BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Basic Artihmetic

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 62
Joined: Fri Dec 21, 2007 2:58 pm
Location: Fremont , California
Thanked: 3 times

by kishore » Sat Jun 21, 2008 6:28 pm
first car sold at = 125/100 * 20,000 = 25000 $

Second car sold at = 80/100 * 20,000 = 16000 $

Selling price of both the cars = 41000 $

Cost Price of both the cars = 40000 $

Since , selling price is less than cost price, its a loss

loss= 41000 - 40000 = 1000 $
Top
Master | Next Rank: 500 Posts
Posts: 124
Joined: Mon May 19, 2008 1:12 pm
Thanked: 3 times

by Nycgrl » Sat Jun 21, 2008 6:30 pm
To solve this question first we have to calculate the cost price of both the cars.

Let the cost price of the care to dealer be X

1st Car X + .25X = 20,000 i.e X = 16000
Now X is cost price we have to calculate the amount he earned when he sold it which is 25% of cost price that is 25% of 16000 = 4000

2nd Car X - .20X = 20,000 i.e X = 25000.
He sold it at 20% loss which is 20% of 25000 = 5000

both the transactions combined he had loss of 1000
Top
Post new topic Post Reply
• Page 1 of 1