Question 1

LEVEL 4

The formula for the area, A, of a triangle is given by

A = \dfrac{1}{2}bh

Rearrange the formula to make b the subject and select the correct answer from the options below.

A: b = \dfrac{2A}{h}

B: b = \dfrac{1}{2}Ah

C: b = \dfrac{A}{2h}

D: b = 2Ah

 

CORRECT ANSWER:   A: b = \dfrac{2A}{h}

 

WORKED SOLUTION:

Firstly, we multiply both sides of the equation by 2 to get

2A = bh

Then we can divide by h to make b the subject. This gives us the answer of

b = \dfrac{2A}{h}.

Question 2

LEVEL 4

Rearrange the following formula to make T the subject.

P = \dfrac{nRT}{V}

 

Select the correct answer from the options below.

A: T = \dfrac{PV}{nR}

B: T = \dfrac{P}{VnR}

C: T = \dfrac{V}{PnR}

D: T = \dfrac{nR}{PV}

 

CORRECT ANSWER:  A: T = \dfrac{PV}{nR}

WORKED SOLUTION:

First, multiply both sides by V.

PV = nRT

Then, dividing both sides by nR makes T the subject, and gives us

\dfrac{PV}{nR} = T.

Question 3

LEVEL 4

The formula for the volume, V, of a sphere is given by

V = \dfrac{4}{3} \pi r^3

Select from the options below the correct rearrangement, for making r the subject of the formula.

A: r = \sqrt[3]{\dfrac{4V}{3 \pi}}

B: r = \sqrt[3]{\dfrac{3V}{4 \pi}}

C: r = \dfrac{4}{3 \pi} \sqrt[3]{V}

D: r = \dfrac{3 \pi}{4} \sqrt[3]{V}

 

CORRECT ANSWER:   B: r = \sqrt[3]{\dfrac{3V}{4 \pi}}

 

WORKED SOLUTION:

First, we divide both sides by \frac{4}{3} to get

\dfrac{3}{4}V = \pi r^3

Then divide both sides by \pi.

\dfrac{3V}{4 \pi} = r^3

Finally, we make r the subject by cube rooting both sides of the equation, which makes

\sqrt[3]{\dfrac{3V}{4 \pi}} = r.

Question 4

LEVEL 4

Which one of the following 4 options is a correct rearrangement of the formula to make g the subject?

T = 2 \pi \sqrt{\dfrac{L}{g}},

Select the correct answer from the list below:

A: g = \dfrac{L^2}{4 \pi^2 T}

B: g = \dfrac{LT^2}{2 \pi^2}

C: g = \dfrac{4 \pi^2 T}{L^2}

D: g = \dfrac{4 \pi^2 L}{T^2}

 

CORRECT ANSWER:    D: g = \dfrac{4 \pi^2 L}{T^2}

 

WORKED SOLUTION:

Firstly, divide by 2 \pi to get

\dfrac{T}{2 \pi} = \sqrt{\dfrac{L}{g}}

Next, square both sides.

\dfrac{T^2}{4 \pi^2} = \dfrac{L}{g}

Then, multiply both sides by g.

\dfrac{gT^2}{4 \pi^2} = L

Now multiply both sides by 4 \pi^2.

gT^2 = 4 \pi^2 L

Finally, dividing by T^2 will make g the subject, and give us the final answer of

g = \dfrac{4 \pi^2 L}{T^2}.

Question 5

LEVEL 4

Which one of the following 4 options is a correct rearrangement of the formula to make y the subject?

x = \sqrt{\dfrac{2y}{3z}}

Select the correct answer from the list below:

A: y = \dfrac{3x^2 z}{2}

B: y = \dfrac{3xz}{2}

C: y = \dfrac{3x^2}{2z}

D: y = \dfrac{2x^2 z}{3}

 

CORRECT ANSWER:    A: y = \dfrac{3x^2 z}{2}

 

WORKED SOLUTION:

Square both sides.

x^2 = {\dfrac{2y}{3z}}

Then, multiply both sides by 3z

3x^2 z = 2y

Now, divide both sides by 2

y = \dfrac{3x^2 z}{2}