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26 May 2017, 04:43
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A certain company sells only cars and trucks, and the average selling price of all the cars and trucks combined is \(18,000. If the average selling price of each car is\)15,500, did the company sell more trucks than cars?
(1) The number of sold cars is 100.
(2) The average price of trucks sold was $21,500.
(1) The number of sold cars is 100.
(2) The average price of trucks sold was $21,500.
26 May 2017, 04:45
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Official Solution:
Such as below, this question is a "2 by 2" question, which is the most common type of question that appears in the current GMAT.
Cars Trucks Number a b The average selling
price for each car c d
In the original condition, there are 4 variables (a, b, c, d) and 2 equations (that the average selling price of all the cars and trucks combined is \(18,000 and that the average selling price of each car is\)15,500). In order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer.
In fact, C is the answer. However, this is also a statistics question, one of the key questions, so you must apply "CMT 4(A: if you get C too easily, consider A or B)". Also, if one condition is a number and the other is a ratio, the ratio beats the number. So, if you take a look at 2), which is a ratio condition, you get \(15,500a+21,500b = 18,000 (a+b), and if you expand this, you get\)15,500a+21,500b = 18,000a+18,000b, and then \(21,500b-18,000b = 18,000a-15,500a, 3,500b = 2,500a, 7b = 5a. Hence, it is always\)b < a$, so yes, it is sufficient. The answer is B. This is a typical 5051-level problem.
Answer: B
Such as below, this question is a "2 by 2" question, which is the most common type of question that appears in the current GMAT.
Cars Trucks Number a b The average selling
price for each car c d
In the original condition, there are 4 variables (a, b, c, d) and 2 equations (that the average selling price of all the cars and trucks combined is \(18,000 and that the average selling price of each car is\)15,500). In order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer.
In fact, C is the answer. However, this is also a statistics question, one of the key questions, so you must apply "CMT 4(A: if you get C too easily, consider A or B)". Also, if one condition is a number and the other is a ratio, the ratio beats the number. So, if you take a look at 2), which is a ratio condition, you get \(15,500a+21,500b = 18,000 (a+b), and if you expand this, you get\)15,500a+21,500b = 18,000a+18,000b, and then \(21,500b-18,000b = 18,000a-15,500a, 3,500b = 2,500a, 7b = 5a. Hence, it is always\)b < a$, so yes, it is sufficient. The answer is B. This is a typical 5051-level problem.
Answer: B
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