Mixtures And Alligations

since you can take any value forget about 5$ and get only 6 and 6.8
(you can only get 6.5 if you get one value bigger and one smaller)
so you can take 0% of 5 and x6 + y6.8, with x + y = 100%
solution is 3/5 and 5/8

x = number of $5 items
y = number of $6 items
z = number of $6.8 items

Sum of number of items * price / number of items = average price
(5x + 6y + 6.8z)/(x + y + z) = 6.5

Multiply both sides by 10 and (x+y+z)

50x + 60y + 68z = 65x + 65y + 65z
3z = 15x + 5y (EQ1)

For EQ1 to work z > x+y

We then examine the constants (3 , 15 and 5)
The common multiples are 15, 30 , 45 etc

Lets start with the 1st multiple:
if leftside of equation = 15, right side is not satisfied (we get either 0 for x and y)

Lets start with the second multiple:
if leftside of equation = 30 (z=10), right side can be satisfied if x=1 and y =2 or viceversa

guerrero wrote:The cost prices of 3 kinds of items are $5, $6 and $6.80, respectively. In what proportion should they be mixed so that the price of the mixture is $6.50 per bag?

a)1:3:10
b)3:6:25
c)3:6:15
d)1:3:25

More than one answer choice works.

Answer choice A: 1:3:10
Total price of the mixture = 1(5) + 3(6) + 10(6.8) = 91.
Number of items in the mixture = 1+3+10 = 14.
Average price = total/number = 91/14 = 6.5.

Answer choice B: 3:6:25
Total price of the mixture = 3(5) + 6(6) + 25(6.8) = 15 + 36 + 170 = 221.
Number of items in the mixture = 3+6+25 = 34.
Average price = total/number = 221/34 = 13/2 = 6.5.

What is the source of this question?

GMATGuruNY wrote:

guerrero wrote:The cost prices of 3 kinds of items are $5, $6 and $6.80, respectively. In what proportion COULD they be mixed so that the price of the mixture is $6.50 per bag?

a)1:3:10
b)3:6:25
c)3:6:15
d)1:3:25

We can plug in the answers, which represent the ratio of the 3 items.

Answer choice A: 1:3:10
Total price of the mixture = 1(5) + 3(6) + 10(6.8) = 82.
Number of items in the mixture = 1+3+10 = 14.
Average price = total/number = 82/14 = 41/7.
Eliminate A.

Answer choice B: 3:6:25
Total price of the mixture = 3(5) + 6(6) + 25(6.8) = 15 + 36 + 170 = 221.
Number of items in the mixture = 3+6+25 = 34.
Average price = total/number = 221/34 = 13/2 = 6.5.
Success!

The correct answer is B.

I think there is an error in the calculation of choice a - see below.
Total price of the mixture = 1(5) + 3(6) + 10(6.8) = 82.

theCEO wrote: I think there is an error in the calculation of choice a - see below.

Thanks for the heads up. I've amended the post.
A and B each offer an acceptable ratio.

One issue here is that an unlimited number of ratios will work.
Combining items costing $5, $6, and $6.80 to form a mixture that costs $6.50 implies the following:
5x + 6y + 6.8z = 6.5(x+y+z)
50x + 60y + 68z = 65x + 65y + 65z
3z = 15x + 5y.

This equation is satisfied by the ratio in A (x=1, y=3, z=10):
3(10) = 15(1) + 5(3)
30 = 30.

This equation is satisfied by the ratio in B (x=3, y=6, z=25):
3(25) = 15(3) + 5(6)
75 = 75.

An infinite number of combinations of x, y and z will satisfy 3z = 15x + 5y.