[GMAT math practice question]
The total price of books A, B, and C is $306. What is the median price of books A, B, and C?
1) The price of book A is $102.
2) The price of book B is $20 more than that of book C.
The total price of books A, B, and C is $306. What is the me
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- Max@Math Revolution
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Let a, b and c be the prices of books A, B and C, respectively.
Since we have 3 variables (a, b and c) and 1 equation, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
The equation given by the original condition is a + b + c = 306.
Conditions 1) & 2):
Since a = 102 and b = c+ 20, we have a + b + c = 102 + c + 20 + c = 2c + 122 = 306 or 2c = 184.
So, c = 92, b = 112 and a = 102.
Thus, the median is 102.
Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
a = 102 is the average price of the books since the total price is 306.
If b < 102, then c > 102, and the median is a = 102.
If b = 102, then c = 102, and the median is 102.
If b > 102, then c < 102, and the median is a = 102.
Thus, the median is 102 in all cases.
Condition 1) is sufficient.
When we have 3 data values, if one value is equal to the average value, then it is the median.
Condition 2)
If a = 102, b = 92 and c = 112, then the median price is 102.
If a = 86, b = 100 and c = 120, then the median price is 100.
Since it doesn't give us a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
Let a, b and c be the prices of books A, B and C, respectively.
Since we have 3 variables (a, b and c) and 1 equation, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
The equation given by the original condition is a + b + c = 306.
Conditions 1) & 2):
Since a = 102 and b = c+ 20, we have a + b + c = 102 + c + 20 + c = 2c + 122 = 306 or 2c = 184.
So, c = 92, b = 112 and a = 102.
Thus, the median is 102.
Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
a = 102 is the average price of the books since the total price is 306.
If b < 102, then c > 102, and the median is a = 102.
If b = 102, then c = 102, and the median is 102.
If b > 102, then c < 102, and the median is a = 102.
Thus, the median is 102 in all cases.
Condition 1) is sufficient.
When we have 3 data values, if one value is equal to the average value, then it is the median.
Condition 2)
If a = 102, b = 92 and c = 112, then the median price is 102.
If a = 86, b = 100 and c = 120, then the median price is 100.
Since it doesn't give us a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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