12 Days of Christmas GMAT Competition - Day 12: Set A consists of 50

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
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Correct answer: D

There are total 50 distinct positive integers.

Average=60
Then positive offset from the mean should equal negative offset from mean.

Statement 1: 10 of the numbers in the set A are greater than 85.

Maximum negative offset that numbers from 35 to 60 can have:
1+2+....25= 325

Let's assume that 10 numbers greater than 85 are 86......95
Minimum positive offset of these 10 numbers: (26+27+...35) = 305

We still have 15 numbers left that can create a minimum offset of: 0+1+...14 (considering numbers are 60,61....)= 105
Minimum positive offset from mean= 305+105=410

Maximum negative offset: 315

Thus there has to be number(s) less than 35 to balance the positive offset from mean.

Thus statement 1 is sufficient.


Statement 2: 8 of the numbers in the set A are greater than 90.

Maximum negative offset that numbers from 35 to 60 can have:
1+2+....25= 325

Minimum positive offset of these 8 numbers: 31+32...38= 276

We still have 17 numbers left that can create a minimum offset of: 0+1+...16 (considering numbers are 60,61....,76)= 136
Minimum positive offset from mean= 276+136= 412

Maximum negative offset: 315

Thus there has to be number(s) less than 35 to balance the positive offset from mean.

Statement 2 is also sufficient.

Thus D

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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.

Solution : lets understand and analyze stem first :
50 distinct integers positive Arithmetic mean = 60 so sum of 50 integers = 3000 , we have to find if any number < 35 y/n



statement 1: 10 numbers in set > 85

SUM of 10 numbers > 86 + 87 + ... 95 we can say > 86 * 10 +( 0+ 1+2+ ...9)
which means sum of 10 numbers > 860 + (10*9/2) which means >905

sum of 40 numbers + sum 10 nos = sum of 50 nos

sum of 10 nos = sum of 50 - sum of 40 nos

3000 - sum of 40 nos > 905

sum of 40 nos < 2095

now lets take minimum sum of 40 nos possible when we have all not less than 35

that sum of 40 nos will be 35 , 36 ...... 74 so this sum will be 35 *40 + (0+1+... 39)= 1400+ 780 = 2180

which is not possible as 2180 > 2095

so we cannot have all nos not less than 35

so statement 1 is sufficient as we have have definite yes

statement 2 :

8 of the numbers in the set A are greater than 90.

sum of 8 nos > 91+ 92 .... 98 = 91* 8+ (0+1....+7) = 728+ 28 = 756


so sum of 42 nos < 2244

so lets see what can be the minimum sum of 42 nos all of them not less than 35

35 +..... 76 = 35 *42 + (0+1+.... 41) = 1470 + 41*42/2 = 1470 +861 = 2331



so 2331 is greater than 2244 which means our supposition is impossible


so we cannot have all nos not less than 35

so statement 2 is sufficient as we have have definite yes

so the answer is d ) each statement alone is sufficienet to answer the question

Given: Set A consists of 50 different positive integers.
Asked: 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?
Sum of all 50 numbers in Set A = 50*60 = 3000

(1) 10 of the numbers in the set A are greater than 85.
Sum of the 10 numbers > 86+87+88+89+90+91+92+93+94+95 = 905
Sum of remaining 40 numbers < 3000 - 905 = 2095
Average of remaining 40 numbers < 2095/40 = 52.375
If we take 20 consecutive numbers above 52 and 20 consecutive numbers below 52. In this case smallest number will be the largest possible.
A = {32,33,34,.....,51,52,53,......72,86,87,......95}
As we see there is a number less than 35 in set A.
SUFFICIENT

(2) 8 of the numbers in the set A are greater than 90.
Sum of the 8 numbers > 91+92+93+94+95+96+97+98 = 756
Sum of remaining 42 numbers < 3000 - 756 = 2244
Average of remaining 42 numbers < 2244/42 = 53.43
If we take 21 consecutive numbers above 53 and 21 consecutive numbers below 53. In this case smallest number will be the largest possible.
A = {32,33,34,.....,51,52,53,54,55......74,91,92,...98}
As we see there is a number less than 35 in set A.
SUFFICIENT

IMO 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?

Given, sum of 50 different positive numbers= 50*60= 3000

Stat1: 10 of the numbers in the set A are greater than 85.
So, minimum sum of numbers greater than 85 will be 86+87+...+95 = 10/2 * (86+95) = 5*181 = 905
Now, sum of 40 different positive numbers= 3000- 905= 2095
So, minimum sum of 40 different positive numbers 35+36...+74 = 40/2 * (35+74) = 20*109 = 2180

But, we have sum available for 40 different positive numbers = 2095, so, there would be number or numbers less than 35. Sufficient.

Stat2: 8 of the numbers in the set A are greater than 90.
So, minimum sum of numbers greater than 90 will be 91+92+...+98 = 8/2 * (91+98) = 4*189 = 756
Now, sum of 42 different positive numbers= 3000- 756= 2244
So, minimum sum of 42 different positive numbers 35+36...+76 = 42/2 * (35+76) = 21*111 = 2331

But, we have sum available for 42 different positive numbers = 2244, so, there would be number or numbers less than 35. Sufficient.

So, I think 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?

Sum =60*50 = 3000

(1) 10 of the numbers in the set A are greater than 85.
If 10 no are more than 85 then min sum =850
Bal = 2150
For no to be above 35 min no to start from 35,36,37......75
Sum = (35+75)/2*40 = 2200 >2150
Hence some of the no has to be under 35
Sufficient

(2) 8 of the numbers in the set A are greater than 90.
Same condition applies
Sufficient
IMO 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?

(1) 10 of the numbers in the set A are greater than 85.
Average: 60, 10 number greater than 85, means there will be 10 distinct number lesser than 45.

As numbers are distinct, the answer is yes - Sufficient

(2) 8 of the numbers in the set A are greater than 90.

means will be at least 8 number lesser than 30 - Sufficient

Correct 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?

Solution -
50*60 = 3000 which is the sum of all the different numbers in the set.

(1) 10 of the numbers in the set A are greater than 85.
Sum of 20 numbers immediately after 85 is 905.
3000 - 905 = 2095
Since the numbers are distinct the sum of remaining 40 can be calculated using Arithmetic Progression formula.
Turns out, some of the numbers are less that 35.
Statement (1) alone is sufficient.

(2) 8 of the numbers in the set A are greater than 90.
Sum of 8 numbers immediately after 90 is 756.
3000 - 756 = 2244
Using Arithmetic Progress formula, it seems that remaining number can be less than 34.
Statement (2) alone is also sufficient.

Answer should be D

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