Frank Brothers Solutions for ICSE Class 9 Mathematics Chapter 6 Changing The Subject Of A Formula

Exercise 6.1

Question 1. Express simple interest \( I \) as a formula where the principal sum is \( A \), the time in years is \( I \), and the rate is \( R \) percent.
Answer: We can represent the simple interest as \( I \). The formula for simple interest is calculated by multiplying the principal sum, the number of years, and the interest rate percentage, then dividing by 100. Let the principal sum be \( A \), the number of years be \( I \), and the rate percentage be \( R \). Based on this information, the equation is:
\( \implies I = \frac{A \times I \times R}{100} \)
In simple words: Write a formula where the simple interest \( I \) is equal to the principal \( A \) times the years \( I \) times the rate \( R \), divided by 100.

Exam Tip: Ensure all variables are clearly defined before formulating the algebraic expression.

Question 2. Write a formula for the volume \( V \), which is equal to one-third of \( \pi \) times the cube of the radius \( r \).
Answer: Let the radius be \( r \). This means the cube of the radius is \( r^3 \). If we take one-third of \( \pi \) multiplied by this cubed radius, we get \( \frac{1}{3}\pi r^3 \). Thus, we can express the volume \( V \) with this equation:
\( \implies V = \frac{1}{3}\pi r^3 \)
In simple words: The volume is one-third multiplied by \( \pi \) and the radius cubed.

Exam Tip: Pay close attention to powers like "cube" (power of 3) versus "square" (power of 2) when setting up formulas.

Question 3. Write a formula to convert centigrade temperature \( C \) into Fahrenheit temperature \( F \), where the Fahrenheit temperature is 32 more than nine-fifths of the centigrade temperature.
Answer: Let the temperature in centigrade be \( C \). Nine-fifths of this temperature is written as \( \frac{9}{5}C \). Adding 32 to this value gives \( \frac{9}{5}C + 32 \). Based on this relationship, the Fahrenheit temperature \( F \) is expressed as:
\( \implies F = \frac{9}{5}C + 32 \)
In simple words: To find the Fahrenheit temperature, multiply the centigrade temperature by nine-fifths and then add thirty-two.

Exam Tip: When a question mentions "more than," it indicates addition, which is usually written after the main term.

Question 4. Write a formula for the mean \( M \) of five numbers \( a, b, c, d, \text{ and } e \).
Answer: The sum of the numbers \( a, b, c, d, \text{ and } e \) is \( a+b+c+d+e \). There are 5 separate values in total. Dividing this total sum by the count of values gives \( \frac{a+b+c+d+e}{5} \). Thus, the mean \( M \) is:
\( \implies M = \frac{a+b+c+d+e}{5} \)
In simple words: Find the average by adding all five numbers together and then dividing that total by five.

Exam Tip: The average or mean is always the sum of all elements divided by the total number of elements.

Question 5. Write a formula relating the focal length \( f \), object distance \( u \), and image distance \( v \), where the reciprocal of the focal length is equal to the sum of the reciprocals of the object distance and image distance.
Answer: Let the distance of the object be \( u \) and the distance of the image be \( v \). The inverse of these distances are \( \frac{1}{u} \) and \( \frac{1}{v} \), respectively. Adding these two fractions gives \( \frac{1}{u} + \frac{1}{v} \). Let the focal length be \( f \), so its inverse is \( \frac{1}{f} \). Therefore, the formula is:
\( \implies \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \)
In simple words: The reciprocal of focal length is found by adding the reciprocals of the object and image distances.

Exam Tip: Make sure to write "reciprocal" as 1 divided by the variable, keeping them in the denominator.

Question 6. Write a formula for the number of diagonals \( d \) drawn from a single vertex of an \( n \)-sided polygon, which is 3 less than the number of sides.
Answer: Let \( n \) represent how many sides the polygon has. If we subtract 3 from this count, we get \( n-3 \). Based on the given details, the number of diagonals \( d \) is:
\( \implies d = n-3 \)
In simple words: Subtract three from the total number of sides to find this value.

Exam Tip: Carefully translate "less 3" or "3 less than" as subtracting 3 from the variable.

Question 7. Write a formula for the area \( A \) of a ring (circular path) with outer radius \( R \) and inner radius \( r \).
Answer: Let the larger outer radius be \( R \) and the smaller inner radius be \( r \). The difference between their squared values is \( R^2 - r^2 \). Multiplying this difference by \( \pi \) gives \( \pi(R^2 - r^2) \). Therefore, the area \( A \) is represented as:
\( \implies A = \pi(R^2 - r^2) \)
In simple words: The area of the ring is \( \pi \) times the difference when you subtract the inner radius squared from the outer radius squared.

Exam Tip: In area problems of rings, always square the radii first before finding their difference, then multiply by pi.

Question 8. The cost price of each article is 30p and the selling price of each article is 20q. Find the total profit \( P \) in rupees on selling 25a articles.
Answer: The cost to buy \( 25a \) items at 30p each is \( 25a \times 30p \). The revenue from selling these \( 25a \) items at 20q each is \( 25a \times 20q \). Since profit is selling price minus cost price, and we need the answer in Rupees (converting paise to rupees by dividing by 100):
\( \implies P = \text{Rs } \frac{25a \times 20q - 25a \times 30p}{100} \)
\( \implies P = \text{Rs } \frac{50a(10q - 15p)}{100} \)
\( \implies P = \text{Rs } \frac{a(10q - 15p)}{2} \)
In simple words: Find the total money made from selling and subtract the cost to buy them. Divide by 100 to change paise into Rupees.

Exam Tip: Remember to divide by 100 to convert prices given in paise into Rupees.

Question 9. Express the total time \( T \) in minutes for a duration of \( x \) hours, \( y \) minutes, and \( z \) seconds.
Answer: We know that 1 hour contains 60 minutes, and 1 minute contains 60 seconds. Therefore, the minutes in \( x \) hours is \( 60x \). The minutes in \( y \) minutes is simply \( y \). To convert \( z \) seconds into minutes, we divide by 60, giving \( \frac{z}{60} \). Adding these values together, the total time \( T \) in minutes is:
\( \implies T = 60x + y + \frac{z}{60} \)
In simple words: Convert hours to minutes by multiplying by 60, keep the minutes as they are, and convert seconds to minutes by dividing by 60. Then add them all.

Exam Tip: When converting smaller units (seconds) to larger units (minutes), divide; when converting larger (hours) to smaller (minutes), multiply.

Question 10. A dozen apples cost Rs. \( x \) and a score of mangoes cost Rs. \( y \). Find the total cost \( C \) of 20 apples and 30 mangoes.
Answer: Since one dozen equals 12 items, 12 apples cost Rs \( x \). This means one apple costs Rs \( \frac{x}{12} \). The price for 20 apples is Rs \( \frac{20x}{12} \). A score represents 20 items, so 20 mangoes cost Rs \( y \). One mango costs Rs \( \frac{y}{20} \). The price for 30 mangoes is Rs \( \frac{30y}{20} \), which simplifies to Rs \( \frac{3y}{2} \). To find the combined cost \( C \):
\( \implies C = \frac{20x}{12} + \frac{3y}{2} \)
\( \implies C = \frac{20x + 18y}{12} \)
\( \implies C = \frac{10x + 9y}{6} \)
In simple words: Find the cost of one apple and one mango first, multiply by the quantities wanted, then add them up using a common denominator.

Exam Tip: Remember the standard conversion terms: 1 dozen is 12, and 1 score is 20.

Exercise 6.2

Question 1. Make \( R \) the subject of the formula: \( A = P\left(1 + \frac{R}{100}\right)^N \)
Answer: Starting with the given formula:
\( \implies A = P\left(1 + \frac{R}{100}\right)^N \)
Divide both sides by \( P \):
\( \implies \frac{A}{P} = \left(1 + \frac{R}{100}\right)^N \)
Take the \( N \)-th root on both sides:
\( \implies \left(\frac{A}{P}\right)^{\frac{1}{N}} = 1 + \frac{R}{100} \)
Subtract 1 from both sides:
\( \implies \left(\frac{A}{P}\right)^{\frac{1}{N}} - 1 = \frac{R}{100} \)
Multiply both sides by 100 to isolate \( R \):
\( \implies 100 \left[\left(\frac{A}{P}\right)^{\frac{1}{N}} - 1\right] = R \)
Rewriting this using radical notation:
\( \implies R = 100 \left[\sqrt[N]{\frac{A}{P}} - 1\right] \)
In simple words: Move \( P \) to the other side, take the \( N \)-th root to remove the power, subtract 1, and then multiply the whole thing by 100 to solve for \( R \).

Exam Tip: When removing a power of \( N \), always take the \( N \)-th root (or raise to power \( \frac{1}{N} \)) on both sides of the equation.

Question 2. Make \( L \) the subject of the formula: \( T = 2\pi \sqrt{\frac{L}{G}} \)
Answer: Given the time period formula:
\( \implies T = 2\pi \sqrt{\frac{L}{G}} \)
Divide both sides by \( 2\pi \):
\( \implies \frac{T}{2\pi} = \sqrt{\frac{L}{G}} \)
Square both sides to eliminate the square root:
\( \implies \left(\frac{T}{2\pi}\right)^2 = \frac{L}{G} \)
Multiply both sides by \( G \):
\( \implies G \left(\frac{T}{2\pi}\right)^2 = L \)
Simplify the squared term:
\( \implies L = \frac{G T^2}{4\pi^2} \)
In simple words: Divide by \( 2\pi \), square everything to get rid of the root, and multiply by \( G \) to leave \( L \) by itself.

Exam Tip: When squaring a fraction like \( \frac{T}{2\pi} \), remember to square both the numerator and every factor in the denominator to get \( 4\pi^2 \).

Question 3. Make \( a \) the subject of the formula: \( S = ut + \frac{1}{2}at^2 \)
Answer: Given the equation of motion:
\( \implies S = ut + \frac{1}{2}at^2 \)
Subtract \( ut \) from both sides:
\( \implies S - ut = \frac{1}{2}at^2 \)
Multiply both sides by 2 to clear the fraction:
\( \implies 2(S - ut) = at^2 \)
Divide both sides by \( t^2 \) to isolate \( a \):
\( \implies \frac{2(S - ut)}{t^2} = a \)
Rearranging the formula:
\( \implies a = \frac{2(S - ut)}{t^2} \)
In simple words: Move the \( ut \) term over, multiply by two to get rid of the fraction, and divide by \( t^2 \) to solve for \( a \).

Exam Tip: Group terms without the target variable on one side first before trying to isolate it.

Question 4. Make \( x \) the subject of the formula: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
Answer: Start with the equation of an ellipse:
\( \implies \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
Isolate the \( x \)-term by subtracting \( \frac{y^2}{b^2} \) from both sides:
\( \implies \frac{x^2}{a^2} = 1 - \frac{y^2}{b^2} \)
Multiply both sides by \( a^2 \):
\( \implies x^2 = a^2 \left(1 - \frac{y^2}{b^2}\right) \)
Combine the terms inside the parentheses into a single fraction:
\( \implies x^2 = a^2 \left(\frac{b^2 - y^2}{b^2}\right) \)
Take the square root of both sides:
\( \implies x = \sqrt{a^2 \left(\frac{b^2 - y^2}{b^2}\right)} \)
Simplify the square root of the perfect squares:
\( \implies x = \frac{a}{b}\sqrt{b^2 - y^2} \)
In simple words: Move the \( y \) fraction to the right side, multiply by \( a^2 \), make a single fraction, and take the square root to get \( x \) by itself.

Exam Tip: Simplify perfect squares like \( a^2 \) and \( b^2 \) out of the square root radical to leave the final answer in its simplest form.

Question 5. Make \( a \) the subject of the formula: \( S = \frac{a(r^n - 1)}{r - 1} \)
Answer: Given the sum of a geometric progression:
\( \implies S = \frac{a(r^n - 1)}{r - 1} \)
Multiply both sides by \( (r - 1) \) to clear the denominator:
\( \implies S(r - 1) = a(r^n - 1) \)
Divide both sides by \( (r^n - 1) \) to isolate \( a \):
\( \implies a = \frac{S(r - 1)}{r^n - 1} \)
In simple words: Multiply by the bottom part of the fraction, then divide by the bracket multiplied with \( a \) to isolate \( a \).

Exam Tip: Treat whole expressions in parentheses as single units when moving them across the equals sign.

Question 6. Make \( r_2 \) the subject of the formula: \( \frac{1}{R} = \frac{1}{r_1} + \frac{1}{r_2} \)
Answer: Given the parallel resistance formula:
\( \implies \frac{1}{R} = \frac{1}{r_1} + \frac{1}{r_2} \)
Isolate the term containing \( r_2 \) on one side:
\( \implies \frac{1}{r_2} = \frac{1}{R} - \frac{1}{r_1} \)
Find a common denominator for the right-hand side:
\( \implies \frac{1}{r_2} = \frac{r_1 - R}{r_1 R} \)
Take the reciprocal of both sides:
\( \implies r_2 = \frac{R r_1}{r_1 - R} \)
In simple words: Keep the term with \( r_2 \) on one side, combine the other two fractions into one, and flip both sides upside down.

Exam Tip: Always combine separate fractions on one side before taking the reciprocal of the entire equation.

Question 7. Make \( a \) the subject of the formula: \( x = \sqrt{\frac{a + b}{a - b}} \)
Answer: Start with the given equation:
\( \implies x = \sqrt{\frac{a + b}{a - b}} \)
Square both sides to remove the radical sign:
\( \implies x^2 = \frac{a + b}{a - b} \)
Cross-multiply by \( (a - b) \):
\( \implies x^2(a - b) = a + b \)
Expand the left-hand side:
\( \implies x^2 a - x^2 b = a + b \)
Gather all terms with \( a \) on the left and other terms on the right:
\( \implies x^2 a - a = b + x^2 b \)
Factor out \( a \) on the left and \( b \) on the right:
\( \implies a(x^2 - 1) = b(x^2 + 1) \)
Divide by \( (x^2 - 1) \) to solve for \( a \):
\( \implies a = \frac{b(x^2 + 1)}{x^2 - 1} \)
In simple words: Square both sides, multiply to get rid of the fraction, group all terms containing \( a \) together, factor \( a \) out, and divide.

Exam Tip: When the target variable appears in multiple terms, collect them all on one side and factor it out.

Question 8. Make \( y \) the subject of the formula: \( w = pq + \frac{1}{2}wy^2 \)
Answer: Given:
\( \implies w = pq + \frac{1}{2}wy^2 \)
Subtract \( pq \) from both sides:
\( \implies w - pq = \frac{1}{2}wy^2 \)
Multiply both sides by 2:
\( \implies 2(w - pq) = wy^2 \)
Divide by \( w \):
\( \implies \frac{2(w - pq)}{w} = y^2 \)
Take the square root of both sides to isolate \( y \):
\( \implies y = \sqrt{\frac{2(w - pq)}{w}} \)
In simple words: Subtract \( pq \), multiply by 2 to clear the fraction, divide by \( w \), and then take the square root of both sides.

Exam Tip: Keep track of lowercase and uppercase letters as they represent different quantities in physics and math.

Question 9. Make \( N \) the subject of the formula: \( I = \frac{NG}{R + Ny} \)
Answer: Given:
\( \implies I = \frac{NG}{R + Ny} \)
Multiply both sides by \( (R + Ny) \):
\( \implies I(R + Ny) = NG \)
Expand the bracket:
\( \implies IR + INy = NG \)
Rearrange to bring all terms containing \( N \) to one side:
\( \implies INy - NG = -IR \)
Factor out \( N \) from the left side:
\( \implies N(Iy - G) = -IR \)
Divide both sides by \( (Iy - G) \):
\( \implies N = \frac{-IR}{Iy - G} \)
Simplify the negative signs:
\( \implies N = \frac{IR}{G - Iy} \)
In simple words: Multiply by the denominator, expand, group all terms with \( N \) on one side, factor out \( N \), and divide to solve.

Exam Tip: Multiplying the numerator and denominator of a fraction by -1 is a clean way to eliminate a leading negative sign.

Question 10. Make \( V \) the subject of the formula: \( K = \frac{1}{2}MV^2 \)
Answer: Start with the kinetic energy formula:
\( \implies K = \frac{1}{2}MV^2 \)
Multiply both sides by 2:
\( \implies 2K = MV^2 \)
Divide both sides by \( M \):
\( \implies \frac{2K}{M} = V^2 \)
Take the square root of both sides to isolate \( V \):
\( \implies V = \sqrt{\frac{2K}{M}} \)
In simple words: Multiply by two, divide by \( M \), and then take the square root of the result.

Exam Tip: Make sure the square root sign covers the entire fraction, including both the numerator and denominator.

Question 11. Make \( d \) the subject of the formula: \( S = \frac{n}{2}\{2a + (n - 1)d\} \)
Answer: Given the sum of an arithmetic progression:
\( \implies S = \frac{n}{2}\{2a + (n - 1)d\} \)
Multiply both sides by 2 and expand the multiplication by \( n \):
\( \implies 2S = 2an + n(n - 1)d \)
Subtract \( 2an \) from both sides:
\( \implies 2S - 2an = n(n - 1)d \)
Factor out 2 on the left side:
\( \implies 2(S - an) = n(n - 1)d \)
Divide both sides by \( n(n - 1) \) to solve for \( d \):
\( \implies d = \frac{2(S - an)}{n(n - 1)} \)
In simple words: Multiply by two to clear the fraction, distribute \( n \), subtract the term without \( d \), and then divide by \( n(n-1) \).

Exam Tip: Expanding terms partially can make it easier to isolate a specific variable like \( d \).

Question 12. Make \( R_2 \) the subject of the formula: \( R^2 = 4\pi(R_1^2 - R_2^2) \)
Answer: Given:
\( \implies R^2 = 4\pi(R_1^2 - R_2^2) \)
Expand the right-hand side:
\( \implies R^2 = 4\pi R_1^2 - 4\pi R_2^2 \)
Rearrange to make the term with \( R_2^2 \) positive:
\( \implies 4\pi R_2^2 = 4\pi R_1^2 - R^2 \)
Divide by \( 4\pi \):
\( \implies R_2^2 = \frac{4\pi R_1^2 - R^2}{4\pi} \)
Take the square root of both sides:
\( \implies R_2 = \sqrt{\frac{4\pi R_1^2 - R^2}{4\pi}} \)
In simple words: Multiply out the bracket, swap the terms to make the target positive, divide by \( 4\pi \), and take the square root.

Exam Tip: When isolating a negative term like \( -4\pi R_2^2 \), add it to both sides first to work with a positive coefficient.

Question 13. Make \( A \) the subject of the formula: \( R = \frac{m_1 B + m_2 A}{m_1 + m_2} \)
Answer: Given:
\( \implies R = \frac{m_1 B + m_2 A}{m_1 + m_2} \)
Multiply both sides by the denominator \( (m_1 + m_2) \):
\( \implies R(m_1 + m_2) = m_1 B + m_2 A \)
Subtract \( m_1 B \) from both sides:
\( \implies R(m_1 + m_2) - m_1 B = m_2 A \)
Divide both sides by \( m_2 \) to isolate \( A \):
\( \implies A = \frac{R(m_1 + m_2) - m_1 B}{m_2} \)
In simple words: Multiply by the entire bottom group, subtract the term with \( B \), and then divide by \( m_2 \).

Exam Tip: Do not expand \( R(m_1 + m_2) \) if the target variable \( A \) is not inside those parentheses.

Question 14. Make \( c \) the subject of the formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Answer: Starting with the quadratic formula:
\( \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Multiply both sides by \( 2a \):
\( \implies 2ax = -b \pm \sqrt{b^2 - 4ac} \)
Add \( b \) to both sides:
\( \implies 2ax + b = \pm \sqrt{b^2 - 4ac} \)
Square both sides to eliminate the square root:
\( \implies (2ax + b)^2 = b^2 - 4ac \)
Rearrange to isolate the term with \( c \):
\( \implies 4ac = b^2 - (2ax + b)^2 \)
Divide by \( 4a \):
\( \implies c = \frac{b^2 - (2ax + b)^2}{4a} \)
In simple words: Multiply by \( 2a \), add \( b \), square both sides, move the \( 4ac \) to make it positive, and divide by \( 4a \).

Exam Tip: Squaring \( \pm \sqrt{X} \) results in just \( X \) since squaring either positive or negative values gives a positive result.

Question 15. Make \( k \) the subject of the formula: \( T = 2\pi \sqrt{\frac{k^2 + h^2}{hg}} \)
Answer: Given:
\( \implies T = 2\pi \sqrt{\frac{k^2 + h^2}{hg}} \)
Divide both sides by \( 2\pi \):
\( \implies \frac{T}{2\pi} = \sqrt{\frac{k^2 + h^2}{hg}} \)
Square both sides to remove the root:
\( \implies \left(\frac{T}{2\pi}\right)^2 = \frac{k^2 + h^2}{hg} \)
Multiply both sides by \( hg \) to clear the fraction:
\( \implies hg \left(\frac{T}{2\pi}\right)^2 = k^2 + h^2 \)
Subtract \( h^2 \) from both sides:
\( \implies hg \left(\frac{T}{2\pi}\right)^2 - h^2 = k^2 \)
Take the square root of both sides to isolate \( k \):
\( \implies k = \sqrt{hg \left(\frac{T}{2\pi}\right)^2 - h^2} \)
Expanding the squared term inside the radical:
\( \implies k = \sqrt{\frac{T^2 hg}{4\pi^2} - h^2} \)
In simple words: Divide by \( 2\pi \), square both sides, multiply by \( hg \), subtract \( h^2 \), and then take the square root.

Exam Tip: When squaring \( \frac{T}{2\pi} \), make sure to distribute the power to both the numerator and the denominator, turning it into \( \frac{T^2}{4\pi^2} \).

Question 16. Substitute \( y = ax + b \) into \( mx + ny = p \) and express \( x \) as the subject of the formula.
Answer: Given the linear equation:
\( \implies mx + ny = p \)
We substitute \( y = ax + b \) into the equation:
\( \implies mx + n(ax + b) = p \)
Expand the term with \( n \):
\( \implies mx + anx + bn = p \)
Group the terms containing \( x \) and factor \( x \) out:
\( \implies x(m + an) + bn = p \)
Subtract \( bn \) from both sides:
\( \implies x(m + an) = p - bn \)
Divide both sides by \( (m + an) \) to solve for \( x \):
\( \implies x = \frac{p - bn}{m + an} \)
In simple words: Replace \( y \) with its expression, multiply through by \( n \), group all terms with \( x \), factor \( x \) out, and divide to isolate it.

Exam Tip: Substituting one equation into another is a standard algebraic method to eliminate a variable.

Question 17. Given \( A = \pi r^2 \) and \( C = 2\pi r \), eliminate \( \pi \) and make \( r \) the subject of the formula.
Answer: We are given two formulas:
\( A = \pi r^2 \) ... (i)
\( C = 2\pi r \) ... (ii)
Divide equation (i) by equation (ii) to eliminate \( \pi \):
\( \implies \frac{A}{C} = \frac{\pi r^2}{2\pi r} \)
Simplifying the right side by cancelling out \( \pi \) and one \( r \):
\( \implies \frac{A}{C} = \frac{r}{2} \)
Multiply both sides by 2 to isolate \( r \):
\( \implies r = \frac{2A}{C} \)
In simple words: Divide the area formula by the circumference formula to get rid of \( \pi \), then multiply by 2 to find \( r \).

Exam Tip: Dividing equations is a highly effective way to eliminate shared constants like \( \pi \).

Question 18. Given \( V = \pi r^2 h \) and \( S = 2\pi r^2 + 2\pi rh \), eliminate \( h \) and express \( V \) in terms of \( S \) and \( r \).
Answer: Let the two formulas be:
\( V = \pi r^2 h \) ... (i)
\( S = 2\pi r^2 + 2\pi rh \) ... (ii)
From equation (ii), we isolate \( h \):
\( \implies 2\pi rh = S - 2\pi r^2 \)
\( \implies h = \frac{S - 2\pi r^2}{2\pi r} \)
Now substitute this expression for \( h \) into equation (i):
\( \implies V = \pi r^2 \left(\frac{S - 2\pi r^2}{2\pi r}\right) \)
Simplify the terms by cancelling \( \pi \) and \( r \):
\( \implies V = r \left(\frac{S - 2\pi r^2}{2}\right) \)
Distribute \( r \) and divide:
\( \implies V = \frac{Sr}{2} - \pi r^3 \)
In simple words: Solve the second equation for \( h \) first, and then plug that value into the first equation to remove \( h \) entirely.

Exam Tip: When substituting, simplify the terms outside the parentheses with the denominator of the fraction to make the final expression cleaner.

Question 19. Make \( x \) the subject of the formula and simplify: \( 3ax + 2b^2 = 3bx + 2a^2 \)
Answer: Given the algebraic equation:
\( \implies 3ax + 2b^2 = 3bx + 2a^2 \)
Rearrange terms to collect all terms with \( x \) on the left:
\( \implies 3ax - 3bx = 2a^2 - 2b^2 \)
Factor out \( x \) on the left:
\( \implies x(3a - 3b) = 2a^2 - 2b^2 \)
Divide both sides to express \( x \):
\( \implies x = \frac{2a^2 - 2b^2}{3a - 3b} \)
Factor the numerator and the denominator:
\( \implies x = \frac{2(a^2 - b^2)}{3(a - b)} \)
Use the difference of squares identity \( a^2 - b^2 = (a - b)(a + b) \):
\( \implies x = \frac{2(a - b)(a + b)}{3(a - b)} \)
Cancel the common factor \( (a - b) \), assuming \( a \neq b \):
\( \implies x = \frac{2(a + b)}{3} \)
In simple words: Group all terms with \( x \) on one side, factor \( x \) out, write as a fraction, and then use the difference of squares rule to cancel out common parts.

Exam Tip: Always look for algebraic identities like the difference of squares to simplify fractional expressions fully.

Question 20. If \( b = \frac{2a}{a - 2} \) and \( c = \frac{4b - 3}{3b + 4} \), express \( c \) in terms of \( a \).
Answer: We are given two equations:
\( b = \frac{2a}{a - 2} \)
\( c = \frac{4b - 3}{3b + 4} \)
Substitute the value of \( b \) into the expression for \( c \):
\( c = \frac{4\left(\frac{2a}{a - 2}\right) - 3}{3\left(\frac{2a}{a - 2}\right) + 4} \)
Simplify the numerators:
\( \implies c = \frac{\frac{8a}{a - 2} - 3}{\frac{6a}{a - 2} + 4} \)
Combine the fractions in the numerator and denominator using a common denominator of \( (a - 2) \):
\( \implies c = \frac{\frac{8a - 3(a - 2)}{a - 2}}{\frac{6a + 4(a - 2)}{a - 2}} \)
Cancel out the common denominator of \( (a - 2) \):
\( \implies c = \frac{8a - 3(a - 2)}{6a + 4(a - 2)} \)
Expand the parentheses:
\( \implies c = \frac{8a - 3a + 6}{6a + 4a - 8} \)
Simplify the terms to get the final expression:
\( \implies c = \frac{5a + 6}{10a - 8} \)
In simple words: Substitute the formula for \( b \) into the equation for \( c \), then simplify the fractions by multiplying through to eliminate the denominators.

Exam Tip: When simplifying complex fractions (fractions within fractions), multiplying both the top and bottom by the common denominator is the quickest way to clear them.

Exercise 6.3

Question 1. In the formula \( R = \frac{h}{2}(a - b) \), make \( h \) the subject. Then find the value of \( h \) when \( R = 108 \), \( a = 16 \), and \( b = 12 \).
Answer: First, let's rearrange the given formula to make \( h \) the subject:
\( R = \frac{h}{2}(a - b) \)
Multiply both sides by 2:
\( \implies 2R = h(a - b) \)
Divide both sides by \( (a - b) \):
\( \implies h = \frac{2R}{a - b} \)
Now substitute the values \( R = 108 \), \( a = 16 \), and \( b = 12 \):
\( \implies h = \frac{2 \times 108}{16 - 12} \)
\( \implies h = \frac{2 \times 108}{4} \)
\( \implies h = 54 \)
In simple words: Solve for \( h \) by multiplying by 2 and dividing by \( a-b \). Then put in the numbers to calculate the final answer.

Exam Tip: Always isolate the required variable first before substituting the numerical values, as this reduces arithmetic errors.

Question 2. In the formula \( v^2 = u^2 + 2as \), make \( s \) the subject. Then find the value of \( s \) when \( u = 3 \), \( a = 2 \), and \( v = 5 \).
Answer: First, rearrange the formula to make \( s \) the subject:
\( v^2 = u^2 + 2as \)
Subtract \( u^2 \) from both sides:
\( \implies v^2 - u^2 = 2as \)
Divide both sides by \( 2a \):
\( \implies s = \frac{v^2 - u^2}{2a} \)
Now substitute \( u = 3 \), \( a = 2 \), and \( v = 5 \) into this formula:
\( \implies s = \frac{5^2 - 3^2}{2 \times 2} \)
\( \implies s = \frac{25 - 9}{4} \)
\( \implies s = \frac{16}{4} \)
\( \implies s = 4 \)
In simple words: Rearrange the formula to isolate \( s \) by subtracting \( u^2 \) and dividing by \( 2a \), then plug in the given values to solve.

Exam Tip: When evaluating powers like \( 5^2 \) and \( 3^2 \), calculate the squares first before performing subtraction.

Question 3. Change the subject of the formula to make \( y \) the subject: \( x = \frac{1-y^2}{1+y^2} \). Hence, find the value of \( y \) when \( x = \frac{3}{5} \).
Answer:
To express \( y \) in terms of \( x \), we rearrange the given formula:
\( x = \frac{1-y^2}{1+y^2} \)
Multiplying both sides by the denominator gives:
\( \implies x(1+y^2) = 1-y^2 \)
\( \implies x+xy^2 = 1-y^2 \)
Now, let us collect the terms containing \( y^2 \) on the left-hand side:
\( \implies xy^2+y^2 = 1-x \)
Factoring out \( y^2 \):
\( \implies y^2(x+1) = 1-x \)
Dividing by \( x+1 \):
\( \implies y^2 = \frac{1-x}{1+x} \)
Taking the positive square root on both sides:
\( \implies y = \sqrt{\frac{1-x}{1+x}} \)

Next, we substitute \( x = \frac{3}{5} \) into this rearranged equation:
\( y = \sqrt{\frac{1-\frac{3}{5}}{1+\frac{3}{5}}} \)
\( \implies y = \sqrt{\frac{\frac{2}{5}}{\frac{8}{5}}} \)
\( \implies y = \sqrt{\frac{2}{8}} \)
\( \implies y = \sqrt{\frac{1}{4}} \)
\( \implies y = \frac{1}{2} \)
In simple words: First, rewrite the equation so that \( y \) is isolated on one side. After that, plug in \( x = \frac{3}{5} \) and calculate the numerical answer.

Exam Tip: Remember to factor out the desired subject variable once all terms containing it are grouped on one side of the equation.

Question 4. Make \( a \) the subject of the formula: \( S = \frac{n}{2}\{2a + (n-1)d\} \). Then find the value of \( a \) when \( S = 50 \), \( n = 10 \), and \( d = 2 \).
Answer:
To express \( a \) in terms of the other variables, we rearrange the formula step-by-step:
\( S = \frac{n}{2}\{2a + (n-1)d\} \)
Multiply by 2 to clear the fraction:
\( \implies 2S = n\{2a + (n-1)d\} \)
Divide both sides by \( n \):
\( \implies \frac{2S}{n} = 2a + (n-1)d \)
Subtract \( (n-1)d \) from both sides:
\( \implies \frac{2S}{n} - (n-1)d = 2a \)
Divide the entire equation by 2:
\( \implies a = \frac{S}{n} - \frac{(n-1)d}{2} \)

Now, substitute the values \( S = 50 \), \( n = 10 \), and \( d = 2 \) into this expression:
\( a = \frac{50}{10} - \frac{(10-1) \times 2}{2} \)
\( \implies a = 5 - \frac{9 \times 2}{2} \)
\( \implies a = 5 - 9 \)
\( \implies a = -4 \)
In simple words: Rearrange the formula to isolate \( a \) on one side. Then, insert the given values for the other variables to compute the numerical result.

Exam Tip: Be careful with the minus signs during final subtraction to ensure you do not make a basic calculation error.

Question 5. Rearrange the formula to make \( x \) the subject: \( a = 1 - \frac{2b}{cx-b} \). Then, calculate \( x \) when \( a = 5 \), \( b = 12 \), and \( c = 2 \).
Answer:
We start by isolating the term with \( x \) in the formula:
\( a = 1 - \frac{2b}{cx-b} \)
Subtract 1 from both sides:
\( \implies a - 1 = -\frac{2b}{cx-b} \)
Multiply both sides by \( (cx-b) \):
\( \implies (a-1)(cx-b) = -2b \)
\( \implies (a-1)(cx-b) + 2b = 0 \)
Expanding the product:
\( \implies acx - ab - cx + b + 2b = 0 \)
\( \implies acx - cx - ab + 3b = 0 \)
Collect the terms with \( x \) on one side and the rest on the other:
\( \implies x(ac-c) = ab - 3b \)
Factorise both sides:
\( \implies xc(a-1) = b(a-3) \)
Divide by \( c(a-1) \) to get \( x \):
\( \implies x = \frac{b(a-3)}{c(a-1)} \)

Now, substitute \( a = 5 \), \( b = 12 \), and \( c = 2 \) into our expression:
\( x = \frac{12(5-3)}{2(5-1)} \)
\( \implies x = \frac{12 \times 2}{2 \times 4} \)
\( \implies x = \frac{24}{8} \)
\( \implies x = 3 \)
In simple words: First, rearrange the equation to solve for \( x \). Then, substitute the values of \( a \), \( b \), and \( c \) to find the numerical value.

Exam Tip: Be methodical when expanding binomial expressions to avoid losing track of signs.

Question 6. Make \( h \) the subject of the formula: \( K = \sqrt{\frac{hg}{d^2} - a^2} \). Then, find the value of \( h \) when \( K = -2 \), \( a = -3 \), \( d = 8 \), and \( g = 32 \).
Answer:
To rearrange the equation to solve for \( h \), we proceed as follows:
\( K = \sqrt{\frac{hg}{d^2} - a^2} \)
Square both sides to eliminate the square root:
\( \implies K^2 = \frac{hg}{d^2} - a^2 \)
Add \( a^2 \) to both sides:
\( \implies K^2 + a^2 = \frac{hg}{d^2} \)
Multiply both sides by \( d^2 \):
\( \implies (K^2 + a^2)d^2 = hg \)
Divide both sides by \( g \):
\( \implies h = \frac{(K^2 + a^2)d^2}{g} \)

Now, we plug in the values \( K = -2 \), \( a = -3 \), \( d = 8 \), and \( g = 32 \) into the formula:
\( h = \frac{((-2)^2 + (-3)^2)(8)^2}{32} \)
\( \implies h = \frac{(4 + 9) \times 64}{32} \)
\( \implies h = \frac{13 \times 64}{32} \)
\( \implies h = 13 \times 2 \)
\( \implies h = 26 \)
In simple words: Square both sides to remove the root, then isolate \( h \). Finally, replace the letters with their given numbers to calculate \( h \).

Exam Tip: Squaring a negative number always yields a positive result, so make sure to write \( (-2)^2 = 4 \) and \( (-3)^2 = 9 \).

Question 7. Express \( x \) as the subject of the formula: \( y = \frac{1-x^2}{1+x^2} \). Then, find the value of \( x \) when \( y = \frac{1}{2} \).
Answer:
To rearrange the formula to make \( x \) the subject:
\( y = \frac{1-x^2}{1+x^2} \)
Multiply both sides by \( 1+x^2 \):
\( \implies y(1+x^2) = 1-x^2 \)
\( \implies y + yx^2 = 1 - x^2 \)
Move all terms with \( x^2 \) to the left side:
\( \implies yx^2 + x^2 = 1 - y \)
Factor out \( x^2 \):
\( \implies x^2(1+y) = 1 - y \)
Divide by \( 1+y \):
\( \implies x^2 = \frac{1-y}{1+y} \)
Take the square root of both sides:
\( \implies x = \sqrt{\frac{1-y}{1+y}} \)

Next, substitute \( y = \frac{1}{2} \) into the rearranged formula:
\( x = \sqrt{\frac{1-\frac{1}{2}}{1+\frac{1}{2}}} \)
\( \implies x = \sqrt{\frac{\frac{1}{2}}{\frac{3}{2}}} \)
\( \implies x = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \)
In simple words: Rearrange the equation to get \( x \) on one side. Then substitute \( y = \frac{1}{2} \) to find the value of \( x \).

Exam Tip: Be precise when simplifying complex fractions inside square roots by canceling out common denominators.

Question 8. Rearrange the formula to make \( y \) the subject: \( \frac{x}{a} + \frac{y}{b} = 1 \). Then find the value of \( y \) when \( a = 2 \), \( b = 8 \), and \( x = 5 \).
Answer:
To rearrange the given equation for \( y \):
\( \frac{x}{a} + \frac{y}{b} = 1 \)
Subtract \( \frac{x}{a} \) from both sides:
\( \implies \frac{y}{b} = 1 - \frac{x}{a} \)
Multiply both sides by \( b \):
\( \implies y = b\left(1 - \frac{x}{a}\right) \)
\( \implies y = b - \frac{b}{a}x \)

Now, we substitute the values \( a = 2 \), \( b = 8 \), and \( x = 5 \):
\( y = 8 - \frac{8}{2} \times 5 \)
\( \implies y = 8 - 4 \times 5 \)
\( \implies y = 8 - 20 \)
\( \implies y = -12 \)
In simple words: Move the \( x \) term to the other side of the equals sign, multiply by \( b \) to get \( y \) alone, and then plug in the numbers.

Exam Tip: Be careful with the order of operations: perform the multiplication before the subtraction.

Question 9. Rearrange the formula to make \( m \) the subject: \( x = \frac{my}{14 - mt} \). Hence, find the value of \( m \) when \( x = 6 \), \( y = 10 \), and \( t = 3 \).
Answer:
To solve for \( m \), we rearrange the formula as follows:
\( x = \frac{my}{14 - mt} \)
Multiply both sides by \( 14 - mt \):
\( \implies (14 - mt)x = my \)
\( \implies 14x - mtx = my \)
Collect all terms containing \( m \) on one side:
\( \implies 14x = my + mtx \)
Factor out \( m \) on the right side:
\( \implies 14x = m(y + tx) \)
Divide by \( y + tx \) to isolate \( m \):
\( \implies m = \frac{14x}{tx + y} \)

Now, substitute the values \( x = 6 \), \( y = 10 \), and \( t = 3 \) into the new formula:
\( m = \frac{14 \times 6}{3 \times 6 + 10} \)
\( \implies m = \frac{84}{18 + 10} \)
\( \implies m = \frac{84}{28} \)
\( \implies m = 3 \)
In simple words: Get the \( m \) terms onto one side, factor \( m \) out, and divide to isolate it. Finally, insert the given numbers to find \( m \).

Exam Tip: Factoring out \( m \) helps to group multiple occurrences of the variable into a single term that can be isolated.

Question 10. Rearrange the formula to make \( I \) the subject: \( M = L + \frac{1}{F}\left\{ \frac{1}{2}N - C \right\} \times I \). Then, find the value of \( I \) when \( M = 44 \), \( L = 20 \), \( F = 15 \), \( N = 50 \), and \( C = 13 \).
Answer:
To rearrange the equation to isolate \( I \):
\( M = L + \frac{1}{F}\left\{ \frac{1}{2}N - C \right\} \times I \)
Subtract \( L \) from both sides:
\( \implies M - L = \frac{1}{F}\left\{ \frac{1}{2}N - C \right\} \times I \)
Multiply both sides by \( F \):
\( \implies F(M - L) = \left\{ \frac{1}{2}N - C \right\} \times I \)
Simplify the expression inside the brackets by finding a common denominator:
\( \implies F(M - L) = \left\{ \frac{N - 2C}{2} \right\} \times I \)
Multiply both sides by 2:
\( \implies 2F(M - L) = (N - 2C) \times I \)
Divide both sides by \( N - 2C \):
\( \implies I = \frac{2F(M - L)}{N - 2C} \)

Now, substitute the values \( M = 44 \), \( L = 20 \), \( F = 15 \), \( N = 50 \), and \( C = 13 \) into the rearranged formula:
\( I = \frac{2 \times 15(44 - 20)}{50 - 2 \times 13} \)
\( \implies I = \frac{30 \times 24}{50 - 26} \)
\( \implies I = \frac{720}{24} \)
\( \implies I = 30 \)
In simple words: First, isolate the term containing \( I \) by moving other variables to the left. Then substitute the values given to calculate \( I \).

Exam Tip: Clear denominators early in multi-step algebraic manipulation to make isolating variables simpler.

Question 11. Make \( g \) the subject of the formula: \( v^2 = u^2 - 2gh \). Then find the value of \( g \) when \( v = 9.8 \), \( u = 41.5 \), and \( h = 25.4 \).
Answer:
To rearrange the formula to make \( g \) the subject:
\( v^2 = u^2 - 2gh \)
Add \( 2gh \) and subtract \( v^2 \) on both sides:
\( \implies 2gh = u^2 - v^2 \)
Divide both sides by \( 2h \):
\( \implies g = \frac{u^2 - v^2}{2h} \)

Next, substitute the values \( v = 9.8 \), \( u = 41.5 \), and \( h = 25.4 \) into the formula:
\( g = \frac{41.5^2 - 9.8^2}{2 \times 25.4} \)
Using the algebraic identity \( a^2 - b^2 = (a+b)(a-b) \):
\( g = \frac{(41.5 + 9.8)(41.5 - 9.8)}{50.8} \)
\( \implies g = \frac{51.3 \times 31.7}{50.8} \)
\( \implies g = \frac{1625.71}{50.8} \)
\( \implies g \approx 32.01 \approx 32 \)
In simple words: Rearrange the formula to solve for \( g \). Use the square difference identity to make the calculation of the squared numbers easier.

Exam Tip: Utilizing the difference of two squares identity \( a^2 - b^2 = (a+b)(a-b) \) is a great way to simplify arithmetic without manual squaring.

Question 12. Make \( f \) the subject of the formula: \( D = \sqrt{\frac{f+p}{f-p}} \). Then find the value of \( f \) when \( D = 13 \) and \( p = 21 \).
Answer:
To rearrange the formula and solve for \( f \):
\( D = \sqrt{\frac{f+p}{f-p}} \)
Square both sides to remove the radical sign:
\( \implies D^2 = \frac{f+p}{f-p} \)
Multiply both sides by \( f-p \):
\( \implies D^2(f-p) = f+p \)
\( \implies D^2f - D^2p = f+p \)
Move all terms with \( f \) to one side and terms with \( p \) to the other:
\( \implies D^2f - f = D^2p + p \)
Factor out \( f \) and \( p \) respectively:
\( \implies f(D^2-1) = p(D^2+1) \)
Divide by \( D^2-1 \):
\( \implies f = \frac{p(D^2+1)}{D^2-1} \)

Now, substitute \( D = 13 \) and \( p = 21 \) into the formula:
\( f = \frac{21(13^2+1)}{13^2-1} \)
\( \implies f = \frac{21 \times 170}{168} \)
Since \( \frac{21}{168} = \frac{1}{8} \):
\( \implies f = \frac{170}{8} \)
\( \implies f = 21.25 \)
In simple words: Square both sides to eliminate the root, group the \( f \) terms together, and factor. Finally, plug in the numbers to find \( f \).

Exam Tip: Look for common factors to simplify the fraction before multiplying larger numbers (like \( 21 \times 170 \n) to save time.

Question 13. Given that \( y = \frac{2z + 1}{2z - 1} \) and \( x = \frac{y + 1}{y - 1} \), express \( z \) in terms of \( x \) and find the value of \( z \) when \( x = 34 \).
Answer:
We can substitute the expression for \( y \) directly into the equation for \( x \):
\( x = \frac{y+1}{y-1} \)
Substitute \( y = \frac{2z+1}{2z-1} \):
\( \implies x = \frac{\left( \frac{2z+1}{2z-1} \right) + 1}{\left( \frac{2z+1}{2z-1} \right) - 1} \)
Multiply both the numerator and the denominator by \( (2z-1) \) to simplify:
\( \implies x = \frac{2z + 1 + 2z - 1}{2z + 1 - (2z - 1)} \)
\( \implies x = \frac{4z}{2} \)
\( \implies x = 2z \)
Rearranging this to solve for \( z \):
\( \implies z = \frac{x}{2} \)

Now, substituting \( x = 34 \), we get:
\( z = \frac{34}{2} = 17 \)
In simple words: Put the value of \( y \) into the second equation, simplify it to get \( x = 2z \), and then solve for \( z \). Finally, divide 34 by 2.

Exam Tip: Substituting one variable into another equation is often much faster than rearranging both equations individually.

Question 14. Rearrange the formula to make \( c \) the subject: \( a = b(1 + ct) \). Then find the value of \( c \) when \( a = 1100 \), \( b = 100 \), and \( t = 4 \).
Answer:
To express \( c \) as the subject of the formula:
\( a = b(1 + ct) \)
Expand the bracket:
\( \implies a = b + bct \)
Subtract \( b \) from both sides:
\( \implies bct = a - b \)
Divide both sides by \( bt \):
\( \implies c = \frac{a - b}{bt} \)

Now, substitute the values \( a = 1100 \), \( b = 100 \), and \( t = 4 \) into this formula:
\( c = \frac{1100 - 100}{100 \times 4} \)
\( \implies c = \frac{1000}{400} \)
\( \implies c = 2.5 \)
In simple words: Multiply out the brackets first, isolate the term with \( c \), and then divide to find \( c \). Put in the given numbers to get the final answer.

Exam Tip: Alternatively, you can divide by \( b \) first to get \( \frac{a}{b} = 1 + ct \), which also leads to the correct result.

Question 15. The volume \( V \) of a cylinder is the product of \( \pi \), the square of the radius \( r \), and the height \( h \). Write the formula for \( V \), make \( r \) the subject of this formula, and calculate \( r \) when \( V = 44\text{ cm}^3 \), \( h = 14\text{ cm} \), and \( \pi = \frac{22}{7} \).
Answer:
The volume \( V \) of a cylinder is represented as:
\( V = \pi r^2 h \)
To solve for \( r \), we rearrange the formula:
\( \implies \frac{V}{\pi h} = r^2 \)
Taking the positive square root of both sides:
\( \implies r = \sqrt{\frac{V}{\pi h}} \)

Now, we substitute \( V = 44 \), \( h = 14 \), and \( \pi = \frac{22}{7} \):
\( r = \sqrt{\frac{44}{\frac{22}{7} \times 14}} \)
Simplify the denominator first:
\( \frac{22}{7} \times 14 = 22 \times 2 = 44 \)
\( \implies r = \sqrt{\frac{44}{44}} \)
\( \implies r = \sqrt{1} = 1\text{ cm} \)
In simple words: Set up the cylinder volume formula, solve for the radius \( r \) by dividing and taking the square root, then plug in the numbers to find the radius.

Exam Tip: Do not forget to state the correct unit (in this case, cm) for the final radius value.

Question 16. The volume \( V \) of a cone is defined as one-third the product of \( \pi \), the square of the base radius \( r \), and the height \( h \). Express \( r \) as the subject of this formula. Hence, find \( r \) when \( V = 1232\text{ cm}^3 \), \( h = 24\text{ cm} \), and \( \pi = \frac{22}{7} \).
Answer:
The volume of a cone is given by:
\( V = \frac{1}{3}\pi r^2 h \)
To solve for \( r \), we rearrange the equation:
\( \implies 3V = \pi r^2 h \)
\( \implies r^2 = \frac{3V}{\pi h} \)
\( \implies r = \sqrt{\frac{3V}{\pi h}} \)

Now, we substitute \( V = 1232 \), \( h = 24 \), and \( \pi = \frac{22}{7} \) into this formula:
\( r = \sqrt{\frac{3 \times 1232}{\frac{22}{7} \times 24}} \)
Simplifying the denominator:
\( \frac{22}{7} \times 24 = \frac{528}{7} \)
Therefore:
\( \implies r = \sqrt{\frac{3696 \times 7}{528}} \)
Since \( \frac{3696}{528} = 7 \):
\( \implies r = \sqrt{7 \times 7} \)
\( \implies r = \sqrt{49} = 7\text{ cm} \)
In simple words: Write the cone volume formula, rearrange it to isolate \( r \) under a square root, then plug in the numbers to calculate the radius.

Exam Tip: Be careful when simplifying complex fractions in calculations to make sure the final square root calculation works out to a whole number where possible.

Question 17. The pressure \( P \) and volume \( V \) of a gas are related by the formula \( PV = C \), where \( C \) is a constant. Given that \( P = 4 \) when \( V = 2\frac{1}{2} \), find the value of \( P \) when \( V = 4 \).
Answer:
First, we convert the mixed fraction to an improper fraction:
\( V = 2\frac{1}{2} = \frac{5}{2} \)
Next, substitute the given values \( P = 4 \) and \( V = \frac{5}{2} \) into the formula to find the constant \( C \):
\( PV = C \)
\( \implies 4 \times \left(\frac{5}{2}\right) = C \)
\( \implies C = 10 \)

Now, we use this constant value of \( C = 10 \) to find the value of \( P \) when \( V = 4 \):
\( PV = C \)
\( \implies P(4) = 10 \)
\( \implies P = \frac{10}{4} \)
\( \implies P = \frac{5}{2} = 2.5 \)
In simple words: First, find the constant \( C \) by multiplying the first set of \( P \) and \( V \). Then, use that constant to find the new \( P \) when the volume changes to 4.

Exam Tip: Ensure that you find the value of the constant \( C \) first before attempting to solve for the unknown variable.

Question 18(a). Make \( m \) the subject of the formula: \( E = \frac{1}{2}mv^2 + mgh \).
Answer:
To express \( m \) in terms of the other variables:
\( E = \frac{1}{2}mv^2 + mgh \)
Factor out \( m \) on the right side of the equation:
\( \implies E = m\left(\frac{1}{2}v^2 + gh\right) \)
Divide both sides by the term in brackets:
\( \implies m = \frac{E}{\frac{1}{2}v^2 + gh} \)
Simplify the complex fraction by writing the denominator with a common base of 2:
\( \implies m = \frac{E}{\frac{v^2 + 2gh}{2}} \)
Multiply the numerator by the reciprocal of the denominator:
\( \implies m = \frac{2E}{v^2 + 2gh} \)
In simple words: Since \( m \) is in both terms on the right, factor it out first. Then, divide both sides to leave \( m \) on its own, and simplify the fraction.

Exam Tip: Factoring the common term \( m \) is the key first step whenever the target variable appears in multiple terms.

Question 18(b). Using the formula derived in part (a), find the value of \( m \) when \( v = 2 \), \( g = 10 \), \( h = 5 \), and \( E = 104 \).
Answer:
Using the rearranged equation for \( m \):
\( m = \frac{2E}{v^2 + 2gh} \)
Substitute the given values \( v = 2 \), \( g = 10 \), \( h = 5 \), and \( E = 104 \):
\( \implies m = \frac{2(104)}{(2)^2 + 2(10)(5)} \)
\( \implies m = \frac{208}{4 + 100} \)
\( \implies m = \frac{208}{104} \)
\( \implies m = 2 \)
In simple words: Put the given values into the formula you found in the first part and do the basic calculation to get the final answer.

Exam Tip: Double check that you multiply \( E \) by 2 in the numerator, as derived in the previous section.

Question 19. Make \( d \) the subject of the formula: \( s = \frac{n}{2}[2a + (n-1)d] \). Then find the value of \( d \) when \( n = 3 \), \( a = n + 1 \), and \( s = 18 \).
Answer:
To rearrange the formula to make \( d \) the subject:
\( s = \frac{n}{2}[2a + (n-1)d] \)
Expand the term with \( a \):
\( \implies s = an + \frac{n(n-1)d}{2} \)
Subtract \( an \) from both sides:
\( \implies s - an = \frac{n(n-1)d}{2} \)
Multiply both sides by 2:
\( \implies 2(s - an) = n(n-1)d \)
Divide both sides by \( n(n-1) \):
\( \implies d = \frac{2(s - an)}{n(n-1)} = \frac{2}{n(n-1)}(s - an) \)

Given \( n = 3 \), we find \( a \):
\( a = n + 1 = 3 + 1 = 4 \)

Now, we substitute \( n = 3 \), \( a = 4 \), and \( s = 18 \) into the rearranged equation:
\( d = \frac{2}{3(3-1)}(18 - 4 \times 3) \)
\( \implies d = \frac{2}{3 \times 2}(18 - 12) \)
\( \implies d = \frac{2}{6}(6) \)
\( \implies d = 2 \)
In simple words: Rearrange the formula to isolate \( d \). Calculate \( a = 4 \) using the given rule, then put all values into the formula to find \( d \).

Exam Tip: Don't forget to calculate \( a \) first before substituting everything else, otherwise you'll be stuck with too many unknown variables.

Question 20. The area \( A \) of the ring-shaped region between two concentric circles with larger radius \( R \) and smaller radius \( r \) is given by the formula \( A = \pi(R^2 - r^2) \). Express \( r \) as the subject of the formula. Then, find \( r \) when \( A = 88\text{ cm}^2 \) and \( R = 8\text{ cm} \) (Take \( \pi = \frac{22}{7} \)).
Answer:
To rearrange the formula for the smaller radius \( r \):
\( A = \pi(R^2 - r^2) \)
Divide both sides by \( \pi \):
\( \implies \frac{A}{\pi} = R^2 - r^2 \)
Add \( r^2 \) and subtract \( \frac{A}{\pi} \) on both sides:
\( \implies r^2 = R^2 - \frac{A}{\pi} \)
Take the positive square root of both sides:
\( \implies r = \sqrt{R^2 - \frac{A}{\pi}} \)

Now, substitute the values \( A = 88 \), \( R = 8 \), and \( \pi = \frac{22}{7} \) into this formula:
\( r = \sqrt{8^2 - \frac{88}{\frac{22}{7}}} \)
Simplify the fraction inside the square root first:
\( \frac{88}{\frac{22}{7}} = 88 \times \frac{7}{22} = 4 \times 7 = 28 \)
Substituting this back:
\( \implies r = \sqrt{64 - 28} \)
\( \implies r = \sqrt{36} \)
\( \implies r = 6\text{ cm} \)
In simple words: Rearrange the formula to make \( r \) the subject. Calculate the division inside the square root first, subtract, and then take the square root.

Exam Tip: Ensure that you carry out the division \( \frac{A}{\pi} \) first before subtracting it from \( R^2 \).