kushal.adhia wrote:The integer n is greater than 7. The average (arithmetic mean) of a group of n numbers is a. When 3 of the numbers are removed, the average of the remaining numbers is b. Which of the following expressions is the average of the 3 numbers removed?
A. (na) / (n-3)
B. {n(a-b)+3b} / 3
C. {n(a-b)+3b} / (n-3)
D. {n(a+b)+3b}/3
E. {n(a-b)} / 3
Try simulating the problem using plugging in: Let's say that n=10, so we have 10 integers. Let's further day that each of the integers is 10, so that the average of the set is a=10 as well.
Now take 3 "tens" out: you are left with 7 tens, and the average of the new set is also b=70/7=10.
So what have we found? If n=10, a=10 then b also equals 10. What did the question ask? what is the average of the 3 removed numbers? we removed 3 tens, so the average is, you guessed it, 10.
All we need to do now is plug in a=b=n=10 into the answer choices, and eliminate those which do not equal 10:
A. (10*10) / (10-3) = 100/7 - not equal to 10. ELIMINATE.
B. {10(10-10)+3*10} / 3 = (0+30)/3 = 10 - possible candidate.
C. {10(10-10)+3*10} / (10-3) = (0 + 30)/7 - not equal to 10. ELIMINATE
D. {10(10+10)+3*10}/3 = (10*20 + 30) / 3 = 230/3 not equal to 10. ELIMINATE
E. {10(10-10)} / 3 = 0/3 = 0 not equal to 10. ELIMINATE.
b is chosen by dint of being the only answer that matches your goal of 10 - since the other answer choices are eliminated, B must be the right answer, and you needn't worry about the algebra as to why.
I will say this - I don't normally like to plug in the same numbers for everything, since it tends to make more than one answer choice match your goal, forcing you to plug in a second set. However, the wealth of variables and complicated answer chocies call for the easiest possible plug in, so I went with 10 overall. If it worked (i.e eliminated all but one), great - if it didn't, at least I've elimianted 2-3 answer choices, leaving only the last 2 to plug in again for.
