I have already finished Calculus II but I go back and practice the basics on Khan Academy. This problem confuses me conceptually every time. I know what the answer is, but I am having a hard time rationalizing the steps. What are the mental steps you take solving this problem? Thanks in advance and please have mercy on me. I am doing this for my own entertainment not for a grade!
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2$\begingroup$ Simply use unitary method. $\endgroup$– Prasun BiswasApr 16, 2015 at 7:10
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1$\begingroup$ This is a division question, specifically $\frac{\frac23}{\frac43} = \frac12$. $\endgroup$– Jeppe Stig NielsenApr 16, 2015 at 20:33
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10 Answers
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If $1$kg of sand fills a fraction $F$, then $4/3$kg of sand must fill a fraction $4/3F$.
But you are given that $4/3F=2/3$.
Solve for $F$
answered Apr 16, 2015 at 7:15
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$\begingroup$ So to be quite sure, the idea is that F = 1KG. The amount of sand I have filling 2/3 of my bucket is 4/3 kg, which is as you say 4/3 of my unit amount. Therefore F = (2/3)(3/4) $\endgroup$ Apr 16, 2015 at 7:25
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6$\begingroup$ @StrykerGraham Actually, to be quite sure, you know that $$\frac{2}{3}\text{ bucket}=\frac{4}{3}kg$$Solve for $kg$, and you get $$1kg=\frac12\text{ bucket}$$Alternatively, in case you want to be more wordy, and less algebra-y, if $4/3kg$ fills $2/3$ of a bucket, then $4kg$ fills two buckets. $\endgroup$– ArthurApr 16, 2015 at 8:13
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Here's a simple elementary approach:
We're told that Caden has 4/3 kg of sand, which fills 2/3 of a bucket.
Three times as much sand would fill three times as many buckets. Thus, if Caden had 4 kg of sand, it would fill 2 buckets.
Therefore, 1 kg of sand would fill half a bucket.
answered Apr 16, 2015 at 13:59
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A picture is worth a thousand words.
This is a drawing of the bucket cut in 6 slices.
The bottom 4 of them are filled with sand.
+------------------------------+
| |
+..... empty .... 1/3 bucket ..+
| |
+------------------------------+
| 1/3 kg sand = 1/6 bucket |
.- +..............................+ -.
/ | 1/3 kg sand = 1/6 bucket | \
| +------------------------------+ |
1 kg sand < | 1/3 kg sand | > 1/2 bucket
| +................ 1/3 bucket ..+ |
\ | 1/3 kg sand | /
`- +------------------------------+ -'
Solution: 1 kg of sand is made up of 3 "slices" of 1/3 kg, each of them filling 1/6 of bucket.
3 * 1/6 = 1/2 bucket
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Let x denote 1 kg od sand y denote 1 bucket $$\frac{4}{3}x=\frac{2}{3}y$$ now $$x=\frac{2}{3}*\frac{3}{4}y=\frac{1}{2}y$$ $$\therefore$$ 1kg of sand fills half a bucket
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In 11th grade chemistry, I learned about conversion factors. I wish I knew about them sooner. You are given this equation:
$\frac{4}{3} S = \frac{2}{3} B$
Rewritten:
$1 = \frac{\frac{2}{3} B}{\frac{4}{3}S}$
Simplified:
$1 = \frac{B}{2S}$
This equation is useful because you can multiply anything by one without changing it. Here's how you'd use this. You have 1 kg of sand which I'd write as $1S$. Let's multiply this by our equation above which I've shown equals one:
$1S * \frac{B}{2S} = \frac{1}{2}B$
Even though we're multiplying by one, the units cancel in such a way that we now have buckets. That's our answer. We're asked for a number of buckets.
Try this with other conversions. Miles to kilometers. Celsius to Kelvin. You don't need to remember that $distance = rate * time$, you just need to know the units and how they relate.
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$\begingroup$ Conversion factors are actually where I first think about going when I approach a problem like this... I just wasn't sure how to put this into the standard "? = known factor (...)" format $\endgroup$ Apr 18, 2015 at 11:40
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'Caden has 4/3 kg of sand which fills 2/3 of his bucket.'
These two, a kilogram and a bucket, are simply two units of measure of sand. The quantities $4/3\,kg$ and $2/3\,bucket$ are simple multiplications of a number and a unit, and they both denote the same amount of sand:
$$4/3\, kg = 2/3\, bucket$$
To find out 'How many buckets will 1kg sand fill?' you simply multiply both sides by an appropriate number so that you get a desired value on the left side:
$$3/4 \times 4/3\,kg = 3/4 \times 2/3\, bucket$$
You can do that, because if two values (LHS and RHS) are equal, and they get multiplied by the same number, the results are also equal. So by associativity you get:
$$(3/4 \times 4/3)\,kg = (3/4 \times 2/3)\, bucket$$ $$1\,kg = (2/4)\, bucket = 1/2\, bucket$$
Solving for $1\,kg$ is quite simple, because we need to multiply by an inverse of the given coefficient. In general case we might want to solve for any other value, say 'how much copper you need to make 3,5 kg of brass if you use 7 kg of copper for 10 kg of brass?' You can write the two equations then:
$$\begin{align} x\,kg_{Cu} & = k \times 3.5\,kg_{brass} \\ 7\,kg_{Cu} & = k \times 10\,kg_{brass} \end{align}$$
where $k$ is a part of Cu in the alloy.
As LHS equals RHS in each equation, you can divide them on both sides and the results are equal, too:
$$\frac {x\,kg_{Cu}}{7\,kg_{Cu}} = \frac {k \times 3.5\,kg_{brass}}{k \times 10\,kg_{brass}} $$ then simplify to get a simple proportion:
$$\frac x7 = \frac {3.5}{10} $$ and finally $$x = \frac {7\times 3.5}{10}$$
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Set up a ratio:
$\frac{4}{3}$ kg per $\frac{2}{3}$ bucket is $\frac{4/3}{2/3} = 2$ kg per bucket
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As the bucket is filled by $2/3$, a half of its actual content is needed to get it filled completely, that is a half of $4/3$ kg.
answered Apr 16, 2015 at 10:33
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It is a simple linear relation.
4/3 kg 2/3 bucket. Add 50 percent or multiply by 3/2
6/3 kg 3/3 bucket = 2 kg 1 bucket
so, 1kg filled in 1/2 or half bucket.
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4/3 Kg fills two thirds of the bucket, e.g. not a filled bucket. 3/3 < 4/3. The number of filled buckets will therefore be none.
