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A used-car dealer sold one car at ...

Expert replies
Source: — Problem Solving |

by MBA.Aspirant » Tue Jun 14, 2011 7:49 pm
1)

He sold a car for 20,000 with a profit of 25%

so 1.25 x = 20,000

x (purchase price) = 16,000 and profit = 20,000- 16,000 = 4000

2nd car sold for 20,000 with a loss of 20%

so 0.8 x = 20000

x = 25000 and loss = 25000 - 20000 = 5000 loss

Total profit/loss = 4000 profit - 5000 loss = 1000 loss

Answer is C
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by MBA.Aspirant » Tue Jun 14, 2011 7:55 pm
2)

18 applicants choose 4

18!/4! 14! = 3060

Answer is E
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by EddieU » Tue Jun 14, 2011 8:15 pm
is the answer for the third question (marked as 4) 2?
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by MBA.Aspirant » Tue Jun 14, 2011 8:30 pm
3)

rate of 6 machines = 1/12

rate of 1 machine = 1/12/6= 1/72


to find out how many machines we need to do the same job in 8 days

1/72 * no. of machines * 8 = 1

no. of machines = 72/8 = 9

so extra machines = 9 -6 = 3 machines

Answer is B

This is an article explaining various ways to solve the same problem: https://www.beatthegmat.com/six-machines ... 52992.html
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by qduong » Wed Jun 15, 2011 7:53 pm
MBA.Aspirant wrote:2)

18 applicants choose 4

18!/4! 14! = 3060

Answer is E
Is there another way to solve this problem because I assume when we use " ! " we really need a calculator ? Thank you.
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by qduong » Thu Jun 16, 2011 10:06 pm
ssiva wrote:The formula is (n!)/[(n-r)! * r!] for a combination problem

18!/[(14!)(4!)] = (18*17*16*15)/(1*2*3*4) = 3060
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