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James Kamau stocks and sells cabbages, oranges and mangoes in his grocery at Kitengela market. On Monday last week, he sold 55 cabbages, 100 oranges and 95 mangoes making a total sale of sh. 1,625. On Tuesday, he sold 60 cabbages, 120 oranges and 80 mangoes making a total sale of sh. 1,580. On Wednesday, he sold 75 cabbages, 150 oranges and 120 mangoes making a total sale of sh. 2,175. He buys these items from a distributor at sh.3, sh.2 and sh.6 for a cabbage, an orange and a mango respectively. Required: a) Three simultaneous equations connecting the number of units sold and total sales. (3 Marks) b) The selling price for each. (9 Marks) c) The profit that James Kamau made on each of the three days and his total profits.​

User Cvarsendan
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a) Let x, y, and z be the selling price of one cabbage, one orange, and one mango, respectively.

From the given data, we can write three simultaneous equations:

Monday: 55x + 100y + 95z = 1625

Tuesday: 60x + 120y + 80z = 1580

Wednesday: 75x + 150y + 120z = 2175

b) To find the selling price of each item, we need to solve the system of equations. We can use any method of solving systems of equations, such as substitution or elimination. Here, we will use the elimination method.

Multiplying the first equation by 6, the second equation by -5, and the third equation by 3, we get:

Monday: 330x + 600y + 570z = 9750

Tuesday: -300x - 600y - 400z = -7900

Wednesday: 225x + 450y + 360z = 6525

Adding all three equations, we get:

255x + 450y + 530z = 8385

Dividing both sides by 5, we get:

51x + 90y + 106z = 1677

Now we can use this equation and any of the original equations to solve for one of the variables. Let's use the first equation:

55x + 100y + 95z = 1625

Multiplying both sides by 106 and subtracting 530 times the first equation from it, we get:

76x + 45z = 43

Solving for x, we get:

x = (43 - 45z)/76

Now we can substitute this value of x into any of the previous equations to solve for y and z. Let's use the third equation:

75x + 150y + 120z = 2175

Substituting x, we get:

75[(43-45z)/76] + 150y + 120z = 2175

Simplifying, we get:

43z/2 - 375/2 + 150y = 825

Solving for y, we get:

y = (825 - 43z/2 + 375/2)/150

Now we can substitute the values of x and y into any of the previous equations to solve for z. Let's use the second equation:

60x + 120y + 80z = 1580

Substituting x and y, we get:

60[(43-45z)/76] + 120[(825-43z/2+375/2)/150] + 80z = 1580

Simplifying, we get:

z = 4.6

Substituting z into the equation for y, we get:

y = 3.45

Substituting z and y into the equation for x, we get:

x = 1.5

Therefore, the selling price for one cabbage is sh. 1.5, for one orange is sh. 3.45, and for one mango is sh. 4.6.

c) The profit that James Kamau made on each of the three days and his total profits:

To calculate the profit, we need to subtract the cost of the items from the revenue generated by selling them.

On Monday:

Cost of cabbages = 55 x 3 = 165 shillings

Cost of oranges = 100 x 2 = 200 shillings

Cost of mangoes = 95 x 6 = 570 shillings

Total cost = 935 shillings

Revenue = 1625 shillings

Profit = Revenue - Cost = 1625 - 935 = 690 shillings

On Tuesday:

Cost of cabbages = 60 x 3 = 180 shillings

Cost of oranges = 120 x 2 = 240 shillings

Cost of mangoes = 80 x 6 = 480 shillings

Total cost = 900 shillings

Revenue = 1580 shillings

Profit = Revenue - Cost = 1580 - 900 = 680 shillings

On Wednesday:

Cost of cabbages = 75 x 3 = 225 shillings

Cost of oranges = 150 x 2 = 300 shillings

Cost of mangoes = 120 x 6 = 720 shillings

Total cost = 1245 shillings

Revenue = 2175 shillings

Profit = Revenue - Cost = 2175 - 1245 = 930 shillings

Total profit over three days:

Profit on Monday + Profit on Tuesday + Profit on Wednesday = 690 + 680 + 930 = 2300 shillings

Therefore, James Kamau made a profit of 690 shillings on Monday, 680 shillings on Tuesday and 930 shillings on Wednesday, with a total profit of 2300 shillings over the three days.

User Franz Becker
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