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Henry made himself a sandwich that was [tex]\frac{11}{12}[/tex] of a foot long. He put pepperoni on [tex]\frac{3}{4}[/tex] of his sandwich. What length of the sandwich did not have pepperoni on it?

A. [tex]\frac{11}{48}[/tex] of a foot
B. [tex]\frac{5}{18}[/tex] of a foot
C. [tex]\frac{11}{16}[/tex] of a foot
D. [tex]\frac{9}{11}[/tex] of a foot


Sagot :

To find the length of Henry's sandwich that did not have pepperoni, we need to go through a few steps:

1. Determine the length of the entire sandwich:
- Henry's sandwich is [tex]\(\frac{11}{12}\)[/tex] of a foot long.

2. Determine the proportion of the sandwich that had pepperoni:
- Henry put pepperoni on [tex]\(\frac{3}{4}\)[/tex] of his sandwich.

3. Calculate the actual length of the sandwich that had pepperoni on it:
- This is done by multiplying the total length of the sandwich by the proportion that had pepperoni:
[tex]\[ \frac{11}{12} \times \frac{3}{4} \][/tex]
- Simplifying this multiplication:
[tex]\[ \frac{11}{12} \times \frac{3}{4} = \frac{11 \times 3}{12 \times 4} = \frac{33}{48} \][/tex]
- Simplifying [tex]\(\frac{33}{48}\)[/tex] (if necessary) we get:
[tex]\[ \frac{33}{48} = \frac{11}{16} \][/tex]
- So, the length of the sandwich with pepperoni is [tex]\(\frac{11}{16}\)[/tex] of a foot.

4. Find the length of the sandwich without pepperoni:
- To get the length without pepperoni, subtract the length of the sandwich with pepperoni from the total length of the sandwich:
[tex]\[ \frac{11}{12} - \frac{11}{16} \][/tex]
- To subtract these fractions, we need a common denominator. The least common multiple of 12 and 16 is 48. Converting both fractions to have this common denominator:
[tex]\[ \frac{11}{12} = \frac{44}{48} \][/tex]
[tex]\[ \frac{11}{16} = \frac{33}{48} \][/tex]
- Now, subtract the fractions:
[tex]\[ \frac{44}{48} - \frac{33}{48} = \frac{44 - 33}{48} = \frac{11}{48} \][/tex]

So, the length of the sandwich that did not have pepperoni is [tex]\(\frac{11}{48}\)[/tex] of a foot.

Among the given options, the correct answer is:

[tex]\(\boxed{\frac{11}{48} \text{ of a foot}}\)[/tex]
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