Place Value Word Problems worksheet with a lightning bolt theme and four math challenges for students.

Problem Analysis:

The worksheet contains several word problems related to place value. Let's solve each problem step by step.

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#### Problem 1:
What number has 7 tens and 3 less ones than tens?

- Step 1: Identify the number of tens.
- The problem states there are 7 tens.
- In place value, 7 tens = \( 7 \times 10 = 70 \).

- Step 2: Determine the number of ones.
- The problem states there are 3 less ones than tens.
- Since there are 7 tens, the number of ones is \( 7 - 3 = 4 \).

- Step 3: Combine the tens and ones to form the number.
- The number is \( 70 + 4 = 74 \).

Answer for Problem 1:
\[
\boxed{74}
\]

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#### Problem 2:
What number has 8 thousands and 4 less hundreds than thousands?

- Step 1: Identify the number of thousands.
- The problem states there are 8 thousands.
- In place value, 8 thousands = \( 8 \times 1000 = 8000 \).

- Step 2: Determine the number of hundreds.
- The problem states there are 4 less hundreds than thousands.
- Since there are 8 thousands, the number of hundreds is \( 8 - 4 = 4 \).
- In place value, 4 hundreds = \( 4 \times 100 = 400 \).

- Step 3: Combine the thousands and hundreds to form the number.
- The number is \( 8000 + 400 = 8400 \).

Answer for Problem 2:
\[
\boxed{8400}
\]

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#### Problem 3:
Jason wrote down a three-digit number and gave the class two clues to guess it.
- Clue A: The hundreds place digit is 8 greater than the ones place digit.
- Clue B: The tens place digit is 4 greater than the ones place digit.
- Clue C: The sum of all three places is 15.

Let the three-digit number be represented as \( \overline{abc} \), where:
- \( a \) is the hundreds place digit,
- \( b \) is the tens place digit,
- \( c \) is the ones place digit.

- Step 1: Translate Clue A into an equation.
- "The hundreds place digit is 8 greater than the ones place digit."
- This gives: \( a = c + 8 \).

- Step 2: Translate Clue B into an equation.
- "The tens place digit is 4 greater than the ones place digit."
- This gives: \( b = c + 4 \).

- Step 3: Translate Clue C into an equation.
- "The sum of all three places is 15."
- This gives: \( a + b + c = 15 \).

- Step 4: Substitute \( a \) and \( b \) from Clues A and B into Clue C.
- From Clue A: \( a = c + 8 \).
- From Clue B: \( b = c + 4 \).
- Substitute into \( a + b + c = 15 \):
\[
(c + 8) + (c + 4) + c = 15
\]
Simplify:
\[
c + 8 + c + 4 + c = 15
\]
\[
3c + 12 = 15
\]
\[
3c = 3
\]
\[
c = 1
\]

- Step 5: Solve for \( a \) and \( b \).
- Using \( a = c + 8 \):
\[
a = 1 + 8 = 9
\]
- Using \( b = c + 4 \):
\[
b = 1 + 4 = 5
\]

- Step 6: Form the three-digit number.
- The number is \( \overline{abc} = 951 \).

Answer for Problem 3:
\[
\boxed{951}
\]

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#### Problem 4:
Michelle typed a 5-digit number on her iPad and gave her friends 3 clues to determine the number.
- Clue A: The highest place, the hundreds place value, and the ones place all have the same digit.
- Clue B: The tens place value is one less than the thousands place.
- Clue C: The hundreds place value is 9 greater than the tens place.

Let the 5-digit number be represented as \( \overline{abcde} \), where:
- \( a \) is the ten-thousands place digit,
- \( b \) is the thousands place digit,
- \( c \) is the hundreds place digit,
- \( d \) is the tens place digit,
- \( e \) is the ones place digit.

- Step 1: Translate Clue A into an equation.
- "The highest place, the hundreds place value, and the ones place all have the same digit."
- This gives: \( a = c = e \).

- Step 2: Translate Clue B into an equation.
- "The tens place value is one less than the thousands place."
- This gives: \( d = b - 1 \).

- Step 3: Translate Clue C into an equation.
- "The hundreds place value is 9 greater than the tens place."
- This gives: \( c = d + 9 \).

- Step 4: Substitute \( c = a \) and \( c = e \) into Clue C.
- From Clue C: \( c = d + 9 \).
- Since \( c = a \), we have \( a = d + 9 \).

- Step 5: Express \( d \) in terms of \( b \) using Clue B.
- From Clue B: \( d = b - 1 \).

- Step 6: Substitute \( d = b - 1 \) into \( a = d + 9 \).
- \( a = (b - 1) + 9 \)
- Simplify:
\[
a = b + 8
\]

- Step 7: Determine possible values for \( b \).
- Since \( a \) is a digit (0–9), and \( a = b + 8 \), \( b \) must be a digit such that \( b + 8 \leq 9 \).
- The only possible value for \( b \) is \( b = 1 \).

- Step 8: Solve for \( a \), \( c \), \( d \), and \( e \).
- If \( b = 1 \):
- \( a = b + 8 = 1 + 8 = 9 \)
- \( d = b - 1 = 1 - 1 = 0 \)
- \( c = d + 9 = 0 + 9 = 9 \)
- \( e = c = 9 \)

- Step 9: Form the 5-digit number.
- The number is \( \overline{abcde} = 91909 \).

Answer for Problem 4:
\[
\boxed{91909}
\]

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Final Answers:

1. \(\boxed{74}\)
2. \(\boxed{8400}\)
3. \(\boxed{951}\)
4. \(\boxed{91909}\)