Word Problems (Grade 10 NSC Matric Mathematics): Revision Notes

Worked Example: Bicycles and Tricycles

Problem: A shop sells bicycles and tricycles. In total there are 7 cycles and 19 wheels. Determine how many of each there are, if a bicycle has two wheels and a tricycle has three wheels.

Solution:

Step 1: Assign variables to unknown quantities

• Let $b$ = number of bicycles
• Let $t$ = number of tricycles

Step 2: Set up the equations

• Total cycles: $b + t = 7$ ... (1)
• Total wheels: $2b + 3t = 19$ ... (2)

Step 3: Rearrange equation (1) and substitute into equation (2)

• From equation (1): $t = 7 - b$
• Substitute into equation (2): $2b + 3(7 - b) = 19$
• Simplify: $2b + 21 - 3b = 19$
• Solve: $-b = -2$, so $b = 2$

Step 4: Calculate the number of tricycles

• $t = 7 - b = 7 - 2 = 5$

Step 5: Check and write final answer

• There are 2 bicycles and 5 tricycles
• Check: $2 + 5 = 7$ cycles ✓ and $2(2) + 3(5) = 4 + 15 = 19$ wheels ✓

Worked Example: Test Marks Problem

Problem: Bongani takes Jane's maths test paper and will not tell her what her mark is. He knows that Jane dislikes word problems so he decides to tease her. Bongani says: "I have 2 marks more than you do and the sum of both our marks is equal to 14. What are our marks?"

Solution:

Step 1: Assign variables

• Let $b$ = Bongani's mark
• Let $j$ = Jane's mark

Step 2: Set up system of equations

• Bongani has 2 more marks than Jane: $b = j + 2$ ... (1)
• Sum of both marks: $b + j = 14$ ... (2)

Step 3: Use equation (1) to express $b$ in terms of $j$

• $b = j + 2$

Step 4: Substitute into equation (2)

• $(j + 2) + j = 14$
• $2j + 2 = 14$

Step 5: Solve for $j$

• $2j = 12$
• $j = 6$

Step 6: Find $b$

• $b = j + 2 = 6 + 2 = 8$

Step 7: Final answer

• Bongani got 8 marks and Jane got 6 marks

Worked Example: Milkshake Prices

Problem: A fruitshake costs R 2,00 more than a chocolate milkshake. If 3 fruitshakes and 5 chocolate milkshakes cost R 78,00, determine the individual prices.

Solution:

Step 1: Assign variables

• Let $x$ = price of chocolate milkshake
• Let $y$ = price of fruitshake

Step 2: Set up equations

• Fruitshake costs R 2,00 more: $y = x + 2$ ... (1)
• Total cost: $3y + 5x = 78$ ... (2)

Step 3: Substitute equation (1) into (2)

• $3(x + 2) + 5x = 78$
• $3x + 6 + 5x = 78$
• $8x = 72$
• $x = 9$

Step 4: Find $y$

• $y = x + 2 = 9 + 2 = 11$

Step 5: Final answer

• One chocolate milkshake costs R 9,00 and one fruitshake costs R 11,00

Worked Example: Consecutive Negative Integers

Problem: The product of two consecutive negative integers is 1122. Find the two integers.

Solution:

Step 1: Assign variables

• Let the first integer be $n$
• Let the second integer be $n + 1$

Step 2: Set up equation

• $n(n + 1) = 1122$

Step 3: Expand and solve

• $n^2 + n = 1122$
• $n^2 + n - 1122 = 0$
• $(n + 34)(n - 33) = 0$
• So $n = -34$ or $n = 33$

Step 4: Apply the constraint

• Since both integers must be negative: $n = -34$
• Second integer: $n + 1 = -34 + 1 = -33$

Step 5: Final answer

• The two consecutive negative integers are -34 and -33