gmat prep ....weird answer ! help !

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 71
Joined: Sun May 18, 2008 11:04 pm
Thanked: 1 times

gmat prep ....weird answer ! help !

by charmaine » Sun Sep 21, 2008 10:26 pm
A used-car dealer sold one car at a profit of 25 percent of the dealer's purchase price for that car and sold another car at a loss of 20 percent of the dealer's purchase price for that car. If the dealer sold each car for $20,000, what was the dealer's total profit or loss, in dollars, for the two transactions combined?

(A) $1,000 profit
(B) $2,000 profit
(C) $1,000 loss
(D) $2,000 loss
(E) $3,334 loss


i keep gettin A but the OA is C
how come ?
thanks :)

Which of the following is equal to the value of 2^5+ 2^5 + 3^5 + 3^5 + 3^5 ?

(A) 5^6
(B) 13^5
(C) 2^6 + 3^6
(D) 2^7 + 3^8
(E) 4^5 + 9^5


the answer for this is C but i think its wrong ???opininons ??
Source: — Problem Solving |

Master | Next Rank: 500 Posts
Posts: 267
Joined: Wed Jul 16, 2008 12:20 am
Thanked: 4 times
Followed by:1 members

by Mani_mba » Sun Sep 21, 2008 11:22 pm
For your first Question,

Purchase price of the car that was sold at 25% profit = 16000
Purchase price of the car that was sold at 20% loss = 25000

Total purchase price = 41000 while sales sum to 40000

The seller incurred a loss of 1000$. So OA is right.

For the second question, it is easy and straight forward,

The question is equivalent to

2*2^5 + 3*3^5 = 2^6 + 3^6.

So option C is right.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

Re: gmat prep ....weird answer ! help !

by lunarpower » Tue Sep 23, 2008 2:31 am
charmaine wrote:A used-car dealer sold one car at a profit of 25 percent of the dealer's purchase price for that car and sold another car at a loss of 20 percent of the dealer's purchase price for that car. If the dealer sold each car for $20,000, what was the dealer's total profit or loss, in dollars, for the two transactions combined?

(A) $1,000 profit
(B) $2,000 profit
(C) $1,000 loss
(D) $2,000 loss
(E) $3,334 loss


i keep gettin A but the OA is C
how come ?
thanks :)
the reason you're getting answer (a) is because you're mis-reading the problem. you're treating the $20,000's as purchase prices, when they are in fact selling prices. in other words, you're increasing $20,000 by 25% to give a false "selling price" of $25,000, and then decreasing $20,000 by 20 percent to give a false "selling price" of $16,000.
you can't do that.
the $20,000's are the selling prices, so you have to work the problem backwards.

first car: let x = purchase price
25% greater than x = $20,000
1.25x = 20,000 (make flash cards for these "percentage multipliers" if you don't know them COLD -- e.g., increase by 25% = multiply by 1.25; decrease by 16% = multiply by 0.84; etc)
x = 20,000/1.25 = 16,000

second car: let x = purchase price
20% less than x = $20,000
0.80x = 20,000
x = 20,000/0.8 = 25,000

so the cars were purchased for 16,000 + 25,000 = $41,000, and sold for only $40,000.
that's a loss of 1 grand

there you go.
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

by lunarpower » Tue Sep 23, 2008 2:41 am
Mani_mba wrote:For the second question, it is easy and straight forward,

The question is equivalent to

2*2^5 + 3*3^5 = 2^6 + 3^6.

So option C is right.
watch it with throwing around words like "easy and straightforward"; obviously the original poster didn't think so.

we're all hearts and flowers here, man. we don't want to call things "easy" if posters don't find them easy.

this is a correct solution; if you don't understand why, then you can write the extra step:
2(2^5) + 3(3^5)
= (2^1)(2^5) + (3^1)(3^5)
add exponents when multiplying --> 2^6 + 3^6

you can also ESTIMATE, or CALCULATE DIRECTLY.

you should make flash cards with powers of 2 and powers of 3, at least up to 2^8 = 256 and 3^6 = 729. if you do so, then you could just add up 32 + 32 + 243 + 243 + 243 = 793, and then add up (c) = 64 + 729 = 793.
(a) has to be wrong because all powers of 5 have a units digit of 5. 793 doesn't end with a five, so there's no need to calculate this.
(b) is ridiculously huge.
(d) is ridiculously huge; if you try to calculate 3^8, you'll notice that it gets out of hand.
(e) is also ridiculously huge.
you can easily do all this within the allotted time, IF YOU START DOING IT RIGHT AWAY.
DO NOT DELIBERATE.

and remember, flash cards are your friends.
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron