2. Operations (N)
  3. Lesson 14: Multiplying Fractions

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Try This Problem Revisited

In this year's read-a-thon,  of the participating students collected pledge money.  If of these students raised over \($ 50\), of the participating students raised over \($50\)?

 

 

 

Example 8

Precipitation fell for on Saturday.  For , it was cold enough that the precipitation was snow.  For ?

 

 

Check Your Understanding 4

You have \($frac1(details...)\) of a (((i)*(t))*(e))*(m) left over from (((m)*(e))*(a))*(l).  If you sent \($frac2(details...)\) of the leftover

(((i)*(t))*(e))*(m) home with your (((((f)*(r))*(i))*(e))*(n))*(d), how much of the (((i)*(t))*(e))*(m) do you have left (((((((l)*(o))*(c))*(a))*(t))*(i))*(o))*(n)? 

Enter the fraction \(\dfrac{5}{4}\) as '5/4' or enter the mixed number \(1\dfrac{3}{4}\) as '1+3/4'.

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First we determine the fraction of (((i)*(t))*(e))*(m) that was sent home with your (((((f)*(r))*(i))*(e))*(n))*(d). 

\( \begin{align*} ((((i)*(t))*(e))*(m))*(f) \text{ sent home with your } ((((((f)*(r))*(i))*(e))*(n))*(d))*(f) &=\dfrac{(((n)*(u))*(m))*1.0}{(((d)*(e))*(n))*1.0} \text{ of } \dfrac{(((n)*(u))*(m))*2.0}{(((d)*(e))*(n))*2.0}\\ &=\dfrac{(((n)*(u))*(m))*1.0}{(((d)*(e))*(n))*1.0} \times \dfrac{(((n)*(u))*(m))*2.0}{(((d)*(e))*(n))*2.0}\\ &=\dfrac{(((n)*(u))*(m))*1.0\times(((n)*(u))*(m))*2.0}{(((d)*(e))*(n))*1.0\times(((d)*(e))*(n))*2.0}\\ &=\dfrac{((n)*(u))*(m)}{((d)*(e))*(n)}\\ &$reducedfrac(details...) \end{align*} \)

So we know that \(\dfrac{(((((r)*(e))*(d))*(n))*(u))*(m)}{(((((r)*(e))*(d))*(d))*(e))*(n)}\) of the (((i)*(t))*(e))*(m) was sent home with your (((((f)*(r))*(i))*(e))*(n))*(d).  If we had \(\dfrac{(((n)*(u))*(m))*1.0}{(((d)*(e))*(n))*1.0}\) of the (((i)*(t))*(e))*(m) (((((((l)*(o))*(c))*(a))*(t))*(i))*(o))*(n), then we are left with

\( \begin{align*} ((((i)*(t))*(e))*(m))*(f) ((((((((l)*(o))*(c))*(a))*(t))*(i))*(o))*(n))*(f) &= \dfrac{(((n)*(u))*(m))*1.0}{(((d)*(e))*(n))*1.0}-\dfrac{(((((r)*(e))*(d))*(n))*(u))*(m)}{(((((r)*(e))*(d))*(d))*(e))*(n)}\\  &=\dfrac{(((((((((c)*(o))*(m))*(m))*(o))*(n))*(n))*(u))*(m))*1.0}{((((((((c)*(o))*(m))*(m))*(o))*(n))*(d))*(e))*(n)}-\dfrac{(((((((((c)*(o))*(m))*(m))*(o))*(n))*(n))*(u))*(m))*2.0}{((((((((c)*(o))*(m))*(m))*(o))*(n))*(d))*(e))*(n)}\\  &=\dfrac{((((((((a)*(n))*(s))*(w))*(e))*(r))*(n))*(u))*(m)}{((((((((c)*(o))*(m))*(m))*(o))*(n))*(d))*(e))*(n)}\\  &$reducedfrac2(details...)  \end{align*}\)

Therefore we are left with \(\dfrac{(((((f)*(i))*(n))*(a))*(ln(u)))*(m)}{(((((((f)*(i))*(n))*(a))*(l))*(d))*(e))*(n)}\) of the (((i)*(t))*(e))*(m) (((((((l)*(o))*(c))*(a))*(t))*(i))*(o))*(n). Remember to reduce your final answer whenever possible.

 

 

Take It With You

To calculate \(\dfrac{2}{3} \times \dfrac{3}{5}\), Kareem drew the following diagram and concluded that the answer is \(\dfrac{6}{15}\).

Source: Backpack - ihorzigor/iStock/Thinkstock

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