a car dealer purchased a group of identical cars for $640,000.after selling all but 2 of the cars at an average profit of $16,000 each, the investment of $640,000 had been regained.how many cars did the dealer purchased?"
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No. of cars purchased by the dealer = n
Cost of each car = $C
Total cost of the cars = Total amount invested by the dealer = $nC --------- (1)
No. of cars sold by the dealer = n - 2
Average profit on each car sold = $p
Average Sale price of each car = $(C + p)
Total amount regained by sale of n - 2 cars = $(n - 2)(C + p) --------- (2)
As per the question : (n - 2)(C + p) = nC ----------- (3)
=> (n - 2)(1 + p/C) = n
=> n / (n - 2) = 1 + p/C ------------- (4)
Given, nC = 640,000 and p = 16,000
p / nC = 1/40
=> p / C = n / 40 -------- (5)
Substituting for p / C in (4) we get
n / (n - 2) = 1 + n/40 = (n + 40) / 40
=> (n + 40)(n -2) = 40n
=> n² + 38n - 80 = 40n
=> n² - 2n - 80 = 0
=> n² - 10n + 8n - 80 = 0
=> n(n - 10) + 8(n - 10) = 0
=> (n - 10)(n + 8) =0
=> n - 10 = 0, or n + 8 = 0
=> n = 10, or n = - 8
Rejecting the negative value, we have
No. of cars purchased by the dealer = n = 10
Cost of each car = $C
Total cost of the cars = Total amount invested by the dealer = $nC --------- (1)
No. of cars sold by the dealer = n - 2
Average profit on each car sold = $p
Average Sale price of each car = $(C + p)
Total amount regained by sale of n - 2 cars = $(n - 2)(C + p) --------- (2)
As per the question : (n - 2)(C + p) = nC ----------- (3)
=> (n - 2)(1 + p/C) = n
=> n / (n - 2) = 1 + p/C ------------- (4)
Given, nC = 640,000 and p = 16,000
p / nC = 1/40
=> p / C = n / 40 -------- (5)
Substituting for p / C in (4) we get
n / (n - 2) = 1 + n/40 = (n + 40) / 40
=> (n + 40)(n -2) = 40n
=> n² + 38n - 80 = 40n
=> n² - 2n - 80 = 0
=> n² - 10n + 8n - 80 = 0
=> n(n - 10) + 8(n - 10) = 0
=> (n - 10)(n + 8) =0
=> n - 10 = 0, or n + 8 = 0
=> n = 10, or n = - 8
Rejecting the negative value, we have
No. of cars purchased by the dealer = n = 10
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The answer to every algebra question begins with "Let x represent ....".
You asked, "How many cars did the dealer purchase?"
So, "Let x represent how many cars the dealer purchased".
Now we re-write all the details using x.
A car dealer purchased a group of identical cars for $640,000.
So the price of each car was ( $640,000 / x ).
After selling all but 2 of the cars...
means after selling (x -2 ) cars
at an average profit of $16,000 each.
Means the profit was ((x-2)*$16,000 )
the investment of $640,000 had been regained.
Means the profit was $640,000
So we know the profit in dollars and we know the profit in terms of x. The two ways of writing the profit are equal. We can write this equation:
$640,000 = ( ( x-2) * $16,000 )
And now we simplify to find x.
Divide both sides of the equation by $16,000
$640,000 / $16000 = ( ( x-2) * $16,000 ) / $16,000
40 = ( x-2 )
Add 2 to both sides of the equation
42 = x
X represents how many cars the dealer purchased. So how many cars the dealer purchased is 42.
You asked, "How many cars did the dealer purchase?"
So, "Let x represent how many cars the dealer purchased".
Now we re-write all the details using x.
A car dealer purchased a group of identical cars for $640,000.
So the price of each car was ( $640,000 / x ).
After selling all but 2 of the cars...
means after selling (x -2 ) cars
at an average profit of $16,000 each.
Means the profit was ((x-2)*$16,000 )
the investment of $640,000 had been regained.
Means the profit was $640,000
So we know the profit in dollars and we know the profit in terms of x. The two ways of writing the profit are equal. We can write this equation:
$640,000 = ( ( x-2) * $16,000 )
And now we simplify to find x.
Divide both sides of the equation by $16,000
$640,000 / $16000 = ( ( x-2) * $16,000 ) / $16,000
40 = ( x-2 )
Add 2 to both sides of the equation
42 = x
X represents how many cars the dealer purchased. So how many cars the dealer purchased is 42.
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The dealer purchased 10 cars at 64,000 each.
when he sold 8 cars at (64,000 + 16,000 profit = 80,000) he had recovered his $640,000
when he sold 8 cars at (64,000 + 16,000 profit = 80,000) he had recovered his $640,000
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42
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42