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Answer:

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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("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:

#("2000 millilitres")/("8 cups")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("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

#("2000 millilitres")/("8 cups") ->(color(red)("2000 millilitres"-:8))/("8 cups"-:8) #

#" "color(red)(uarr)#

#color(red)("The bit in red is where the shortcut comes from")#

#=(250" millilitres")/(1" cup")#

So 1 cup holds 250 millilitres

Answer:

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

#1/3-:3/9=1#

Explanation:

A fractions structure is such that you have:

#("count")/("size indicator")" "->" "("numerator")/("denominator")#

You can not #ul("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.
.......................................................................................................................
#color(blue)("Using first principles")#

Initial condition: #-> 1/3-:3/9#

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

#color(green)( [1/3 color(red)(xx1)] -:3/9" "->" "[1/3color(red)(xx3/3)]-:3/9 )#

#" "[3/9]-:3/9 #

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

#" "3-:3=1#

#color(green)(" Try this with different values. It really does work!")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Short cut method")#

#" "(color(green)(1))/(color(brown)(3))-:(color(green)(3))/(color(brown)(9))#

Turn the divisor upside down and multiply:

#" "(color(green)(1))/(color(brown)(3))xx(color(brown)(9))/(color(green)(3))#

This is the same as:

#" "color(green)(1/3)xxcolor(brown)(9/3)#

#" "color(green)(1/3)" "xx" "color(brown)(9/3)#

#" "uarr" "uarr#

#"Division of counts conversion factor"#

#" "=1#

Answer:

#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:

#(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#

Answer:

#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.

Answer:

#-3/2," "-9/8," "0," "2/5," "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#

#2/5 = 112/280#

#0 =0/280#

#-3/2 =( -420)/280" "larr# the smallest

#-9/8 = (-315)/280#

#8/7 = 320/280" "larr# the greatest

Now you compare the numerators.

Remember to use the numbers in their original form.

Answer:

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?

Let's add first.

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

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

#(-3)+(-9)+6+12+(-5)#
#(-3)+(-9)+18+(-5)#
#(-12)+18+(-5)#
#(-12)+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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