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A dealer has a lot that can hold 30 vehicles. In this lot, there are two available models, A and B. The dealer normally sells at least twice as many model A cars as model B cars. If the dealer makes a profit of $1300 on model A cars and $1700 on model B cars, how many of each car should the dealer have in the lot given that the total profit for the sale of both cars is $43,000?

a. 20 model A
10 model B

b. 10 model A
20 model B

c. 5 model A
25 model B

d. 16 model A
14 model B


i would appreciate if you explained!

User Obed
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1 Answer

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The answer is a. 20 model A, 10 model B

a - the number of model A vehicles
b - the number of model B vehicles

A dealer has a lot that can hold 30 vehicles:
a + b = 30

The dealer makes a profit of $1300 on model A cars: 1300a
The dealer makes a profit of $1700 on model B cars: 1700b
The total profit for the sale of both cars is $43,000:
1300a + 1700b = 43000

The system of two equations:
a + b = 30
1300a + 1700b = 43000
________
Express the first equation in the term of a:
a = 30 - b
1300a + 1700b = 43000
_________
Substitute a from the first equation into the second one:
1300(30 - b) + 1700b = 43000
39000 - 1300b + 1700b = 43000
39000 + 400b = 43000
400b = 43000 - 39000
400b = 4000
b = 4000/400
b = 10

Since a + b = 30 and b = 10, then
a + 10 = 30
a = 30 - 10
a = 20
User Vinay Nagaraj
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