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A grocer wants to mix two kinds of coffee. One kind sells for \$1.45 per pound, and other sells for \$2.20 per pound. He wants to mix a total of 21 pounds and sell it for \$1.70Per pound. How many pounds of each kind should he use in the new mix? (ROund off the answers to the  nearest hundredth)

Detailed Solution:

Given that there are two kinds of coffee.

Let us say that  “X pounds” of coffee is sold at \$1.45 per pound.

Therefore the  cost  of  the “X pounds” of coffee is \$1.45*X

The other kind of coffee “Y” pounds” is sold at \$2.20 per pound.

Therefore the cost of “Y pounds” of coffee would be \$2.20*Y.

Since there are two variables X and Y, we need to frame two equations in X and Y to solve them.

Given that the grocer wants to mix two kinds of coffee and the total weight of the mixture is 21 pounds.

Hence the first  equation can be

X+Y = 21——–>Equation1

We are also given that the total cost of the mixture is \$1.70 per pound.

The cost  of  21 pounds of the  mixture would be 21*1.7=35.7

Hence the  second equation would be

1.45X+2.20Y=35.7 ——–> Equation 2

Solving the equations 1 and 2:

X+Y = 21–>Equation1

1.45X+2.20Y=35.7 –> Equation 2

From the  equation1, we can say that  Y = 21-X.

Let us substitute this new value of Y = 21-x in the  second  equation 1.45X+2.20Y =1.7

1.45X+2.20(21-x) =35.7


-0.75X =35.7-46.2


Therefore X= \dfrac {-10.5}{-0.75}

X=14 pounds

Therefore the  value  of  Y would be Y= 21-X=21-14

Y= 7 Pounds

Therefore, the grocer needs to mix 14 pounds of coffee which sells for  \$1.45 and 7pounds of coffee which sells for  \$ 2.20 pounds.


We can cross-check the answers we got for the X =14 and  Y =7 values.

When the values of X and Y are plugged back into the original equations, they must hold good.

Let us take up  the  equation1



21=21.Therefore, the solutions we  arrived work for the Equation1

Let us cross-check with the equation2.

1.45X+2.2Y =35.7

1.45(14)+2.2(7) =35.7


Therefore, the solutions we arrived work for the Equation2 as well.

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