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12 Nov 2021, 03:51
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Set A consists of 50 different positive integers. If the average (arithmetic mean) of the numbers in the set is 60, is any of the numbers in the set A less than 35?
(1) 10 of the numbers in the set A are greater than 85.
(2) 8 of the numbers in the set A are greater than 90.
(1) 10 of the numbers in the set A are greater than 85.
(2) 8 of the numbers in the set A are greater than 90.
12 Nov 2021, 03:51
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Official Solution:
Set A consists of 50 different positive integers. If the average (arithmetic mean) of the numbers in the set is 60, is any of the numbers in the set A less than 35?
First of all, notice that the average of 60 means that the sum of the numbers is \(50*60 = 3,000\).
(1) 10 of the numbers in the set A are greater than 85.
Let's see whether we can have a set with no number less than 35. In this case, since all numbers will be more than or equal to 35, we should try to minimize all of them for the sum not to exceed 3,000.
The minimum sum of the 10 numbers mentioned in this statement would be \(86+87+...+95=\frac{86+95}{2}*10=905\)
The remaining 40 numbers must total \(3,000-905=2,095\)
The minimum sum of the remaining 40 numbers, if there is no number less than 35, is \(35+36+...+74=\frac{35+74}{2}*40=2,180\), which is more that 2,095.
So, we cannot have a set with no number less than 35. Sufficient.
(2) 8 of the numbers in the set A are greater than 90.
Again, let's see whether we can have a set with no number less than 35.
The minimum sum of the 8 numbers mentioned in this statement would be \(91+92+...+98=\frac{91+98}{2}*8=756\)
The remaining 40 numbers must total \(3,000-756=2,244\)
The minimum sum of the remaining 42 numbers, if there is no number less than 35, is \(35+36+...+76=\frac{35+76}{2}*42=2,331\), which is more that 2,244.
So, we cannot have a set with no number less than 35. Sufficient.
Answer: D
Set A consists of 50 different positive integers. If the average (arithmetic mean) of the numbers in the set is 60, is any of the numbers in the set A less than 35?
First of all, notice that the average of 60 means that the sum of the numbers is \(50*60 = 3,000\).
(1) 10 of the numbers in the set A are greater than 85.
Let's see whether we can have a set with no number less than 35. In this case, since all numbers will be more than or equal to 35, we should try to minimize all of them for the sum not to exceed 3,000.
The minimum sum of the 10 numbers mentioned in this statement would be \(86+87+...+95=\frac{86+95}{2}*10=905\)
The remaining 40 numbers must total \(3,000-905=2,095\)
The minimum sum of the remaining 40 numbers, if there is no number less than 35, is \(35+36+...+74=\frac{35+74}{2}*40=2,180\), which is more that 2,095.
So, we cannot have a set with no number less than 35. Sufficient.
(2) 8 of the numbers in the set A are greater than 90.
Again, let's see whether we can have a set with no number less than 35.
The minimum sum of the 8 numbers mentioned in this statement would be \(91+92+...+98=\frac{91+98}{2}*8=756\)
The remaining 40 numbers must total \(3,000-756=2,244\)
The minimum sum of the remaining 42 numbers, if there is no number less than 35, is \(35+36+...+76=\frac{35+76}{2}*42=2,331\), which is more that 2,244.
So, we cannot have a set with no number less than 35. Sufficient.
Answer: D
15 Nov 2025, 14:26
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Bunuel
Cant we just adjust the terms, like 34 and 75 for maintaining the sum same and also having a set of numbers with less than 35. Please correct me if my understanding is wrong, I think it should be E.
Cant we just adjust the terms, like 34 and 75 for maintaining the sum same and also having a set of numbers with less than 35. Please correct me if my understanding is wrong, I think it should be E.
Bunuel
