Problem Solving

For the past n days, the arithmetic mean daily production of a company was 50 units. If today's production of 90 units raises the average to 55 units per day. What is the value of n?

Can someone explain this please?

Hi Pearson,

Here it goes.

The sum of daily productions for the past n days is 50*n.
Now today's production is 90. If we add this to the total sum and recalculate the average (including one more day that is today) we have an avg of 55 (as given in the question). So we can set up the equation as

(50n+90)/(n+1)=55

If you solve for n, the answer you get is 7.

Let me know if you understood this.

For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n ?

(A) 30
(B) 18
(C) 10
(D) 9
(E) 7

This is a weighted average/mixture question.

Average for the first n days = 50.
Average for today = 90.
Average for the MIXTURE of all the days = 55.

To determine the value of n, use ALLIGATION.

Step 1: Plot the 3 averages on a number line, with the average for the first n days and that for today on the ends and the average for the mixture in the middle.
50----------------55----------------90

Step 2: Calculate the distances between the averages.
50--------5-------55--------35------90

Step 3: Determine the ratio in the mixture.
The required ratio of 50-unit days to 90-unit days is equal to the RECIPROCAL of the distances in red.
(50-unit days) : (90-unit days) = 35:5 = 7:1.

Thus, for every 1 day with 90 units, there must be 7 days with 50 units.
Thus, today's production of 90 units requires n=7 days with 50 units.

The correct answer is E.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html