Home / iGCSE Mathematics (0580) : C2.1Substitute numbers for words and letters in formulae. iGCSE Style Questions Paper 3

iGCSE Mathematics (0580) : C2.1Substitute numbers for words and letters in formulae. iGCSE Style Questions Paper 3

Question

(a) \( T = P + 5 – 3Q \)

Find the value of \( T \) when \( P = 6 \) and \( Q = 8 \).

(b) Simplify.

\( 3a – b + 7a + 2b – 4a \)

(c) Multiply out.

\( 5(2x – 3y) \)

(d) Solve.

\( 5x – 1 = 3x + 1 + 9 \)

(e) Make \( t \) the subject of the formula \( p = 5t – 3 \).

(f) Entry to a castle costs \( \$x \) for an adult and \( \$y \) for a child.

Entry for 2 adults and 3 children costs \( \$15.00 \).
Entry for 3 adults and 5 children costs \( \$23.50 \).

Write down a pair of simultaneous equations to show this information and solve them to find the value of \( x \) and the value of \( y \).

▶️ Answer/Explanation
Solution

(a) Ans: -13

Substitute \( P = 6 \) and \( Q = 8 \) into \( T = P + 5 – 3Q \).

\( T = 6 + 5 – 24 = -13 \).

(b) Ans: \( 6a + b \)

Combine like terms: \( (3a + 7a – 4a) + (-b + 2b) = 6a + b \).

(c) Ans: \( 10x – 15y \)

Expand using the distributive property: \( 5 \times 2x – 5 \times 3y = 10x – 15y \).

(d) Ans: \( x = 5.5 \)

Simplify and solve: \( 5x – 1 = 3x + 10 \) → \( 2x = 11 \) → \( x = 5.5 \).

(e) Ans: \( t = \frac{p + 3}{5} \)

Rearrange the equation: \( p + 3 = 5t \) → \( t = \frac{p + 3}{5} \).

(f) Ans: \( x = 4.5 \), \( y = 2 \)

Form equations: \( 2x + 3y = 15 \) and \( 3x + 5y = 23.5 \).

Solve by elimination: Multiply first by 3 and second by 2, then subtract to find \( y = 2 \).

Substitute \( y = 2 \) into the first equation to find \( x = 4.5 \).

Question

(a) Simplify. \( a + 4a – 3a \)

(b) Simplify. \( 8b – 4 \times 7b \)

(c) 

The perimeter of this shape is equal to the perimeter of a square. Find an expression for the length of one side of the square. Give your answer in its simplest form.

(d) Victoria buys 5 cups of tea and 4 cakes for \$15.69.
Isabella buys 3 cups of tea and 7 cakes for \$17.97.

Write down a pair of simultaneous equations and solve them to find the cost of one cup of tea and the cost of one cake.

▶️ Answer/Explanation
Solution

(a) Ans: \(2a\)

Combine like terms: \(a + 4a = 5a\), then \(5a – 3a = 2a\).

(b) Ans: \(-20b\)

First multiply: \(4 \times 7b = 28b\), then subtract from \(8b\) to get \(-20b\).

(c) Ans: \(6x + 3\)

Add all sides: \(3x – 9 + 4x + 3 + x + 7 + 9x + 8 = 17x + 9\). Divide by 4 for square side: \(\frac{17x + 9}{4}\) simplifies to \(6x + 3\).

(d) Ans: Tea \$1.65, Cake \$1.86

Set up equations: \(5t + 4c = 15.69\) and \(3t + 7c = 17.97\). Solve by elimination: multiply first by 3, second by 5, then subtract. This gives \(c = 1.86\), substitute back to find \(t = 1.65\).

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