A restaurant has one type of milk that has 2% fat and another that has 7% fat. How many quarts of each type does the restaurant need to make 40 quarts of a milk mixture that is 5% fat?

let the no. of quarts required of 2% fat milk be "x" and no. of quarts required of 7% fat milk be "y"

as per problem, Total no. of quarts should be 40, so, x+y=40.....(1)

Now, the total percentage of fat should be 5% so, 5% of 40 = 2% of x + 7 % of y 5*40/100 =2*x/100+7*y/100 2 = 2x/100+7y/100 or, 2x+7y =200 or,2x+7y=200...(2)

now from equation (1), x+y=40 x=40-y

plugging the value of x in (2), we get 2(40-y)+7y=200 80-2y+7y=200 5y=120 y=24

and x = 40-24 = 16

so, 16 quarts of 2% fat milk and 24 quarts of 7% milk is required.

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