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Re: A company purchased at least 10 units of a certain product from each o [#permalink]
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Cost X = 100 + 20x10 + 5 (X-10)

Cost Y = 150 + 10x10 + 10 (Y-10)


Average = (Cost X + Cost Y)/ (X+Y) = 20
450 + 5X + 10Y = 20X + 20Y
3X + 2Y = 90
X +2/3*Y = 30
=> Y must be divisible by 3 = 15, 27, 30

Y= 15; X = 20
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A company purchased at least 10 units of a certain product from each o [#permalink]
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chetan2u wrote:
A company purchased at least 10 units of a certain product from each of Suppliers X and Y. Supplier X charged a $100 fixed cost, $20 per unit for each of the first 10 units, and $5 per unit for each unit purchased in excess of 10 units. Supplier Y charged a $150 fixed cost, $15 per unit for each of the first 10 units, and $10 per unit for each unit purchased in excess of 10 units. Including the fixed costs, the combined average cost for the units was $20 per unit.

Select a number of units purchased from Supplier X and a number of units purchased from Supplier Y that are jointly consistent with the given information. Make only two selections, one in each column.



 

We know that at least 10 units were bought from each supplier. 
Cost of first 10 units from X = 100 + 20*10 = 300
Cost of first 10 units from Y = 150 + 15*10 = 300

So on average cost of first 20 units was $30 per unit. Now additional units are bought at cheaper price of $5 per unit or $10 per unit. The average price of all these additional units would be between $5 and $10 say it is $n. But the overall average is $20 which is closer to the $30 price. Hence number of additional units bought must be less than 20.

Of the given options, there are only  2 possible options - 15 and 20 (additional units 15) or 15 and 22 (additional units 17). In all other cases, additional units will be more than 20 which is not acceptable. We will check for one of them. If that doesn't work, we know that the other one will. 

Try additional units 15. Then the weights are 15 and 20 which means that 

\(\frac{3}{4} = \frac{(30 - 20)}{(20 - n)}\)
\(x = $\frac{20}{3}\)

This is the average of $5 and $10.

\(\frac{w1}{w2} = \frac{(10 - 20/3)}{(20/3 - 5)} = 2:1\)

Hence 10 additional units were bought at $5 and 5 additional units were bought at $10. So 20 units were bought from X and 15 were bought from Y.

ANSWER: 20 from X and 15 from Y

Note that the solution looks long with multiple calculations but while solving it, you can do it directly on a single number line (using scale method). It happens much faster. 

Here is the video solution of this question: 
https://youtu.be/waWVH6zPj-4

For scale method of weighted averages, check: 
https://www.youtube.com/watch?v=_GOAU7moZ2Q
 

­
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A company purchased at least 10 units of a certain product from each o [#permalink]
­There can be an alternate method which i used : 

For supplier X : 100 (fixed) + 20 (10) + x = 100 + 200 + y = 300 + x
For Y : 150 (fixed) + 15 (10) +  y = 150 + 150 + y = 300 + y

Now we know average unit price is 20 $  i.e,  300 + x + 300 + y / no. of items = 20
This also gives an estimation that total qty. combining x and y should be greater than 30.
I started to look for combination of 35 and just verified where i can get 20$ unit price.

Prepared additional table for additional prices (beyond 10 qty.) below 



300 + (50) + 300 + (50) / 15 + 20 = 700/35 = 20

P.S : This method can be inefficient if table have more fields but considering the current question, using above it can be solved in lesser time.
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A company purchased at least 10 units of a certain product from each o [#permalink]
­3x+2y=40 where x and y are the excess units of X and Y
x = (40-2y)/3

Is x is an integer, 40-2y should be a multiple of 3..

y=5, x=10 ; total 15, 20
y=8, x= 8 ; total 18 ,26
y=11, x=6.. you get the pattern.. y=5,8,11,14.. x=10,8,6,4,2...



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A company purchased at least 10 units of a certain product from each o [#permalink]
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