Question

Can you please help me solve this problem? I tried multiplication, tu then I read the question and realized that I had no idea how to solve it. ☹️

Answer 1

3 pounds, 2 pounds, 0.60 pounds

Explanation:

This can be solved using division.
The problem can be restated as: For a given amount of money how much candies can you buy.

For (a)
`7 / 3.50 = 2`
We have 2 "$3.50" in "$7" and we know that for every $3.50 we have, we can buy 1 pound of candies.
Therefore, for $7, we can buy 2 pounds of candies.

The same can be done for (b) and (c)

For (b)
`10.50 / 3.50 = 3`

For (c)
`2.10 / 3.50 = 0.60`

Answer 2

See a solution process below:

Explanation:

We can solve this using ratios:

`1"lb = $3.50` or `1"lb"/($3.50)`

**So, to find how much candy we can buy for $7 we can solve:#

`x/($7.00) = (1"lb")/($3.50)`

`color(red)($)color(red)(7.00) xx x/($7.00) = color(red)($)color(red)(7.00) xx (1"lb")/($3.50)`

`cancel(color(red)($)color(red)(7.00)) xx x/color(red)(cancel(color(black)($7.00))) = cancel(color(red)($))color(red)(7.00) xx (1"lb")/(color(red)(cancel(color(black)($)))3.50)`

`x = (color(red)(7.00)"lb")/3.50`

`x = (cancel(color(red)(7.00))2"lb")/color(red)(cancel(color(black)(3.50)))`

#x = 2"lb"

Therefore you can buy 2 pounds of candy for $7.00

**To find how much candy we can buy for $10.50 we can solve:#

`x/($10.50) = (1"lb")/($3.50)`

`color(red)($)color(red)(10.50) xx x/($10.50) = color(red)($)color(red)(10.50) xx (1"lb")/($3.50)`

`cancel(color(red)($)color(red)(10.50)) xx x/color(red)(cancel(color(black)($10.50))) = cancel(color(red)($))color(red)(10.50) xx (1"lb")/(color(red)(cancel(color(black)($)))3.50)`

`x = (color(red)(10.50)"lb")/3.50`

`x = (cancel(color(red)(10.50))3"lb")/color(red)(cancel(color(black)(3.50)))`

#x = 3"lb"

Therefore you can buy 3 pounds of candy for $10.50

**To find how much candy we can buy for $2.10 we can solve:#

`x/($2.10) = (1"lb")/($3.50)`

`color(red)($)color(red)(2.10) xx x/($2.10) = color(red)($)color(red)(2.10) xx (1"lb")/($3.50)`

`cancel(color(red)($)color(red)(2.10)) xx x/color(red)(cancel(color(black)($2.10))) = cancel(color(red)($))color(red)(2.10) xx (1"lb")/(color(red)(cancel(color(black)($)))3.50)`

`x = (color(red)(2.10)"lb")/3.50`

`x = 0.60"lb"`

Therefore you can buy 0.6 or `3/5` pounds of candy for $2.10

Answer 3

You use division.

Explanation:

For `$7`:

`($7.00)/($3.50"/pound") = 2" pounds"`

For `$10.50`:

`($10.50)/($3.50"/pound") = 3" pounds"`

For `$2.10`:

`($2.10)/($3.50"/pound") = 0.6" pounds"`

Answer 4

`color(blue)2` pounds for $7

`color(blue)3` pounds for $10.50

`color(blue)(3/5` pound for $2.10

Explanation:

The weight to cost ratio stays the same;
so `(1" pound")/($3.50)` is a constant.

a. The number of pounds bought for `$7` must be equal to this constant:
`color(white)("XXX")("number of pounds bought for "$7)/($7)=(1" pound")/($3.50)`

after multiplying both sides by `$7`
`color(white)("XXX")"number of pounds bought for " $7=(1" pound")/($3.50)xx($7.00)/color(white)("x")`

`color(white)("XXXXXXXXXXXXXXXXXXXXXX")=2" pounds"`

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Similarly
`color(white)("XXX")"number of pound bought for "$10.50 = (1" pound")/($3.50)xx($10.50)/color(white)("x")`

`color(white)("XXXXXXXXXXXXXXXXXXXXXXX")=3" pounds"`

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and
`color(white)("XXX")"number of pound bought for "$2.10 = (1" pound")/($3.50)xx($2.0)/color(white)("x")`

`color(white)("XXXXXXXXXXXXXXXXXXXXXXX")=210/350" pound"`

`color(white)("XXXXXXXXXXXXXXXXXXXXXXX")=3/5" pound"`

Answer 5

Slightly different presentation of the solution. First part is using first principles. It shows where the shortcut method of cross multiply comes from.

Explanation:

We are given the initial condition: `("weight in pounds") /("cost")= (1lb)/($3.50)`

Dropping the units of measurement `->1/3.50`

`color(magenta)("First part - "ul("in detail")". The other parts will not be detailed")`

Let the unknown value be `x` pounds

`1/3.50=x/7`

To find the value of `x` we need to change the `3.50` from `1/3.50` into 7

So we can do this: `3.50xx7/3.50=7`
Also, if you multiply by 1 you do not change the value

`color(green)([1/3.50color(red)(xx1)])=x/7`

but we can write 1 as `(7-:3.5)/(7-:3.5)`

`color(green)([1/3.50color(red)(xx (7-:3.5)/(7-:3.5))])=x/7`

`color(green)([ (color(green)(1)color(red)(xx7-:3.5))/(color(green)(3.5)color(red)(-:3.5xx7))] )= x/7 color(white)("ddd")`

`color(white)("d")color(green)([ (color(green)(1)color(red)(xx7-:3.5))/7] )= x/7 color(white)("ddd")`Notice the top is the same as `1/3.5xx7`

which is where your shortcut of cross multiply comes from.

Comparing just the tops of the ratios (numerators) we have

`color(white)("dddddddd")color(magenta)(ul(bar(| color(white)(2/2) x=7-:3.5 = 2color(white)(2/2)|)`

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`color(magenta)("Second part - shortcut method")`

Given `(1lb)/($3.50)= x/($10.50)`

Cross multiply

`1lbxx($10.50)/($3.50)=x`

`1lbxx(cancel($)color(white)("d")cancel(10.50)^3)/(cancel($)color(white)("d")cancel(3.50)^1)=x = 3`

Did you know that you can cancel out units of measurement the way I have?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(magenta)("Third part - shortcut method")`

Given `(1lb)/($3.50)= x/($2.10)`

`x=(1lbxxcancel($)2.10)/(cancel($)3.50) = 3/5lb -=6/10lb = 0.6lb`