souvik101990 wrote:
List M consists of 50 decimals, each of which has a value between 1 and 10 and has two non-zero digits after the decimal place (e.g. 5.68 could be a number in List M). The sum of the 50 decimals is S. The truncated sum of the 50 decimals, T, is defined as follows. Each decimal in List M is rounded down to the nearest integer (e.g. 5.68 would be rounded down to 5); T is the sum of the resulting integers. If S - T is x percent of T, which of the following is a possible value of x?
I. 2%
II. 34%
III. 99%
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III
Let's examine the
EXTREME CASESS - T = x percent of T
So, S - T = (x/100)T
Divide both sides by to get: (S - T)/T = x/100
Multiply both sides by 100 to get:
x = 100(S - T)/T First, we we'll MINIMIZE the value of x by minimizing the value of S - T and maximizing the value of T.
This occurs when list M = {9.11, 9.11, 9.11, 9.11, 9.11, 9.11, . . . . .9.11, 9.11}
So, S = (50)(9.11)
And T = (50)(9)
So, S - T = (50)(9.11) - (50)(9)
= (50)(9.11 - 9)
= (50)(0.11)
Plug these values into the
above equation to get x =100(50)(0.11)/(50)(9)
= 100(0.11)/9
= 11/9
≈1.2222...
So, the MINIMUM value of x is approximately 1.22%First, we we'll MAXIMIZE the value of x by maximizing the value of S - T and minimizing the value of T.
This occurs when list M = {1.99, 1.99, 1.99, 1.99, 1.99, . . . 1.99, 1.99}
So, S = (50)(1.99)
And T = (50)(1)
So, S - T = (50)(1.99) - (50)(1)
= (50)(1.99 - 1)
= (50)(0.99)
Plug these values into the
above equation to get x =100(50)(0.99)/(50)(1)
= 100(0.99)/1
= 99
So, the MAXIMUM value of x is 99%Combine the results to get:
1.22% < x ≤
99%All three values (2%, 34% and 99%) fall within this range of x-values.
Answer: E
Cheers,
Brent
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