Select the correct answer.

Robert is a 15-year-old boy. His recommended dietary intake for vitamin B12, C, and E is [tex]2.4 \mu g / \text{day}[/tex], [tex]75 \text{mg} / \text{day}[/tex], and [tex]15 \text{mg} / \text{day}[/tex], respectively. He eats a salmon fillet with one cup of boiled green beans and half a cup of fresh sliced strawberries for lunch. If the nutrient values of his meal components are as shown below, for which vitamins did Robert meet at least [tex]75\%[/tex] of his recommended daily allowance?

\begin{tabular}{|l|l|l|l|}
\hline
& Vitamin B12 & Vitamin C & Vitamin E \\
\hline
salmon fillet [tex]$(124 \text{g})$[/tex] & [tex]$5.87 \mu g$[/tex] & [tex]$0.0 \text{mg}$[/tex] & [tex]$0.60 \text{mg}$[/tex] \\
\hline
boiled green beans [tex]$(1 \text{cup})$[/tex] & [tex]$0.0 \mu g$[/tex] & [tex]$12.1 \text{mg}$[/tex] & [tex]$0.57 \text{mg}$[/tex] \\
\hline
strawberries, sliced [tex]$(1/2 \text{cup})$[/tex] & [tex]$0.0 \mu g$[/tex] & [tex]$49.0 \text{mg}$[/tex] & [tex]$0.42 \text{mg}$[/tex] \\
\hline
\end{tabular}

A. vitamin B12, vitamin C, and vitamin E

B. vitamin B12 and vitamin C only

C. vitamin B12 and vitamin E only

D. vitamin C and vitamin E only

E. vitamin B12 only



Answer :

Let's determine whether Robert meets at least 75% of the recommended daily intake for each vitamin based on his lunch components: salmon filet, boiled green beans, and sliced strawberries.

1. Vitamin B12:
- Salmon filet: [tex]\(5.87\, \mu g\)[/tex]
- Boiled green beans: [tex]\(0.0\, \mu g\)[/tex]
- Sliced strawberries: [tex]\(0.0\, \mu g\)[/tex]

Total Vitamin B12 intake:
[tex]\[ 5.87\, \mu g + 0.0\, \mu g + 0.0\, \mu g = 5.87\, \mu g \][/tex]

Percentage of the recommended intake:
[tex]\[ \frac{5.87\, \mu g}{2.4\, \mu g} \times 100 = 244.583\% \][/tex]

Since 244.583% is greater than 75%, Robert meets the 75% criteria for Vitamin B12.

2. Vitamin C:
- Salmon filet: [tex]\(0.0\, mg\)[/tex]
- Boiled green beans: [tex]\(12.1\, mg\)[/tex]
- Sliced strawberries: [tex]\(49.0\, mg\)[/tex]

Total Vitamin C intake:
[tex]\[ 0.0\, mg + 12.1\, mg + 49.0\, mg = 61.1\, mg \][/tex]

Percentage of the recommended intake:
[tex]\[ \frac{61.1\, mg}{75\, mg} \times 100 = 81.467\% \][/tex]

Since 81.467% is greater than 75%, Robert meets the 75% criteria for Vitamin C.

3. Vitamin E:
- Salmon filet: [tex]\(0.60\, mg\)[/tex]
- Boiled green beans: [tex]\(0.57\, mg\)[/tex]
- Sliced strawberries: [tex]\(0.42\, mg\)[/tex]

Total Vitamin E intake:
[tex]\[ 0.60\, mg + 0.57\, mg + 0.42\, mg = 1.59\, mg \][/tex]

Percentage of the recommended intake:
[tex]\[ \frac{1.59\, mg}{15\, mg} \times 100 = 10.6\% \][/tex]

Since 10.6% is less than 75%, Robert does not meet the 75% criteria for Vitamin E.

Based on the calculations, Robert meets at least 75% of the recommended daily intake for Vitamin B12 and Vitamin C only.

Therefore, the correct answer is:
B. vitamin B12 and vitamin C only