Change of Subject Formula

Whenever the subject of a formula is to be changed, the intention is to make the value becoming the subject stand alone and remain on the left hand side of the expression. The methods used to effect this change may be any basic mathematical principles like multiplication, addition, subtraction, division, finding square roots, etc.

Whatever operation that is used to perform the change must be applied to both sides of the expression (formula).

Examples;

1. Make u the subject of the formula; v² = u² + 2as.

Solution

Step 1

Rearrange by bringing “u²” to the LHS of the expression and carrying v² to the RHS of the expression. The reason for this is because you have been instructed to make “u” the subject of the formula according to the question.

-u² = 2as – v²

Note;

Whenever you carry figures from one side of an expression to another, the sign changes to the opposite.

Step 2

Multiply both sides by -1 in order to eliminate the – (minus sign);

u² = 2as + v²

Step 3

Take square root of both sides;

2. Change the subject of the formula below to h;

Solution

Step 1

The target value here is “h”. So in order to make “h” subject, you need to remove the square root by taking square of both sides;

Step 2

Cross multiply both sides;

2Qh = 100p²

Step 3

Divide both sides by 2Q;

Final answer will be;

3. Express S in terms of T, X and K in the formula below;

Solution

Step 1

Remove the square root by taking square of both sides;

Step 2

Split the RHS expression apart to make it easier for you to solve. See below;

Step 3

Expand (2X) ² and remove the square root;

Note; raise to power of 2 can easily cancel square root. This is why: (2 × ½ gives 1 where ½ is the mathematical interpretation of square root)

Step 4

Cross multiply both sides;

4SX² X 1 = T² X K

4SX² = KT²

Step 5

Divide both sides by 4X² because you need “S” to be alone;

Lastly, we need to find S when T = 8, X = 2 and K = 3. The only way to do this is by substituting for T, X and K respectively in our new formula.

S = 3 × 4
Note; 16 goes in 64 gives 4

S = 12. (Answer)

Practice these questions below;

1. Make d the subject of the formula below; 

2. Make m the subject of the expression below;

3. Make v the subject of the relation;

4. Given that v² = u² + 2as, express a in terms of u, v and s. Hence, find the value of a when u = 15, v = 20 and s = 5.

5. An expression is given as

Make R the subject of the formula and find R when K = 20, q = 15 and D = 6.

Ratio and Percentage

A ratio is a numerical method of comparing two quantities of the same type, such as sum of money, length, weight, ages, marks, etc…

As ratio is a comparison of relative magnitude of two quantities, it may not necessarily contain units. It may be in fraction or separated by column in between them. For example; ^P/_Q or P:Q which is pronounced as P ratio Q. A ratio is always numbers.

The following rules should be implemented when dealing with ratios.

Rule 1; If ^a/_b and ^c/_d are two different ratios. Let ^a/_b = ^c/_d

Therefore, ad = bc (when cross multiplied)

Rule 2; If ^a/_b = ^c/_d

Therefore, ^b/_a = ^d/_c (alternating the ratios)

Rule 3; If ^a/_b = ^c/_d

Therefore, ^a/_c = ^b/_d (inverting the ratios)

Examples;

1. In a certain class, the ratio of boys to girls is 2:5. If there are 40 boys, find how many girls are there.

Solution

Let the no of girls be x

Then 2:5 = 40:x

²/₅ = ⁴⁰/_x (cross multiply)

2x = 40 X 5

2x = 200 (divide both sides by 2)

x = ²⁰⁰/₂

Therefore, x = 100.

This means that there are 100 girls in the class.

2. Which ratio is greater, 6:8 or 12:22?

Solution

6:8 is the same as ⁶/₈.

Divide 6 by 8 and the answer is 0.75.

While 12:22 is the same as ¹²/₂₂.

Divide 12 by 22 and the answer is 0.545.

Which is bigger between 0.75 and 0.545?

Obviously it is 0.75.

Therefore this means that 6:8 is greater than 12:22.

 

3. Decrease #110.81 in the ratio 4:7.

Solution

In order to decrease #110.81, multiply it by ⁴/₇.

#110.81 X ⁴/₇

This can be written as; 

= #63.92 (final answer)

Practice these questions below;

1. The ratio of the circumference of a circle to its diameter is 22:7. What is the circumference of a circle of diameter 15.6m?

2. In each class, find which of the two ratios is greater;

(a) 16 : 7  or  17 : 6

(b) 2.5g : 2kg  or  0.4kg : 300kg

(c) #1.60 : #4  or  #6 : #11.

3. Divide 490 in the ratio 2:5:7.

4. In preparing a recipe for cake, the ratio of the flour to sugar is 40:3. Find the required amount of flour to 18kg of sugar.

5. A worker’s income is increased in the ratio 36:30. Find the increase percent.

FRACTIONS

A fraction is part of a whole number which contains a numerator and denominator. A numerator is the figure placed on top while the denominator is the figure placed below. There are three types of fractions which are;

1. Proper Fraction; This is a fraction which has its numerator lower than the denominator E.g. ½, ²/₅, ⁷/₉, etc.

2. Improper Fraction; This has its numerator higher (bigger) than the denominator. E.g.  ³/₂, ⁴/₃, ⁹/₅, etc.

3. Mixed Fractions; This consists of a whole number and a proper fraction written together. E.g. 1¹/₂, 2³/₄, 5⁶/₇, etc.

Solving any question related to fractions shouldn’t be difficult if students can master the term BODMAS. BODMAS stands for the order which arithmetic are carried out, which are as follows;

B; Bracket

O; Of

D; Division

M; Multiplication

A; Addition

S; Subtraction

Example 1; Simplify 5²/₃ ÷ (23²/₅ of 3¹/₄ – 22)

Note;

Before applying the BODMAS rule to this question, you will have to convert all the mixed fractions to proper fractions. You can do these by multiplying the whole number with first the denominator and second add the result with the numerator and finally dividing your total result with the denominator. (This is the format of conversion; Whole Number × Denominator + Numerator ÷ Denominator).

= ¹⁷/₃ ÷ (¹¹⁷/₅ of ¹³/₄ – 22)

Applying the rule means you’ll first deal with the ones in bracket where “of” stands for multiplication.

= ¹⁷/₃ ÷ (¹¹⁷/₅ × ¹³/₄ – 22)

= ¹⁷/₃ ÷ (¹⁵²¹/₂₀ – 22)

Find the L.C.M of the fractions in bracket;

= ¹⁷/₃ ÷ (¹⁵²¹/₂₀ – ⁴⁴⁰/₂₀)

= ¹⁷/₃ ÷ (¹⁰⁸¹/₂₀)

Remove the bracket;

= ¹⁷/₃ ÷ ¹⁰⁸¹/₂₀

Note;

It is impossible to divide two fractions, so you’ll have to change the mathematical sign to multiplication (It’s a rule in mathematics) which makes the fraction to the right take an inverse form.

= ¹⁷/₃ × ²⁰/₁₀₈₁

= ³⁴⁰/₃₂₄₃ (Answer).

Example 2; Evaluate

Convert all mixed fractions to proper fractions;

 

 

Find the L.C.M of both fractions above and below;

 

= ¹¹/₄ ÷ ¹¹/₈ (Change division sign to multiplication sign just like example 1 above)

= ¹¹/₄ × ⁸/₁₁ (11 cancels 11 while 4 in 8 gives 2)

= 2 (Answer).

Practice these questions below;

Simplify the following;

1. (a) 2³/₄ + 3²/₅ – 1¹/₂

    (b) 2¹/₆ + (3³/₅ ÷ 1¹/₈)

    (c) (2⁵/₆ – 3¹/₂) ÷ 1¹/₂ of 5²/₃

2. Simplify this fraction; 

3. (a) ⁵/₉ ÷ (1³/₈ – ¹/₃)

    (b)  5¹/₃ ÷ (4⁵/₆ – 3¹/₅)

    (c) 2¹/₂ ÷ (2¹/₄ ÷ 4¹/₃)

 

4. Simplify this fraction below;