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

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

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New post 02 Feb 2019, 16:22
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A
B
C
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E

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
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Re: A car dealer sells only two models of cars. Alpha 9 and Beta 10. Was t  [#permalink]

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New post 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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