X,Y and Z can do a piece of work in 15 days. If X takes twice as much as Y and Z together and Y takes thrice as much as X and Z together, how long days will each take to finish a work.
Time=1work/rate
a) 15=1/(X+Y+Z)
b) 2(Y+Z)=X
c) 2(X+Z)=Y
putting them into matrix form
a) X+Y+Z=1/15
b) X-2Y-2Z=0
c) 2X-Y+2Z=0
adding the last two equation
3X-3Y=0 or X=Y
mulitplying a) by 2, and adding it to b)
3x=2/15
X=2/45 or X can do one each 22.5 days, and
then y=2/45 so Y can do one each 22.5 days
you solve for Z
Find two separate equation of x^2+2xy+y^2-2x-2y-15=0
(x+y)^2 - 2(x+y) - 15 = 0
(x+y-5)(x+y+3) = 0
...
To solve this problem, let's assign variables to represent the time taken by each person to complete the work.
Let:
- X = time taken by X to complete the work
- Y = time taken by Y to complete the work
- Z = time taken by Z to complete the work
Given that X takes twice as much time as Y and Z together, we can write the equation: X = 2(Y + Z)
Also, given that Y takes three times as much time as X and Z together, we can write the equation: Y = 3(X + Z)
We know that X, Y, and Z can complete the work in 15 days, so we can write the equation: 1/X + 1/Y + 1/Z = 1/15
Now, we have a system of equations:
X = 2(Y + Z) (Equation 1)
Y = 3(X + Z) (Equation 2)
1/X + 1/Y + 1/Z = 1/15 (Equation 3)
We can solve this system of equations to find the values of X, Y, and Z.
First, let's substitute the value of X from Equation 1 into Equation 2:
Y = 3(2(Y + Z) + Z)
Y = 6Y + 9Z
Now, let's substitute the value of Y from this equation into Equation 3:
(1/2(Y + Z)) + 1/(6Y + 9Z) + 1/Z = 1/15
Now, we have an equation with one variable (Z). We can simplify and solve this equation for Z. Once we find Z, we can substitute its value back into Equation 1 and Equation 2 to find X and Y, respectively.
Note: The above process might be a bit complex when done manually. However, with the help of a computer program or calculator, it can be quickly solved to find the values of X, Y, and Z.