Anaira Mitch wrote:A set consists of n consecutive integers in which the smallest term is 1. What is the value of n?
(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.
(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.
Let's take each statement one by one.
(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.
Say the number removed is x.
Thus, x + sum of the remaining (n - 1) numbers = Sum of n numbers
=> x + 15(x - 1) = n(n + 1)/2
2x = n^2 - 29n +30 ---(1)
Insufficient.
(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.
Say the number removed is y.
As with Statement 1, similarly, we have
=> y + 16(x - 1) = n(n + 1)/2
2y = n^2 - 31n +32 ---(2)
Insufficient.
(1) and (2) together
Here, if we can find a relationship between x and y, we will get a solution.
Let's understand an average of consecutive numbers.
Say the consecutive numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 10, and 11. The average of these 11 numbers = (1 + 11)/2 = 6.
Case 1: Say we removed 11, the greatest number, then the average = (66 - 11)/10 = 5.5. Average reduced by 0.5.
Case 2: Say we removed 1, the smallest number, then the average = (66 - 1)/10 = 6.5. Average increased by 0.5.
Maximum possible effect on average upon removing any numbers = 0.5 + 0.5 = 1.
Since as per statement 1, the new average is 15 and as per statement 2, the new average is 16, the effect on average = 16 - 15 = 1, Maximum possible change. Thus, one of the numbers removed must be the greatest among them, i.e., n and the other must be the smallest, i.e., 1.
Or, x = n and y = 1.
From eqn 2, we have
2y = n^2 - 31n +32
2*1 = n^2 - 31n +32
n^2 - 31n + 30 = 0
=> n = 30. Sufficient.
The correct answer:
C
Hope this helps!
-Jay
_________________
Manhattan Review Test Prep
Locations:
Manhattan Review Himayatnagar |
Free GMAT Practice Test |
GRE Prep Jayanagar |
Madhapur GRE Coaching | and many more...
Schedule your free consultation with an experienced GMAT Prep Advisor!
Click here.
