The cost of 100 fruits is rupees 500. Cost of 1 watermelon ,1 guava,1 banana respectively rupees 50,10 and 1 find the no.of fruit of each type?
50w+10g+b=500
w+g+b=100
I'd set w and then work with the others
w=1;
10g+b = 450
g+b = 99
g=39, b=60
Just lucky, I guess; not all selections would come up with integer values.
Set up a system of equations:
50w+10g+b=500; w+g+b=100
Solve the right equation:
b=100-w-g
Substitute:
50w+10g+100-w-g=500
49w+9g+100=500
49w+9g=400
W and G must be whole numbers, so if w=0, then the closest integer is 44
If g=0, closest integer is 8
I'm seeing no solution mathwise. You have to guess and check.
To find the number of each type of fruit, we can set up a system of equations.
Let's assume:
Let x be the number of watermelons.
Let y be the number of guavas.
Let z be the number of bananas.
Given the cost of each fruit, we can write the following equations:
50x + 10y + z = 500 (equation 1)
x + y + z = 100 (equation 2)
To solve this system of equations, we can use a method called substitution.
1. Start by solving equation 2 for one variable.
z = 100 - x - y (equation 3)
2. Substitute equation 3 into equation 1.
50x + 10y + (100 - x - y) = 500
Simplifying the equation gives:
49x + 9y = 400 (equation 4)
3. We can simplify equation 4 further by dividing by 49.
x + (9/49)y = 400/49
4. To eliminate the fraction, we can multiply the equation by 49.
49x + 9y = 400
Multiply both sides of equation 4 by 49:
49x + (9/49) * 49y = (400/49) * 49
Simplifying the equation gives:
49x + 9y = 400 (equation 5)
5. Now we can solve the system of equations equation 5 and equation 2.
Equation 5: 49x + 9y = 400 (equation 5)
Equation 2: x + y + z = 100 (equation 2)
By solving these equations, you can find the values of x, y, and z, which represent the number of watermelons, guavas, and bananas, respectively.