Math Snap

STEP 1

Assumptions1. The fraction of students who got an A is $\frac{1}{3}$
. The fraction of students who got a B is $\frac{}{5}$
3. The fractions represent distinct groups of students, i.e., a student cannot get both an A and a B.

STEP 2

To find the fraction of students who got either an A or a B, we need to add the fraction of students who got an A to the fraction of students who got a B.
$$Fraction\, of\, students\, who\, got\, A\, or\, B = Fraction\, of\, students\, who\, got\, A + Fraction\, of\, students\, who\, got\, B$$

STEP 3

Now, plug in the given values for the fraction of students who got an A and the fraction of students who got a B.
$$Fraction\, of\, students\, who\, got\, A\, or\, B = \frac{1}{3} + \frac{2}{5}$$

STEP 4

To add these fractions, we need to find a common denominator. The least common denominator (LCD) of3 and is15.

STEP 5

Convert the fractions to have the common denominator.
$$\frac{1}{3} = \frac{1 \times5}{3 \times5} = \frac{5}{15}$$$$\frac{2}{5} = \frac{2 \times3}{5 \times3} = \frac{}{15}$$

STEP 6

Now, add the two fractions.
$$Fraction\, of\, students\, who\, got\, A\, or\, B = \frac{5}{15} + \frac{6}{15}$$

Calculate the sum.
$$Fraction\, of\, students\, who\, got\, A\, or\, B = \frac{5 +6}{15} = \frac{11}{15}$$So, $\frac{11}{15}$ of the students got either an A or a B.