GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 28 Jan 2020, 19:15

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

### Show Tags

07 Nov 2019, 03:04
00:00

Difficulty:

35% (medium)

Question Stats:

66% (01:47) correct 34% (01:39) wrong based on 53 sessions

### HideShow timer Statistics

A and B brought a radio each for the same price, $z. A and B sold their radio for$x and $y respectively. If x>y>z, A’s profit is what percent greater than B’s? (1) x – y = 20 (2) x – z = 40 Are You Up For the Challenge: 700 Level Questions _________________ Math Expert Joined: 02 Aug 2009 Posts: 8327 A and B brought a radio each for the same price,$z. A and B sold thei  [#permalink]

### Show Tags

07 Nov 2019, 21:14
Bunuel wrote:
A and B brought a radio each for the same price, $z. A and B sold their radio for$x and $y respectively. If x>y>z, A’s profit is what percent greater than B’s? (1) x – y = 20 (2) x – z = 40 Are You Up For the Challenge: 700 Level Questions A's profit = $$\frac{x-z}{z}$$, and B's profit = $$\frac{y-z}{z}$$ % change of A over B = $$\frac{A's.profit-B's.profit}{b"s.profit}*100...........\frac{\frac{x-z}{z}-\frac{y-z}{z}}{\frac{y-z}{z}}*100=\frac{\frac{x-y}{z}}{\frac{y-z}{z}}*100=\frac{x-y}{y-z}*100$$... Hence, we require values of x, y and z OR values of x-y and y-z We have some fractions for A’s and B’s profit. When we substitute it in the formula $$\frac{\frac{(x-z)}{z}-\frac{(y-z)}{z}}{(\frac{y-z}{z})}=\frac{\frac{(x-z-(y-z))}{z}}{\frac{((y-z)}{z)}}=\frac{(x-z-y+z)}{y-z}=\frac{(x-y)}{(y-z)}$$ Now we know x-y and y-z can be found from the two statements. Subtract statement I from ii $$(x-z)-(x-y)=40-20=20....y-z=20$$ Statement I and II give us these values when combined Answer = $$\frac{20}{20}*100=100%$$ y-z can be found by subtracting Statement I from statement II C _________________ Intern Joined: 15 Jun 2015 Posts: 2 Re: A and B brought a radio each for the same price,$z. A and B sold thei  [#permalink]

### Show Tags

25 Nov 2019, 22:17
chetan2u : I have a doubt, the stmt. 2 says "x-z = 40", I'm wondering how was it substituted into "y-z", please help me with this as I'm stuck at this point when I solved it myself.
Intern
Joined: 24 Sep 2019
Posts: 26

### Show Tags

29 Nov 2019, 09:57
1
pkky wrote:
chetan2u : I have a doubt, the stmt. 2 says "x-z = 40", I'm wondering how was it substituted into "y-z", please help me with this as I'm stuck at this point when I solved it myself.

Hi

We have some fractions for A’s and B’s profit. When we substitute it in the formula
$$\frac{\frac{(x-z)}{z}-\frac{(y-z)}{z}}{(\frac{y-z}{z})}=\frac{\frac{(x-z-(y-z))}{z}}{\frac{((y-z)}{z)}}=\frac{(x-z-y+z)}{y-z}=\frac{(x-y)}{(y-z)}$$
Now we know x-y and y-z can be found from the two statements.
Subtract statement I from ii
$$(x-z)-(x-y)=40-20=20....y-z=20$$
_________________