Problem Analysis and Solutions
The provided image contains a set of math problems from a practice test for the IAAT™ Exam. Below, I will solve each problem step by step.
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####
Problem 1:
If \( n + 9 = 13 - 10 \), then what is the value of \( n \)?
Solution:
1. Simplify the right-hand side of the equation:
\[
13 - 10 = 3
\]
So the equation becomes:
\[
n + 9 = 3
\]
2. Solve for \( n \) by isolating it:
\[
n = 3 - 9
\]
\[
n = -6
\]
Answer:
\[
\boxed{A}
\]
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####
Problem 2:
\( m \) is a whole number. If you subtract 2 from \( m \), which expression is correct?
Solution:
Subtracting 2 from \( m \) gives:
\[
m - 2
\]
Answer:
\[
\boxed{D}
\]
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####
Problem 3:
If \( 6a + 3 = 21 \) and \( 6b + 3 = 15 \), then what can we determine about \( b \)?
Solution:
1. Solve for \( a \) in the first equation:
\[
6a + 3 = 21
\]
Subtract 3 from both sides:
\[
6a = 18
\]
Divide by 6:
\[
a = 3
\]
2. Solve for \( b \) in the second equation:
\[
6b + 3 = 15
\]
Subtract 3 from both sides:
\[
6b = 12
\]
Divide by 6:
\[
b = 2
\]
3. Compare \( a \) and \( b \):
\[
a = 3 \quad \text{and} \quad b = 2
\]
Therefore, \( a > b \).
Answer:
\[
\boxed{D}
\]
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####
Problem 4:
If \( m = 7 \), then \( 4m \equiv ? \)
Solution:
Substitute \( m = 7 \) into the expression \( 4m \):
\[
4m = 4 \times 7 = 28
\]
Answer:
\[
\boxed{C}
\]
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####
Problem 5:
Solve for \( t \) when \( t + 9 = 7 + 16 \).
Solution:
1. Simplify the right-hand side of the equation:
\[
7 + 16 = 23
\]
So the equation becomes:
\[
t + 9 = 23
\]
2. Solve for \( t \) by isolating it:
\[
t = 23 - 9
\]
\[
t = 14
\]
Answer:
\[
\boxed{B}
\]
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####
Problem 6:
Solve for \( b \) when \( 43 = b + 13 \).
Solution:
1. Isolate \( b \) by subtracting 13 from both sides:
\[
43 - 13 = b
\]
\[
b = 30
\]
Answer:
\[
\boxed{B}
\]
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####
Problem 7:
Simplify \( \frac{5a - 15b}{5} \).
Solution:
Factor out 5 from the numerator:
\[
\frac{5a - 15b}{5} = \frac{5(a - 3b)}{5}
\]
Cancel the 5 in the numerator and denominator:
\[
a - 3b
\]
Answer:
\[
\boxed{B}
\]
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####
Problem 8:
If \( q = 6r \), then \( 6r - 3 = ? \)
Solution:
1. Substitute \( q = 6r \) into the expression \( 6r - 3 \):
\[
6r - 3
\]
2. Since \( q = 6r \), replace \( 6r \) with \( q \):
\[
6r - 3 = q - 3
\]
Answer:
\[
\boxed{A}
\]
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Final Answers:
1. \(\boxed{A}\)
2. \(\boxed{D}\)
3. \(\boxed{D}\)
4. \(\boxed{C}\)
5. \(\boxed{B}\)
6. \(\boxed{B}\)
7. \(\boxed{B}\)
8. \(\boxed{A}\)
Parent Tip: Review the logic above to help your child master the concept of algebra 1 practice test.
