can anybody tell me the expression for marginal revenue ??
if u need any other info do let me know..i know nothing about it so may be i m asking an incomplt question.
@freebull said:can anybody tell me the expression for marginal revenue ??if u need any other info do let me know..i know nothing about it so may be i m asking an incomplt question.
you could have googled..
see this(if it helps)--->http://en.wikipedia.org/wiki/Marginal_revenue
@MikelArteta said:If x, y, and z are positive real numbers, for what ratio of the values of y and z is the value of ( x/y + z/12x + 4y/x + x/3z) the minimum?Ans is 1/4.Found this one in the prep section. Could anyone post the solution. Thanks
Let x/y = a & z/x = b
Therefore, x/y + z/12x + 4y/x + x/3z = (a+ 4/a) + (1/12)(b +4/b)
Minimum value occurs when (a+4/a) and (b + 4/b) are minimum.
Minimum value of (a + 4/a) and (b+ 4/b) occur at a=2 and b=2.(since a and b are positive we do not consider negative values)
Therefore, x/y + z/12x + 4y/x + x/3z = (a+ 4/a) + (1/12)(b +4/b)
Minimum value occurs when (a+4/a) and (b + 4/b) are minimum.
Minimum value of (a + 4/a) and (b+ 4/b) occur at a=2 and b=2.(since a and b are positive we do not consider negative values)
(x/y + z/12x + 4y/x + x/3z) is at a minimum when x/y =2 and z/x =2.
y : z = (y/x)*(x/z) =(1/2)*(1/2) = 1: 4
@freebull said:can anybody tell me the expression for marginal revenue ??if u need any other info do let me know..i know nothing about it so may be i m asking an incomplt question.
Do you have any question on marginal revenue ?
@MikelArteta said:If x, y, and z are positive real numbers, for what ratio of the values of y and z is the value of ( x/y + z/12x + 4y/x + x/3z) the minimum?Ans is 1/4.Found this one in the prep section. Could anyone post the solution. Thanks
for minimum value of expression,
x/y= z/12x=4y/x=x/3z
==> x/y=4y/x=> x/y=2
x/y= z/12x=4y/x=x/3z
==> x/y=4y/x=> x/y=2
And, z/12x=x/3z
=> z=2x
==> y/z=1/2*y/x=1/2*1/2=1/4
@deedeedudu said:9No is 27
Answer is correct.
Share your approach. Then I'll share mine. I found this question on difficult side :neutral:
@vijay_chandola said:'X' is the largest sum of rupees which can never be paid using any number of coins of denominations Rs.4, Rs.8, Rs.13 and Rs.18. What is the sum of the digits of 'X'?(a) 9 (b) 10 (c) 11 (d) None of these
Is there a general approach to such questions, but anyways here I did this:
Forgetting 13 for a while, I used 4,8,18 and realized that I can create only even numbers
=> 4,8,12,16,18,20,22,24,26... -> alternate even other than the missing 14 in between
Now using 13, I can create all multiples of 13, and also numbers which can be formed by adding 13 to each of the even numbers .. but the missing even number i.e. 14 + 13 = 27 will be missing..
=> 27 is the largest number which cannot be created => Sum of digits = 9 .. ?
@vijay_chandola said:'X' is the largest sum of rupees which can never be paid using any number of coins of denominations Rs.4, Rs.8, Rs.13 and Rs.18. What is the sum of the digits of 'X'?(a) 9 (b) 10 (c) 11 (d) None of these
8+4k - 24,28,32..
13+4k - 26,30,34..
18+13+4k - 35,39,43..
8+13+4k-25,29,33..
so from 29 all numbers are covered
checking back 27 is max
so 9
13+4k - 26,30,34..
18+13+4k - 35,39,43..
8+13+4k-25,29,33..
so from 29 all numbers are covered
checking back 27 is max
so 9
@soham2208 said:Is there a general approach to such questions, but anyways here I did this:Forgetting 13 for a while, I used 4,8,18 and realized that I can create only even numbers=> 4,8,12,16,18,20,22,24,26... -> alternate even other than the missing 14 in betweenNow using 13, I can create all multiples of 13, and also numbers which can be formed by adding 13 to each of the even numbers .. but the missing even number i.e. 14 + 13 = 27 will be missing..=> 27 is the largest number which cannot be created => Sum of digits = 9 .. ?
Yeah, you might wanna refer to McNugget problem.
@vijay_chandola said:Answer is correct.Share your approach. Then I'll share mine. I found this question on difficult side
Min no of the type 4a+3 that can be made by adding given nos is
31
Hence 27 is the required no :)
_/\_ @vijay_chandola sir
31
Hence 27 is the required no :)
_/\_ @vijay_chandola sir
http://en.wikipedia.org/wiki/Coin_problem#McNugget_numbers
@soham2208 said:Is there a general approach to such questions, but anyways here I did this:Forgetting 13 for a while, I used 4,8,18 and realized that I can create only even numbers=> 4,8,12,16,18,20,22,24,26... -> alternate even other than the missing 14 in betweenNow using 13, I can create all multiples of 13, and also numbers which can be formed by adding 13 to each of the even numbers .. but the missing even number i.e. 14 + 13 = 27 will be missing..=> 27 is the largest number which cannot be created => Sum of digits = 9 .. ?
@rachit_28 said:Do you have any question on marginal revenue ?
see the attachement plz !!
@Anivesh90 said:you could have googled..see this(if it helps)--->en.wikipedia.org/wiki/Marginal...
instead of scanning whole lot of unwanted theory i wanted just the relevant part..tht's why i asked it here.
otherwise i too know for wht purpose google has been put up there.
hope i didn't waste much of ur precious time.
thanks anyway !!
@vijay_chandola said:'X' is the largest sum of rupees which can never be paid using any number of coins of denominations Rs.4, Rs.8, Rs.13 and Rs.18. What is the sum of the digits of 'X'?(a) 9 (b) 10 (c) 11 (d) None of these
OA: a) 9
Solution:
For these kind of questions, form the numbers in form of 4K or 4K +1 or 4K + 2 or 4K + 3
(as all numbers are of the form 4K or 4K +1 or 4K + 2 or 4K + 3)
Coins available: 4, 8, 13 and 18
1) We can make every number of the form 4K
2) We can make every number of the form 4K + 1 starting with 13
3) We can make every number of the form 4K + 2 starting with 18
4) We can make every number of the form 4K + 3 starting with (18 + 13) = 31
Hence, largest number should be less than 31, which is 27.
@vijay_chandola said:'X' is the largest sum of rupees which can never be paid using any number of coins of denominations Rs.4, Rs.8, Rs.13 and Rs.18. What is the sum of the digits of 'X'?(a) 9 (b) 10 (c) 11 (d) None of these
Clearly all no.s of type 4k can be formed.
After 13, all no.s of the form 4k+1 can be formed.
After 18 all no.s of the form 4k+2 can be formed.
After 39 all no.s of the form 4k+3 can be formed.
Now check for no.s of the form 4k+3 less than 39.
35 = 13+18+4
31 = 13+18
27 cant be formed by any means. hence ans is 27
After 13, all no.s of the form 4k+1 can be formed.
After 18 all no.s of the form 4k+2 can be formed.
After 39 all no.s of the form 4k+3 can be formed.
Now check for no.s of the form 4k+3 less than 39.
35 = 13+18+4
31 = 13+18
27 cant be formed by any means. hence ans is 27
I think, marginal revenue = 0 =>Change in revenue with addition of new people per trip is zero
So, if g(x) = Number of people * Fare = x*(54 - x/32)^2
g'(x) = 0
=> (54-x/32)^2 - (x/16)*(54-x/32) = 0 => x = 1728(rejected) or 54-x/32 - 2x/32 = 0
=> 3x = 32*54 => x = 576 ..
So, is it 576 ?