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two old cars

by sanju09 » Wed Mar 02, 2011 12:35 am
Ian purchased two old cars for $5000. He sold the first at 12 percent loss and the second at a gain of 8 percent. In this bargain he neither lost nor gained anything. Which of the following is the selling price of one of the two cars?
(A) $3500
(B) $3260
(C) $1850
(D) $1760
(E) $1500



[spoiler]made up[/spoiler]
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by HSPA » Wed Mar 02, 2011 12:53 am
Before selling : 8:12 == 2000, 3000

If 2000 element is sold at 12% loss i.e. 1760 and 3000 element is sold at 8% profit i.e. 3240 then resulting gain is zero.

Ans : 1760 or 3240
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by gmatmachoman » Wed Mar 02, 2011 1:15 am
As I am always a fan of Sanju,i wish i can pen down my solution.

X +Y = 5000

0.88X +1.08Y= 5000 ( accounting for the loss in X & profit in Y)

Solve for X & Y

Now the stem asks for selling price ( tricky part)

SO calculate either of .88X or 1.08Y. Look for options

I get1760 as one of the given options
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by sanju09 » Wed Mar 02, 2011 1:33 am
gmatmachoman wrote:As I am always a fan of Sanju,i wish i can pen down my solution.

X +Y = 5000

0.88X +1.08Y= 5000 ( accounting for the loss in X & profit in Y)

Solve for X & Y

Now the stem asks for selling price ( tricky part)

SO calculate either of .88X or 1.08Y. Look for options

I get1760 as one of the given options
Thanks Govind!!

Actually whenever we trade two things at par by incurring loss on one and a gain on the other, then, mathematically, the loss happens to be equal to the gain made. If the car of loss cost him $x, then the car of gain must have cost him $(5000 - x), such that

0.12 x = 0.08 (5000 - x)

or 0.2 x = 400

or x = 2000

A loss of 12 percent brings it down to [spoiler]$1760


D

NB: Such kind of questions can be cracked in many other quicker and successful ways.[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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by ankur.agrawal » Wed Mar 02, 2011 1:55 am
Wow Sanju. This method will save a lot of time.:)
sanju09 wrote:
gmatmachoman wrote:As I am always a fan of Sanju,i wish i can pen down my solution.

X +Y = 5000

0.88X +1.08Y= 5000 ( accounting for the loss in X & profit in Y)

Solve for X & Y

Now the stem asks for selling price ( tricky part)

SO calculate either of .88X or 1.08Y. Look for options

I get1760 as one of the given options
Thanks Govind!!

Actually whenever we trade two things at par by incurring loss on one and a gain on the other, then, mathematically, the loss happens to be equal to the gain made. If the car of loss cost him $x, then the car of gain must have cost him $(5000 - x), such that

0.12 x = 0.08 (5000 - x)

or 0.2 x = 400

or x = 2000

A loss of 12 percent brings it down to [spoiler]$1760


D

NB: Such kind of questions can be cracked in many other quicker and successful ways.[/spoiler]
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by BarryLi » Wed Mar 02, 2011 2:09 am
sanju your solution is clever. I definitely would do the question the same way gmatmachoman before reading your solution.
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