I am bogged down by the ask of the question. how to do it within 2 mins?
List T consist of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum of the 30 decimals, E, is defined as follows. Each decimal in T whose tenths digit is even is rounded up to the nearest integer, and each decimal in T whose tenths digits is odd is rounded down to the nearest integer. If 1/3 of the decimals in T have a tenths digit that is even, which of the following is a possible value of E – S ?
I. -16
II. 6
III. 10
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
Hi Kukrejaria,
https://crackverbal.com/forum/threads/questions/ps-og-q218/
You can use the above the link, where this question has been already discussed.
Besides the best way to solve this question is Plugging in,
Whenever there is a bigger range, try to pick a smaller range and solve the question.
Here,
Let’s say instead of 30 decimals, Lets work for 3 decimals.
Let say T consists of 3 decimals,
In third of decimals are even (tenths place is even), so one is even and two are odd(means tenths position is odd)
T = {1.1, 1.1, 1.2}
So sum, S = 3.5
It’s about finding the Estimated sum, then S value Maximum and S value minimum.
For Estimated Sum(E),
So say for an example say if,
Decimal value is 1.2 is rounded to 2. If it is 1.1 it is rounded to 1.
So therefore, here E = 4.
Since the question is asking for all possible value. Let find out minimum and maximum values.
S max = {1.9 + 1.9 + 1.8} = 5.6 making the tenth position as max as possible,
S min = {1.1 + 1.001 + 1.002} = 3.1(approximate) making the tenth position as min as possible,( minimum value for an even and odd in tenths place )
E-S min = 4 – 3.1 = 0.9
E-S max = 4 – 5.6 = -1.6
This is for 3 values, then for ten values it would be range, 9 to 16.
Thus, the min/max of E is -16 and 6, so I , II apply.
III statement is beyond the range.
So answer is C. I and II only.
Hope this helps
