a merchant buys 2 quantities of rice at rs.60 and rs.40 per kg respectively.he pays at total of rs.10000. on the other hand he had paid a flat rate of rs.50 per kg. he would have paid 10% more. how many kg of each type of rice did he buy?
let the amount of rice at rs.60 be x kg
let the amount of rice at rs.40 be y kg
60x + 40y = 10000 ---> 3x+2y = 500
50(x+y) = 11000 -----> x+y = 220 or y = 220-x
sub back into the 1st
3x + 2(220-x) = 500
3x + 440 - 2x = 500
x = 60 , then y = 220 - 60 = 160
check:
60(60) + 160(40) = 10000 , check!
50(220) = 11000 , check!
To solve this problem, let's assume that the merchant bought x kg of rice at Rs. 60 per kg and y kg of rice at Rs. 40 per kg.
According to the given information, the merchant paid a total of Rs. 10000 for these two quantities of rice.
So, the first equation we can form is:
60x + 40y = 10000 ----(1)
Now, let's consider the second scenario where the merchant paid a flat rate of Rs. 50 per kg for the rice. In this case, the total cost would have been 10% more than Rs. 10000.
So, the second equation we can form is:
50(x + y) = 1.10(10000)
50x + 50y = 11000 ----(2)
Now we have a system of linear equations with two variables, x and y. We can solve this system to find the values of x and y.
To solve the system of equations (1) and (2), we can use any method such as substitution or elimination. Here, I will use the elimination method to solve it.
First, let's multiply equation (2) by 2 to eliminate the y term:
2(50x + 50y) = 2(11000)
100x + 100y = 22000 ----(3)
Now, subtract equation (1) from equation (3) to eliminate the y term:
(100x + 100y) - (60x + 40y) = 22000 - 10000
40x + 60y = 12000
Simplify the above equation:
2x + 3y = 600 ----(4)
Now we have a new equation (4) with two variables, x and y.
Next, we can use equation (1) and equation (4) to solve for x and y.
Solving equation (1) and equation (4) simultaneously will give us the values of x and y.
Let's solve these equations:
60x + 40y = 10000 ---(1)
2x + 3y = 600 ---(4)
Now we can solve these two equations using a method like substitution or elimination to find the values of x and y.