A greengrocer was selling apples at a penny each, chickoos at 2 for a penny…
2026
A greengrocer was selling apples at a penny each, chickoos at 2 for a penny and peanuts at 3 for a penny. A father spent 7 pennies and got the same amount of each type of fruit for each of his three children (each child receiving at least one of every kind). What did each child get?
- A.
1 apple, 2 chickoos, 2 peanuts
- B.
1 apple, 3 chickoos, 2 peanuts
- C.
1 apple, 2 chickoos, 1 peanut
- D.
1 apple, 1 chickoo, 1 peanut
Attempted by 435 students.
Show answer & explanation
Correct answer: C
CONCEPT: When one fixed amount of money is shared so that every recipient gets an identical bundle of priced items, the situation becomes a single linear equation in the per-person counts. Multiply each per-person count by the number of people and by its unit price, set the sum equal to the total spent, and look for non-negative whole-number solutions. If several integer solutions exist, the conditions stated in words select the intended one.
APPLICATION:
Let a, b and c be the apples, chickoos and peanuts each child receives. For 3 children the father buys 3a apples, 3b chickoos and 3c peanuts.
Unit prices: an apple costs 1 penny, a chickoo ½ penny (2 for a penny) and a peanut ⅓ penny (3 for a penny). Total cost: 3a×1 + 3b×½ + 3c×⅓ = 7.
Simplify to 3a + 1.5b + c = 7, then multiply through by 2 to clear the half: 6a + 3b + 2c = 14.
Search non-negative integers. Taking a = 1 gives 3b + 2c = 8, which is satisfied by b = 2 and c = 1.
So each child gets 1 apple, 2 chickoos and 1 peanut.
CROSS-CHECK:
Totals: 3 apples (3 pennies) + 6 chickoos (3 pennies) + 3 peanuts (1 penny) = 7 pennies. ✓
Other integer solutions of 6a + 3b + 2c = 14, such as a = 1, b = 0, c = 4 or a = 2, b = 0, c = 1, leave a child with none of some item. The condition that each child gets some of every type makes 1 apple, 2 chickoos, 1 peanut the intended share.