A car dealership sold two cars: the first car at a 10% profit and the second car at a 10% loss, which gave them an overall profit margin of 5% from these two sales. If the dealership's total profit was $1000, what was the sale price of each car?
$5,000 and $1,000
$9,000 and $5,000
$11,000 and $9,000
$15,000 and $5,000
$20,000 and $10,000
Tough One
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IMO D
1000 dollars is 5% so total sale price must be 20,000 dollars
so must be one of C and D
use the answers
D- $15,000 and $5,000
10 % profit on first=1500 dollars
10% loss on 2nd=500dollars
total profit =1000 dollars
must be the answer
try C it doesn't satisfy
1000 dollars is 5% so total sale price must be 20,000 dollars
so must be one of C and D
use the answers
D- $15,000 and $5,000
10 % profit on first=1500 dollars
10% loss on 2nd=500dollars
total profit =1000 dollars
must be the answer
try C it doesn't satisfy
The powers of two are bloody impolite!!
IMO D
I used a diff approach
Lets say the cost price of cars = A (car1) & B (car2)
Selling price Car1 = 1.1 * A
Selling price Car2 = 0.9 * B
Profit = 5% = 5/100 = [(1.1A+0.9B) - (A+B)]/(A+B)
5/100 = 0.1 (A-B)/A+B ==> A = 3B --- (1)
Profit = 1000 = 0.1(A-B) --- (2)
From (1) and (2)
==> B = 5000
A = 15000
I used a diff approach
Lets say the cost price of cars = A (car1) & B (car2)
Selling price Car1 = 1.1 * A
Selling price Car2 = 0.9 * B
Profit = 5% = 5/100 = [(1.1A+0.9B) - (A+B)]/(A+B)
5/100 = 0.1 (A-B)/A+B ==> A = 3B --- (1)
Profit = 1000 = 0.1(A-B) --- (2)
From (1) and (2)
==> B = 5000
A = 15000
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Cost Price = X , Sales Price = 1.1X
Cost Price = y, Sales Price = 0.9Y
1.1X + 0.9Y - (X+Y) = 1000
1.1X + 0.9Y = X + Y + 0.05(X+Y)
solve the 2 eqn give X = 3Y
Y = 5000
X =15000
Cost Price = y, Sales Price = 0.9Y
1.1X + 0.9Y - (X+Y) = 1000
1.1X + 0.9Y = X + Y + 0.05(X+Y)
solve the 2 eqn give X = 3Y
Y = 5000
X =15000