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Since the beginning of algebra, I have always been terrible at word problems. I don't know why. I have this word problem here and would love for someone to explain in large depth how to solve these word problems. Thanks.

A pen floating in the weightlessness of space is 30 inches above the floor of the space capsule and is rising at 1.5 inches per second. In how many seconds will it reach the 6-foot-tall ceiling?

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Ted: I rolled back your question to the original one, as Dave invested a considerable amount of time in answering that question (and did so very nicely, I think). Feel free to ask the other question as a new question. –  t.b. Jul 13 '11 at 15:24
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2 Answers

Note: The problem I discuss below is not the problem that appeared at this site when I first read the problem statement. Apparently the statement of the problem was changed while I was writing my answer, which I think is still sufficient for the original poster to make use of.

The problem asks for how long it will take an object to travel a certain distance (not given explicitly, but we can find this distance) when traveling at a constant speed. So this is a

distance = (rate)(time)

$d = rt$

problem. We are given that $r = 1.5$ in/sec. Using this information gives us

$d = (2.5)t$

We are asked to find the value of $t$, so it's not likely to be among the given information (except maybe for trick questions). That leaves $d$. Can you find the value of $d$? Draw two horizontal lines to indicate the floor and ceiling, and label the distance between the parallel lines as 6 ft. Now draw a line segment straight up from the floor (i.e. the lower parallel line) and stopping somewhere between the parallel lines, and label its length 30 in. Then $d$ is the distance from the top of the line segment to the upper parallel line. Do you see that this distance is 6 ft $-$ 30 in? (If you have already traveled 30 inches into a trip having a total distance of 6 feet, then the distance that remains to be traveled is 6 ft $-$ 30 in.) Since the units of rate (i.e. speed) are inches/second, $d$ needs to be in inches and $t$ needs to be in seconds.

$d =$ 6 ft $-$ 30 in

$d =$ (6 ft)(12 in/ft) $-$ 30 in

$d =$ 72 in $-$ 30 in

$d = 42$ in

Putting this into $d = (2.5)t$ gives

$42 = (2.5)t$

Dividing both sides by 2.5 gives

$t = 42 \div 2.5$

$t =$ 16.8 sec

If you have to do this without a calculator, note that 2.5 = 5/2, so the computation becomes

$42 \div \frac{5}{2} = \frac{42}{1} \cdot \frac{2}{5} = \frac{84}{5} = \frac{80}{5} + \frac{4}{5} = 16 + 0.8 = 16.8$

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This seems a nice answer to a different question –  Ross Millikan Jul 13 '11 at 15:21
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@Ross: The OP changed the question while this answer was written. –  t.b. Jul 13 '11 at 15:24
    
OK, I have deleted my answer to the interim problem. –  Ross Millikan Jul 13 '11 at 15:27
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It is good for you to first identify different categories of word problems. Following are few that I encounter a lot of times

    1. Speed or velocity Related Problems

    2. Mixture Problems

    3. Proportional related Problems

and list goes on.

When you are able to successfully Identify these problems then it would be much easier for you to come up with strategy to solve them.Following is a link worth visiting.

http://www.themathpage.com/alg/word-problems.htm

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