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Jamb mathematics tutorial: variation

Hi guys. Today we’ll be looking to completely exhaust variation based on jamb mathematics syllabus.

If you feel you need a good textbook instead, see the recommended textbook jamb asked students to read for mathematics.

However, for those who feel they have completely exhausted this topic, click here to see tough questions on variation

Having said all that, let’s proceed.

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Objectives of jamb mathematics variation

Candidates should be able to:

i. solve problems involving direct, inverse, joint and partial variations.

ii. Solve problems on percentage increase and decrease in variation.

Jamb mathematics variation content

(a) Direct variation.

(b) Inverse variation.

(d) Partial variation.

(e) Percentage increase and decrease.

1.1 variation

In everyday life, we come across relationships between quantities. These relationships can be considered in terms of varying one quantity and knowing how the other quantities vary. For example, the variation in the cost of items and their quantities; the variation in the distance travelled by the car and the time taken; the variation in the amount of money a school receive and the population of the school etc.

Basically, there are four types of variation, namely

Direct variation
Indirect (inverse) variation
Joint variation
Partial variation

1.2 Direct variation

If a person buys a packet of biscuit for # $100\cdot00$ , then two packets will be # $200\cdot00$ , three packets will be # $300\cdot00$ and so on. The number of packet(s) can be said to be directly proportional to the price. For all the case, the value for $\frac{price}{number of packets}=$ # $100\cdot00$ . This is called the proportionality constant often represented as ‘K’ or ‘C’.

$x\propto y$ means x varies as y or x is directly proportional to y. Removing the proportionality sign $(\propto)$ and replacing it with an equality sign will give $x=ky$ where k which is the proportionality constant equals $\frac{x}{y}$

Example 1
If $1m$ of wire has a mass of $xg$ , what will be the mass of 25m of the same wire?

Solution
Length of wire varies directly as its mass

$1m$ of wire $=xg\\25m$ of wire $=(25\times x)g\\=25xg$

Example 2
If $a\propto b$ and $a=2\cdot4$ when $b=3$ , find the relationship between a and b.

Solution
Given: $a\propto b\\a=kb\cdots$ eq(i) Where k is a constant

$a=2\cdot 4$ when $b=3$

Substitute for a and b in eq(i)

2\cdot4=3k\\k=0\cdot8

Thus $a=0\cdot8b$ is the formula which connects a and b

Example 3
If $p\propto q$ and $p=4\cdot5$ when $q=12$ , find

The relationship between p and q
p when $q=16$
q when $p=2.4$

Solution
Given: $p\propto q\\p=kq\cdots$ eq(i) where k is constant

$p=4\cdot 5$ when $q=12$

Substituting for p and q in eq(i)

4\cdot5=12k\\k=\frac{4\cdot 5}{12}\\k=\frac{3}{8}

1. $p=\frac{3}{8}q$

2. when $q=16\\p=\frac{3}{8}\times16\\\therefore p=6$

3. when $p=2\cdot 4\\2\cdot4=\frac{3}{8}q\\8\times 2\cdot4=3q\\\therefore q=6\cdot 4$

1.3 Inverse variation

This is the opposite of direct variation. It has to do with the quantity of something reducing as the quantity of another is increasing. For example, if a car travels a certain distance; the greater its average speed, the less time it will take.

If $a\propto \frac{1}{b}$ , we say a varies inversely as b and the proportionality constant is given as $k=ab$ from $a=\frac{k}{b}$ .

Example 4
If $x\propto \frac{1}{y}$ and $x=6$ when $y=3$ . Find the formula that connects x and y

Solution
Given: $x\propto \frac{1}{y}$ , where k is a constant

$x=k\times \frac{1}{y}$ , where k is a constant

Thus $x=\frac{k}{y} \cdots$ eq(i)

$x=6$ when $y=3$

6=\frac{k}{3}

k=18

$x=\frac{18}{y}$ is the formula that connects x and y

Example 5}
If $\theta \propto \frac{1}{n}$ and $\theta =30°$ when $n=4$ , find

$\theta$ when $n=12$
n when $\theta=8$

Solution
Given: $\theta \propto \frac{1}{n}\\\theta=\frac{k}{n}$ where k is a constant

$\theta=90°$ when $n=4\\90=\frac{k}{4}\\k=360\\\theta=\frac{360}{n}$

1. When $n=12\\\theta=\frac{360}{12}\\\therefore \theta=30°$

2. When $\theta=8°\\8=\frac{360}{n}\\n=\frac{360}{8}\\\therefore n=45$

1.4 Joint variation

When a quantity varies as more than one quantity varies, it is termed joint variation. For example, the area of a plane sheet of paper varies as the length and breadth vary i.e the area is directly proportional to the length and breadth of the sheet of paper written as $a\propto lb$ . There are other cases, either inversely proportional to two quantities or directly proportional to one and inversely proportional to another.

Example 6
y varies jointly as x and z. When $x=2$ and $z=3$ , $y=30$

Find the equation relating x,y and z
Find y when $x=4$ and $z=6$

Solution
Given: $y\propto xz\\y=kxz$ where k is a constant

When $x=2$ , $z=3$ and $y=30\\30=k(2)(3)\\k=5$

1. y=5xz

2. When $x=4$ and $z=6\\y=5\times4\times6\\\therefore y=120$

Example 7
y varies inversely as $x^2$ and x varies directly as $z^2$ . Find the relationship between y and z, given that C is a constant

Solution
$y\propto \frac{1}{x^2}\\\therefore y=\frac{A}{x^2}\cdots$ eq(i) where A is a constant

$x\propto z^2\\x=Bz^2\cdots$ eq(i) where B is a constant

from eq(i), $x=\sqrt{\frac{A}{y}}$

from eq(ii), $x=Bz^2$

equating x

\sqrt{\frac{A}{y}}=Bz^2

squaring both sides

\frac{A}{y}=B^2z^4\\y=\frac{A}{B^2z^4}

Remember A and B are constants, so $\frac{A}{B^4}$ can be represented by another constant C

\therefore y=\frac{C}{z^4}

i.e y varies inversely as $z^4$

1.5 Partial variation

When a quantity has a relationship with other quantities and some of the quantities varies and others are kept constant, it is termed partial variation i.e partly varies and partly constant. For example, when a tailor makes a dress, the total cost depends on two things; first the cost of the cloth, secondly the amount of time it takes to make the dress.

The cost of the cloth is constant but the time taken to make the dress may vary. A simple dress will take a short time to make; a dress with a difficult pattern will take a long time. We can say that the cost is partly constant and partly varies with the amount of time taken. In algebraic form, $c=A+Kt$ where c is the cost; t is the time taken; A and K ae constants.

Example 8
R is partly constant and partly varies with E. When $R=530$ , $E=1600$ , and When $R=730$ , $E=3600$

Find the formula which connects R and E
Find R when $E=1300$

Solution
Given: $R\propto C+KE$ where C and K are constants

when $R=530$ , $E=1600\\530=C+1600K\cdots$ eq(i)

when $R=730$ , $E=3600\\730=C + 3600K\cdots$ eq(ii)

subtracting eq(ii) from eq(iii)

200=2000K\\K=\frac{1}{10}

Substitute $k=\frac{1}{10}$ in eq(ii)

530=C+1600(\frac{1}{10})\\530=C+160\\C=370

1. $R=370+\frac{1}{10}E$

2. when $E=1300\\R=370+\frac{1300}{10}\\R=370+130\\\therefore R=500$

Example 9
x is partly constant and partly varies as y. When $y=5$ , $x=7$ and when $y=7$ , $x=8$ . Find

The law of the variation
when $y=11$

Solution
$x=C+Ky$ where C and K are constants

when $x=7$ , $y=5\\7=C+5K\cdots$ eq(i)

$x=8$ , $y=7\\8=C+7K\cdots$ eq(ii)

subtract eq(ii) from eq(i)

1=2K\\K=\frac{1}{2}

substitute $k=\frac{1}{2}$ into eq(ii)

7=C+5(\frac{1}{2})\\7=C+2\cdot 5\\C=4\cdot 5

1. $x=4\cdot 5+\frac{y}{2}\\2x=9+y$

2. $y=11\\2x=9+11\\2x=20\\\therefore x=10$

1.6 Percentage increase and decrease

In direct variation, as a quantity increases by a certain per cent, another quantity increases by the same percent and for inverse variation, as a quantity increases by a given percentage, the other quantity does not reduce by equal percentage; same applies to joint variation.

Example 10
If x varies directly as y, by what amount will x increase or decrease if y increases by $20$ %

Solution
Given: $x\propto y\\x=Ky$

at $x=100$ %, $y=100$ %

$100$ % $=k\times 100$ %

$k=1\\\therefore x=y$ {the relationship connecting x and y}

$y=120$ % { $100$ % $+20$ %}

then $x=120$ % since $x=y\\\therefore$ x increases by $20$ % too

Example 11
If x varies inversely as y, by what amount will x increase or decrease if y increases by $20$ %

Solution
Given: $x\propto \frac{1}{y}\\x=\frac{K}{y}$

at $x=100$ %, $y=100$ %

$100$ % $=\frac{K}{100\%}\\k=1$

note: $100$ % $\times 100$ % $=\frac{100}{100}\times \frac{100}{100}=1\\x=\frac{1}{y}$ {relationship between x and y}

when y increases by $20$ %

$x=\frac{1}{120\%}\\120$ % $=\frac{120}{100}=1\cdot20\\x=\frac{1}{1\cdot 20}\\x=0\cdot 8333\\\therefore x=83\cdot3$ % { $0\cdot 8333 \times 100$ %}

$100$ % $-83\cdot3$ % $=16\cdot7$ %

$\therefore$ x reduces by $16\cdot7$ % as y increases by $20$ %

In all conditions, $k=1$ because all quantities are $100$ % at first. We’ll skip looking for k henceforth

Example 12
If x varies jointly as y and z, find the percentage increase or decrease in x if y increases by $16$ % and z decreases by $16$ %

Solution
Given: $x\propto yz\\x=kyz$

remember, $k=1\\\therefore x=yz$ {relationship between x,y,z}

an increase by $16$ % makes $y=116$ %, and a decrease by $16$ % makes $z=84$ %

$x= 1\cdot 16\times 0\cdot84\\x=0\cdot 9744$ i.e $x=97\cdot44$ %

$100$ % $-97\cdot44$ % $=2\cdot56$ %

$\therefore x$ decreases by $2\cdot56$ %

For solutions to more than 20 selected tough jamb past questions on this topic, click here

Click here to see the official jamb mathematics syllabus for yourself

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