Nice challenge

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Nice challenge

by adam15 » Fri Dec 04, 2009 8:11 pm
Archie, Betty and Coach bought a radio. Archie paid the least amount of money, Coach paid twice as much as Archie, and Betty paid the most. Archie paid seventy dollars less than the difference between what Betty and Coach paid. If Archie had paid twice as much as he did pay, Betty and Coach would have each paid ten dollars less. Which of the following equals the amount Betty paid?


40

90

130

140

Insufficient information

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by Stuart@KaplanGMAT » Fri Dec 04, 2009 11:31 pm
adam15 wrote:Archie, Betty and Coach bought a radio. Archie paid the least amount of money, Coach paid twice as much as Archie, and Betty paid the most. Archie paid seventy dollars less than the difference between what Betty and Coach paid. If Archie had paid twice as much as he did pay, Betty and Coach would have each paid ten dollars less. Which of the following equals the amount Betty paid?


40

90

130

140

Insufficient information
The question seems a bit ambiguous, but I'm going to interpret it as the three of them are splitting the cost of one radio.

With that in mind, we have:

A+B+C = R

C = 2A

A = (B - C) - 70

A = 20 (this is really the key to the question and comes from the "If Archie had paid twice as much as he did pay, Betty and Coach would have each paid ten dollars less." part of the problem - see below for full translation)

and 0 < A < B < C

We know that A=20

Therefore:

C = 2(20) = 40

and:

20 = (B - 40) - 70

20 = B - 110

130 = B... done!

* * *

Back to that key deduction that A=20.

We know that:
If Archie had paid twice as much as he did pay, Betty and Coach would have each paid ten dollars less.
Well, if Betty and Coach each pay 10 dollars less, that's a total of 20 extra dollars that A is paying. We can derive the following equation:

(New amount Archie paying) - (Old amount Archie was paying) = 20

Archie normally pays A dollars, so if he's paying twice as much, he's paying 2A dollars. Accordingly:

New amount = 2A
Old amount = A

and:

2A - A = 20

A = 20
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by enniguy » Sat Dec 05, 2009 7:01 am
This is how I did it.
From sentence 1:
A + B + C = Radio

From sentence 2:
C=2A

From sentence 3:
A = B - C - 70

From sentence 4:
2A + (B-10) + (C-10) = Radio.

From 4 and 1:
2A + (B-10) + (C-10) = A + B + C

You get A = 20.
So, from sentence 2:
C = 40.
From sentence 3:
B = 130.

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by adam15 » Sat Dec 05, 2009 9:37 am
this is the right answer thank you.
I used backsolving .
by the writing the second equation.
as follow:
A=(B-70)/3
an check for the answer choice and check for a multiple of 3

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by Stuart@KaplanGMAT » Sat Dec 05, 2009 10:32 am
adam15 wrote:this is the right answer thank you.
I used backsolving .
by the writing the second equation.
as follow:
A=(B-70)/3
an check for the answer choice and check for a multiple of 3
Backsolving would normally be great for a question like this; however, choice E ("insufficient information") makes backsolving more dangerous, since that choice makes it possible that there's more than one solution.

So, for example, plugging in 130 works, but so might another choice (or another number not even listed).
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by adam15 » Sat Dec 05, 2009 4:09 pm
Thank you Stuart for your remark.