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**mathsmypassion****Member**- Registered: 2008-12-01
- Posts: 33

Two candles, same height, are lit at the same time. The first takes 4 hours to burn and the second takes 3 hours. When is the first candle twice the height of the second one?

I need a solution that doesn't use equation. Thanks.

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**janetpc2008****Member**- Registered: 2009-02-03
- Posts: 7

TAKE A PIECE OF GRAPH PAPER. Draw each candle on the paper 60 units high.

The 4hr candle will burn 15 units per hour. - mark these on the graph as hours 0,1,2,3,4. 0 at top, 4 at bottom. You can think of each unit as 4 minutes (4x15=60min)

The 3hr candle will burn 20 units per hour -mark them also from 0 to 3. You can think of each unit here as 3 minutes (3x20=60min)

You can see that at hour 2 the 4hr candle is at 30 units up, and the 3hr candle is at 20 units up. It is just after this hour that the 3hr candle will be half the height of the 4hr candle.

NOW - burn each candle for 12minutes, which is 3 units on the 4hr candle, getting to 27 units, AND 4 units on the 3hr candle, getting to 16 units. This is not quite half (16/27).

Burn each candle AGAIN for 12 minutes. The 4hr candle goes to 24 units, and the 3hr candle goes to 12. Now we are at half (12/24).

The time elapsed is 2hrs plus 12minutes plus 12 minutes.

I hope this is clear enough without showing you my drawing.

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**mathsmypassion****Member**- Registered: 2008-12-01
- Posts: 33

Yes, it is clear. Many thanks. What made you choose 60 as units high? I realized that 12 minutes would be a good time unit after I solved the problem using equations- but saying that 12 comes from 3 times 4 is not right.

12 minutes is actually 1/5 of one hour. What's the trick I miss?

*Last edited by mathsmypassion (2009-02-12 16:00:22)*

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**janetpc2008****Member**- Registered: 2009-02-03
- Posts: 7

It was a bit of hit-and-miss, but it seemed the easiest way to explain.

We could use the 30 high column also:

On the 30 high column, we would need to move 4 units on the 3hr candle (4x6min) and 3 units on the 4hr candle (3x8min).

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**mathsmypassion****Member**- Registered: 2008-12-01
- Posts: 33

Thank you very much.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Just like to point out that a solution that **does** use equation would be far simpler.

Let the length of the candles be units. Then the first candle burns at the rate of units per hour and the second burns at the rate of units per hour.

After

hours, the length of the first candle is units and the length of the second candle is units.Solving gives

hours, i.e. 2 hours 24 minutes.Im not sure why a solution not using equation should be needed, really.

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**mathsmypassion****Member**- Registered: 2008-12-01
- Posts: 33

My daughter's maths teacher apparently showed them a solution without solving equation. My daughter said she understood his explanation (kind of short in fact) but she couldn't reproduce it for me. In my opinion, trying to solve this problem without equation is against the nature. But I cannot say I know everything. So I agree 100% with you Jane, and thanks for reply.

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**Monox D. I-Fly****Member**- Registered: 2015-12-02
- Posts: 981

mathsmypassion wrote:

My daughter's maths teacher apparently showed them a solution without solving equation. My daughter said she understood his explanation (kind of short in fact) but she couldn't reproduce it for me. In my opinion, trying to solve this problem without equation is against the nature. But I cannot say I know everything. So I agree 100% with you Jane, and thanks for reply.

Probably her teacher just did what janetpc2008 suggested: TAKE A PIECE OF GRAPH PAPER. As a math person myself, I am also glad whenever my students solve problems that way. It shows that they understand the initiate concept rather than merely memorizing exhausting formulas.

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