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.

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.