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17 Jul 2010, 12:41
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At a certain supplier, a machine of type A costs $20,000 and a machine of type B costs$50,000. Each machine can be purchased by making a 20 percent down payment and repaying the remainder of the cost and the finance charges over a period of time.If the finance charges= 40 percent of the rremainderof the cost, how much less would 2 machines of type A cost than 1 machine of type B?

A. $10,000 B.$11,200
C. $12,000 D.$12,800
E. $13,200 [Reveal] Spoiler: OA Last edited by Bunuel on 30 Mar 2012, 10:39, edited 1 time in total. Edited the question Math Expert Joined: 02 Sep 2009 Posts: 35220 Followers: 6615 Kudos [?]: 85238 [2] , given: 10231 Re: Problem Solving [#permalink] ### Show Tags 17 Jul 2010, 16:50 2 This post received KUDOS Expert's post EnterMatrix wrote: At a certain supplier, a machine of type A costs$20,000 and a machine of type B costs $50,000. Each machine can be purchased by making a 20 percent down payment and repaying the remainder of the cost and the finance charges over a period of time.If the finance charges= 40 percent of the rremainderof the cost, how much less would 2 machines of type A cost than 1 machine of type B? a.$10,000
b. $11,200 c.$12,000
d. $12,800 e.$13,200

1 machine of type B will cost: 20% down payment of 50,000 = 10,000 plus remaining sum (50,000-10,000=40,000) with 40% of finance charges 40,000*1.4=56,000 --> 10,000+56,000=66,000;

2 machine of type A will cost: 20% down payment of 2*20,000 = 8,000 plus remaining sum (40,000-8,000=32,000) with 40% of finance charges 32,000*1.4=44,800 --> 8,000+44,800=52,800;

Difference = 66,000 - 52,800 = 13,200.

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Re: Problem Solving [#permalink]

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17 Jul 2010, 18:10
i got the same answer E as bunuel did...what's the official answer?
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Re: Problem Solving [#permalink]

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18 Jul 2010, 02:20
OA is E...was just considering a 40% of the remaining cost instead of the 1.4 factor..so was stuck!! Thanks...
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Re: Tough GMAT prep PS [#permalink]

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21 Feb 2011, 05:43
2 A type machines cost: 40000.
20% down payment: 40000/5 = 8000
Remainder: 40000-8000=32000
Financial charges on remainder=32000*40/100=12800
Total money spent for 2 type A's = 12800+8000+32000=52800

1 B type machine cost: 50000
20% down payment: 50000/5=10000
Remainder: 50000-10000=40000
Financial charges on remainder=40000*40/100=16000
Total money spent for 1 type B = 16000+10000+40000=66000

Difference: 66000-52800=13200

Ans: "E"
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Re: Tough GMAT prep PS [#permalink]

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21 Feb 2011, 19:09
mydreambschool wrote:
Hi,

Can someone please explain tough PS question from gmat prep.

Please find screenshot attached in the word doc file.

Many thanks.

The cost of 2 machines of type A is 40,000 while the cost of one machine of type B is 50,000.
The down payment and rate of interest are the same for both the cases. So all expenses will be same for both the cases except for the extra 10,000 to be paid in case of machine B and extra interest calculated as 40% of 8000 = 3200 (because out of 10,000, 20% down payment is 2000 and rest 8000 is the loaned amount)
So we need to pay a total of 13,200 extra in case of machine B.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 21 Mar 2010 Posts: 314 Followers: 5 Kudos [?]: 30 [0], given: 33 Re: Problem Solving [#permalink] ### Show Tags 22 Feb 2011, 12:58 E as well. Although, is it ok to consider total cost at the end of payment period when there is no specific mention abt it? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6960 Location: Pune, India Followers: 2024 Kudos [?]: 12704 [0], given: 221 Re: Problem Solving [#permalink] ### Show Tags 22 Feb 2011, 14:24 mbafall2011 wrote: E as well. Although, is it ok to consider total cost at the end of payment period when there is no specific mention abt it? Yes, it will be the cost at the end of the period. The question stem says "how much less would 2 machines of type A cost than 1 machine of type B?" Consider this: I want to buy a car - either a Camry or an Accord Camry down payment 20%, 6% interest on rest Accord down payment 15%, 6% interest on rest price of Camry $$, Price of Accord$$$ etc etc....
How much less would an Accord cost as compared to Camry?

It has to be the overall cost.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 29 Nov 2012 Posts: 900 Followers: 14 Kudos [?]: 947 [0], given: 543 Re: At a certain supplier, a machine of type A costs$20,000 and [#permalink]

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27 Jun 2013, 01:27
Why do we multiply by 1.4 and not .4.. Please explain
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17 Jul 2013, 00:59
Down payment for A 20000 * 20/100 = 4000

Remaining is 20000- 4000 = 16000

finance charge for A is 40/100 * 16000 = 6400

Total cost is ( 16000 + 4000 + 6400) = 26400

Since we have 2 machines total cost is 26400 * 2 = 52800

Now for B

Down payment is 50000 * 20/100 = 10000

Remainder is (50000 - 10000) 40000

Finance is on the remainder so 40000 * 40/100 = 16000

Total cost for B is ( 40000 + 10000 + 16000) = 66000

66000 - 52800 = 13200

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07 Jul 2015, 18:31
Cost of Machine = Down Payment + Remainder + Remainder*Finance Percent = Machine Cost + Remainder*Finance Percent
A=2E4+.8*2E4*.4 = 1.32(2E4)
B=5E4 + .8*5E4*.4 = 1.32(5E4)
B-2A=13200

Once you model the Cost of the Machine, the problem is just arithmetic.
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Re: At a certain supplier, a machine of type A costs $20,000 and [#permalink] ### Show Tags 09 Jul 2016, 02:22 EnterMatrix wrote: At a certain supplier, a machine of type A costs$20,000 and a machine of type B costs $50,000. Each machine can be purchased by making a 20 percent down payment and repaying the remainder of the cost and the finance charges over a period of time.If the finance charges= 40 percent of the remainder of the cost, how much less would 2 machines of type A cost than 1 machine of type B? A.$10,000
B. $11,200 C.$12,000
D. $12,800 E.$13,200

Okay..I tried to reduce the amount of calculations..here's me take..
After all..how much do we have to pay?

20% down payment done..
we're left with 80%..on which 40% interest is accrued..in other words..112%

Total amount to be paid..in each case..is 112% + 20% = 132% of the actual value

So..
1.32(50000 - 2*20000) = 13200 (E).

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