A clock store sold a certain ....

[factor26]

A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time?
a $270
b $250
c $240
d $220
e $200

answer is [/spoiler]a[spoiler] can someone please help me solve? [/spoiler]

[vaibhavgupta]

A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time?
a $270
b $250
c $240
d $220
e $200

answer is [/spoiler]a[spoiler] can someone please help me solve? [/spoiler]

let the original price be 100 now he sold it off at 120 to collector. collector resold at 60. now it will be sold at 108 (80% profit).
40- when 100
100- 100/40*100= 250

now 20% add... 300. resale at half 150.. then 180/100*150= 270

Hence

A

[tpr-becky]

Take this question one step at a time through the algebra (Plug in the answers would work but be complicated)

First let P be the original price of the store sold it for a 20% profit. this is best calculated by muliplying P by 120%. On the GMAT, since you don't have a calcualtor it is best to keep everything in fractions so:

P(120/100) = 12p/10

Then the store bought it back and 1/2 that price so: (12p/10)(1/2) = 6p/10

then they sold it again at a profit of 80% over the last price = (6p/10)(180/100) = 6(18)(p)/100

Then it tells you that the difference between the buy back price and the old price was 100.
So p - [(6p)/10)] = 100 or 10p - 6p = 1000 or 4p = 1000 thus p = 250

Now you have to find the second sale price, put the 250 in for p. and you get 6(18)(250)/100.

You can cancel before you multiply: 3(18)(25)/5 = 3(18)5) = 270

[GMATGuruNY]

A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time?
a $270
b $250
c $240
d $220
e $200

answer is [/spoiler]a[spoiler] can someone please help me solve? [/spoiler]

We can guess and check.

Let the original cost = 100.
Sold at a 20% profit = 120.
Buy-back price = (1/2)*120 = 60.
Original cost - the buy-back price = 100-60 = 40.

Too small: the problem states that the difference between the original cost and the buy-back price is 100.
Since 100/40 = 5/2, all of the values determined above -- including the buy-back price -- must be multiplied by a factor of 5/2.

Thus:
Actual buy-back price = (5/2)*60 = 150.
Resold at an 80% profit = 150 + .8*(150) = 270.

The correct answer is A.

[factor26]

thank you all for your help. Would it be easier to plug in the answers choices to find the answer?

Question: how much was the clock sold for the second time?

A - 270 -- second time the goods sold at an 80% premium over cost ... thus 1.80(x)=270, x or buy back price here is $150.

orig cost = buy back cost + $100
orig cost = 150 + 100 = $250

orig sale price = 1.20 (250)
orig sale price = $300.00

every parameter is satisfied here thus the answer is a .... seems like a little overkill here though ... thanks guys for making it easier!!!