A company sells a container of mixed nuts that is 50% peanuts and another container that is 80% peanuts. How many cups of each mixture would be needed to make 105 cups that is 60% peanuts? Help me to understand how to solve problems like these, please.

Let n be the amount of 60% peanuts. Then the amount of 12% peanuts is 12-n. So:
.6n+.12(12-n)=.4(12)

.48n+1.44=4.8
.48n=3.36
n=7 cups 60% peanuts
12-n=5 cups 12% peanuts. ☺☺☺☺

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Cetacea Cetacea · Answer 2 · 2022-01-11T12:09:13+01:00

The number of 50% peanut cup = 70.

The number of 80% peanut cup = 35.

Let the number of 50% peanut cup = x

The number of 80% peanut cup = y.

So, total number of cups: x + y = 105 or y = 105 - x ......(1)

Total amount of peanut is:

50% of x + 80% of y = 60% of 105

[tex]0.50x+0.80y=0.60(105)\\0.50x+0.80y=63[/tex]

Plug y = 105 - x and solve for x.

[tex]0.50x+0.80y=63\\0.50x+0.80(105-x)=63\\0.50x+84-0.80x=63\\-0.30x=-21\\x=70[/tex]

So, y = 105 - 70 = 35.

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