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Featured 3 months ago

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

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:

The change required is that we have millilitres and 1 cup

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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:

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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

So 1 cup holds 250 millilitres

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.

A fractions structure is such that you have:

You can not

.......................................................................................................................

The shortcut method adopts a sort of 'indirect division' in that first it divides the counts then multiplies by a conversion factor.

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Initial condition:

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

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Turn the divisor upside down and multiply:

This is the same as:

Featured 6 days ago

Use PEMDAS to help with the order of operations:

Pfor Parentheses

Efor Exponents

MDfor Multiplication and Division (left to right)

ASfor Addition and Subtraction (left to right)

So we find:

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

#=(4xx4(10)-3(9)+2)-:3+(7)(7)-2xx2+(7)(7)+1#

#=(4xx40-27+2)-:3+49-2xx2+49+1#

#=(160-27+2)-:3+49-2xx2+49+1#

#=(133+2)-:3+49-2xx2+49+1#

#=135-:3+49-2xx2+49+1#

#=45+49-4+49+1#

#=94-4+49+1#

#=90+49+1#

#=139+1#

#=140#

Featured 3 weeks ago

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

Now the

So first is to simplify:

Subtract

Remember how

Featured 2 weeks ago

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

Now you compare the numerators.

Remember to use the numbers in their original form.

Featured yesterday

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

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

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

Let's add first.

What about if we turn it into a purely addition problem?

What about if we go from last to first?

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

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