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Tom, Jane, and Sue each purchased a new house. The average [#permalink]
 09 Oct 2009, 23:27
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17% (00:42) wrong based on 7 sessions
Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was $120,000. What was the median price of the three houses? (1) The price of Tom’s house was $110,000. (2) The price of Jane’s house was $120,000.
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reply2spg wrote: Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was $120,000. What was the median price of the three houses? (1) The price of Tom’s house was $110,000. (2) The price of Jane’s house was $120,000. We have three prices: a, b and c. (a+b+c)/3=120 The median price would be: the second biggest. a<=b<=c --> median price b. (1) One of the prices is 110, less than average of 120. It's possible 110 to be a or b price, so insufficient. (2) One of the prices is 120 equals to average. It must be the b price, as it's not possible this price to be lowest or highest because it's equals to the average, only 2 cases a<120<c or a=b=c=120 Answer B.
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good explanation ,thaank you very much
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kirankp wrote: Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was $120,000. What was the median price of the three houses?
(1) The price of Tom’s house was $110,000.
(2) The price of Jane’s house was $120,000. Did not you get B? If one of the house is equal to mean, then it is the median because other 2 houses (both) cannot be > 120,000 or < 120,000. The wrost case, is one is < 120,000 and the other is >120,000. So B is suff....
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Mean and one of the values out of 3 values are same, hence median has to be equal to mean.
Ans. is B.
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Bunuel wrote: We have three prices: a, b and c. (a+b+c)/3=120
The median price would be: the second biggest.
a<=b<=c --> median price b.
(1) One of the prices is 110, less than average of 120. It's possible 110 to be the a or b price, so insufficient.
(2) One of the prices is 120 equals to average. It must be the b price, as it's not possible this price to be lowest or highest because it's equals to the average, only 2 cases a<120<c or a=b=c=120
Answer B. if one of the price is 110, aka not avg 120, wouldn't it be "a"? Since there are 3 numbers, average is 120, anything less than average will be "a"? How can 110 possibly be b?
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catty2004 wrote: Bunuel wrote: We have three prices: a, b and c. (a+b+c)/3=120
The median price would be: the second biggest.
a<=b<=c --> median price b.
(1) One of the prices is 110, less than average of 120. It's possible 110 to be the a or b price, so insufficient.
(2) One of the prices is 120 equals to average. It must be the b price, as it's not possible this price to be lowest or highest because it's equals to the average, only 2 cases a<120<c or a=b=c=120
Answer B. if one of the price is 110, aka not avg 120, wouldn't it be "a"? Since there are 3 numbers, average is 120, anything less than average will be "a"? How can 110 possibly be b? Try to construct different scenarios, you'll see that it's not that hard: Sum is 120*3=360. 110+120+130=360; 100+ 110+150=360.
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Bunuel wrote: reply2spg wrote: Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was $120,000. What was the median price of the three houses? (1) The price of Tom’s house was $110,000. (2) The price of Jane’s house was $120,000. We have three prices: a, b and c. (a+b+c)/3=120 The median price would be: the second biggest. a<=b<=c --> median price b. (1) One of the prices is 110, less than average of 120. It's possible 110 to be a or b price, so insufficient. (2) One of the prices is 120 equals to average. It must be the b price, as it's not possible this price to be lowest or highest because it's equals to the average, only 2 cases a<120<c or a=b=c=120 Answer B. You have no idea how long I've been looking for a more well explained answer, thank you so much!!! Everyone else justs calculates the 360-250 and shows examples, this Issa much more intuitive when you pair it with a bell curve, thank you thank you thank uou
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kalpeshchopada7 wrote: Mean and one of the values out of 3 values are same, hence median has to be equal to mean.
Ans. is B. Is the above a rule general to all sets? Could someone answer this please
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By defintion, Median divides the distribution of values such that exactly half lie below the median and half above the median. example, given 2,5,9,50 (notice that to calculate median, the values must first be arranged in ascending order), the median is (5+9)/2=7. meaning below 7 like two values and above 7 lie two values. By contrast, Mean or average doesn't necessarily do that. It is affected by the magnitude of each value. if one value is extreme, the mean or average shifts towards that extremity. In the above example, the average is about 16. half of the numbers aren't less than this 16, there are three numbers less than 16. similarly, half aren't above 16, only one i.e. 50 is above 16. When you are told that there are only three values and one of them is actually the mean or the average, what's that really saying? if all three numbers are different, you know that the average has to fall somewhere between the smallest and the biggest number. and we are given that this middle number is 120,000 and also happens to be the mean. it doesn't matter now what the other two numbers are. this becomes an evenly spaced set of three numbers. the smaller the smallest number, the larger the largest number has to be to keep the average 120,000 constant. So as a rule, you can remember that for any evenly spaced set, the mean is always equal to the median. example, 2,4,6 or 10,20,30,40. eaakbari wrote: kalpeshchopada7 wrote: Mean and one of the values out of 3 values are same, hence median has to be equal to mean.
Ans. is B. Is the above a rule general to all sets? Could someone answer this please |
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koisun wrote: By defintion, Median divides the distribution of values such that exactly half lie below the median and half above the median. example, given 2,5,9,50 (notice that to calculate median, the values must first be arranged in ascending order), the median is (5+9)/2=7. meaning below 7 like two values and above 7 lie two values. By contrast, Mean or average doesn't necessarily do that. It is affected by the magnitude of each value. if one value is extreme, the mean or average shifts towards that extremity. In the above example, the average is about 16. half of the numbers aren't less than this 16, there are three numbers less than 16. similarly, half aren't above 16, only one i.e. 50 is above 16. When you are told that there are only three values and one of them is actually the mean or the average, what's that really saying? if all three numbers are different, you know that the average has to fall somewhere between the smallest and the biggest number. and we are given that this middle number is 120,000 and also happens to be the mean. it doesn't matter now what the other two numbers are. this becomes an evenly spaced set of three numbers. the smaller the smallest number, the larger the largest number has to be to keep the average 120,000 constant. So as a rule, you can remember that for any evenly spaced set, the mean is always equal to the median. example, 2,4,6 or 10,20,30,40. eaakbari wrote: kalpeshchopada7 wrote: Mean and one of the values out of 3 values are same, hence median has to be equal to mean.
Ans. is B. Is the above a rule general to all sets? Could someone answer this please Thanks for the explanation. So I can generalize that if a number in a set is equal to the mean, the set is an evenly spaced set and hence the mean = median. ??? Awaiting your views?
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Yes it is given that 120 is the mean. In statement 2 it says one of the numbers is 120. You can memorize it as a rule or test it on any 3 nos. The median has to be 120. Read bunuel's post too.
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