Rearranging and substituting: do as directed

Exam papers often give a for­mula with an instruc­tion such as "make $t$ the sub­ject, then find $t$ when $v = 30$, $u = 10$ and $a = 4$". Ques­tions of this kind are usu­ally headed "do as directed": you must do exactly what is asked, in the order asked. This les­son puts the whole chap­ter together: rearrange first, then sub­sti­tute, then check that the answer makes sense.

Two ways to find an unknown#

Sup­pose you know every quan­tity in a for­mula except one. There are two ways to find it.

• Sub­sti­tute first, then solve. Put the known num­bers in, which gives a lin­ear equa­tion in the unknown, and solve it.
• Rearrange first, then sub­sti­tute. Make the unknown the sub­ject, then put the num­bers in.

Both give the same answer. Rear­rang­ing first is bet­ter when you will use the for­mula many times, or when the ques­tion tells you to; sub­sti­tut­ing first is often quicker for a sin­gle cal­cu­la­tion.

Exam­ple 1: both ways#

The area of a trapez­ium is $\displaystyle A = \frac{1}{2}(a + b)h$. Find $h$ when $A = 60$, $a = 7$ and $b = 5$.

Sub­sti­tut­ing first:

$$\displaystyle \begin{aligned} 60 &= \frac{1}{2}(7 + 5)h \\ 60 &= 6h \\ h &= 10 \end{aligned}$$

Rear­rang­ing first, using $\displaystyle h = \frac{2A}{a + b}$ from ear­lier in the chap­ter:

$$\displaystyle h = \frac{2 \times 60}{7 + 5} = \frac{120}{12} = 10$$

The same answer, as it must be.

Rearrange, then sub­sti­tute#

Exam­ple 2#

Make $t$ the sub­ject of $v = u + at$, and find $t$ when $v = 30$, $u = 10$ and $a = 4$.

$$\displaystyle \begin{aligned} v - u &= at \\ t &= \frac{v - u}{a} \end{aligned}$$

Now sub­sti­tute:

$$\displaystyle t = \frac{30 - 10}{4} = \frac{20}{4} = 5$$

Exam­ple 3: body tem­per­a­ture#

Make $C$ the sub­ject of $\displaystyle F = \frac{9C}{5} + 32$, and change $98.6$ degrees Fahren­heit to Cel­sius.

From the third les­son of this chap­ter, $\displaystyle C = \frac{5(F - 32)}{9}$. Then

$$\displaystyle C = \frac{5(98.6 - 32)}{9} = \frac{5 \times 66.6}{9} = \frac{333}{9} = 37$$

So $98.6$ degrees Fahren­heit is $37$ degrees Cel­sius, the nor­mal tem­per­a­ture of the human body.

Exam­ple 4: rate of inter­est#

Sita invested ₹7500 for $3$ years and earned ₹1350 sim­ple inter­est. Make $R$ the sub­ject of $\displaystyle I = \frac{PRT}{100}$ and find the rate.

$$\displaystyle R = \frac{100I}{PT} = \frac{100 \times 1350}{7500 \times 3} = \frac{135000}{22500} = 6$$

The rate is $6$ per cent a year.

Exam­ple 5: radius of a tank#

A cylin­dri­cal water tank holds $1540$ cubic m and is $10$ m tall. Find its radius, using $V = \pi r^{2}h$ and $\displaystyle \pi = \frac{22}{7}$.

From the fourth les­son of this chap­ter, $\displaystyle r = \sqrt{\frac{V}{\pi h}}$. First work out the frac­tion under the root:

$$\displaystyle \frac{V}{\pi h} = \frac{1540}{\frac{22}{7} \times 10} = 1540 \times \frac{7}{220} = 49$$

So $r = \sqrt{49} = 7$. The radius is $7$ m.

Exam­ple 6: radius from cir­cum­fer­ence#

Make $r$ the sub­ject of $C = 2\pi r$, and find the radius of a wheel whose cir­cum­fer­ence is $88$ cm. Take $\displaystyle \pi = \frac{22}{7}$.

$$\displaystyle r = \frac{C}{2\pi} = 88 \div \frac{44}{7} = 88 \times \frac{7}{44} = 14$$

The radius is $14$ cm.

When you need the for­mula many times#

Exam­ple 7#

An auto charges ₹30 plus ₹15 a kilo­me­tre, so the fare is $F = 30 + 15k$. Three pas­sen­gers paid ₹90, ₹165 and ₹255. How far did each travel?

Rearrange once:

$$\displaystyle \begin{aligned} F - 30 &= 15k \\ k &= \frac{F - 30}{15} \end{aligned}$$

Then use it three times:

They trav­elled $4$ km, $9$ km and $15$ km. Rear­rang­ing once saved solv­ing three sep­a­rate equa­tions.

Does the answer make sense?#

After sub­sti­tut­ing, ask whether the answer is sen­si­ble. A length, a time or a mass must be pos­i­tive. A num­ber of peo­ple must be a whole num­ber. And look out for val­ues a for­mula can­not take:

• In $\displaystyle x = \frac{3y + 2}{y - 1}$, $y$ can­not be $1$, because the denom­i­na­tor would be zero.
• In $\displaystyle r = \sqrt{\frac{A}{\pi}}$, $A$ can­not be neg­a­tive, because a neg­a­tive num­ber has no real square root. That is no sur­prise: an area is never neg­a­tive.

Prac­tice#

Do as directed. Take $\displaystyle \pi = \frac{22}{7}$ where needed.

1. Make $r$ the sub­ject of $C = 2\pi r$, and find $r$ when $C = 132$.
2. Make $b$ the sub­ject of $P = 2(l + b)$, and find $b$ when $P = 50$ and $l = 14$.
3. Make $T$ the sub­ject of $\displaystyle I = \frac{PRT}{100}$, and find $T$ when $I = 2160$, $P = 9000$ and $R = 8$.
4. Make $F$ the sub­ject of $\displaystyle C = \frac{5(F - 32)}{9}$, and find $F$ when $C = 25$.
5. Make $h$ the sub­ject of $V = \pi r^{2}h$, and find $h$ when $V = 924$ and $r = 7$.
6. Make $a$ the sub­ject of $v^{2} = u^{2} + 2as$, and find $a$ when $v = 20$, $u = 10$ and $s = 50$.
7. Make $s$ the sub­ject of $A = s^{2}$, and find $s$ when $A = 2.25$.
8. Make $u$ the sub­ject of $v = u + at$, and find $u$ when $v = 18$, $a = 2.5$ and $t = 4$.
9. Make $x$ the sub­ject of $\displaystyle y = \frac{x + 2}{x - 3}$, and find $x$ when $y = 6$.
10. Make $R$ the sub­ject of $\displaystyle \frac{1}{R} = \frac{1}{a} + \frac{1}{b}$, and find $R$ when $a = 6$ and $b = 3$.
11. Make $n$ the sub­ject of $\displaystyle S = \frac{n}{2}(a + l)$, and find $n$ when $S = 210$, $a = 3$ and $l = 39$.

Answers#