Number System - Questions & Discussions

MBA CAT 2022
964
x + y + z = 17

no of positive integral solution : 16 C 2 = 16 * 15/2 = 120

but x,y,z
x = x + 10

x + y + z = 7

no of solution = 6 C 2 = 15

so we get 120 - (3*15) = 75

Dear Naga25French,
Doesnt the case given by allow value of x to be zero as well....?
I am kinda of confused....? With this concept...
965
coeficient of X^17 in (x+x^2+..x^9)(1+x+..x^9)^2
=x(1-x^9)(1-x^10)^2*(1-x)^-3
= (x-x^10)(1-2x^10)*(1-x)^-3
= x(1-x)^-3 - 2x^11(1-x)^-3 - x^10(1-x)^-3
= 16+3-1C2 - 2* 6+3-1C2 - 7+3-1C2
= 153 - 56 - 36 = 61...
wats the OA???

puys pls can anyone tell wats the O.A for this question???
966
q)How many 3 digit number has its sum of the digits equal to 17.
(Solution other than conventional way is needed)


Say Number is xyz

x>=1, y>=0 z>=0

here let x=9-t,y=9-u and z=9-v

since x,y,z
(9-t)+(9-u)+(9-v)= 17

t+u+v = 10

12C2 ways =66 ways

but here t cannot be 9 since then x will be 0

when t=9

u+v= 1

so 2 solns exist u=1, v=0 or u=0,v=1

so total solns = 66-2 =64


Wat is the OA??
967
Say Number is xyz

x>=1, y>=0 z>=0

here let x=9-t,y=9-u and z=9-v

since x,y,z
(9-t)+(9-u)+(9-v)= 17

t+u+v = 10

12C2 ways =66 ways

but here t cannot be 9 since then x will be 0

when t=9

u+v= 1

so 2 solns exist u=1, v=0 or u=0,v=1

so total solns = 66-2 =64


Wat is the OA??

dude here whn t+u+v = 10
u will have three cases when one of them is 10 and the other two will be 0 hence
the ans shud be 64-3 = 61...wat say:)
968
x + y + z = 17

no of positive integral solution : 16 C 2 = 16 * 15/2 = 120

but x,y,z
x = x + 10
x + y + z = 7

no of solution = 6 C 2 = 15

so we get 120 - (3*15) = 75



Hi,
Can u explain the part in bold..

I think if we have x+y+z the we introduce a dummy variable
x+y+z+w=10

No of solutions..
9C3=84

So total number of solutions
120-84=36.
969
Dear Naga25French,
Doesnt the case given by allow value of x to be zero as well....?
I am kinda of confused....? With this concept...



x cant be zero because we need a 3 digit number...
970
Say Number is xyz

x>=1, y>=0 z>=0

here let x=9-t,y=9-u and z=9-v

since x,y,z
(9-t)+(9-u)+(9-v)= 17

t+u+v = 10

12C2 ways =66 ways

but here t cannot be 9 since then x will be 0

when t=9

u+v= 1

so 2 solns exist u=1, v=0 or u=0,v=1

so total solns = 66-2 =64


Wat is the OA??

dude here whn t+u+v = 10
u will have three cases when one of them is 10 and the other two will be 0 hence
the ans shud be 64-3 = 61...wat say:)


Dude...I have a doubt here when i put x = 9-t i guess it means t can vary from 0 to 9 only otherwise if t = 10, x becomes -ve

same is the case wid y and z.
971
Dude...I have a doubt here when i put x = 9-t i guess it means t can vary from 0 to 9 only otherwise if t = 10, x becomes -ve

same is the case wid y and z.


Hi Guys sorry to disappoint you all but i think there is more than 61 or 64 ways possible...
Lets see the question again
Ques says a 3 digit a number say xyz should have sum of its digits so x+y+Z
Lets keep this ques aside for a sec
lets say question was x+y+z sol:- here xyz to be a 3 digit number atleast has to be 1
x (1 given to x) + y + z now x + y + z + D = 8
now this we can get in 11 C 3 way = 11*10*9/3*2= 165

So for sum of digits less than or equal to 8 we get 165 .so for somthin like 17 it has to be more than 165 hope this helps
Unfortunately i am finding hard to arrive at a solution for this
972

If x+y+z=17 then the answer is 62
but x+y+z

973
Dude...I have a doubt here when i put x = 9-t i guess it means t can vary from 0 to 9 only otherwise if t = 10, x becomes -ve

same is the case wid y and z.

its not like tht all tht matters is ur second eqn...t+u+v = 10
here u will have three possibilites like i said in which one will be 10 rt? so if u substitute tht into the first eqn one of the variable will become negative, hence u need to subtract those three cases. but i need to knw the OA..
974
If x+y+z=17 then the answer is 62
but x+y+z
its no where given tht we need to find for x+y+z
975
976

Sorry aryan its wrong... cause u are neglecting then condition when x y z is greater than 10

977
NoRuLeS Says
Sorry aryan its wrong... cause u are neglecting then condition when x y z is greater than 10

sorry corrected my mistake

x=1, y+z=16 (y=7,8,9) solutions 3
x=2, y+z=15 solutions 4
....
....
x=8 y+z=9 solutions 10
x=9 y+z=8 solutions 9

total 3+4+.......+10+9=61
978

Hello Puys, Try solving this question ??
1 : Garibchand decided to sell 89 articles (of same kind). After achieving break even (on exact no. of articles), he decided to give 10% discount and when he achieved exact 10% profit, he decided to give 20% discount. Garibchand sold all articles with 20% net profit. Had he not given discount, he would have made profit of

(1)23.6% (2) 25.3% (3) 27.8% (4) 29.2% (5) 31.4%

979
Hello Puys, Try solving this question ??
1 : Garibchand decided to sell 89 articles (of same kind). After achieving break even (on exact no. of articles), he decided to give 10% discount and when he achieved exact 10% profit, he decided to give 20% discount. Garibchand sold all articles with 20% net profit. Had he not given discount, he would have made profit of

(1)23.6% (2) 25.3% (3) 27.8% (4) 29.2% (5) 31.4%

I think the answer should be 23.6%, although I can't do this without using a calculator. I solved it by using options and worked with the one which gave me an integer value (as is required in the question).

So, if we suppose that the actual price of 89 articles is Rs. 100 each, then total comes to Rs. 8900. So if we work with 23.6%, that means the price that he started off with was Rs. 123.6.

Thus, no. of articles sold to breakeven = 8900/123.6 = 72 (You won't get an integer value by using any other option)

Then he decides to give a 10% discount, i.e. price = 123.6*0.9 = 111.24
Till he achieved 10% profit, i.e. a further income of = 8900*0.1 = 890

Thus, no. of articles sold = 890/111.24 = 8

Lastly, he gives a 20% discount, i.e. price = 123.6*0.8 = 98.88
He achieved a total net profit of 20%, thus further income = 8900*0.1 = 890 (Only multiplying by 0.1 and NOT 0.2 because we have already included 890 in the previous part, when he gave 10% discount)

Thus, no. of articles sold = 890/98.88 = 9

Thus, total articles sold = 89 😃
980
Hello Puys, Try solving this question ??
1 : Garibchand decided to sell 89 articles (of same kind). After achieving break even (on exact no. of articles), he decided to give 10% discount and when he achieved exact 10% profit, he decided to give 20% discount. Garibchand sold all articles with 20% net profit. Had he not given discount, he would have made profit of
(1)23.6% (2) 25.3% (3) 27.8% (4) 29.2% (5) 31.4%


SP -> a
CP -> b
no of articles at a = x
no of articles at 0.9 a = y
no of articles at 0.8 a = 89-x-y
xa = 89b ---1
0.9ay = 8.9b ---2
(89-x-y)(0.8 )a = 8.9b ---3
we need to find (a-b)/b
solving 1,2,3
(a-b)/b= 23.6%
981

Can someone please explain how to go about in the following type of questions :

Q> Which of these is greater:

a) 200^300 or 300^200 or 400^150
b) 5^100 or 2^200
c) 10^20 or 40^10

I did solve these by making the factorizing the number and making all the base numbers equal, but this is not true in all the case. I would really appreciate if someone provided a clear cut approach for these types of questions.

982
Can someone please explain how to go about in the following type of questions :

Q> Which of these is greater:

a) 200^300 or 300^200 or 400^150
b) 5^100 or 2^200
c) 10^20 or 40^10

I did solve these by making the factorizing the number and making all the base numbers equal, but this is not true in all the case. I would really appreciate if someone provided a clear cut approach for these types of questions.


a) 200^300 = (200^6)^50
300^200 = (300^4)^50
400^150 = (400^3)^50

Obviously 200^6 > 300^4 > 400^3
Result follows.

b) 2^200 = 4^100
c) 10^20 = 100^10 > 40^10
983

41^36) mod 7*11=?

(41^36) mod 7 = 1
(41^36) mod 11 = 3

so from here also remainder = 36

PLZ ANYONE EXPLAIN ME HW THIS 36 CAME...