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## Kamal has to fill the same measuring cup 8 times to get 2 liters of water. How many milliliters of water does the cup hold?

Tony B
Featured 3 months ago

Further explanation (expansion) on the solution by SS. Showing where his shortcut method comes from.

#### Explanation:

Although it does not look as though people are using ratio when solving this type of question they are.

Initial condition expressed as a ratio in fraction form gives:

#("2 Litres")/("8 cups")" "# this is stating that for every 8 cups we have 2 litres

The change required is that we have millilitres and 1 cup
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Dealing with the millilitres first}}$

Known that 1 Litre = 1000 millilitres

So 2 Litres = 2000 millilitres

The actual volume has not changed, just the units of measurement. So we now have:

$\left(\text{2000 millilitres")/("8 cups}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Dealing with the 8 cups}}$

As this is a ratio we need to maintain proportionality.

So if we reduce the count of cups we need to reduce the count of millilitres by an appropriate a mount.

To change 8 cups into 1 cup divide by 8. For multiply or divide in ratios what you do the bottom you do to the top.

Divide top and bottom by 8

$\left(\text{2000 millilitres")/("8 cups") ->(color(red)("2000 millilitres"-:8))/("8 cups} \div 8\right)$

$\text{ } \textcolor{red}{\uparrow}$

$\textcolor{red}{\text{The bit in red is where the shortcut comes from}}$

$= \left(250 \text{ millilitres")/(1" cup}\right)$

So 1 cup holds 250 millilitres

## What is #\frac { \frac { 1} { 3} } { \frac { 3} { 9} }#?

Tony B
Featured 1 month ago

A different approach. Tends not to be used as the shortcut method is very much faster. Explained why the shortcut method works.

$\frac{1}{3} \div \frac{3}{9} = 1$

#### Explanation:

A fractions structure is such that you have:

$\left(\text{count")/("size indicator")" "->" "("numerator")/("denominator}\right)$

You can not $\underline{\text{directly divide the counts}}$ unless the size indicators are the same.

.......................................................................................................................
The shortcut method adopts a sort of 'indirect division' in that first it divides the counts then multiplies by a conversion factor.
.......................................................................................................................
$\textcolor{b l u e}{\text{Using first principles}}$

Initial condition: $\to \frac{1}{3} \div \frac{3}{9}$

Multiply by 1 and you do not change the value of the fraction. However, 1 comes in many forms.

$\textcolor{g r e e n}{\left[\frac{1}{3} \textcolor{red}{\times 1}\right] \div \frac{3}{9} \text{ "->" } \left[\frac{1}{3} \textcolor{red}{\times \frac{3}{3}}\right] \div \frac{3}{9}}$

$\text{ } \left[\frac{3}{9}\right] \div \frac{3}{9}$

Now the 'size indicators' (denominators) are the same you can directly divide the counts;

$\text{ } 3 \div 3 = 1$

$\textcolor{g r e e n}{\text{ Try this with different values. It really does work!}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Short cut method}}$

$\text{ } \frac{\textcolor{g r e e n}{1}}{\textcolor{b r o w n}{3}} \div \frac{\textcolor{g r e e n}{3}}{\textcolor{b r o w n}{9}}$

Turn the divisor upside down and multiply:

$\text{ } \frac{\textcolor{g r e e n}{1}}{\textcolor{b r o w n}{3}} \times \frac{\textcolor{b r o w n}{9}}{\textcolor{g r e e n}{3}}$

This is the same as:

$\text{ } \textcolor{g r e e n}{\frac{1}{3}} \times \textcolor{b r o w n}{\frac{9}{3}}$

$\text{ "color(green)(1/3)" "xx" } \textcolor{b r o w n}{\frac{9}{3}}$

$\text{ "uarr" } \uparrow$

$\text{Division of counts conversion factor}$

$\text{ } = 1$

## #(4xx4(6+4)-3(8+1)+2)-:3+(5+2)(5+2)-2xx2+(42-:6)(42-:6)+1# ?

George C.
Featured 6 days ago

$140$

#### Explanation:

Use PEMDAS to help with the order of operations:

P for Parentheses

E for Exponents

MD for Multiplication and Division (left to right)

AS for Addition and Subtraction (left to right)

So we find:

$\left(4 \times 4 \left(6 + 4\right) - 3 \left(8 + 1\right) + 2\right) \div 3 + \left(5 + 2\right) \left(5 + 2\right) - 2 \times 2 + \left(42 \div 6\right) \left(42 \div 6\right) + 1$

$= \left(4 \times 4 \left(10\right) - 3 \left(9\right) + 2\right) \div 3 + \left(7\right) \left(7\right) - 2 \times 2 + \left(7\right) \left(7\right) + 1$

$= \left(4 \times 40 - 27 + 2\right) \div 3 + 49 - 2 \times 2 + 49 + 1$

$= \left(160 - 27 + 2\right) \div 3 + 49 - 2 \times 2 + 49 + 1$

$= \left(133 + 2\right) \div 3 + 49 - 2 \times 2 + 49 + 1$

$= 135 \div 3 + 49 - 2 \times 2 + 49 + 1$

$= 45 + 49 - 4 + 49 + 1$

$= 94 - 4 + 49 + 1$

$= 90 + 49 + 1$

$= 139 + 1$

$= 140$

## At Westbrook Elementary School, 96 students signed up for movie-making classes after school. There were 23 third graders and 36 fourth graders. The rest were fifth graders. How many fifth graders signed up for the class?

David S.
Featured 3 weeks ago

$37$

#### Explanation:

I think here the best way to set this problem up is by making an equation. You know that all of the students is equal to $96$, so we can put that on one side. On the other side, we can add up values.

$23 + 36 + x = 96$

Now the $x$ is there because we don't yet know how many 5th graders there are. When we simplify and solve, we'll find a number that can fit in place of the $x$.

So first is to simplify:

$59 + x = 96$

Subtract $59$ from both sides to cancel that term

$x = 37$

Remember how $x$ represented how many 5th graders there were in the original equation? Well now we have our answer. There are $37$ 5th graders in the class.

## How do you reorder these numbers from least to greatest: #2/5, 0, -3/2, -9/8, 8/7#?

EZ as pi
Featured 2 weeks ago

$- \frac{3}{2} , \text{ "-9/8," "0," "2/5," } \frac{8}{7}$

#### Explanation:

Another method of comparing these numbers is to convert them to fractions which have a common denominator. IN this way you will avoid working with recurring decimals.
Note that some of the fractions are negative.

The LCM is $280$

$\frac{2}{5} = \frac{112}{280}$

$0 = \frac{0}{280}$

$- \frac{3}{2} = \frac{- 420}{280} \text{ } \leftarrow$ the smallest

$- \frac{9}{8} = \frac{- 315}{280}$

$\frac{8}{7} = \frac{320}{280} \text{ } \leftarrow$ the greatest

Now you compare the numerators.

Remember to use the numbers in their original form.

## How do you evaluate #-3- 9+ 6+ 12- 5#?

Gimpy C.
Featured yesterday

Conventionally, left to right but it doesn't matter.

#### Explanation:

We have a lot of freedom with this problem. Just do one thing at a time, and you should get it.

$- 3 - 9 + 6 + 12 - 5$
$- 12 + 6 + 12 - 5$
$- 6 + 12 - 5$
$6 - 5$
$1$

We might ask ourselves though. Can we do it in any other order?

$- 3 - 9 + 6 + 12 - 5$
$- 3 - 9 + 18 - 5$
$- 12 + 18 - 5$
$6 - 5$
$1$

$\left(- 3\right) + \left(- 9\right) + 6 + 12 + \left(- 5\right)$
$\left(- 3\right) + \left(- 9\right) + 18 + \left(- 5\right)$
$\left(- 12\right) + 18 + \left(- 5\right)$
$\left(- 12\right) + 13$
$1$

What about if we go from last to first?

$- 3 - 9 + 6 + 12 - 5$
$- 3 - 9 + 6 + 7$
$- 3 - 9 + 13$
$- 3 + 4$
$1$

The point is that you just do it. You can't go wrong. Addition and subtraction have the same precedence.

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