Exponential functions Practice Questions (Grade 11)

Warm-upeasy

An account starts with $\text{\char36}500$ and earns $20\%$ interest compounded yearly.

Choose the function model that correctly calculates the amount in the account after $3$ years.

The amount after $t$ years is $A(t) = P(1 + r)^{t}\textsf{.}$ Here $P = 500\textsf{,}$ $r = 0.20\textsf{,}$ and $t = 3\textsf{,}$ so $A(3) = 500(1.20)^{3} = 864\textsf{.}$

Mid-gameeasy

An $80$ -milligram sample of a radioactive isotope has a half-life of $4$ years.

Choose the function model that correctly calculates the amount remaining after $12$ years.

The remaining amount after $t$ years is $A(t) = 80\left(\dfrac{1}{2}\right)^{\textstyle\frac{t}{h}}\textsf{,}$ where $h$ is the half-life. Here $t = 12$ and $h = 4\textsf{,}$ so $A(12) = 80\left(\dfrac{1}{2}\right)^{\textstyle\frac{12}{4}} = 80\left(\dfrac{1}{2}\right)^{3} = 10\textsf{.}$

Mid-gameeasy

An account starts with $\text{\char36}100$ and earns $10\%$ interest compounded yearly.

Choose the function model that correctly calculates the amount in the account after $2$ years.

The amount after $t$ years is $A(t) = P(1 + r)^{t}\textsf{.}$ Here $P = 100\textsf{,}$ $r = 0.10\textsf{,}$ and $t = 2\textsf{,}$ so $A(2) = 100(1.10)^{2} = 121\textsf{.}$

Mid-gameeasy

A $128$ -milligram sample of a radioactive isotope has a half-life of $2$ years.

Choose the function model that correctly calculates the amount remaining after $6$ years.

The remaining amount after $t$ years is $A(t) = 128\left(\dfrac{1}{2}\right)^{\textstyle\frac{t}{h}}\textsf{,}$ where $h$ is the half-life. Here $t = 6$ and $h = 2\textsf{,}$ so $A(6) = 128\left(\dfrac{1}{2}\right)^{\textstyle\frac{6}{2}} = 128\left(\dfrac{1}{2}\right)^{3} = 16\textsf{.}$

Mid-gameeasy

An account starts with $\text{\char36}2{,}000$ and earns $10\%$ interest compounded yearly.

Choose the function model that correctly calculates the amount in the account after $3$ years.

The amount after $t$ years is $A(t) = P(1 + r)^{t}\textsf{.}$ Here $P = 2{,}000\textsf{,}$ $r = 0.10\textsf{,}$ and $t = 3\textsf{,}$ so $A(3) = 2{,}000(1.10)^{3} = 2{,}662\textsf{.}$

Mid-gameeasy

An $80$ -milligram sample of a radioactive isotope has a half-life of $5$ years.

Choose the function model that correctly calculates the amount remaining after $10$ years.

The remaining amount after $t$ years is $A(t) = 80\left(\dfrac{1}{2}\right)^{\textstyle\frac{t}{h}}\textsf{,}$ where $h$ is the half-life. Here $t = 10$ and $h = 5\textsf{,}$ so $A(10) = 80\left(\dfrac{1}{2}\right)^{\textstyle\frac{10}{5}} = 80\left(\dfrac{1}{2}\right)^{2} = 20\textsf{.}$

Mid-gameeasy

An account starts with $\text{\char36}256$ and earns $25\%$ interest compounded yearly.

Choose the function model that correctly calculates the amount in the account after $3$ years.

The amount after $t$ years is $A(t) = P(1 + r)^{t}\textsf{.}$ Here $P = 256\textsf{,}$ $r = 0.25\textsf{,}$ and $t = 3\textsf{,}$ so $A(3) = 256(1.25)^{3} = 500\textsf{.}$

Buzzer beatereasy

A $32$ -milligram sample of a radioactive isotope has a half-life of $1$ year.

Choose the function model that correctly calculates the amount remaining after $4$ years.

The remaining amount after $t$ years is $A(t) = 32\left(\dfrac{1}{2}\right)^{\textstyle\frac{t}{h}}\textsf{,}$ where $h$ is the half-life. Here $t = 4$ and $h = 1\textsf{,}$ so $A(4) = 32\left(\dfrac{1}{2}\right)^{\textstyle\frac{4}{1}} = 32\left(\dfrac{1}{2}\right)^{4} = 2\textsf{.}$

Exponential functions. Game on.

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