# Cutting cake into 5 equal pieces

If a cake is cut into $5$ equal pieces, each piece would be $80$ grams heavier than when the cake is cut into $7$ equal pieces. How heavy is the cake?

How would I solve this problem? Do I have to try to find an algebraic expression for this? $5x = 7y + 400$?

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I thought the question was: how to cut a cake in 5 equal pieces? That is a hard problem.. –  Fabio F. Jun 19 '12 at 20:30
@FabioF. I also thought so, and can suggest sticking 50 candles into it and then cutting it so that each piece contains 10 candles. –  texnic Jun 19 '12 at 20:56
@texnic Guys really? Cutting a cake into 5 equal pieces is 360/5. –  user34059 Jun 20 '12 at 6:11
@KirillFuchs That would admittedly improve precision, however I haven't seen a cake that would accommodate 360 candles yet... –  texnic Jun 20 '12 at 6:43
Where did 400 come from? If you had not added 400, then you would have written one of the equations from the sentence 5x = 7y. The other is x=y+80. Therefore … –  richard Jun 20 '12 at 12:09

The first step is to turn the word problem into an equation; one-fifth of the cake is $80$ grams heavier than one-seventh of the cake, so one-fifth of the cake equals one-seventh of the cake plus 80. "The cake" (specifically its mass) is $x$, and we can work from there:

$$\dfrac{x}{5} = \dfrac{x}{7}+80$$

$$\dfrac{x}{5} - \dfrac{x}{7} = 80$$

Here comes the clever bit; multiply each fraction by a form of one that will give both fractions the same denominator: $$\dfrac{7}{7}\cdot \dfrac{x}{5} - \dfrac{5}{5}\cdot\dfrac{x}{7} = 80$$

$$\dfrac{7x}{35} - \dfrac{5x}{35} = 80$$

$$\dfrac{2x}{35} = 80$$

$$2x = 2800$$

$$x = 1400$$

You can check your answer by plugging it into the original equation; if the two sides are indeed equal the answer is correct:

$$\dfrac{1400}{5} = \dfrac{1400}{7} + 80$$

$$280 = 200 + 80$$

$$\ \ \ \ \ \ \ \ \ \ \ \ 280 = 280 \ \ \text{<-- yay!}$$

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Much quicker to multiply both sides of the original equation by $35$. The result is $7x=5x+2800$. You immediately get to your sixth equation. For the life of me, I don’t know why the high schools in the US don’t emphasize this approach. –  Lubin Aug 14 '12 at 1:22
Because the point of the exercise is usually learning to cross-multiply fractions. Once you know how that works, you can easily see why simply multiplying everything by the LCM does the same trick. I remember going through a lot of really tedious math in prealgebra only to have the teacher show us the shortcut the next day. –  KeithS Sep 7 '12 at 15:06

If cut it into 5 pieces, but decide you wanted 7 instead, so you cut away 80 grams from each slice, and make two (not so pretty) pieces from that. Now you have 7 pieces of equal size, which means you've got 400g evenly divided among the two ugly out of the seven pieces, so the whole cake weighted 1400g.

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I admit my answer wasn't of the highest educational value, nor was it particulary well written, gramatically. I am surprised so many of you liked it. –  Arthur Jun 20 '12 at 9:11

The natural algebraic approach is to give the unknown weight of the cake a name, $x$, and then translate the information we have into an equation: $$\frac{x}{5} = \frac{x}{7}+80$$ From there it is just algebra: multiply everything by 5 and then 7 to clear denominators, rearrange to separate the multiples of $x$ from the constants, solve.

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Let $w$ be the weight of the cake in grams.

If you cut the cake into $5$, one slice would weigh $\frac{w}{5}$ grams. If it was cut into $7$, it would weigh $\frac{w}{7}$ grams. We know

$$\frac{w}{5}=\frac{w}{7}+80$$

Solve for $w$ (the weight of the cake in grams).

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The way to deal with such a question is often to break it down into pieces, and write an equation for each piece.

So suppose the whole cake is $c$ grams (c for cake, name the thing you want to find)

Then one seventh of the cake is $s=\frac c 7$ (s for seven)

And one fifth of the cake is $f=\frac c 5$ (f for five)

And you also have the equation $f=s+80$

That deals with all the information you have been given in an orderly fashion - except it involves a lot of extra variable names, and you only want to know about $c$. You can take any of the three equations and use the others to substitute out $f$ and $s$ to give you a single equation in $c$ alone, which you should be able to solve. In fact the third one is easiest, even though it doesn't involve $c$ at all at the moment. But I'll leave that to you.

I really wanted to point out that it is sometimes easier to write down a number of very simple equations, and solve these, than to try to get directly to a single equation, when it is easy to make a mistake.

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+1 For an excellent point, sir. –  Cameron Buie Jun 19 '12 at 17:34
@Steven if your read nothing else of this answer, note what Mark has to say in his final sentence. Absorbing this way of will be far more profitable for you in the long run than any direct answers to your questions. –  Drew Christianson Jun 19 '12 at 17:40
Have you ever tried solving "Brian is twice as old as Max was when he was eight years older than Samantha when John was born ..." type puzzles? Organising the information in this very simple way is a great help in such situations when they get your brain spinning. –  Mark Bennet Jun 19 '12 at 17:54

Cut the cake into $35$ (that is, $5\times 7$) equal pieces. Let the weight of each piece be $p$. Dividing the cake between $5$ people means each person gets $7p$. Dividing between $7$ people means each person gets $5p$.

So $7p-5p=80$, and therefore $p=40$, and the whole cake weighs $(35)(40)$.

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Let the weight of cake be $35t$ grams.

So, the weight of each piece when the cake was cut into $7$ pieces $= \dfrac{35t}{7} = 5t.$
Also, the weight of each piece when the cake was cut into $5$ pieces = $\dfrac{35}{5} = 7t.$

$$7t = 5t + 80$$ $$7t - 5t = 80$$ $$2t = 80$$ $$t = 40$$

So, the weight of the cake is $35 \cdot 40 = 1400$ grams.

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$$\frac{w}{5}=\frac{w}{7}+80$$

Multiply both sides by 35

$$7w=5w+80\cdot 35$$

substracting both sides by $$2w$$

$$2w=80\cdot 35$$

dividing both sides by 2

$$w=40\cdot 35$$

So weight of cake is 1400

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my answer is the most boring but the most standard isn't it? –  Jim Thio Jun 20 '12 at 2:03

Suppose original weigth of cake is $7x.$ When it is cut into $5$ pieces, each piece weighs $80$ grams more than when it cut into $7$ pieces. So, the equations to model the problem would be

$$5(x+80) = 7x$$ $$5x + 400 = 7x$$ $$x = 200$$

So original weight of cake is $7x = 7 \cdot 200 = 1400$ grams

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Run-on equations like that are really hard to read. –  Henning Makholm Jun 19 '12 at 22:38