Let's solve each problem step by step.
---
Problem 1:
I have eight in the ones place. I am greater than 70 but less than 89. What number am I?
- The number has an
8 in the ones place.
- The number is between
70 and
89 (exclusive).
The possible numbers are:
- 78 (since it is the only number between 70 and 89 that ends with 8).
Thus, the number is:
\[
\boxed{78}
\]
---
Problem 2:
I have a two in the ones place. I am less than 13 but greater than 11. What number am I?
- The number has a
2 in the ones place.
- The number is between
11 and
13 (exclusive).
The only number that fits these criteria is:
- 12 (since it is the only number between 11 and 13 that ends with 2).
Thus, the number is:
\[
\boxed{12}
\]
---
Problem 3:
I have a nine in the ones place. I am less than 50 but greater than 47. What number am I?
- The number has a
9 in the ones place.
- The number is between
47 and
50 (exclusive).
The only number that fits these criteria is:
- 49 (since it is the only number between 47 and 50 that ends with 9).
Thus, the number is:
\[
\boxed{49}
\]
---
Problem 4:
I am an even number with two digits. I have two in the ones place and six in the tens place. What number am I?
- The number has a
6 in the tens place.
- The number has a
2 in the ones place.
- The number must be
even (which is already satisfied since the ones digit is 2).
Combining these, the number is:
- 62.
Thus, the number is:
\[
\boxed{62}
\]
---
Problem 5:
I have two digits with an odd number in the ones place. When you add the two digits I am made up of, the sum is 5. What number am I?
- The number has
two digits.
- The ones digit is
odd.
- The sum of the two digits is
5.
Let the number be represented as \(10a + b\), where:
- \(a\) is the tens digit,
- \(b\) is the ones digit.
From the problem:
1. \(b\) is odd.
2. \(a + b = 5\).
Since \(b\) is odd, the possible values for \(b\) are: 1, 3, or 5.
#### Case 1: \(b = 1\)
- \(a + 1 = 5 \implies a = 4\).
- The number is \(41\).
#### Case 2: \(b = 3\)
- \(a + 3 = 5 \implies a = 2\).
- The number is \(23\).
#### Case 3: \(b = 5\)
- \(a + 5 = 5 \implies a = 0\).
- This is not a valid two-digit number.
Thus, the possible numbers are \(41\) and \(23\). However, the problem does not specify additional constraints to choose between them. Assuming the problem intends a single solution, we can present both possibilities, but typically, the first valid solution is preferred.
Thus, the number is:
\[
\boxed{41}
\]
---
Final Answers:
1. \(\boxed{78}\)
2. \(\boxed{12}\)
3. \(\boxed{49}\)
4. \(\boxed{62}\)
5. \(\boxed{41}\)
Parent Tip: Review the logic above to help your child master the concept of place value worksheet word problems.
