Green tea word problem

Photo credit: leojmelsrub

Ahmed sent me in this kind of word problem:


A tea producer want to market mixed green tea leaves at $14 per pound. how many pounds of high mountain green tea leaves worth $20 per pound must be mixed with 90 pounds of regular green tea leaved worth $10 per pound?

I have solve many problems like this one on my blog, but it never hurts to solve some more. This can be solved with algebra, using a chart. I've done that before for similar problems... so if you are reading this, and you feel a bit "rusty" in this area, try to make the chart yourself first, before you read further!

For the chart, we also need to choose a variable or several. In this case it is easy:  the unknown is obviously what is asked, or the amount of high mountain green tea. Note also that the cost is always the price per pound times the amount.

mountain green   regular green  the mixture
 tea leaves  tea leaves
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amount |   x   90   90 + x
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cost   |  20x   $900   14(90 + x)
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The chart is ready. Its purpose is to help us write an EQUATION of some sort which will solve x.

So where can we find something equals something? It comes from the cost. The COST of mountain green tea + the COST of regular green tea = COST of the mixture.

20x + 900 = 14(90 + x)

20x + 900 = 1260 + 14x

6x = 360

x = 60 

Now, let's check. That's always the last step in solving equations.

We need 60 pounds of mountain green tea leaves mixed with 90 pounds of regular green tea leaves. The mixture will weigh 150 pounds. The cost of mountain green tea leaves will be $1200, the cost of regular green tea will be $900, and the total cost will be $2100. Calculating cost per pound: $2100 / 150 lb = $14 per pound, so it checks.

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Two problems about fractional parts

I have 2 questions on fractions which I can't solve.

There were 3/5 as many adults as children on a bus. At the next bus stop, 6 adults and 6 children boarded the bus. As a result, there were 2/3 as many adults as children on the bus. How many people were on the bus at first?

A solution using bar diagrams (Singapore style):

children |----|----|----|----|----|

adults   |----|----|----|

Then we have:

children |----|----|----|----|----| +6

adults   |----|----|----| +6

Here look at the difference of children and adults. There are two "blocks" more children than adults. We know the number of adults is 2/3 of the number of children... therefore the DIFFERENCE of two blocks must be 1/3 of the children.

Now look:

children |----|----|----|----|----| +6

Two of those blocks is 1/3 of the total... the other two blocks is another 1/3 of the total... so |----| +6 or one block and 6 must be 1/3 of the total.
This means the +6 must be one block. Or, one block = 6. This now solves the problem, because originally we had 8 "blocks" of people, or 48 people.


Solution with algebra:
Let c be the number of children at first. There were (3/5)c adults at first.

Then 6 more adults and 6 more children come in, so now we have c + 6 children, and (3/5)c + 6 adults.

Now, the amount of adults = (2/3) of the amount of children.

(3/5)c + 6 = (2/3)(c + 6)

As this equation has fractions, let's multiply both sides by 15 to start.

9c + 90 = 10(c + 6)

9c + 90 = 10c + 60

30 = c

There were 30 children and 18 adults at first, or 48 total people.

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2/5 of the counters in a box were red and the rest were blue. After putting another 48 blue counters into a box, 3/4 of the counters were blue. How many counters were in the box at first?

A solution using bar diagrams (Singapore style):

blue |----|----|----|

red  |----|----|

Then:

blue |----|----|----| + 48

red  |----|----| 

...and now 3/4 of the counters are blue. So, |----|----| (the red ones) or two blocks is 1/4 of the counters. Therefore there are a total of 8 blocks of counters. but I drew five blocks and 48. So, 48 must be worth three "blocks", or one block is 16.

At first we had five "blocks" of counters, or 5 x 16 = 80 counters.

CHECK:
80 counters... 32 red and 48 blue.
Add 48 blue. Now we have 32 red and 96 blue, a total of 128.
Reds are 32/128 = 8/32 = 1/4. That checks.

A solution using algebra:
At first, we have (2/5)c counters red and (3/5)c counters blue. ONce we hadd 48 blue counters, we have (2/5)c red counters and (3/5)c + 48 blue counters, and c + 48 total counters. It says now 3/4 of the counters are blue. We write that as an equation:

(3/4) of all counters = number of blue counters

(3/4)(c + 48) = (3/5)c + 48 Multiply this by 20.

15(c + 48) = 12c + 960

15c + 720 = 12c + 960

3c = 240

c = 80

There were originally 80 counters.