A factory has ordered 58 wheels for the small tempos that they make. Each tempo has 3 wheels. In how many tempos can they fix the wheels?
Given: 58 wheels, 3 wheels per tempo
Formula: Number of tempos = Total wheels ÷ Wheels per tempo
Calculation: Using partial quotient method: 30 wheels = 10 tempos (58 - 30 = 28 wheels left) 15 wheels = 5 tempos (28 - 15 = 13 wheels left) 9 wheels = 3 tempos (13 - 9 = 4 wheels left) 3 wheels = 1 tempo (4 - 3 = 1 wheel left) Total tempos = 10 + 5 + 3 + 1 = 19 tempos
Answer: The factory can fix wheels in 19 tempos, with 1 wheel left over.
Explanation
The chapter introduces division using the partial quotient method. For 58 ÷ 3: Take out groups of 10s, 5s, or other easy multiples. 30 wheels = 10 tempos (58-30=28 left). 15 wheels = 5 tempos (28-15=13 left). 9 wheels = 3 tempos (13-9=4 left). 3 wheels = 1 tempo (4-3=1 left). Total tempos = 10+5+3+1 = 19, with 1 wheel remaining.
Solution Steps
Step 1: 58 wheels ÷ 3 wheels per tempo
Step 2: 30 wheels = 10 tempos (58-30=28 wheels left)
Step 3: 15 wheels = 5 tempos (28-15=13 wheels left)
Step 4: 9 wheels = 3 tempos (13-9=4 wheels left)
Step 5: 3 wheels = 1 tempo (4-3=1 wheel left)
Step 6: Total tempos = 10+5+3+1 = 19, remainder = 1 wheel