Blend Coffee Beans
Blend Coffee Beans
Mary Wart blends coffee for Tast-Delight. She needs to prepare 170pounds?
of blended coffee beans selling for $4.56 per pound. She plans to do this by blending together a high-quality bean costing $5.50 per pound and a cheaper bean at $3.50 per pound. To the nearest pound, find how much high-quality coffee bean and how much cheaper coffee bean she should blend.
She should blend ? lbs. of high quality beans.
She should blend ? lbs. of cheaper beans.
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"Mixture" Word Problems
Mixture problems involve creating a mixture from two or more things, and then determining some quality (percentage, price, etc) of the resulting mixture. For example:
Your school is holding a "family friendly" event this weekend. Students have been pre-selling tickets to the event; adult tickets are $5.00, and child tickets (for six-years-old and under) are $2.50. From past experience, you expect about 13,000 people to attend the event. But this is the first year in which tickets prices have been reduced for the younger children, so you really don't know how many child tickets and how many adult tickets you can expect to sell. Your boss wants you to estimate the expected ticket revenue. You decide to use the information from the pre-sold tickets to estimate the ratio of adults to children, and figure the expected revenue from this information.
You consult with your student ticket-sellers, and discover that they have not been keeping track of how many child tickets they have sold. The tickets are identical, until the ticket-seller punches a hole in the ticket, indicating that it is a child ticket. But they don't remember how many holes they've punched. They only know that they've sold 548 tickets for $2460. How much revenue from each of child and adult tickets can you expect?
To solve this, we need to figure out the ratio of tickets that have already been sold. If we work methodically, we can find the answer.
Let A stand for the number of adult tickets pre-sold, and C stand for the child tickets pre-sold. Then A + C = 548. Also, since each adult ticket cost $5.00, then ($5.00)A stands for the revenue brought in from the adult tickets pre-sold; likewise, ($2.50)C stands for the revenue brought in from the child tickets. Then ($5.00)A + ($2.50)C = $2460. But we can only solve an equation with one variable, not two. So look again at that first equation. If A + C = 548, then A = 548 – C (or C = 548 – A; it doesn't matter at this stage which variable you solve for). Organizing this information in a grid, we get:
tickets sold $/ticket total $
adult 548 – C $5 $5(548 – C)
child C $2.50 $2.50C
total 548 --- $2460
From the last column, we get (total $ from the adult tickets) plus (total $ from the child tickets) is (the total $ so far), or, as an equation:
($5.00)(548 – C) + ($2.50)C = $2460
$2740 – ($5.00)C + ($2.50)C = $2460
$2740 – ($2.50)C = $2460
–($2.50)C = –$280
C = –$280/–$2.50 = 112
Then 112 child tickets were pre-sold, and A = 548 – 112 = 436 adult tickets were sold. (Using "A" and "C" for our variables, instead of "x" and "y", was helpful, because the variables suggested what they stood for. We knew instantly that "C = 112" meant "112 child tickets". This is a useful technique.)
Will It Blend - Coffee Beans.avi
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