Math

QuestionA fruit stand needs to set prices: $10\$ 10 for 4 apples & 4 oranges, $12\$ 12 for 6 apples & 6 oranges. Unique prices? A) Yes; $1.00\$ 1.00 apple, $1.50\$ 1.50 orange. B) Yes; $1.00\$ 1.00 apple, $1.00\$ 1.00 orange. C) No; many solutions. D) No; no solution.

Studdy Solution

STEP 1

Assumptions1. The cost of4 apples and4 oranges is 10.Thecostof6applesand6orangesis10. The cost of6 apples and6 oranges is 123. We assume that the cost of each fruit is constant, i.e., the cost of an apple or an orange does not change with the quantity bought.
4. We are looking for a unique solution for the price of an apple and an orange.

STEP 2

Let's denote the price of an apple as aa and the price of an orange as oo. We can then write the two given conditions as a system of linear equations.
4a+4o=104a +4o =106a+6o=126a +6o =12

STEP 3

We can simplify these equations by dividing the first equation by and the second equation by6.
a+o=2.5a + o =2.5a+o=2a + o =2

STEP 4

Now we can see that the two equations are contradictory. The price of an apple plus the price of an orange cannot be both 2.and2. and 2 at the same time.Therefore, the system of equations has no solution. The answer is D) No; the system has no solution.

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