GKT101: General Knowledge for Teachers – Math

1. A. $h = 12$

Let's plug in $h=12$ and see if the inequality is true.

$\begin{aligned} 6 & \geq \frac{h}{2} \\ 6 & \stackrel {?}{\geq} \frac{12}{2} \\ 6 & \geq 6 \end{aligned}$

Yes, $h =12$ is a solution!

Now let's try $h =14$.

$\begin{aligned} &6 \geq \frac{h}{2} \\ &6 \stackrel {?}{\geq} \frac{14}{2} \\ &6 \not {\geq} 7 \end{aligned}$

No, $h=14$ is not a solution.

Let's try $h =16$.

$\begin{aligned} 6 & \geq \frac{h}{2} \\ 6 & \stackrel {?}{\geq} \frac{16}{2} \\ 6 & \not{\geq} 8 \end{aligned}$

No, $h = 16$ is not a solution.

The following $h$-value satisfies the inequality $6 \geq \frac{h}{2}$:

• $h = 12$

2. B.$z =9$, C. $z =10$

Let's plug in $z=8$ and see if the inequality is true.

$\begin{aligned} &7 < \frac{z}{2}+3 \\ &7  \stackrel {?}{ < } \frac{8}{2} + 3 \\ &7 \stackrel{?}{ < } 4+3 \\ &7 \nless 7 \\ \end{aligned}$

No, $z = 8$ is not a solution.

Now let's try $z=9$.

$\begin{aligned} &7 < \frac{z}{2}+3 \\ &7 \stackrel {?}{ < } \frac{9}{2} + 3 \\ &7 \stackrel{?}{ < } 4.5 +3 \\ &7 < 7.5 \\ \end{aligned}$

Yes, $z =9$ is a solution!

Let's try $z =10$.

$\begin{aligned} &7 < \frac{z}{2}+3 \\ &7 \stackrel {?}{ < } \frac{10}{2} + 3 \\ &7 \stackrel{?}{ < } 5 +3 \\ &7  < 7.5 \\ \end{aligned}$

Yes, $z =10$ is a solution!

The following $z$-values satisfy the inequality $7 < \frac{z}{2}+3$:

• $z =9$
• $z= 10$

3. A. $f=4$

Let's plug in $f=4$ and see if the inequality is true.

$\begin{aligned} f+8 & \leq 12 \\ 4+8 & \stackrel{?}{\leq} 12 \\ 12 & \leq 12 \end{aligned}$

Yes, $f=4$ is a solution.

Now let's try $f=5$.

$\begin{aligned} f+8 & \leq 12 \\ 5+8 & \stackrel {?} { \leq } 12 \\ 13 & \leq 12 \end{aligned}$

No, $f=5$ is not a solution.

Let's try $f=6$.

$\begin{aligned} f+8 & \leq 12 \\ 6+8 & \stackrel {?} { \leq } 12 \\ 14 & \leq 12 \end{aligned}$

No, $f=6$ is not a solution.

The following $f$-values satisfy the inequality $f+8 \leq 12:$

• $f=4$

4. A. $n=10$

Let's plug in $n=10$ and see if the inequality is true.

$\begin{aligned} 3 n-7 & < 26 \\ 3(10)-7 & \stackrel{?}{ < } 26 \\ 30-7 & \stackrel{?}{ < } 26 \\ 23 & < 26 \end{aligned}$

Yes, $n=10$ is a solution!

Now let's try $n=11$.

$\begin{aligned} 3 n-7 & < 26 \\ 3(11)-7 & \stackrel{?}{ < } 26 \\ 33-7 & \stackrel{?}{ < }26 \\ 23 & \nless 26 \end{aligned}$

No, $n=11$ is not a solution.

Let's try $n=12$.

$\begin{aligned} 3 n-7 & < 26 \\ 3(12)-7 & \stackrel{?}{ < } 26 \\ 36-7 & \stackrel{?}{ < } 26 \\ 29 & \nless 26 \end{aligned}$

No, $n=12$ is not a solution.

The following $n$-values satisfy the inequality $3 n-7 < 26$:

• $n=10$