Divide numbers ending in zeroes: word problems

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:

  1. Write the division problem: ( 120 \div 6 )
  2. Ignore the zero and divide the remaining numbers:
  • Remove the zero from 120, leaving 12.
  • Divide 12 by 6.
  • ( 12 \div 6 = 2 )
  1. Add the zero back to the quotient:
  • Since we divided 12 by 6 and got 2, add the zero back to get 20.
  1. 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:

  1. Write the division problem: ( 500 \div 10 )
  2. 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 )
  1. 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:

  1. Write the division problem: ( 2,400 \div 30 )
  2. 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.
  1. Divide the simplified numbers:
  • ( 240 \div 3 )
  • Divide 240 by 3:
    • 240 divided by 3 is 80.
  1. 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:

  1. Write the division problem: ( 6,000 \div 15 )
  2. 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 )
  1. Divide the simplified numbers:
  • Now, we have ( 1,200 \div 3 ).
  • Divide 1,200 by 3:
    • 1,200 divided by 3 is 400.
  1. 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 â‚¹36,000 to buy new light bulbs for the street lamps. If each light bulb costs â‚¹6,

Divide the total amount of money by the cost of each light bulb. Divide â‚¹36,000 by â‚¹6.

First divide 36 by 6.

36 ÷ 6 = 6

Now count the zeroes.

36,000 Ã· 6 = ?

There are 3 zeroes. Write 3 zeroes in the answer.

₹36,000 Ã· â‚¹6 = 6,000

The city will be able to buy 6,000 light bulbs.

🔔 A real estate agent has â‚¹4,800 to spend on newspaper ads. If each ad costs â‚¹8,

Divide the amount of money by the cost of each newspaper ad. Divide â‚¹4,800 by â‚¹8.

First, divide 48 by 8.

48 ÷ 8 = 6

Now count the zeroes.

4,800 Ã· 8 = ?

There are 2 zeroes. Write 2 zeroes in the answer.

₹4,800 Ã· â‚¹8 = 600

The real estate agent will be able to buy 600 ads.

🔔 Monroe County has â‚¹8,000 to buy new street signs. If each sign costs â‚¹2,000,

Divide the amount of money by the cost of each street sign. Divide â‚¹8,000 by â‚¹2,000.

First, divide â‚¹8 by â‚¹2.

₹8 ÷ â‚¹2 = 4

Now count the zeroes in each number.

₹8,000 Ã· â‚¹2,000 = ?

There are the same number of blue zeroes in â‚¹8,000 as in â‚¹2,000.

₹8,000 Ã· â‚¹2,000 = 4

Monroe County will be able to buy 4 new street signs.

let’s practice: