Snipped....

sowmyanarayanan Says

please explain the red part!!!!

See.. If there are 'n' people and 3 vehicles; each person has 3 ways to choose vehicle.

So, 3*3*3*... n times = 3^n total ways
But, no vehicle should be empty. A vehicle will be empty when all 'n' people choose only 1 or 2 vehicles. We need to remove such cases.
Now, 2 vehicles can be chosen in 3C2 ways i.e. 3 ways.

And each person has 2 choices now.

So, total 3C2 * (2*2*2*.. n times) = 3 * 2^n ways
But, this case of two vehicles will also include when all people choose only 1 vehicle. They can pick one vehicle in 2 ways (any one of 2). So, such cases will be 3C2 * 2 = 6
But, we know that selecting only 1 vehicle can be done in 3 ways. So, we have overcounted 3 ways.
So, in all we should remove *3*2^n - 3) cases.

So, total cases = 3^n - (3*2^n - 3)
= 3^n - 3*2^n + 3

sowmyanarayanan Says

please explain the red part!!!!

Its inclusion-exclusion principle.
The red part has been elaborated by subha bhai couple of posts previous to Think bhai's post ...

if we have 'n' people going in 3 vehicles so that each vehicale has at least 1 person; then it can be done in:

3^n - 3*2^n + 3
=> 3^6 - 3*2^6 + 3
= 729 - 192 + 3
= 540

please explain the red part!!!!

Ooo maaai GAWD !!!



Abe kya kar raha hai.. thik se dekh .. mock ka question hai ye to ..
Vehicles are different ..

it would be like
1,1,4 - 6C4 * 2C1 * 3!/2 = 90
1,2,3 = 6C3*3C2 * 3! = 360
2,2,2 = 6C2*4C2 = 90



Nachiket bhai .. salaam !!!

Was feeling left out on this thread with you guys not around .. :)

vehicle are different to maine bi maana tha glati se shakal same kar di thi har ek ki...


u know early morning!!! :)

BHAI log iska koi formulae hai kya.
If there are 'n' points and 'm' are collinear, how many traingles can be formed

m*C(n-m,2) + (n-m)*C(m,2)

C() is combination..

ksagarwa Says

On a Christmas day, Arjun goes to his school for a party where he wants to carry a colorful square box full of chocolates for his friends. After arranging the chocolates in the box he finds that he is still left with 67 chocolates with him. Now in order to be able to carry all the remaining chocolates as well, he plans to increase the size of the box in such a way so as increase one chocolate along both dimensions of the square, but then discovered that he still has space for 32 chocolates. Therefore, the actual number of chocolates that he has originally planned to carry with him is:

It should be 2468.

We have square of n*n size. Then it becomes (n+1)*(n+1) and can accomodate 67+32 more chocolates.

=> Difference between two consecutive sqaures is 99
=> Numbers are 49 and 50 ... (99-1)/2 and (99+1)/2

So, we have 49^2 + 67 = 50^2 - 32 = 2468 chocolates

BHAI log iska koi formulae hai kya.
If there are 'n' points and 'm' are collinear, how many traingles can be formed

Bhai iska formula to yahi hai..
nC3 - mC3
Aur agar st lines poocha hai toh
nC2 - mC2 + 1

SIX people have to attend a wedding in three diff vehicles,each of which can accomodate atmost six people.in how many ways can they go,so that they use all the three vehicles???
an easier approach plzz...anyone????
OA is 540

The No of people in vehicles can be 1,1,4 or 1,2,3 0r 2,2,3
For case 1 1 4
6c4*2c1*3!/2!=90
For case 1 2 3
6c3*3c2*3! =360
For case 2 2 2
6c2*4c2*3!/3! = 90
total = 540

BHAI log iska koi formulae hai kya.
If there are 'n' points and 'm' are collinear, how many traingles can be formed

SIX people have to attend a wedding in three diff vehicles,each of which can accomodate atmost six people.in how many ways can they go,so that they use all the three vehicles???
an easier approach plzz...anyone????
OA is 540

If we have 'n' people going in 3 vehicles so that each vehicale has at least 1 person; then it can be done in:

3^n - 3*2^n + 3
=> 3^6 - 3*2^6 + 3
= 729 - 192 + 3
= 540

SIX people have to attend a wedding in three diff vehicles,each of which can accomodate atmost six people.in how many ways can they go,so that they use all the three vehicles???
an easier approach plzz...anyone????
OA is 540

1,2,3

2,2,2

1,1,4

1,2,3--- C(6,3)*C(3,2) * 3!
2,2,2 ---C(6,2)*C(4,2)
1,1,4----C(6,4)*C(2,1) *3!/2!

Add them all ---540

GUYS SOME ONE GIVE THE OA FOR THE CHISTMAS CHOCOLATE BOX PROBLEM, FROM THE PREVIOUS PAGE!!!


i am solving it like this:
n - current side of the box
a - amount he wants to take

(1) => n^2 = a - 67
(2) => ( n + 1 )^2 = a + 32

n^2 = 2401

so a = 2468

OA PLEEEAAASSSSEEE!!!!

eqn of RC = 9y + 2x = 18
BP => y + 2x = 6

they intersect at point X (9/4,3/2)

so slope = 2/3

Ooo maaai GAWD !!!

x+y+z=6

x'+y'+z'=3

5c2=10 ways...

Abe kya kar raha hai.. thik se dekh .. mock ka question hai ye to ..
Vehicles are different ..

it would be like
1,1,4 - 6C4 * 2C1 * 3!/2 = 90
1,2,3 = 6C3*3C2 * 3! = 360
2,2,2 = 6C2*4C2 = 90

My take is 2/3

Just drop a perpendicular from point X on x-axis. Say, the point is Z.
So, triangle AZX and CAB are similar.

So, slope of AX = XZ/AZ = AB/AC = 6/9 = 2/3

Nachiket bhai .. salaam !!!

Was feeling left out on this thread with you guys not around .. :)

SIX people have to attend a wedding in three diff vehicles,each of which can accomodate atmost six people.in how many ways can they go,so that they use all the three vehicles???
an easier approach plzz...anyone????
OA is 540

One person has 3 choices so 3^6
Now where we have few cases when one or two cars are empty.
For one car empty-
so we need to deduct 3*2^6
but here we have cases which have two cars empty those cases are 3 in real but here we have counted them 6 times. (2^6 has 2 such cases when all six goes to a single car and 3*2=6)
so our answer is 3^6-3*2^6+3=729-192+3=732-192=540

Triangle ABC has vertices A (0, 0), B (0, 6) and C (9, 0). The points P and Q lie on side AC such that
AP = PQ = QC. Similarly, the points R and S lie on side AB such that AR = RS = SB. If the line
segments PB and RC intersect at X, then the slope of the line AX is .......... ??

My take is 2/3

Just drop a perpendicular from point X on x-axis. Say, the point is Z.
So, triangle AZX and CAB are similar.

So, slope of AX = XZ/AZ = AB/AC = 6/9 = 2/3

SIX people have to attend a wedding in three diff vehicles,each of which can accomodate atmost six people.in how many ways can they go,so that they use all the three vehicles???
an easier approach plzz...anyone????
OA is 540

x+y+z=6

1,1,4=6c1*5c1*3=90
1,2,3,=360 ways
2,2,2=90 ways

so 540..

The lengths of the sides of an acute angled triangle are 5 , x and 12 , where 5 < x < 12, . how many integral values of x are possible?

1
0
6
4

5 + x > 12
x > 7

and,
12 + 5 > x
x < 17

25 + x^2 > 144
x^2 > 119
x > 10

So 1 value is possible namely 11..

Triangle ABC has vertices A (0, 0), B (0, 6) and C (9, 0). The points P and Q lie on side AC such that
AP = PQ = QC. Similarly, the points R and S lie on side AB such that AR = RS = SB. If the line
segments PB and RC intersect at X, then the slope of the line AX is .......... ??

eqn of RC = 9y + 2x = 18
BP => y + 2x = 6

they intersect at point X (9/4,3/2)

so slope = 2/3

pkaman Says

for acute, a^2 + b^2 > c^2

what about obtuse pankaj bhai ?

SIX people have to attend a wedding in three diff vehicles,each of which can accomodate atmost six people.in how many ways can they go,so that they use all the three vehicles???
an easier approach plzz...anyone????
OA is 540

Is thr any way to solve equations with 4 variables/unknowns like this??

A +B + C + D = 1
8A +4B + 2C + D = 3
27A+9B + 3C + D = 7
64A+16B+ 4C + D = 14

Subtract 1st eq from 2nd , 2nd from 3rd , 3rd from 4th
You will get 3 equations , 3 variables

Now subtract 1from 2 , 2nd from 3
You will get 2 equations with 2 variables

Get the values of the variables.

Put it in the remaining two equations and find the other two variables

explain pankaj bhai.
'

@ sowmyanarayana - no,the answer is 1.
p.s my cat is on the 16th.

for acute, a^2 + b^2 > c^2

Only 11 is possible.
So, 1

explain pankaj bhai.
'

@ sowmyanarayana - no,the answer is 1.









p.s my cat is on the 16th.

The lengths of the sides of an acute angled triangle are 5 , x and 12 , where 5 < x < 12, . how many integral values of x are possible?

1
0
6
4

x value can be 8, 9, 10, 11
so 4
am i right?

The lengths of the sides of an acute angled triangle are 5 , x and 12 , where 5 < x < 12, . how many integral values of x are possible?

1
0
6
4

Only 11 is possible.
So, 1

Triangle ABC has vertices A (0, 0), B (0, 6) and C (9, 0). The points P and Q lie on side AC such that
AP = PQ = QC. Similarly, the points R and S lie on side AB such that AR = RS = SB. If the line
segments PB and RC intersect at X, then the slope of the line AX is .......... ??

so by the points given the following can be derived:
P(0,3)
Q(0,6)
R(2,0)
S(4,0)

find the eqns of the line BP and CR which are 9x + 2y = 18 and x + 2y = 6
solve them to gewt point X i.e. (1.2, 2.4)

Now use the slope formula tan-1(y2 -y1 / x2 - x1 ) for the points A and X
=tan-1(2.4/1.2)
=tan-1(2)
= ( i dont know the value )

The lengths of the sides of an acute angled triangle are 5 , x and 12 , where 5 < x < 12, . how many integral values of x are possible?

1
0
6
4

For the problem of chocolates you can first find the difference as 67+32=99
Now x+y=99,x-y=1 so x=50
answer=50^2-32=2468


Triangle ABC has vertices A (0, 0), B (0, 6) and C (9, 0). The points P and Q lie on side AC such that
AP = PQ = QC. Similarly, the points R and S lie on side AB such that AR = RS = SB. If the line
segments PB and RC intersect at X, then the slope of the line AX is .......... ??

Won't solve the entire prob subha bhai.. at office ..
AP = PQ = QC
So, coordinates of P = (3,0)
AR = RS = SB
So, R = (0,2)
Now, we have coordinates for P,B. We can get equation for that and similarly for R,C.

So, we just need to find the intersection point of these two lines ...

Is thr any way to solve equations with 4 variables/unknowns like this??

A +B + C + D = 1
8A +4B + 2C + D = 3
27A+9B + 3C + D = 7
64A+16B+ 4C + D = 14

For the problem of chocolates you can first find the difference as 67+32=99
Now x+y=99,x-y=1 so x=50
answer=50^2-32=2468
Triangle ABC has vertices A (0, 0), B (0, 6) and C (9, 0). The points P and Q lie on side AC such that
AP = PQ = QC. Similarly, the points R and S lie on side AB such that AR = RS = SB. If the line
segments PB and RC intersect at X, then the slope of the line AX is .......... ??

R (0,2)
P (3,0)
Now we need to take out point X
For line PB
=> y - 0 = -6/3 * (x - 3)
=> y = -2x + 6 --- (i)
For line RC
=> y - 2 = 2/(-9) * (x - 0)
=> -9y + 18 = 2x
=> 9y = 18 - 2x --- (ii)
To find out point X solve both those eq (i) and (ii)
x = 2.25, y= 1.5
Slope of AX = 1.5 - 0 / 2.25 - 0 = 1.5/2.25 = 0.66

Three Professors Dr. Gupta ,Dr Sharma and Dr. Singh are evaluating answer scripts of a subject .Dr Gupta is 40% more efficient than Dr.Sharma ,who is 20 % more efficient than Dr.Singh .Dr Gupta takes 10 days less than Dr Sharma to complete the evaluation work Dr. Gupta starts the evaluation work and works for 10 days and then Dr. Sharma takes over Dr. Sharma evaluates for next 15 days and then stops .In how many days ,Dr Singh can complete the remaining evaluation work ??
(A)7.2
(b)9.5
(C)11.5
(D) none of the

plzz post the approach !!!

For the problem of chocolates you can first find the difference as 67+32=99
Now x+y=99,x-y=1 so x=50
answer=50^2-32=2468


Triangle ABC has vertices A (0, 0), B (0, 6) and C (9, 0). The points P and Q lie on side AC such that
AP = PQ = QC. Similarly, the points R and S lie on side AB such that AR = RS = SB. If the line
segments PB and RC intersect at X, then the slope of the line AX is .......... ??

Getting 3/2
Method :

Form the eq of line CR and PB
Found intersection point X

now slope of AX = y2-y1/x2-x1
I m getting 3/2 if there is not calc mistake

EDIT : Taken B and C at the wrong ordinate.
Should be then 2/3

Pradeep borrowed Rs 12.600 at 10% rate of compound interest If this amount has to be repaid in two equal annual instalments find the value of each instalment.

with approach....

formula for equal installment under compound interest is:
(P.r)/(100.(1-(100/(100+r))^n))

= ( 12600 x 10 ) / ( 100 x ( 1 - ( 100 / ( 100 + 10 ) )^ 2))
= 7260

kindly correct me if i am wrong. and OA please!!

HARIV Says

It ll be 1/4 of the original triangle

hi hariv!!!

is this a property???

if yes can u elaborate?

rohitpal Says

yeh plus 99 kyon liya hain bhai tumne? is it bcoz of 67 + 32 ?

Thats correct Rohit bhai..
Kyoki 67 adjust karne ke baad bhi 32 jagah aur bacha hua tha..
so 67 + 32 = 97

P.S How goes ur prep for CAT 2011?

For the problem of chocolates you can first find the difference as 67+32=99
Now x+y=99,x-y=1 so x=50
answer=50^2-32=2468


Triangle ABC has vertices A (0, 0), B (0, 6) and C (9, 0). The points P and Q lie on side AC such that
AP = PQ = QC. Similarly, the points R and S lie on side AB such that AR = RS = SB. If the line
segments PB and RC intersect at X, then the slope of the line AX is .......... ??

x*x=z
(x+1)*(x+1)=z+99
==>x^2+x+x+1=z+99
==>2x=98
==>x=49
he planned to carry 49*49+67 = 2468

yeh plus 99 kyon liya hain bhai tumne? is it bcoz of 67 + 32 ?

ksagarwa Says

On a Christmas day, Arjun goes to his school for a party where he wants to carry a colorful square box full of chocolates for his friends. After arranging the chocolates in the box he finds that he is still left with 67 chocolates with him. Now in order to be able to carry all the remaining chocolates as well, he plans to increase the size of the box in such a way so as increase one chocolate along both dimensions of the square, but then discovered that he still has space for 32 chocolates. Therefore, the actual number of chocolates that he has originally planned to carry with him is:

x*x=z
(x+1)*(x+1)=z+99
==>x^2+x+x+1=z+99
==>2x=98
==>x=49
he planned to carry 49*49+67 = 2468

ksagarwa Says

On a Christmas day, Arjun goes to his school for a party where he wants to carry a colorful square box full of chocolates for his friends. After arranging the chocolates in the box he finds that he is still left with 67 chocolates with him. Now in order to be able to carry all the remaining chocolates as well, he plans to increase the size of the box in such a way so as increase one chocolate along both dimensions of the square, but then discovered that he still has space for 32 chocolates. Therefore, the actual number of chocolates that he has originally planned to carry with him is:

49^2 = 2401
50^2 = 2500

So, initially, he planned to carry 2401+67 chocolates in a 49*49 box.
On increasing dimensions, he gets a 50*50 box with 2500 spaces.
So, spaces left = 2500 - 2468 = 32.

i have one question...
suppose three co-ordinates are given... and they form a triangle.... now their mid points are joined to form another triangle..... we need to find the area of the triangle formed by joining mid points of the sides...

how do we find it????
is it just half of the area of the bigger triangle???

It ll be 1/4 of the original triangle

i have one question...
suppose three co-ordinates are given... and they form a triangle.... now their mid points are joined to form another triangle..... we need to find the area of the triangle formed by joining mid points of the sides...

how do we find it????
is it just half of the area of the bigger triangle???

100 = 2^2*5^2
number of odd factors = 3
number of ways it can be written as sum of consecutive natural numbers = 3-1=2

Thala, How did you come up with the bold part? Pl elaborate.

Snipped....

PLEASE ANSWER!!

In a certain examination paper, taken by a total of 36000 students, there are q questions. For p = 1, 2 .. q, there are 2^(q - p) students who answered p or more questions wrongly. The total number of wrong answers is 65,535.

I understood the solution till..
students with 1 or more wrong answers = 2^(q-1)
students with 2 or more wrong answers = 2^(q-2) and so on

Hence, students with exactly 1 wrong answer = 2^(q-1) - 2^(q-2) = 2^(q-2) students with exactly 2 wrong answers = 2^(q-2) - 2^(q-3) = 2^(q-3)
.
.
students with exactly q wrong answers = ??? (in solution its given 2^0)

So, No. of WRONG ANSWERS is summations of
(students with exactly p wrong ans)*p
= 1*2^(q-2) + 2*2^(q-3)+ 3*2^(q-4).....+ q*2^0
= 2^(q-1)+ 2^(q-2)+ 2^(q-3).....+1 How??


ANSWER =16

for such type of sums,direct formula....2^n -1=65355
=>2^n=65356
=>n=16

Pradeep borrowed Rs 12.600 at 10% rate of compound interest If this amount has to be repaid in two equal annual instalments find the value of each instalment.

with approach....

Here i think borrowed amount is 12,600:sneaky:
then ans shud be 6000/11 = 5454.5....

let the amount of installment is A and borrowed amount is B...then

A+A^2 = 12600..

WHAT'S THE OA YAAR

On a Christmas day, Arjun goes to his school for a party where he wants to carry a colorful square box full of chocolates for his friends. After arranging the chocolates in the box he finds that he is still left with 67 chocolates with him. Now in order to be able to carry all the remaining chocolates as well, he plans to increase the size of the box in such a way so as increase one chocolate along both dimensions of the square, but then discovered that he still has space for 32 chocolates. Therefore, the actual number of chocolates that he has originally planned to carry with him is:

100 = 2^2*5^2
number of odd factors = 3
number of ways it can be written as sum of consecutive natural numbers = 3-1=2

could you plz elaborate. I mean is there any formula.

sone ka time

nitish narang Says

How many ways are there to represent 100 as the sum of consecutive natural numbers?

100 = 2^2*5^2
number of odd factors = 3
number of ways it can be written as sum of consecutive natural numbers = 3-1=2

How many ways are there to represent 100 as the sum of consecutive natural numbers?

Pradeep borrowed Rs 12.600 at 10% rate of compound interest If this amount has to be repaid in two equal annual instalments find the value of each instalment.

with approach....

Sahil Kataria Says

10 ways? For what? And how?

bdf ko select kar liya

but 5 more ayenge ab jo bi aye

b x d y f z

ab wo x,y,z ki positions hi fill karenge na
x+y+z=5

now x,y>0

x'+y'+z=3

so 5c2 or 10..

hope u got it...

like in prev ques (abcd wala) you did 7c4

so, in this ques after x'+y'+z=3
remaining are b,d and f, so it should be 5c3(i know it will be 10)

n yar in prev ques what 5! n 4c2 for?
plz bare with me. i'm a newbie pagal..

Sahil Kataria Says

10 ways? For what? And how?

bdf ko select kar liya

but 5 more ayenge ab jo bi aye

b x d y f z

ab wo x,y,z ki positions hi fill karenge na
x+y+z=5

now x,y>0

x'+y'+z=3

so 5c2 or 10..

hope u got it...

lo approach
bdf teeno ayenge hi ayenge

so b x d y f z

x+y+z=5

so 10 ways and for bfd 10 again

so 20

20*5!*4c2=14400

10 ways? For what? And how?

Sahil Kataria Says

What is b',c' n d' and how is a+b'+c'+d'+e=3

b'=b +1
c'=c+1
d'=d+1

as they cannot be zero!!!


@pkbhai: bhai kyu lae re ho garib ki!!!

Please solve this one...

Ther r 10 stations between A and B, both exclusive, along a railway track. In how many ways can a train halt at the stations if there is a total of 4 halts(6, if A and B are included) and none of the 4 halts r at consecutive stations?

yaar abhi 5-6 page pehle hi kara 35 ayega...

lo copy paste bi kar deta hu


its just like picking 4 books out of 10..such that no two are consecutive

a st1 b st2 c st3 d st4 e

a+b+c+d+e=6

b,c,d>0

a+b'+c'+d'+e=3

7c4

so 35 ways...

What is b',c' n d' and how is a+b'+c'+d'+e=3

souravsaha008 Says

Can you please explain it again? I was not able to get it.

bhai kahan doubt hai isme

souravsaha008 Says

Can you please explain it again? I was not able to get it.

see 4 stations were to be picked from 10 such that no 2 are consecutive



so if i pick any 4

st1 st2 st3 st4

a represent no of station before st 1
b before st2 but after st1

so on..

a st1 b st2 c st3 d st4

b,c,d>0 since no 2 can be consecutive

a+b+c+d+e=6

a+b'+c'+d'+e=3

7c3 ways
so 35

jo samaj ni aata do highlight...

oh galat question kar diya!!!

lo approach
bdf teeno ayenge hi ayenge

so b x d y f z

x+y+z=5

so 10 ways and for bfd 10 again

so 20

20*5!*4c2=14400

Can you please explain it again? I was not able to get it.