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GRE Math Challenge #138-In a certain store, computer X costs
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Updated on: 30 Jan 2019, 11:51
Updated on: 30 Jan 2019, 11:51
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Question Stats:
73% (01:16) correct
26% (03:03) wrong based on 178 sessions
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In a certain store, computer X costs 30 percent more than computer Y, and computer Y costs 30 percent more than computer Z.
Quantity A:
Quantity B:
1) Quantity A is greater.
2) Quantity B is greater.
3) Both Quantities are Equal
4) Cannot be determined
Quantity A |
Quantity B |
The cost of computer X minus the cost of computer Y |
The cost of computer Y minus the cost of computer Z |
Quantity A:
Quantity B:
1) Quantity A is greater.
2) Quantity B is greater.
3) Both Quantities are Equal
4) Cannot be determined
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: GRE Math Challenge #138
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16 May 2015, 14:37
16 May 2015, 14:37
2
sandy wrote:
In a certain store, computer X costs 30 percent more than computer Y, and computer Y costs 30 percent more than computer Z.
Quantity A: The cost of computer X minus the cost of computer Y.
Quantity B: The cost of computer Y minus the cost of computer Z.
Quantity A: The cost of computer X minus the cost of computer Y.
Quantity B: The cost of computer Y minus the cost of computer Z.
Let's assign some easy-to-work-with values.
Let computer Z cost $100
Computer Y costs 30 percent more than computer Z
30% of $100 = $30
So, computer Y costs $130
Vomputer X costs 30 percent more than computer Y
30% of $130 = $39
So, computer X costs $169
We get:
Quantity A: The cost of computer X minus the cost of computer Y = $169 - $130 = $39
Quantity B: The cost of computer Y minus the cost of computer Z = $130 - $100 = $30
Answer:
Show: ::
A
Cheers,
Brent
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: GRE Math Challenge #138-In a certain store, computer X costs
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24 Jan 2019, 06:42
24 Jan 2019, 06:42
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sandy wrote:
In a certain store, computer X costs 30 percent more than computer Y, and computer Y costs 30 percent more than computer Z.
Quantity A: The cost of computer X minus the cost of computer Y.
Quantity B: The cost of computer Y minus the cost of computer Z.
Quantity A: The cost of computer X minus the cost of computer Y.
Quantity B: The cost of computer Y minus the cost of computer Z.
Here's another approach:
Computer Y costs 30 percent more than computer Z.
Let z = cost of computer Z
So, cost of computer Y = z + (30% of z)
= z + 0.3z
= 1.3z
ASIDE
So, increasing a value by 30% is the same as multiplying that value by 1.3
Similarly, increasing a value by 20% is the same as multiplying that value by 1.2
And increasing a value by 78% is the same as multiplying that value by 1.78
etc
Computer X costs 30 percent more than computer Y
The cost of computer Y = 1.3z
So, cost of computer X = (1.3)(1.3z)
= 1.69z
SUMMARY:
Cost of computer X = 1.69z
Cost of computer Y = 1.3z
Cost of computer Z = z
We get:
QUANTITY A: 1.69z - 1.3z
QUANTITY B: 1.3z - z
Simplify to get:
QUANTITY A: 0.39z
QUANTITY B: 0.3z
Since z must be POSITIVE, we can safely divide both sides by z to get:
QUANTITY A: 0.39
QUANTITY B: 0.3
Answer: A
Cheers,
Brent
