Need expert help with my solution

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voodoo_child: Master | Next Rank: 500 Posts; Posts: 405; Joined: Thu Feb 10, 2011 1:44 amFollowed by: 1 member; Thanked: 3 times

by voodoo_child » Wed Apr 11, 2012 4:42 am

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

OA - D

S1 = selling price of car1
S2 = selling price of car2

S1 = 1.1C1; S2=0.9C2 ......(i)

S1 + S2 = 1.05(C1+C2) ...(ii)

Profit = $1000 = (S1+S2) - (C1+C2) = (1/20)*(C1+C2).......(iii)
Therefore, C1+C2 = 20K ......(iv)

Now, I can also solve (i) and (ii):
1.1C1 + 0.9C2 = 1.05C1 + 1.05C2 => C1 = 3C2 .......(v)

Using (iv) and (v), C1=5K and C2 = 15K

S1 =5K(1.1) = 5.5K
S2 = 13.5K.

Something is wrong. Can someone please help me?

NB: I know another easy method to solve this by using backtrace. However, I am not sure what's wrong with above equations...Please help me

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Source: Beat The GMAT — Problem Solving | Wed Apr 11, 2012 4:42 am

neelgandham: Community Manager; Posts: 1060; Joined: Fri May 13, 2011 6:46 amFollowed by: 52 members; Location: Utrecht, The Netherlands; Thanked: 318 times

by neelgandham » Wed Apr 11, 2012 5:07 am

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?

voodoo_child - You are correct! I did a quick 'google' and found that the question is
what was the cost price of each car?
and not
what was the sale price of each car?

Correction
C1 = 3C2
C1+C2 = 20000
C1 = 15000 and C2 = 5000
S1 = 15000*(1.1) = 16500
S2 = 5000*(0.9) = 4500

Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/

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Shalabh's Quants: Master | Next Rank: 500 Posts; Posts: 134; Joined: Fri Apr 06, 2012 3:11 amFollowed by: 5 members; Thanked: 35 times

by Shalabh's Quants » Wed Apr 11, 2012 6:12 am

voodoo_child wrote: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

OA - D

S1 = selling price of car1
S2 = selling price of car2

S1 = 1.1C1; S2=0.9C2 ......(i)

S1 + S2 = 1.05(C1+C2) ...(ii)

Profit = $1000 = (S1+S2) - (C1+C2) = (1/20)*(C1+C2).......(iii)
Therefore, C1+C2 = 20K ......(iv)

Now, I can also solve (i) and (ii):
1.1C1 + 0.9C2 = 1.05C1 + 1.05C2 => C1 = 3C2 .......(v)

Using (iv) and (v), C1=5K and C2 = 15K

S1 =5K(1.1) = 5.5K
S2 = 13.5K.

Something is wrong. Can someone please help me?

NB: I know another easy method to solve this by using backtrace. However, I am not sure what's wrong with above equations...Please help me

Well i recommend a little easier approach.

Lets go by Options...

Since First car is to make money and second to loose. Right!

A & B cannot be answers as their prices are too low to make 10% as 1000.

C can also not be answer, as 10% of 11000 = 1100 and loss on second car 10% of 9000 = -900 will not make 1000 overall gain.

Answer should be D.

10% of 15000 - 10% of 5000 = 1000, which is also 5% of (15000+5000).

Option E also gives $ 1000/- profit, but not 5% overall profit, hence rejected.

Shalabh Jain,
e-GMAT Instructor

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voodoo_child: Master | Next Rank: 500 Posts; Posts: 405; Joined: Thu Feb 10, 2011 1:44 amFollowed by: 1 member; Thanked: 3 times

by voodoo_child » Wed Apr 11, 2012 8:16 am

Shalabh's Quants wrote:
voodoo_child wrote: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

OA - D

S1 = selling price of car1
S2 = selling price of car2

S1 = 1.1C1; S2=0.9C2 ......(i)

S1 + S2 = 1.05(C1+C2) ...(ii)

Profit = $1000 = (S1+S2) - (C1+C2) = (1/20)*(C1+C2).......(iii)
Therefore, C1+C2 = 20K ......(iv)

Now, I can also solve (i) and (ii):
1.1C1 + 0.9C2 = 1.05C1 + 1.05C2 => C1 = 3C2 .......(v)

Using (iv) and (v), C1=5K and C2 = 15K

S1 =5K(1.1) = 5.5K
S2 = 13.5K.

Something is wrong. Can someone please help me?

NB: I know another easy method to solve this by using backtrace. However, I am not sure what's wrong with above equations...Please help me
Well i recommend a little easier approach.

Lets go by Options...

Since First car is to make money and second to loose. Right!

A & B cannot be answers as their prices are too low to make 10% as 1000.

C can also not be answer, as 10% of 11000 = 1100 and loss on second car 10% of 9000 = -900 will not make 1000 overall gain.

Answer should be D.

10% of 15000 - 10% of 5000 = 1000, which is also 5% of (15000+5000).

Option E also gives $ 1000/- profit, but not 5% overall profit, hence rejected.

Thanks but what's wrong with my method? Do you mind highlighting it?

Thanks

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Shalabh's Quants: Master | Next Rank: 500 Posts; Posts: 134; Joined: Fri Apr 06, 2012 3:11 amFollowed by: 5 members; Thanked: 35 times

by Shalabh's Quants » Wed Apr 11, 2012 10:44 am

voodoo_child wrote:
Shalabh's Quants wrote:
voodoo_child wrote: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

OA - D

S1 = selling price of car1
S2 = selling price of car2

S1 = 1.1C1; S2=0.9C2 ......(i)

S1 + S2 = 1.05(C1+C2) ...(ii)

Profit = $1000 = (S1+S2) - (C1+C2) = (1/20)*(C1+C2).......(iii)
Therefore, C1+C2 = 20K ......(iv)

Now, I can also solve (i) and (ii):
1.1C1 + 0.9C2 = 1.05C1 + 1.05C2 => C1 = 3C2 .......(v)

Using (iv) and (v), C1=5K and C2 = 15K

S1 =5K(1.1) = 5.5K
S2 = 13.5K.

Something is wrong. Can someone please help me?

NB: I know another easy method to solve this by using backtrace. However, I am not sure what's wrong with above equations...Please help me
Well i recommend a little easier approach.

Lets go by Options...

Since First car is to make money and second to loose. Right!

A & B cannot be answers as their prices are too low to make 10% as 1000.

C can also not be answer, as 10% of 11000 = 1100 and loss on second car 10% of 9000 = -900 will not make 1000 overall gain.

Answer should be D.

10% of 15000 - 10% of 5000 = 1000, which is also 5% of (15000+5000).

Option E also gives $ 1000/- profit, but not 5% overall profit, hence rejected.
Thanks but what's wrong with my method? Do you mind highlighting it?

Thanks

There is nothing wrong in your method. You have also correctly calculated C1= 5000/- and C2 as 15000/-.
This question in fact asks for Cost Price. neelgandham also suggests it. You have probably wrongly typed the question as
'What was the sale Price of each car?'

If you still insist that Question asks for Sale price, then you are correct.

Shalabh Jain,
e-GMAT Instructor

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