MATH SOLVE

3 months ago

Q:
# Mia bought 10 1/9 lb of flour. She used 2 3/4 lb of flour to bake a banana cake and some to bake a chocolate cake. After baking two cakes, she had 3 5/6 lb of flour left. How much flour did she use to bake the chocolate cake?

Accepted Solution

A:

We know that she started with 10 1/9 pounds and after the two cakes she was left with 3 5/6. So if we subtract what she started with minus what she is left with we will get what she used.

This is easiest done by converting the mixed numbers into improper fractions, finding a common denominator and subtracting (subtract the numerators and keep the denominator) as follows:

[tex]10 \frac{1}{9}-3 \frac{5}{6}= \frac{91}{9}- \frac{23}{6} =\frac{182}{18}- \frac{69}{18}= \frac{113}{18} [/tex]

That's how much flour she spent on both cakes. If we subtract from this what she spent on the banana cake 2 3/4 we will be left with what she used for the chocolate cake.

This is found as follows: [tex] \frac{113}{18} -2 \frac{3}{4} = \frac{113}{18} - \frac{11}{4} = \frac{226}{36} - \frac{99}{36} = \frac{127}{36}=3 \frac{19}{36} [/tex]

She used [tex]3 \frac{19}{36} [/tex] pounds of flour on the chocolate cake.

This is easiest done by converting the mixed numbers into improper fractions, finding a common denominator and subtracting (subtract the numerators and keep the denominator) as follows:

[tex]10 \frac{1}{9}-3 \frac{5}{6}= \frac{91}{9}- \frac{23}{6} =\frac{182}{18}- \frac{69}{18}= \frac{113}{18} [/tex]

That's how much flour she spent on both cakes. If we subtract from this what she spent on the banana cake 2 3/4 we will be left with what she used for the chocolate cake.

This is found as follows: [tex] \frac{113}{18} -2 \frac{3}{4} = \frac{113}{18} - \frac{11}{4} = \frac{226}{36} - \frac{99}{36} = \frac{127}{36}=3 \frac{19}{36} [/tex]

She used [tex]3 \frac{19}{36} [/tex] pounds of flour on the chocolate cake.