Divide numbers ending in zeroes: word problems
key points
Let’s go through the process with step-by-step explanations:
Problem 1: Sharing Chocolates
Problem: Sarah has 120 chocolates. She wants to divide them equally among 6 friends. How many chocolates will each friend get?
Steps:
- Write the division problem: ( 120 \div 6 )
- Ignore the zero and divide the remaining numbers:
- Remove the zero from 120, leaving 12.
- Divide 12 by 6.
- ( 12 \div 6 = 2 )
- Add the zero back to the quotient:
- Since we divided 12 by 6 and got 2, add the zero back to get 20.
- Conclusion: Each friend will get 20 chocolates.
Problem 2: Distributing Pencils
Problem: There are 500 pencils in a box. A teacher wants to distribute them equally among 10 students. How many pencils will each student get?
Steps:
- Write the division problem: ( 500 \div 10 )
- Ignore the zeroes and divide the remaining numbers:
- Remove one zero from each number, leaving 50 and 1.
- Divide 50 by 1.
- ( 50 \div 1 = 50 )
- Conclusion: Each student will get 50 pencils.
Problem 3: Packing Books
Problem: A bookstore packs 2,400 books equally into 30 boxes. How many books will be in each box?
Steps:
- Write the division problem: ( 2,400 \div 30 )
- Simplify by removing one zero from both the dividend and the divisor:
- Remove one zero from 2,400, leaving 240.
- Remove one zero from 30, leaving 3.
- Divide the simplified numbers:
- ( 240 \div 3 )
- Divide 240 by 3:
- 240 divided by 3 is 80.
- Conclusion: Each box will contain 80 books.
Problem 4: Dividing Apples
Problem: A farmer has 6,000 apples. He wants to pack them into 15 crates equally. How many apples will each crate have?
Steps:
- Write the division problem: ( 6,000 \div 15 )
- Simplify the division problem:
- Notice that 6,000 and 15 can both be simplified.
- Divide 6,000 by 15:
- First, simplify by removing a common factor. In this case, we can divide both by 5 to make it easier:
- ( 6,000 \div 5 = 1,200 )
- ( 15 \div 5 = 3 )
- Divide the simplified numbers:
- Now, we have ( 1,200 \div 3 ).
- Divide 1,200 by 3:
- 1,200 divided by 3 is 400.
- Conclusion: Each crate will contain 400 apples.
Recap:
- For each division problem involving numbers ending in zeroes, we simplified the problem by removing the zeroes first.
- Then, we performed the division on the simplified numbers.
- Finally, we added back the appropriate number of zeroes to the quotient, if necessary.
These steps should help 6th grade students understand how to handle division problems involving numbers ending in zeroes.
🔔 A city has ₹36,000 to buy new light bulbs for the street lamps. If each light bulb costs ₹6,
how many new light bulbs will the city be able to buy?
Divide the total amount of money by the cost of each light bulb. Divide ₹36,000 by ₹6.
First divide 36 by 6.
36 ÷ 6 = 6
Now count the zeroes.
36,000 ÷ 6 = ?
There are 3 zeroes. Write 3 zeroes in the answer.
₹36,000 ÷ ₹6 = 6,000
The city will be able to buy 6,000 light bulbs.
🔔 A real estate agent has ₹4,800 to spend on newspaper ads. If each ad costs ₹8,
how many ads will the real estate agent be able to buy?
Divide the amount of money by the cost of each newspaper ad. Divide ₹4,800 by ₹8.
First, divide 48 by 8.
48 ÷ 8 = 6
Now count the zeroes.
4,800 ÷ 8 = ?
There are 2 zeroes. Write 2 zeroes in the answer.
₹4,800 ÷ ₹8 = 600
The real estate agent will be able to buy 600 ads.
🔔 Monroe County has ₹8,000 to buy new street signs. If each sign costs ₹2,000,
how many new street signs will the county be able to buy?
Divide the amount of money by the cost of each street sign. Divide ₹8,000 by ₹2,000.
First, divide ₹8 by ₹2.
₹8 ÷ ₹2 = 4
Now count the zeroes in each number.
₹8,000 ÷ ₹2,000 = ?
There are the same number of blue zeroes in ₹8,000 as in ₹2,000.
₹8,000 ÷ ₹2,000 = 4
Monroe County will be able to buy 4 new street signs.
let’s practice: