Angadbir Says

http://www.pagalguy.com/forum/quanti...ml#post2841386 (Official Quant Thread for CAT 2011 )
post it here

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http://www.pagalguy.com/forum/quantitative-questions-and-answers/69715-official-quant-thread-cat-2011-a.html#post2841386

Will miss you

please continue here :

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Praveena.T Says

Sandeep is planning to attend wedding.He starts at a specific time and specific speed so as to be there at the exact time.he was driving for exactly 1 hour when his car had a flat tyre.He changed the tyre in 10 minutes.He started driving the remaining distance @ 40 km/h.he was late for the wedding by 30 minutes.If the puncture had happened 25 km later,he would be late for wedding by 20 min.what is the total dist. travelled by him?

erorcrept Says

anybody for this?

The time difference of 10 mins was created by the distance of 25 km which Sandeep traveled @ v (normal speed) in first case, and 40kmph in second case.
Hence, 25/40 -25/v = 10/60
Or, v = 600/11

Again, Let x = distance to be traveled by Sandeep after the flat tyre.
Since he also used 10 mins for repairs, in which he traveled a distance of '0', the actual time of delay due to slower travel by Sandeep (@ 40kmph) = 20 mins, in first case.
Therefore, x/(600/11) - x/4 = 20/60
Or, x = 50
Total distance traveled by Sandeep = (1)*600/11 + 50 = 104(6/11) = 104.5 km

600/11 + 50 = 104.54

from given conditions ,we can deduce that he traveled exactly twice at a slow speed when it got punctured

so after tire burst in case 1 25+25 = 50 km still to go

now
for 50 km at 40 km / hr he takes
50/40*60 = 75 min..

and he is 20 min late ..
so if he was traveling at original speed
he must take 55 min ..

or speed 50/55*60 or 600/11

for 1 hr driving at 600/11 ...

i got the same

Trichi Says

could you pls explain the bold part..

souravsaha008 Says

why are we adding 1 at the end of the result?

mytotemspins Says

sir can you explain the bolded part a little more.. why did you consider the case when all are equal and why are you diving it by 6+9+1..?

Total cases = 190

Now, since boxes are also identical (1, 2, 17) and (1, 17, 2) are identical.
That means each combination has to be counted just once.

Now, when all x, y, z are different, then these case are counted 6 times.
When two are same, these cases are counted three times
When all are same, its counted just once

So, number of cases when all are equal = 1
Number of cases when two are equal = 9*3 (as x = y = 0 to 9, but when its 6, all are equal)
Number of cases when all are different = 190 - 27 - 1 = 162

So, require answer will be 162/6 + 27/3 + 1 = 37

a + b + c = 18
C(20, 18 ) = 190 ways

When all are equal - 1 case
When exactly two are equal - 9*3 cases

So, required number of cases = (190 - 27 - 1)/6 + 9 + 1 = 37


could you pls explain the bold part..

all can be equal in only one way - 6, 6, 6
when two are equal then they can be chosen in following 9 ways-
0 0 18
1 1 16
..
8 8 2
9 9 0.
each of the above 9 can be rearranged in 3!/2! = 3 ways,so total 3x9 =27

erorcrept Says

anybody for this?

600/11 + 50 = 104.54

from given conditions ,we can deduce that he traveled exactly twice at a slow speed when it got punctured

so after tire burst in case 1 25+25 = 50 km still to go

now
for 50 km at 40 km / hr he takes
50/40*60 = 75 min..

and he is 20 min late ..
so if he was traveling at original speed
he must take 55 min ..

or speed 50/55*60 or 600/11

for 1 hr driving at 600/11 ...

a + b + c = 18
C(20, 18 ) = 190 ways

When all are equal - 1 case
When exactly two are equal - 9*3 cases

So, required number of cases = (190 - 27 - 1)/6 + 9 + 1 = 37


There is one more case - not delivering any letter to any house.

sir can you explain the bolded part a little more.. why did you consider the case when all are equal and why are you diving it by 6+9+1..?

a + b + c = 18
C(20, 18 ) = 190 ways

When all are equal - 1 case
When exactly two are equal - 9*3 cases

So, required number of cases = (190 - 27 - 1)/6 + 9 + 1 = 37

why are we adding 1 at the end of the result?

a + b + c = 18
C(20, 18 ) = 190 ways

When all are equal - 1 case
When exactly two are equal - 9*3 cases

So, required number of cases = (190 - 27 - 1)/6 + 9 + 1 = 37


could you pls explain the bold part..

vikasneuer Says

Its 10080.

Yeah it must

Cases are

1234569=7! Cases
1234587=7! Cases

Total cases are 10080

Q. In how many ways, 18 identical chocolates can be distributed among 3 identical boxes?
Options :
(a) 25
(b) 210
(c) 105
(d) 37
please answer with explanation.

both identical and that means just count the number of ways !

dont think of selection or arrangement ..

a + b + c = 18

list out cases .. it comes to 37

100 % sure these kind of problems will not come in cat

Q. In how many ways, 18 identical chocolates can be distributed among 3 identical boxes?
Options :
(a) 25
(b) 210
(c) 105
(d) 37
please answer with explanation.

I did it the same way as chill sir...
Anyways, other longg way...The combination will be
0,0,18 - 1 way
0,x,y ; x + y = 18; 9 ways (x,y

0)
1,x,y ; x + y = 17; Ways = 8
2,x,y ; x + y = 16; Ways = 7
.....
6,6,6 ; 1 way
Ways = 1+2+4+5+7+8+9+1 = 37

jain4444 Says

Is it 15120 ?

Its 10080.

Q. In how many ways, 18 identical chocolates can be distributed among 3 identical boxes?
Options :
(a) 25
(b) 210
(c) 105
(d) 37
please answer with explanation.

a + b + c = 18
C(20, 18 ) = 190 ways

When all are equal - 1 case
When exactly two are equal - 9*3 cases

So, required number of cases = (190 - 27 - 1)/6 + 9 + 1 = 37

naga/chill/Mn/Ravi Teja/MaskedMenace/Pkaman... or anyone who is online...

can u explain this to me?

There is one more case - not delivering any letter to any house.

naga/chill/Mn/Ravi Teja/MaskedMenace/Pkaman... or anyone who is online...

can u explain this to me?

Check this out...:)

erorcrept Says

anybody for this?

vt =d

v + 40 x t-2/3 = d

v + 40 x t -1/2 - 25/v = d

solving v=150
t=d/150
d= 370 x 5 / 11 ??

---wrong analysis--

---snipped.. query already answered ---

above question:

First box - 0 choc
possible combination in others (considering all boxes are identical)
0/18, 1/17, 2/16........ 9/9 = 10

first box - 1 choc (all possible comb of one of the boxes having one chocolate are over in previous case, so we start with 1/16 in other boxes)
1/16.... 8/9 = 8

first box - 2 chocs
2/14, 3/13, 4/12............8/8 = 7

this goes on till first box has 6 chocs.

therefore, total ways = 10+8+7+5+4+2+1 = 37

In how many ways can 30 identical balls be distributed among 7 distinct bags (numbered from 1 to 7), such that each bag has at least 1 ball and no two bags have the same number of balls?
a)5040
b)10080
c)15120
d)20160

Is it 15120 ?

Q. In how many ways, 18 identical chocolates can be distributed among 3 identical boxes?
Options :
(a) 25
(b) 210
(c) 105
(d) 37
please answer with explanation.

bolded part....

when n=2, arent there two ways of delivering a letter : delivering to first house or deliverign to second house.

similarly, for n=3:
3 ways when one letter is delivered
1 way when 2 letters are delivered = total 4 ways....??

am i thinking too much??

Can you clarify one thing.
e.g., no. of houses = 2, i.e., A and B
So, the letter can be delivered either to A or to B. So, won't that be 2 ways, not 1 ?

my mistake,
f(1) should have been 2 (Either the letter is delivered or it is not)

and f(2) should have been 3

f(3) = 5.. so on.

and then f(19) = 10946

But, note that here we have included cases, where no letters are delivered as well.

In how many ways can 30 identical balls be distributed among 7 distinct bags (numbered from 1 to 7), such that each bag has at least 1 ball and no two bags have the same number of balls?
a)5040
b)10080
c)15120
d)20160

snipped...... wrong approach.

1. 2 balls in the bag having 6 balls will yield .... 1 2 3 4 5 8 7
3. 1 ball in the bag having 6 balls, and 1 ball in the bag having 7 balls will yield 1 2 3 4 5 7 8
both are the same set no real diff ....
so only 2 ways so 10080

Realized already

Praveena.T Says

Sandeep is planning to attend wedding.He starts at a specific time and specific speed so as to be there at the exact time.he was driving for exactly 1 hour when his car had a flat tyre.He changed the tyre in 10 minutes.He started driving the remaining distance @ 40 km/h.he was late for the wedding by 30 minutes.If the puncture had happened 25 km later,he would be late for wedding by 20 min.what is the total dist. travelled by him?

anybody for this?

We can have 1,2,3,4,5,6,7 balls in 7 bags.
Rest of the two balls can be added as :
1. 2 balls in the bag having 6 balls
2. 2 balls in the bag having 7 balls
3. 1 ball in the bag having 6 balls, and 1 ball in the bag having 7 balls.
Total = 2 ways (3. and 1. are same!)
Now, the ways of choosing 7 bags having distinct number of balls = 7!
And we also have three ways of putting in the rest two balls.
Thus, total number of ways = 7!*2 = 10080

1. 2 balls in the bag having 6 balls will yield .... 1 2 3 4 5 8 7
3. 1 ball in the bag having 6 balls, and 1 ball in the bag having 7 balls will yield 1 2 3 4 5 7 8
both are the same set no real diff ....
so only 2 ways so 10080

Co-prime is always relative ( see the CO )

But as its not given,we'll assume all number that do not have common HCF
= 1+2+3+5+7+11+13 =42

14 = 7^1 *2^1

Now, Number of co primes less than is 14 *(1-1/2)(1-1/7) = 6

And sum of all numbers co prime and less that 14 is 14 * 6/2 = 42.

And numbers are 1+3+5+9+11+13 = 42

In how many ways can 30 identical balls be distributed among 7 distinct bags (numbered from 1 to 7), such that each bag has at least 1 ball and no two bags have the same number of balls?
a)5040
b)10080
c)15120
d)20160

Sum of 1 to 7 is 28, so only possibilities are:-
(1, 2, 3, 4, 5, 7, 8 ) - 7!
(1, 2, 3, 4, 5, 6, 9) - 7!

So, 2*7! = 10080 cases

In how many ways can 30 identical balls be distributed among 7 distinct bags (numbered from 1 to 7), such that each bag has at least 1 ball and no two bags have the same number of balls?
a)5040
b)10080
c)15120
d)20160

We can have 1,2,3,4,5,6,7 balls in 7 bags.
Rest of the two balls can be added as :
1. 2 balls in the bag having 6 balls
2. 2 balls in the bag having 7 balls
3. 1 ball in the bag having 6 balls, and 1 ball in the bag having 7 balls.
Total = 2 ways (3. and 1. are same!)
Now, the ways of choosing 7 bags having distinct number of balls = 7!
And we also have two ways of putting in the rest two balls.
Thus, total number of ways = 7!*2 = 10080

In how many ways can 30 identical balls be distributed among 7 distinct bags (numbered from 1 to 7), such that each bag has at least 1 ball and no two bags have the same number of balls?
a)5040
b)10080
c)15120
d)20160

1 ball each so 23 balls left

0 1 2 3 4 5 6 in each bag adds up to 21
so 0 1 2 3 4 5 8
or 0 1 2 3 4 6 7
2x7! is the ans = 10080 ans

If letter to 1 house is delivered, then the house around it does not get a letter delivered on that day.

Suppose no. of houses, n = 1
No. of ways = f(1) = 1

Suppose no. of houses, n = 2
No. of ways = f(2) = 1

n = 3
No. of ways = f(3) = 2

n=4,
No. of ways
If first house is not delivered (it becomes similar to case where n=3), then total ways = f(3) = 2
If first house is deliverd (it becomes similar to case where n=2), then total ways = f(2) = 1

Total f(4) = f(3) + f(2)

Similarly, We will be able to observe, that its a fibonacci series, where f(n+2) = f(n+1) + f(n)
Ans = 19th term of the series = 10946

Can you clarify one thing.
e.g., no. of houses = 2, i.e., A and B
So, the letter can be delivered either to A or to B. So, won't that be 2 ways, not 1 ?

If letter to 1 house is delivered, then the house around it does not get a letter delivered on that day.

Suppose no. of houses, n = 1
No. of ways = f(1) = 1

Suppose no. of houses, n = 2
No. of ways = f(2) = 1

n = 3
No. of ways = f(3) = 2

n=4,
No. of ways
If first house is not delivered (it becomes similar to case where n=3), then total ways = f(3) = 2
If first house is deliverd (it becomes similar to case where n=2), then total ways = f(2) = 1

Total f(4) = f(3) + f(2)

Similarly, We will be able to observe, that its a fibonacci series, where f(n+2) = f(n+1) + f(n)
Ans = 19th term of the series = 10946

bolded part....

when n=2, arent there two ways of delivering a letter : delivering to first house or deliverign to second house.

similarly, for n=3:
3 ways when one letter is delivered
1 way when 2 letters are delivered = total 4 ways....??


am i thinking too much??

In how many ways can 30 identical balls be distributed among 7 distinct bags (numbered from 1 to 7), such that each bag has at least 1 ball and no two bags have the same number of balls?
a)5040
b)10080
c)15120
d)20160

hi5blast Says

can you please explain it

Naga bhai, please explain
1. how u converted this into the fibonacci series.
2. how did u calculate 19th term of the series?

Wont it matter if the letter is first delivered at 1st house, then at 4th and then at 7....??

1) if u look carefully , we are jus finding the number of patterns ..so fibonacci is way to go

2) no other choice adding

took me some 2 min

2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946

see i am jus calculating for minimum possible number .. lowest is 2 and next lowest is 3 ..

i will give small eg .. say u need to form four digit numbers with only digits 1 or 2 such that there are no consecutive 1's?

jus calculate manually :

1212
2121
2221
2122
2212
1222
2222

by fibonacci series way :

2, 3, 5, 8 --> 8 ways

Praveena.T Says

Sandeep is planning to attend wedding.He starts at a specific time and specific speed so as to be there at the exact time.he was driving for exactly 1 hour when his car had a flat tyre.He changed the tyre in 10 minutes.He started driving the remaining distance @ 40 km/h.he was late for the wedding by 30 minutes.If the puncture had happened 25 km later,he would be late for wedding by 20 min.what is the total dist. travelled by him?

is it approx 104.5 kms?????

20 + 2a = angle ACB + angle BCD = angle ACD
x +2b = angle DBC + angle ABC = angle ABD

now ACD and ABD are opposite angles of cyclic quadrilateral
2a and 2b are angles BCD and ABC , not in the figure but can be drawn

I think u didn't get my question. My question is that opposite angles of a cyclic quad are supplementary not equal.. so how did u equate them?

typical application of fibonacci series ..

if first house is delivered with a letter , then third house can get letter ..

3 - 1 = 2

next value is 3 ..

so we need to find 19th fibonacci number for series 2,3 which is 10946

yup, this series is applicable only....when we consider that case also when no house will get letter but in ques its given that postman deliver letter everyday

warriorashu Says

in how many ways the letters of the word AFLATOON be arranged if the consonants and vowels must occupy alternate places?

_ _ _ _ _ _ _ _ - Taking 1st place as consonant (F,L,T,N). So, 1,3,5,7 will be a consonant. => 4! ways. remaining 4 places can be filled (A,A,O,O) in 4!/2!*2! ways = 6 ways. total 24*6 = 144 ways.

Now, when 1,3,5,7 places is a vowel, it can be filled in 4!/2!*2! ways = 6 ways
the consonants will be fillled in 4! ways = 24 => total = 24*6 = 144
=> total ways = 144+144 = 288 ways