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A car dealer sells only two models of cars. Alpha 9 and Beta 10. Was t

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Manager
Joined: 07 Jun 2018
Posts: 51
Location: United States
A car dealer sells only two models of cars. Alpha 9 and Beta 10. Was t  [#permalink]

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02 Feb 2019, 16:22
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Difficulty:

65% (hard)

Question Stats:

50% (02:05) correct 50% (01:55) wrong based on 42 sessions

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A car dealer sells only two models of cars. Alpha 9 and Beta 10. Was the revenue from the sales of Beta 10 model greater than 90 percent of the car dealer's total revenue?

1. The dealer sells Beta 10 model for ten times as much as he sells the Alpha 9 model

2. The dealer sells nine Beta 10 models for every two Alpha 9 model
Manager
Joined: 07 Aug 2017
Posts: 86
Location: India
GPA: 4
WE: Information Technology (Consulting)
Re: A car dealer sells only two models of cars. Alpha 9 and Beta 10. Was t  [#permalink]

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03 Feb 2019, 00:31
Jazzmin wrote:
A car dealer sells only two models of cars. Alpha 9 and Beta 10. Was the revenue from the sales of Beta 10 model greater than 90 percent of the car dealer's total revenue?

1. The dealer sells Beta 10 model for ten times as much as he sells the Alpha 9 model

2. The dealer sells nine Beta 10 models for every two Alpha 9 model

Consider, A and B as revenue from the sales of Alpha 9 and Beta 10. Therefore, total sales T = A + B

$$A_C$$ = Cost of Alpha 9
$$B_C$$ = Cost of Beta 10
$$A_N$$ = Number of Alpha 9 sold
$$B_N$$ = Number of Beta 10 sold

$$T= A+B = A_C*A_N+B_C*B_N$$

Statement 1:
$$B_C=10*A_C$$
$$A_C=\frac{B_C}{10}$$

Not sufficient, since we don't know how many individual cars were sold.

Statement 2:
$$\frac{A_N}{B_N}=\frac{2}{9}$$
$$A_N=\frac{2}{9}*B_N$$
Not sufficient, since the cost of individual models is not known.

Combining 1 & 2:
$$T = A_C*A_N+B_C*B_N$$
$$T= \frac{B_C}{10}*\frac{2}{9}B_N+B_C*B_N$$
$$T= B_C*B_N(1/45+1)$$
$$T= B*46/45$$
$$B= T*45/46=0.97*T$$
Thus, B > 0.90T

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Re: A car dealer sells only two models of cars. Alpha 9 and Beta 10. Was t   [#permalink] 03 Feb 2019, 00:31
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