1. **Problem 12:** 5 boxes each containing 48 marbles were repacked into 8 smaller boxes. How many marbles were in each of the 8 boxes? 2. **Problem 13:** Kay has 12 peanuts. Alan has four times as many. How many peanuts do they have altogether? 3. **Problem 14:** A library charges 50¢ a day when returning books after the return date. A student borrowed 3 books and paid 9.00 in late fees. How many extra days did the student have the books? --- ### Problem 12 Solution: 1. Total marbles initially = number of boxes × marbles per box = $5 \times 48$ 2. Calculate total marbles: $$5 \times 48 = 240$$ 3. These 240 marbles are repacked into 8 smaller boxes equally, so marbles per smaller box = $$\frac{240}{8}$$ 4. Simplify the fraction: $$\frac{\cancel{240}}{\cancel{8}} = 30$$ 5. So, each smaller box contains **30** marbles. --- ### Problem 13 Solution: 1. Kay has 12 peanuts. 2. Alan has four times as many peanuts as Kay, so Alan has $$4 \times 12 = 48$$ peanuts. 3. Total peanuts = Kay's peanuts + Alan's peanuts = $$12 + 48 = 60$$ 4. So, altogether they have **60** peanuts. --- ### Problem 14 Solution: 1. The library charges 50¢ per day per book for late returns. 2. The student borrowed 3 books and paid 9.00 in late fees. 3. Let $d$ be the number of extra days the student had the books. 4. Total late fee = number of books × daily charge × days late $$9.00 = 3 \times 0.50 \times d$$ 5. Simplify the right side: $$9.00 = 1.5 \times d$$ 6. Solve for $d$: $$d = \frac{9.00}{1.5}$$ 7. Simplify the fraction: $$d = \frac{\cancel{9.00}}{\cancel{1.5}} = 6$$ 8. So, the student had the books **6** extra days. --- **Final answers:** - Problem 12: 30 marbles per smaller box - Problem 13: 60 peanuts altogether - Problem 14: 6 extra days