Susan bought #2 3/4# pounds of potato salad at $5.60 a pound. How much did she spend?

2 Answers
May 11, 2017

See a solution process below:

Explanation:

The formula for solving this problem is:

#c = p xx a# Where:

#c# is the total cost, what we will be solving for

#p# is the price of the item, #($5.60)/(lb)# for this problem.

#a# is the amount purchased. #2 3/4 lb# for this problem.

Substituting gives:

#c = ($5.60)/(lb) xx 2 3/4 lb#

First, we can cancel common terms:

#c = ($5.60)/color(red)(cancel(color(black)(lb))) xx 2 3/4 color(red)(cancel(color(black)(lb)))#

#c = $5.60 xx 2 3/4#

Next, we can convert the mixed fraction to an improper fraction:

#c = $5.60 xx (2 + 3/4)#

#c = $5.60 xx ((4/4 xx 2) + 3/4)#

#c = $5.60 xx (8/4 + 3/4)#

#c = $5.60 xx 11/4#

Now, we can multiply the two terms to calculate #c#:

#c = ($61.60)/4#

#c = $15.40#

Susan spent $15.40

May 11, 2017

Susan spent #2 3/4 \times $5.60=$11/4\times 56/10=$(11\times 14)/10=$ 15.40#.

Explanation:

The question is meant to test how well you can convert between all sorts of ways to represent fractions.

We start off by converting everything to improper fractions.

#2 3/4=2 + 3/4=(2\times 4 + 3)/4=(8 + 3)/4=11/4#

Hence, we have #11/4# pounds of potato salad.

Now, #$5.60=$560/100=$56/10#

(Note that we could simplify the fraction even further at this point, but since we are going to do that at the end anyway, we decide to keep the #10# in the denominator for ease of calculation)

Then comes a simple application of the Unitary Method:

#1# pound of potato salad costs #$56/10#
Therefore, #11/4# pounds of potato salad costs #$11/4 \times 56/10=$ (11 \times 14)/10=$154/10=$15.40#.

(We cancel #56# by #4# to get #14#. Then we multiply #11# by #14# to get #154#. Finally we divide by #10# and write the answer with #2# places of decimal, as we normally do for units of currency)

Note : All the calculations can also be done by converting everything to decimals, but more often than not, the methods becomes cumbersome and lengthy. Also, you may encounter non-terminating fractions like #1/3#. Therefore, it is recommended to convert everything to improper fractions and perform short and easy calculations as and when necessary.