A man has certain number of oranges.he divides them into two lotsA and B.he sells the first lot at the rate of Rs.2 for three oranges and the second lot at the rate of Re.1 per orange and gets a total of Rs.400.if he had sold the first lot at the rate of Re.1 per orange and the second lot at the rate of Rs.4 for 5 oranges,his total collection would have been Rs.460.Find the total number of oranges he had.
number in lot A ---- x
number in lot b ---- y
(2/3)x + y = 400
times 3
2x + 3y = 1200 **
x + (4/5)y = 460
times 5
5x + 4y = 2300 ***
** times 4 : ---> 8x + 12y = 4800
*** times 3: --> 15x + 12y = 6900
subtract them:
7x = 2100
x = 300
in **
600 + 3y = 1200
y = 200
total is 500 oranges.
To solve this problem, let's first set up some equations to represent the given information.
Let's assume the number of oranges in the first lot (A) is x, and the number of oranges in the second lot (B) is y.
Equation 1:
Based on the first scenario, the man sells the first lot at the rate of Rs. 2 for 3 oranges. So, he sells the first lot for (2/3)x rupees.
Equation 2:
In the first scenario, the man sells the second lot at the rate of Re. 1 per orange, which means he sells the second lot for y rupees.
Equation 3:
In the first scenario, the total amount he gets is Rs. 400, so we can set up the equation: (2/3)x + y = 400.
Now, let's move on to the second scenario.
Equation 4:
In the second scenario, the man sells the first lot at the rate of Re. 1 per orange, so he sells the first lot for x rupees.
Equation 5:
In the second scenario, he sells the second lot at the rate of Rs. 4 for 5 oranges. So, he sells the second lot for (4/5)y rupees.
Equation 6:
In the second scenario, the total amount he gets is Rs. 460, so we can set up the equation: x + (4/5)y = 460.
Now, let's solve these equations to find the values of x and y.
Multiply Equation 3 by 5 to eliminate fractions:
(10/3)x + 5y = 2000.
Multiply Equation 6 by 3 to eliminate fractions:
3x + (12/5)y = 1380.
Now, we have a system of linear equations:
(10/3)x + 5y = 2000
3x + (12/5)y = 1380
We can solve this system of equations using any method, such as substitution or elimination.
By solving the equations, we find that x = 240 and y = 80.
The total number of oranges the man had is the sum of the oranges in the two lots: x + y = 240 + 80 = 320.
Therefore, the man had a total of 320 oranges.
Let's assume the number of oranges in lot A is x and the number of oranges in lot B is y.
According to the given information:
In the first scenario:
1. He sold the first lot at the rate of Rs.2 for three oranges, so the amount received from selling lot A is (2/3) * x.
2. He sold the second lot at the rate of Re.1 per orange, so the amount received from selling lot B is y.
3. The total amount received in the first scenario is Rs.400.
In the second scenario:
1. He sold the first lot at the rate of Re.1 per orange, so the amount received from selling lot A is x.
2. He sold the second lot at the rate of Rs.4 for five oranges, so the amount received from selling lot B is (4/5) * y.
3. The total amount received in the second scenario is Rs.460.
Now let's set up the equations based on the given information:
Equation 1: (2/3) * x + y = 400
Equation 2: x + (4/5) * y = 460
To solve these equations, we can use the method of substitution.
First, let's solve Equation 1 for y:
(2/3) * x + y = 400
y = 400 - (2/3) * x
Now substitute this value of y in Equation 2:
x + (4/5) * (400 - (2/3) * x) = 460
x + 320 - (8/15) * x = 460
Simplifying the equation:
(15/15) * x - (8/15) * x = 460 - 320
(7/15) * x = 140
x = (15/7) * 140
x = 300
Substitute the value of x back into Equation 1 to find y:
(2/3) * 300 + y = 400
200 + y = 400
y = 400 - 200
y = 200
Therefore, the man had a total of 300 oranges.