JAMB 2019 Mathematics · Question 11

A bricklayer charges ₦1,500 per day for himself and ₦500 per day for his assistant. If a two bedroom flat was…

Question

A bricklayer charges ₦1,500 per day for himself and ₦500 per day for his assistant. If a two bedroom flat was built for ₦95,000 and the bricklayer worked 10 days more than his assistant, how much did the assistant receive?

  1. A.N20,000
  2. B.N28,000
  3. C.N31,200
  4. D.N41,000

Answer: A. N20,000

Explanation written with AI and checked against the answer key

1. Analyze the Question

We need to find the total amount paid to the assistant. We are given:

  • Bricklayer's daily rate: ₦1,500
  • Assistant's daily rate: ₦500
  • Total cost of the project: ₦95,000
  • Relationship between days worked: The bricklayer worked 10 days more than the assistant.
2. Explain the Concept

This is an algebraic word problem. We can represent the number of days the assistant worked as a variable (xx). Since the bricklayer worked 10 days more, his days will be (x+10)(x + 10). We then set up an equation where the sum of the payments to both equals the total cost.

3. Show the Solution
  • Let xx = number of days the assistant worked.
  • Then, x+10x + 10 = number of days the bricklayer worked.

Calculate the total cost: (Bricklayer's rate × Bricklayer's days) + (Assistant's rate × Assistant's days) = Total Cost 1,500(x+10)+500(x)=95,0001,500(x + 10) + 500(x) = 95,000

Expand and solve for xx: 1,500x+15,000+500x=95,0001,500x + 15,000 + 500x = 95,000 2,000x+15,000=95,0002,000x + 15,000 = 95,000

Subtract 15,000 from both sides: 2,000x=95,000−15,0002,000x = 95,000 - 15,000 2,000x=80,0002,000x = 80,000

Divide by 2,000: x=80,0002,000x = \frac{80,000}{2,000} x=40x = 40 days

The assistant worked for 40 days.

Calculate the assistant's total pay: Assistant's pay = (Assistant's daily rate) × (Assistant's days) Assistant's pay = 500×40500 \times 40 Assistant's pay = ₦20,000

4. Identify the Answer

Comparing our result to the given options, the correct choice is A.

5. Exam Tip

When solving "work and pay" problems, always define your variable based on the person who worked the least amount of time. This usually makes the addition in your equation simpler and reduces the chance of making a sign error.

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