Question 1207335: Hi
Mark bought a number of pens at an average cost of $2.50 He bought another pen for $8.50 and the average cost of the pens became $4.50. How many pens did he buy altogether.
Thanks
Found 3 solutions by mananth, greenestamps, josgarithmetic:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! Mark bought a number of pens at an average cost of $2.50 He bought another pen for $8.50 and the average cost of the pens became $4.50. How many pens did he buy altogether.
Let Mark buy x pens
Average cost of pen $2.50
Total paid by him for pens 2.50x
Another pen he bought for $8.50
Total cost of all pens = 2.50x +8.50
The average cost now became $4.50
Average cost of all pens = total purchase cost /Total number of pens
4.50 = (2.50x+8.50)/(x+1)
4.50*(x+1)= 2.50x +8.50
4.50x +4.50 = 2.50x +8.50
4.50x-2.50x = 8.50-4.50
2x = 4
x=2
Altogether he bought 3 pens.
Check
2*2.50 +8.50=13.50
number of pens 3
13.50/3 = 4.50
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
Here is a very unorthodox method for solving a problem like this.... A bit difficult to learn; but if you can learn it, it is much faster than the formal algebraic solution method. Of course, if a formal algebraic solution is needed, this won't help you.
We can treat this as a "mixture" problem -- we are "mixing" pens that cost $2.50 each with pens that cost $8.50 each to get a "mixture" of pens that cost an average of $4.50 each.
For an unorthodox method for solving that problem, look at the three numbers 2.50, 4.50, and 8.50 (on a number line, if it helps) and observe/calculate that 4.50 is 1/3 of the way from 2.50 to 8.50 (from 2.50 to 8.50 is a difference of 6; from 2.50 to 4.50 is a difference of 2; 2/6 = 1/3).
That means 1/3 of the pens were the more expensive ones. Then, since there was one $8.50 pen, the total number of pens was 3.
ANSWER: 3 pens (1 at $8.50 and 2 at $2.50 each)
CHECK:
$8.50 + 2($2.50) = $13.50
3($4.50) = $13.50
Answer by josgarithmetic(39616)  (Show Source):
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