Two airplanes leave an airport at the same time and travel in opposite directions. One plane travels 95 km per hour faster than
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Answer:
Step-by-step explanation:
Step 1 is to collect the information in a table format. The attached Table1 summarizes what we know in columns 1 - 3. Columns 4 - 6 show the data we're about to calculate, so ignore them for the moment.
.I use A for the price of one apple and C for the price of 1 chocolate bar. We know how many of each are purchased by both Mary and Bill, and the total each paid.
The last column (6) shows the price for each, which comes from the derivation presented below.
We can set the information in columns 1 - 3 into equations:
Mary: 3A+1C = 2.70 [This states that Mary bought 3 apples at a price of A for each apple, and 1 chocolate bar at a price of C per bar.]
Bill: 1A + 4C = 6.40 [In the same fashion, this summarizes how Bill parted with his $6.40]
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Now let's rearrange either equation to solve for C. I'll use Mary's equation:
3A+1C = 2.70
1C = 2.70 - 3A
This tells us that the price of 1 chocolate bar is $2.70 minus the cost of 3 apples. Interesting, but we need the price of a chocolate bar to determine an exact amount.
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Now take Bill's equation and substitute the value we just derived for the price of a chocolate bar <u>C</u> = (2.70-3A).
1A + 4<u>C</u> = 6.40
1A + 4(2.70 - 3A) = 6.40
1A + 10.80 - 12A = 6.40
-11A = -4.40
A = 0.40 This is the price for 1 apple. This is the value in the last column (6) of attached Table1.
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Now we can use the value of $0.40 in either equation to find the price of a chocolate bar. I'll use the rearranged equation we developed for Mary:
1C = 2.70 - 3A
1C = 2.70 - 3(0.40) [A = 0.40 from the above work]
1C = 2.70 - 1.20
1C = 1.50
The price of 1 chocolate bar is $1.50. Pricey, and bad for health. But very tasty.
Columns 4 and 5 are the total costs for the apples and bars bought by each, and their sums add up to the amounts we were given at the start of the problem. So our prices must be correct.
