The School Tuck Shop Change
A tuck shop at a primary school starts the term with 480 oranges and 360 apples. On Monday it sells 125 oranges and 90 apples. On Tuesday a delivery of 200 oranges and 150 apples arrives, and the shop sells 160 oranges and 110 apples. On Wednesday the shop wants to pack fruit into bags for a class trip. Each bag must contain the same number of oranges and the same number of apples, and every bag must be identical. The shopkeeper wants to use all the remaining oranges and all the remaining apples, with none left over. She also wants the greatest possible number of bags.
1. Work out how many oranges and how many apples remain at the end of Tuesday.
Read the worked answer
Oranges: start with 480, sell 125, so 480 − 125 = 355. A delivery adds 200, giving 555. Then sell 160, so 555 − 160 = 395 oranges remain.
Apples: start with 360, sell 90, so 360 − 90 = 270. A delivery adds 150, giving 420. Then sell 110, so 420 − 110 = 310 apples remain.
So at the end of Tuesday there are 395 oranges and 310 apples.
2. Find the greatest number of identical bags the shopkeeper can make using all the remaining fruit, and state how many oranges and apples go in each bag.
Read the worked answer
We need the greatest common divisor of 395 and 310.
395 − 310 = 85. Now find GCD(310, 85). 310 = 3 × 85 + 55. 85 = 1 × 55 + 30. 55 = 1 × 30 + 25. 30 = 1 × 25 + 5. 25 = 5 × 5 + 0. So the GCD is 5.
Check: 395 ÷ 5 = 79 and 310 ÷ 5 = 62. Both divide exactly.
So the greatest number of identical bags is 5, with 79 oranges and 62 apples in each bag.
3. The shopkeeper wonders whether she could instead make 10 identical bags using all the fruit. Explain, using divisibility, why this is or is not possible.
Read the worked answer
For 10 identical bags using all the fruit, both 395 and 310 would need to be divisible by 10.
A number is divisible by 10 only if its last digit is 0. The last digit of 395 is 5, and the last digit of 310 is 0. Since 395 is not divisible by 10 (395 ÷ 10 = 39 remainder 5), the oranges cannot be shared equally among 10 bags with none left over.
So 10 bags is not possible. The greatest number that works for both is 5, as found earlier.