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AQA GCSE · Question 14 · Algebra
4 chocolate bars and 3 packets of mints cost £4.70.
5 chocolate bars and 1 packet of mints cost £4.50.
Work out the cost of a chocolate bar and the cost of a packet of mints.
4 chocolate bars and 3 packets of mints cost £4.70.
5 chocolate bars and 1 packet of mints cost £4.50.
Work out the cost of a chocolate bar and the cost of a packet of mints.
How to approach this question
1. Define two variables, for example, `c` for the cost of a chocolate bar and `m` for the cost of mints.
2. Write down two simultaneous equations based on the information given. It's often easier to work in pence to avoid decimals (4c + 3m = 470).
3. Solve the simultaneous equations. You can use either the elimination or substitution method.
- For elimination, you could multiply the second equation by 3 to make the `m` terms equal, then subtract one equation from the other.
- For substitution, rearrange the second equation to find `m` in terms of `c`, then substitute this into the first equation.
4. Once you find the value of one variable, substitute it back into one of the original equations to find the other.
5. Write down your final answers clearly for both items.
Full Answer
Let c be the cost of a chocolate bar and m be the cost of a packet of mints.
From the problem, we can form two simultaneous equations:
1) 4c + 3m = 4.70
2) 5c + m = 4.50
It is easiest to use the substitution method here. Rearrange equation (2) to make m the subject:
m = 4.50 - 5c
Now substitute this expression for m into equation (1):
4c + 3(4.50 - 5c) = 4.70
4c + 13.50 - 15c = 4.70
-11c + 13.50 = 4.70
13.50 - 4.70 = 11c
8.80 = 11c
c = 8.80 / 11
c = 0.80
So, a chocolate bar costs £0.80 (or 80p).
Now substitute the value of c back into the rearranged equation (2) to find m:
m = 4.50 - 5(0.80)
m = 4.50 - 4.00
m = 0.50
So, a packet of mints costs £0.50 (or 50p).
Check with equation (1):
4(0.80) + 3(0.50) = 3.20 + 1.50 = 4.70. Correct.
Chocolate bar: £0.80
Packet of mints: £0.50
This is a word problem that can be solved by setting up and solving a pair of simultaneous linear equations.
Let `c` be the cost of one chocolate bar in pounds.
Let `m` be the cost of one packet of mints in pounds.
From the information given, we can write two equations:
Equation (1): 4c + 3m = 4.70
Equation (2): 5c + 1m = 4.50
We can solve these using the elimination method. To eliminate `m`, we can multiply Equation (2) by 3:
Equation (3): 3 × (5c + m) = 3 × 4.50 => 15c + 3m = 13.50
Now we subtract Equation (1) from Equation (3):
(15c + 3m) - (4c + 3m) = 13.50 - 4.70
11c = 8.80
c = 8.80 / 11
c = 0.80
So, a chocolate bar costs £0.80.
Now substitute c = 0.80 into Equation (2) to find `m`:
5(0.80) + m = 4.50
4.00 + m = 4.50
m = 4.50 - 4.00
m = 0.50
So, a packet of mints costs £0.50.
Final Answer:
Chocolate bar = £0.80
Packet of mints = £0.50
Common mistakes
✗ Setting up the equations incorrectly.
✗ Making arithmetic errors, especially when multiplying or subtracting equations involving decimals.
✗ Finding the value for one variable but forgetting to find the second.
✗ Mixing up which variable represents which item in the final answer.
Practice the full AQA GCSE Maths Higher Tier Paper 1 Non-Calculator
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