motion2020 wrote:
If the mean of set A {918,273,645,331; 918,273,645,320; 918,273,645,301; 918,273,645,421; x} is 918,273,645,321, what is the value of x?
(A) 918,273,645,221
(B) 918,273,645,231
(C) 918,273,645,232
(D) 918,273,645,241
(E) 918,273,645,242
Please do not use excell, or online calculator. This question is supposed to test not your calculator skills, GRE calculator will not fit this number BTW
918,273,645,331; 918,273,645,320; 918,273,645,301; 918,273,645,421; x
For simplicity, let us assume 918,273,645 as \(A\)
Now the numbers are A,331; A,320; A,301; A,421; x
Approach 1:Given; \(\frac{A,331 + A,320 + A,301 + A,421 + x}{5} = A,321\)
\(A,331 + A,320 + A,301 + A,421 + x = 5(A,321)\)
\(A,1373 +x = A,1605\)
\(x = A,1605 - A,1373\)
\(x = A,232\)
Approach 2: Using a Number line
Since, the mean is A,321 - it should lie in the centre
Calculate the difference between the numbers
A,301 = 20 to left
A,320 = 1 to left
A,331, = 10 to right
A,421 = 100 to right
Overall chance = 89 to right
To maintain the balance, the new number should be 89 to the left i.e. A,321 - A,089 = A,232
Hence, option C
Attachments
If the mean of set A.png [ 5.42 KiB | Viewed 1398 times ]