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Alex73 [517]
2 years ago
9

Ice-Cream Palace needed 6 gallons of milk today to make their daily special. They had 6 1⁄2 quarts of skim milk and 1 pint of wh

ole milk. How many pints of milk did they still need to buy?
Mathematics
2 answers:
Elza [17]2 years ago
5 0
First, we should convert all of the measurements into one unit.

Let's convert everything into fluid ounces since it's the smallest unit in the American Liquid Measurement system.

They had 6 1/2 quarts of skim milk.

6 1/2 quarts=208 fluid ounces.  

They had 1 pint of whole milk, which is equal to 16 fluid ounces.

Next, we add both numbers 208, and 16, together.

208+16=224.

So far Ice-Cream Palace has 224 fluid ounces of milk.

Next, we'll convert 6 gallons into fluid ounces.

6 gallons=768 fluid ounces.

Next, we subtract 224 from 768.

768-224=544.

That means they need to buy 544 ounces of milk.

All we have to do now is convert 544 into pints.

544 ounces= 34 pints.

They still need to buy 34 pints of milk.

I hope you have a wonderful day! :D

crimeas [40]2 years ago
5 0
Okay, so 2 cups is a pint, two pints is a quart and four quarts is a gallon.

We need to know how many more pints we need to get to 6 gallons, so I'm going to convert it all to pints, 6 1/2 quarts is 13 pints.


All together we have 14 pints, and you need 8 pints for a gallon, so in 6 gallons we need 48. 48-14=34. 

We need 34 more pints of milk.
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Step-by-step explanation:

Let the amount of apples bought be x

Amount of strawberries bought be y and;

amount of oranges bought be z

If we buy 2 apples, 3 boxes of strawberry, and 4 oranges, the fruit would cost $15.30, then this will be expressed as;

2x+3y+4z = 15.30 ........... 1

If we buy 1 box of strawberry, 4 apples, and 2 oranges, the fruit would cost $10.90, this will be expressed as:

4x+y+2z = 10.90 ......... 2

If we buy 1 orange, 5 apples and 2 boxes of strawberry, the fruit would cost $13.70, this is expressed as:

5x+2y+x = 13.70 .......... 3

Solving the three equations simultaneously:

2x+3y+4z = 15.30 ........... 1

4x+y+2z = 10.90 ......... 2

5x+2y+z = 13.70 .......... 3

Reduce the number of equations

multiply equation 1 by 2 and subtract from equation 2

equation 1 * 2 will give:

4x+6y+8z = 30.6 ........... 4

eqn (4)-eqn(2)

6y-y + 8z-2z = 30.6-10.90

5y+6z = 19.7 ......... 5

Also multiply equation 2 by 5 and eqn 5 by 4 and subtract from each other.

eqn(2)* 5 will give:

20x+5y+10z = 54.5 .......... *

eqn(3) * 4 will give

20x+8y+4z = 54.8.......**

Subtsrct ** from *

8y-3y+(4z-10z)= 54.8-54.5

5y-6z = 0.3 .................. 6

Equate 5 and 6 and solve:

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subtract:

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