a man has one kind of coffee at 'a' paise per kg and another at 'b' paise per kg.how much of each must he take to form a mixture of (a-b)kg, which he can sell at 'c' paise per kg without loss.
I think you must mean a+b kg, since you're taking some of each.
value of x kg of a is ax
value of y kg of b is by
value of mixture is thus ax+by
to sell at c paise/kg,
ax+by = (x+y)*c
now, we only have one equation, so it's hard to pin down both x and y. However, we can let x=1 without loss of generality, because we can then figure y as a multiple of x.
So, letting x=1,
a+by = (y+1)c
a+by-cy = c
y(b-c) = c-a
y = (c-a)/(b-c)
This will be positive, since c is between a and b.
So, if a=5 and b=10, and c=7, then we have x=1, y=2/3
check: 1*5 + (2/3)*10 = 35/3 = (1 + 2/3)*7
So, as long as y = 2x/3, the equation still holds.
Hmmm. I see I was too general. We need x=a and y=b
a^2 + b^2 = (a+b)c
so, if we let a=1, then
1+b^2 = (1+b)c
1+b^2 = c + bc
b^2 - cb + (1-c) = 0
b = 1/2 (c+√(c^2+4c-4))
If we have a=a, instead of a=1, then
b = a/2 (c+√(c^2+4c-4))
check: if we have a=1 and b=1+√2 = 2.8, then we can sell a+b = 3.8kg at $2.
1*1 + (1+√2)(1+√2) = 4+2√2
(1 + 1+2√2)(2) = 4+2√2
You cannot in general pick any random values for a and b and expect to make it fit any given c. You can pick a and b and figure c, or you can pick a and c and calculate b.
To determine how much of each kind of coffee the man must take to form a mixture, we can use the concept of weighted averages.
Let's assume the man needs to form a mixture of (a-b) kg of coffee. We'll use the following variables:
- x: amount of coffee at 'a' paise per kg
- y: amount of coffee at 'b' paise per kg
- c: selling price in paise per kg
First, let's establish the total cost of the mixture and the total weight of the mixture.
The total cost of all the coffee at 'a' paise per kg would be x multiplied by 'a' paise. Therefore, the cost of x kg of coffee at 'a' paise per kg is x * a paise.
Similarly, the total cost of all the coffee at 'b' paise per kg would be y multiplied by 'b' paise. Therefore, the cost of y kg of coffee at 'b' paise per kg is y * b paise.
Since the man wants to sell the mixture without incurring any loss, the selling price per kg should cover the total cost. So, the selling price should be equal to the total cost divided by the total weight (a-b) kg.
The total cost of the mixture would be x * a paise + y * b paise.
The total weight of the mixture is the sum of x kg and y kg, so it's x + y kg.
Setting up the equation:
c paise/kg = (x * a paise + y * b paise) / (x + y) kg
To solve for x and y, we need to solve this equation.
Note: The question does not provide the values for a, b, and c, so to find specific values of x and y, we'd need to know these variables.
However, using this equation, you can now substitute the values of a, b, and c to find the specific values of x and y that would result in a mixture that can be sold at 'c' paise without any loss.