values ?
Could you explain it in details ?

q.) if denotes the greatest integer less than or equal to a, find the number of possible values of x (x is an integer, 0 < x < 1000) such that
+ + = 31x/30
a.) 31...........b.) 32............c.) 33............d.) 34..............e.) none of these

is the oa=33

Q.) 2010 coins, each of value 1, 2 or 3 are arranged in a row such that between any two coins of value 1 there is at least one coin, between any two of value 2 there are at least two coins, and between any two of value 3 there are at least three coins. The maximum number of coins of value 3 that can be in the row?

a.)502............b.)670............c.)503.............d.)669..............e.)501

The probability of a missile hitting a bridge is 1/2 and it takes 3 missiles to destroy a bridge.what is the min. No of missiles to be fired so that the probability of the bridge being destroyed exceeds .99?
1.7
2.8
3.14
4.15

Q.) If denotes the greatest integer less than or equal to a, find the number of possible values of x (x is an integer, 0 < x < 1000) such that
+ + = 31x/30
a.) 31...........b.) 32............c.) 33............d.) 34..............e.) None of these

ACE bhai why you counted 4 seated car wala cases twice

by taking 2 males and 2 females in 4 seated car once and multiply it by 3!/2! we have all those in which 2m+2f seated in 1st,2nd and 3rd cars then why again you considered that case ?

See.. in first one, we have one man each in a car of 4. So, there we need to group women as (3, 3) and hence, we need a division by 2!.

In second case, we have two men in a car of 4.
That car can be chosen in 3 ways.
Now, other cars will have 3 women or 2 women.
So, car with 3 women can be chosen in 2 ways.

Here, as remaining two cars have different number of women; we should not divide by 2.

So, it will be:
Select 2 men out of 4 => 4C2 ways
Select a car for these two men => 3 ways
Select a car for 3 women => 2 ways
Select 2 women for the car of two men => 7C2 ways
Select 3 women for the car with 3 women => 5C3 ways
Other women will go in remaining car
And remaining two men can choose car in 2*1 ways

=> Total ways = 4C2 * 3 * 2 * 7C2 * 5C3 * 2 = 15120

It seems, you missed the point of two women teams of (3, 2) in other cars to be taken in 2*1 ways.

What is OA bs0409 bhai? Who knows, I can be wrong as well. Things are never easy with those many number of women involved

bs0409 Says

Anybody for this one.............:banghead:

first lets select the car i which there will be 2 males n the two males who'll be together=3*4C2
now other two males can be seated in two cars is 2 ways
so as for males being seated
no. of ways=3*4C2*2

now the females

here one car will be carrying 3 people now it can be the car with 2 males or other two. So, three cases are possible out of which 2 will be same

Case-1(3 person car is with 2 males)
now we have to select a female out of 7=7C1
and for other two we have to select 3 out of 6=6C3
total=7C1*6C3

case-2(3 person car is other car)
for car with 2 males we'll select 2 females=7C2
for 3person car females would be=5C2
there would be another case like this so multiply by 2

so total 3*2*4C2

damn yaar nice one

Case 2: Two men in a car of 4
These two men can be selected in 4C2 ways and they can pick a car in 3 ways. Remaining men can pick cars in 2*1 ways.
Now, we have 7 women and (2, 3, 2) posistions in cars of capacity (4, 4, 3) resp.
So, first car can have women in 7C2 ways. Second car can have them in 5C3 ways and remaining two will go to third car.
One more point here that the cars which have 1 man each, have capacity of 4 and 3. So, the car with capacity of 4 can be chosen in 2 ways. Other will have capacity of 1.

So, total = 4C2 * 3 * 2*1 * 7C2 * 5C3 * 2*1 ways = 15120 ways

In all, we have (5040+15120) ways = 20160 ways

male = 4
female = 7

when 2 males in 3 seated car :-
4 4 3 = 4C1*7C3*3C1*4C3*2C2*1C1*3!/2! = 5040

when 2 males in 4 seated car :-
4 4 3 = 4C2*7C2*2C1*5C3*1C1*2C2*3!/2! = 7560

total = 5040+7560 = 12600 ???

ACE bhai why you counted 4 seated car wala cases twice

by taking 2 males and 2 females in 4 seated car once and multiply it by 3!/2! we have all those in which 2m+2f seated in 1st,2nd and 3rd cars then why again you considered that case ?

An institute conducts 32 tests.
no. of tests attempted by students are as follows:
bobby - 16
emran - 18
govinda - 20

The no. of tests written by more than one students is atleast :
1)8
2)11
3)13
4)16
5)15

I + 2II + 3III = 16+18+20 = 54
I + II + III = 32

=> II + 2III = 54 - 32 = 22

Now we have to maximize III in order to minimize at least one
=> II = 0 and III = 11

So, 11 is the answer

an institute conducts 32 tests.
no. Of tests attempted by students are as follows:
bobby - 16
emran - 18
govinda - 20
the no. Of tests written by more than one students is atleast :
1)8
2)11
3)13
4)16
5)15

ans is 2
the three together wrote 54 tests.
Hence min no of tests written by 2 or more students=54-32=22/2=11

An institute conducts 32 tests.
no. of tests attempted by students are as follows:
bobby - 16
emran - 18
govinda - 20

The no. of tests written by more than one students is atleast :
1)8
2)11
3)13
4)16
5)15

My take is 11.

54 - 32 = 22.
Say all 3 gave these 22 tests. So for 1 we have to divide by 2. 22/2 = 11.

An institute conducts 32 tests.
no. of tests attempted by students are as follows:
bobby - 16
emran - 18
govinda - 20
The no. of tests written by more than one students is atleast :
1)8
2)11
3)13
4)16
5)15

It should be 11

We have tests written by all 3 = 16 + 18 + 20 = 54
Total tests = 32
=> Extra tests = (54 - 32) = 22

Now, we will have try to maximize tests given by all three.
Now, we have overcounted 22. So, say all three gave a particualr test. We will count it thrice instead of once. So, two overcounts.

So, 22 corresponds to 11 overcounts.
=> 11 tests written by all 3. And other tests written by only one of them.

bs0409 Says

There are 11 members in a family out of which there are 4 males and remaining females. The family has hired three cars for a trip to zoo. The members are to be seated in the cars in such a way that there are not more than four members in one car and there is at least one male in each car. How many different ways can the members travel?

male = 4
female = 7

when 2 males in 3 seated car :-
4 4 3 = 4C1*7C3*3C1*4C3*2C2*1C1*3!/2! = 5040

when 2 males in 4 seated car :-
4 4 3 = 4C2*7C2*2C1*5C3*1C1*2C2*3!/2! = 7560

total = 5040+7560 = 12600 ???

A train starts from Delhi at a : b oclock (i.e. b minutes after a oclock). It reaches Chandigarh on the same day at b : c oclock after taking exactly c hours and a minutes. How many different values of a are possible? All the times are given in 24-hour clock format.

60b + c - 60a - b = 60c + a.
=> a = 59(b-c)/61.
So only possible is when b = c, ie, a = 0. Otherwise it would be next day.

An institute conducts 32 tests.
no. of tests attempted by students are as follows:
bobby - 16
emran - 18
govinda - 20

The no. of tests written by more than one students is atleast :
1)8
2)11
3)13
4)16
5)15

bs0409 Says

There are 11 members in a family out of which there are 4 males and remaining females. The family has hired three cars for a trip to zoo. The members are to be seated in the cars in such a way that there are not more than four members in one car and there is at least one male in each car. How many different ways can the members travel?

My take is 20160

Number of people in car can be (4, 4, 3) only.
Now, we have 4 males for 3 cars. So, it has to be (2, 1, 1).
Now, we get two cases: 2 men in a car of 3 and 2 men in a car of 4.

Case 1: Two men in a car of 3
These two men can be selected in 4C2 ways. Then, they can pick a car in 3 ways. Remaining two men can pick cars in 2*1 ways.
Now, we have 7 women and (1, 3, 3) psostions in car.
One lady can be chosen in 7 ways. Second car can have ladies in 6C3 ways while other car will have remaining ladies.

So, total = 4C2 * 3 * 2*1 * 7 * 6C3 ways = 5040 ways

Case 2: Two men in a car of 4
These two men can be selected in 4C2 ways and they can pick a car in 3 ways. Remaining men can pick cars in 2*1 ways.
Now, we have 7 women and (2, 3, 2) posistions in cars of capacity (4, 4, 3) resp.
So, first car can have women in 7C2 ways. Second car can have them in 5C3 ways and remaining two will go to third car.
One more point here that the cars which have 1 man each, have capacity of 4 and 3. So, the car with capacity of 4 can be chosen in 2 ways. Other will have capacity of 1.

So, total = 4C2 * 3 * 2*1 * 7C2 * 5C3 * 2*1 ways = 15120 ways

In all, we have (5040+15120) ways = 20160 ways

Find the wrong number in the series. OA hai lekin method nahi pata

Oa is option 3.........

For how many positive integers x between 1 and 1000, both inclusive, is 4x^6 + x^3 + 5 is divisible by 7?
a. 0
b. 4
c. 11
d. 36

E(7) = 6

Case 1: When x is co-prime to 7
a.) 4x^6 + x^3 + 5 MOD(7) = x^3 + 2 MOD(7)

x^3 will leave remainder either 1 or -1 with 7
In any case x^3 + 2 will leave remainder 1 or 3

=> Not divisible by 7
Case 2: When x is not co-prime to 7

=> 4x^6 + x^3 + 5 will leave remainder 5
=> Not divisible by 7

So the expression is never divisible by 7


Pranaam Puylog ____/\____

bs0409 Says

There are 11 members in a family out of which there are 4 males and remaining females. The family has hired three cars for a trip to zoo. The members are to be seated in the cars in such a way that there are not more than four members in one car and there is at least one male in each car. How many different ways can the members travel?

Anybody for this one.............:banghead:

underlined part thoda samjhaoge.....

e.g. 15^3 leaves remainder 1 when divided by 7

Bhai sorry, I just read my post, it sounded ambiguous. What i meant was. All cubes leave a remainder of +1, -1 or 0 with 7.

Case 1 -> Remainder is 1.
4*1 + 1 + 5 = 10.
10 mod 7 = 3.

Case 2 -> Remainder is -1.
4*1 + (-1) + 5 = 9.
9 mod 7 = 2.

Case 3 -> Remainder is 0.
4*0 + 0 + 5 = 5.
5 mod 7 = 5.

Hope it is clear now.

Oa is option 3.........

For how many positive integers x between 1 and 1000, both inclusive, is 4x^6 + x^3 + 5 is divisible by 7?
a. 0
b. 4
c. 11
d. 36

A)0????
@SUMEET-YES,HAVE PUT ON THE SPECS

My take is 0.

x^3 will always leave a remainder of -1 with 7.
and x^6,ie, (x^3)^2 mod 7 wld be +1.
so, x^6 + x^3 + 5 = 5 (mod 7).

samajh gaya ab thanks Enceladus

My take is 0.

x^3 will always leave a remainder of -1 with 7.
and x^6,ie, (x^3)^2 mod 7 wld be +1.
so, x^6 + x^3 + 5 = 5 (mod 7).

underlined part thoda samjhaoge.....

e.g. 15^3 leaves remainder 1 when divided by 7


A train starts from Delhi at a : b o'clock (i.e. 'b' minutes after a o'clock). It reaches Chandigarh on the same day at b : c o'clock after taking exactly 'c' hours and 'a' minutes. How many different values of a are possible? All the times are given in 24-hour clock format.

For how many positive integers x between 1 and 1000, both inclusive, is 4x^6 + x^3 + 5 is divisible by 7?
a. 0
b. 4
c. 11
d. 36

My take is 0.

x^3 will always leave a remainder of -1 with 7.
and x^6,ie, (x^3)^2 mod 7 wld be +1.
so, x^6 + x^3 + 5 = 5 (mod 7).

but sir there is 121 and 484 as well
for 484 we get 2 as false rem(484)/13=3 for 121 all r false

u overlooked that question says digit at ten's place should be 7
so it can't be 484 or 121

hota hai kabhi kabhi

but sir there is 121 and 484 as well
for 484 we get 2 as false rem(484)/13=3 for 121 all r false!

sorry!gone blind

jain4444 Says

sood bold part samjhaiyo !!

Koi bi element subset mein hoga ya toh ni hoga. So har 10 element ke pas 2 options hai. Hence 2^10.

(A7A)^17 is a perfect square means A7A is also a prfct sq.
26^2 is 676 (remember this).
676 = 26*26 which is divisible by 13.
So 3 is false.

but sir there is 121 and 484 as well
for 484 we get 2 as false rem(484)/13=3 for 121 all r false!

Square can be 676. Just out of memory as it is better to keep all sqaures till 30 in mind.

(A7A)^17 = (A7A)^16 * (A7A)
=> (A7A) is a perfect sqaure

=> Remainder by 13 is 0 and not 3.

pratyasha1990 Says

my take is opt2

(A7A)^17 is a perfect square means A7A is also a prfct sq.
26^2 is 676 (remember this).
676 = 26*26 which is divisible by 13.
So 3 is false.

Oa is option 3.........

For how many positive integers x between 1 and 1000, both inclusive, is 4x^6 + x^3 + 5 is divisible by 7?
a. 0
b. 4
c. 11
d. 36

N = (A7A)^17 is a perfect square. Which of the following statement is FALSE?
(1) A is an even digit.
(2) A is divisible by 3
(3) When N is divided by 13 we get remainder 3.
(4) None of these

(A7A)^17 is a perfect square means A7A is also a prfct sq.
26^2 is 676 (remember this).
676 = 26*26 which is divisible by 13.
So 3 is false.

OA is A < B.....

I think one of the ways to be sure is to take the first 3-4 terms and test.

N = (A7A)^17 is a perfect square. Which of the following statement is FALSE?
(1) A is an even digit.
(2) A is divisible by 3
(3) When N is divided by 13 we get remainder 3.
(4) None of these

my take is opt2

I remember this question from last year. Naga sir ne solution post kia tha I guess.
Sum of all ten terms = 55.
55/2 = 27.xx
Total sets possible = 2^10.
Half of these will have sum > 27.
So, answer = 2^10/2 = 2^9.

Vivek bhai ne yaad dilaya.

sood bold part samjhaiyo !!

There are 11 members in a family out of which there are 4 males and remaining females. The family has hired three cars for a trip to zoo. The members are to be seated in the cars in such a way that there are not more than four members in one car and there is at least one male in each car. How many different ways can the members travel?

N = (A7A)^17 is a perfect square. Which of the following statement is FALSE?
(1) A is an even digit.
(2) A is divisible by 3
(3) When N is divided by 13 we get remainder 3.
(4) None of these

Square can be 676. Just out of memory as it is better to keep all sqaures till 30 in mind.

(A7A)^17 = (A7A)^16 * (A7A)
=> (A7A) is a perfect sqaure

=> Remainder by 13 is 0 and not 3.

B>A.

Every term is same in both numbers, but B mein ek step jayda hai. So B is greater ofcrs.

my take A>B


@ BS0409 i am 10 passout

viveksingh25 Says

bro i think b me ek term jyada he bt that is in denominator which will render it a lesser value.....

Is it A>B not sure


difference is in last term
A has 2007+1/2008 and has 2007+1/(2007+ 1/2008+2009^-1)
2008+2009^-1 > 2008
so A will have 2007+greater and B will have 2007+less
Now in A 2006+1/(2007+greater) in B 2006+1/(2007+less)
for A 2006+less for B 2006+more
so it will keep on alternating its position
for odd no. like 2007 A>B
for even no. like 2006 AB

my take is A> B
_/\_ Nachiket bhai,sood bhai,bs409 bhai,vivek bhai, aur wo jinka naam yaad nahi

Enceladus Says

yaar but jo lcm lia hai wo bi toh inverse hoke upar jayega.

OA is A < B.....

I think one of the ways to be sure is to take the first 3-4 terms and test.

N = (A7A)^17 is a perfect square. Which of the following statement is FALSE?
(1) A is an even digit.
(2) A is divisible by 3
(3) When N is divided by 13 we get remainder 3.
(4) None of these

viveksingh25 Says

nachiket bhai....last year u posted some shortcut way to find last two digit of a certain no. raised to some power...us post ka link agar he to post kar do bhai....thanks in advance...

Here you go:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/73031-official-quant-thread-cat-2011-a-38.html#post2993035

@ Vivek bhai.....
No one can be trusted right now. Any sort of absentism from the forum will arouse suspicion.....After all V-day is coming........:p:p:p


A watch showed 10 minutes past 6 o clock on Thursday morning when the correct time was 6 o clock. It loses uniformly and was observed to be 15 minutes slow at 8 o clock. on Saturday morning. when did the watch show correct time?

50 hrs = 25 minutes

so , after 20 hrs it shows correct time i.e. 2 AM friday

Aah.. There are two possible arrangements here.
See.. You can select 5 places in 6C5 ways. Events can occur in 5! ways.
And places can be in arranged in 5! ways as well.

For examples, we have places (P1, P2, P3, P4, P5, P6) out of which we choose say (P1 to P5). And we have 5 events (E1, E2, E3, E4, E5).

So, P1 to P5 can be sequenced n 5! ways.
And then, (E1 to E5) can come in 5! ways.

Like, first place can be P4 and first event can be E4. Or first place can be P4 with first event being E5.

So, both events and places need to be premuted.
That will make it 5!*5! for events and places instead of just a single 5!.



Here, first position cannot be chosen in 6 ways. When we have circle (or in general a regular polygon for that matter), we have cyclic order. We need to break this loop. So, we fix a position.
Like, (a, b) in row can be (ab, ba) but in circle (ab, ba) comes out to be same.

Hence, first place to be chosen should not be 6. We have to fix it to some place. And then others can be numbered as (P1 to P5) with respect to fixed postion.

As you started with 6; you got 144.
Hope this is clear :)

nachiket bhai....last year u posted some shortcut way to find last two digit of a certain no. raised to some power...us post ka link agar he to post kar do bhai....thanks in advance...

viveksingh25 Says

bro i think b me ek term jyada he bt that is in denominator which will render it a lesser value.....

yaar but jo lcm lia hai wo bi toh inverse hoke upar jayega.

yes.
You are right......

Which year you graduated from JU?



Choose one answer.
a. A > B
b. A < B
c. A = B
d. A - B = 2009
e. B - A = 2009

my take is A> B
_/\_ Nachiket bhai,sood bhai,bs409 bhai,vivek bhai, aur wo jinka naam yaad nahi

yes.
You are right......

Which year you graduated from JU?



Choose one answer.
a. A > B
b. A < B
c. A = B
d. A - B = 2009
e. B - A = 2009

Is it A>B not sure


difference is in last term
A has 2007+1/2008 and has 2007+1/(2007+ 1/2008+2009^-1)
2008+2009^-1 > 2008
so A will have 2007+greater and B will have 2007+less
Now in A 2006+1/(2007+greater) in B 2006+1/(2007+less)
for A 2006+less for B 2006+more
so it will keep on alternating its position
for odd no. like 2007 A>B
for even no. like 2006 AB

B>A.

Every term is same in both numbers, but B mein ek step jayda hai. So B is greater ofcrs.

bro i think b me ek term jyada he bt that is in denominator which will render it a lesser value.....

I remember this question from last year. Naga sir ne solution post kia tha I guess.
Sum of all ten terms = 55.
55/2 = 27.xx
Total sets possible = 2^10.
Half of these will have sum > 27.
So, answer = 2^10/2 = 2^9.

Vivek bhai ne yaad dilaya.

thoda detail mein explain kar do. and how can u say "Half of these will have sum > 27."just bcoz 27.5 is half of the 55??

yes.
You are right......

Which year you graduated from JU?



Choose one answer.
a. A > B
b. A < B
c. A = B
d. A - B = 2009
e. B - A = 2009

my take A>B


@ BS0409 i am 10 passout

yes.
You are right......

Which year you graduated from JU?



Choose one answer.
a. A > B
b. A < B
c. A = B
d. A - B = 2009
e. B - A = 2009

B>A.

Every term is same in both numbers, but B mein ek step jayda hai. So B is greater ofcrs.

I understood what you were doing here but I could not understand one thing. The total number of ways in which 5 events can occur at 6 places = 6C5 * 5! = 6*5! (Here I havent considered circular arrangement, which would eventually reduce the number of events). So, after implementing a condition, how can it be 14*5! i.e. a greater number of ways.

Aah.. There are two possible arrangements here.
See.. You can select 5 places in 6C5 ways. Events can occur in 5! ways.
And places can be in arranged in 5! ways as well.

For examples, we have places (P1, P2, P3, P4, P5, P6) out of which we choose say (P1 to P5). And we have 5 events (E1, E2, E3, E4, E5).

So, P1 to P5 can be sequenced n 5! ways.
And then, (E1 to E5) can come in 5! ways.

Like, first place can be P4 and first event can be E4. Or first place can be P4 with first event being E5.

So, both events and places need to be premuted.
That will make it 5!*5! for events and places instead of just a single 5!.

Secondly, I approached it in the following way.
Lets start off with events numbered as E1, E2, E3, E4 and E5.
E1 can occur at any of the 6 places.
E2 can then occur at only 3 points. (one point has been used by E1 and the neighboring points have been locked out.)
E3 can then happen at only 2 points. (two points have been used by E1 and E2 and 2 points adjacent to E2 have been locked out.)
E4 can then occur at only 1 or 2 points depending upon the fact whether E3 occurred next to E1 or not.
Also, E5 would be left with 1 or 2 points in case E4 could occur at 2 or 1 points respectively.

If my description isnt clear, then kindly try to understand the same using a diagram.

The total number of ways = 6*3*2*2*1 when E3 occurs next to E1 and
number of ways = 6*3*2*1*2 in case E3 doesnt occur next to E1.
So, the total number of ways = 2* (6*3*2*2*1) = 144.

What is the flaw in my approach?

Here, first position cannot be chosen in 6 ways. When we have circle (or in general a regular polygon for that matter), we have cyclic order. We need to break this loop. So, we fix a position.
Like, (a, b) in row can be (ab, ba) but in circle (ab, ba) comes out to be same.

Hence, first place to be chosen should not be 6. We have to fix it to some place. And then others can be numbered as (P1 to P5) with respect to fixed postion.

As you started with 6; you got 144.
Hope this is clear

at 2 o clock on friday morning.

in 50 hrs clock loses 25 min
so to lose 10 min it takes 20 hrs...

bhai is the answer to that subset wala question 2^9

total no. of subsets = 2^10
half of it would be greater than 27..


@ bs...aapki baat to sahi he..valentine day around every things need to b seen in suspicious ways.....and more so if seen from the perspective of some one from kolkata..after all mai v wahi ka graduate hu....did my engineering from jadavpur university ... :):):):cheers:

yes.
You are right......

Which year you graduated from JU?



Choose one answer.
a. A > B
b. A < B
c. A = B
d. A - B = 2009
e. B - A = 2009

Set S= {1,2,3,4,5,6,7,8,9,10}
then how many sunsets of S are possible that have the sum of the elements greater than 27 ?

I remember this question from last year. Naga sir ne solution post kia tha I guess.
Sum of all ten terms = 55.
55/2 = 27.xx
Total sets possible = 2^10.
Half of these will have sum > 27.
So, answer = 2^10/2 = 2^9.

Vivek bhai ne yaad dilaya.

@ Vivek bhai.....
No one can be trusted right now. Any sort of absentism from the forum will arouse suspicion.....After all V-day is coming........:p:p:p


A watch showed 10 minutes past 6 o clock on Thursday morning when the correct time was 6 o clock. It loses uniformly and was observed to be 15 minutes slow at 8 o clock. on Saturday morning. when did the watch show correct time?

at 2 o clock on friday morning.

in 50 hrs clock loses 25 min
so to lose 10 min it takes 20 hrs...

bhai is the answer to that subset wala question 2^9

total no. of subsets = 2^10
half of it would be greater than 27..


@ bs...aapki baat to sahi he..valentine day around every things need to b seen in suspicious ways.....and more so if seen from the perspective of some one from kolkata..after all mai v wahi ka graduate hu....did my engineering from jadavpur university ... :):):):cheers:

A watch showed 10 minutes past 6 o clock on Thursday morning when the correct time was 6 o clock. It loses uniformly and was observed to be 15 minutes slow at 8 o clock. on Saturday morning. when did the watch show correct time?

It loses 25 mins in 24*2 + 2 = 50 hours.
So it lost those extra 10 mins in -> 50*10/25 = 20 hrs.
=> Correct time wld be when the clock shows - 2 am on saturday morning.

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@BS bhai - haan yaar 124 ya 248 se equate toh kara hi ni meine

Q) There are 6 points on a circle. five events occur at 5 of these points ,one at each point. in how many ways can they occur such that no two successive events occur at adjacent points?

I think this should be 14*5!

First, we can select 5 positions out of 6 on that circle. As it is circular arrangement, we need to fix a position. Say, we fix it and name it position 1.

Then, we have 2-3-4-5-6 as other places of circle with 2 and 6 next to position 1. All we need to do is select 4 more positions. This is like making 4 digit numbers using digits (2, 3, 4, 5 and 6) such that no two consecutive digits are adjecent; and numbers do not start with 2 or 6.

Starting with 3: 3 _ _ _
Second position can be taken by 5 or 6. When taken by 5; we can have 14 or 26 in the end.
Second position when taken by 6; we can have 24, 42 or 25 as last two digits.
=> Total 5 numbers for 3 at position 2.
=> Total 5 numbers for 5 at position 2 as we have symmetry.

Starting with 4:
We can have seocnd position as 2 or 6. With 2; we can have last two digits as 53 or 63. Two cases and two more symmetrical cases for with 6.
=> Total 4 cases

So, in all we have (5+5+4) i.e. 14 ways of selecting 5 positions.

And then, we can place 5 events in 5! positions for each of selected combination.

So, in all 14*5! ways.

I understood what you were doing here but I could not understand one thing. The total number of ways in which 5 events can occur at 6 places = 6C5 * 5! = 6*5! (Here I havent considered circular arrangement, which would eventually reduce the number of events). So, after implementing a condition, how can it be 14*5! i.e. a greater number of ways.

Secondly, I approached it in the following way.
Lets start off with events numbered as E1, E2, E3, E4 and E5.
E1 can occur at any of the 6 places.
E2 can then occur at only 3 points. (one point has been used by E1 and the neighboring points have been locked out.)
E3 can then happen at only 2 points. (two points have been used by E1 and E2 and 2 points adjacent to E2 have been locked out.)
E4 can then occur at only 1 or 2 points depending upon the fact whether E3 occurred next to E1 or not.
Also, E5 would be left with 1 or 2 points in case E4 could occur at 2 or 1 points respectively.

If my description isnt clear, then kindly try to understand the same using a diagram.

The total number of ways = 6*3*2*2*1 when E3 occurs next to E1 and
number of ways = 6*3*2*1*2 in case E3 doesnt occur next to E1.
So, the total number of ways = 2* (6*3*2*2*1) = 144.

What is the flaw in my approach?

Enceladus Says

Bhaiyo meine yahan kya bewakoofi kari hai?? pls batana. :banghead:

The equation should be 64 + 8x + y = 81 + 9y + x=124

solving u will get x=7,y=4

x+y=11

bs bhai.... sunday koo ranchi chala gaya tha dosto k sath outing pe....kal ka pura din court k chakkar kat reha tha..income tax saving k liye ek affidevit banwana tha.......aisi koi chizz nai hai jo pg se impoertant ho bhai

@ Vivek bhai.....
No one can be trusted right now. Any sort of absentism from the forum will arouse suspicion.....After all V-day is coming........:p:p:p


A watch showed 10 minutes past 6 o clock on Thursday morning when the correct time was 6 o clock. It loses uniformly and was observed to be 15 minutes slow at 8 o clock. on Saturday morning. when did the watch show correct time?

Is it 11..........??

The no. is 124
124 in base 8=174
124 in base 9=147

my take - 11

possible three digit no. can b 124 or 248

later on 248 can b neglected

starting with 124
124 = 8^2+8x+y
so, 8x+y=60 (for 248 this equation would have been 8x+y = 188 which is nt possible, hence nglected)

124 = 9^2+9y+x

so 9y+x = 43

so x=7,y=4

64 + 8x + y = 81 + 9y + x.
=> 8y - 7x = 17.
=> y = 3, x = 1. and y = 6, x = 2. ans so on......
Bhai CBD hai kya?? ya kuch galat kar ra hun mein. :-/

Bhaiyo meine yahan kya bewakoofi kari hai?? pls batana.

Oa for the first one is 4
Second one has OA 11/3

A watch which gains uniformly was observed to be 1 minute slow at 8 am on a day. at 6 pm on the same day it was 1 minute fast. at what time did the watch show the correct time?

my take 1pm

it gain 2 min in 10 hrs
so it will show correct tym when it gains 1 min ie after 5 hrs @ 1pm



bro... kindly post the approach for that tap wala question.....

Can u post the solution for the 2nd ques?? 11/3 wala??

For this question. The clock gains 2 mins in 10 hours. Hence it wld gain 1 min in 5 hrs. So the correct time is at 1pm.

The question is from Nishit Sinha. I dont have solution . U can show your approach. I am personally getting 4/3 which is not in the options.......


Set S= {1,2,3,4,5,6,7,8,9,10}
then how many sunsets of S are possible that have the sum of the elements greater than 27 ?

Oa for the first one is 4
Second one has OA 11/3

A watch which gains uniformly was observed to be 1 minute slow at 8 am on a day. at 6 pm on the same day it was 1 minute fast. at what time did the watch show the correct time?

Can u post the solution for the 2nd ques?? 11/3 wala??

For this question. The clock gains 2 mins in 10 hours. Hence it wld gain 1 min in 5 hrs. So the correct time is at 1pm.

My take is 4.

Let the capacity of the tank be 106 ltrs.
=> We need to fill 60lts in 1 min to be able to fill 106lts in 106 secs.
So, (106/3) + (106/8 ) + (106/15) + (106/24) = 60.06 ~ 60lts.
Hence, 4.

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Jain ke liye har ques asaan hai. Jain \m/

1) my take 3

(1/3+1/8+1/15)*106/60>1

so pump no. 2,3,4 if opened can fill the tank with in 106 sec


2) % of work done by respective pumps per min is
(33.33+8.5+6.6 + 4.16 + 2.7 + 2.1+1.xx +1.xx+1.0x + 0.xx+0.xx+.....so on )

this adds to b nearly 66%

so it suld require 3/2 min to complete the job....
but no such option:-o


bs bhai.... sunday koo ranchi chala gaya tha dosto k sath outing pe....kal ka pura din court k chakkar kat reha tha..income tax saving k liye ek affidevit banwana tha.......aisi koi chizz nai hai jo pg se impoertant ho bhai

Vivek bhai, the bold part is coming out to be lesser than 1. so ek pipe aur on karni padegi.
and jara 2nd wale ki thodi approach samjao. mere toh dimaag mein hi ni aa ra.

120 ltrs of tank capacity

2nd = 120/3 = 40
3rd = 120/8 = 15
4th = 120/15 = 8
5th = 120/24 = 5

40+15+8+5 = 68(1+76/100) = 119.68 approx = 120

so , we need 4 pipes and 5 if we count 1st one also

i am not sure enough for 2nd part.....

but still jo maine kara he i am illustrating..

pump 2nd work/per min 1/3 = 33.33%
pump 3rd work /min = 1/8= 12.5%
pump 4th work/min= 1/15 = 6.66%

so on......

if added willcome nearly to 66% work is done in one min.

so complete the job ie for 100% completion of work 3/2 min required

Oa for the first one is 4
Second one has OA 11/3

A watch which gains uniformly was observed to be 1 minute slow at 8 am on a day. at 6 pm on the same day it was 1 minute fast. at what time did the watch show the correct time?