EquationsMCQMTP May 20, ICAI SMQuestion 1022 of 221
All Questions

The wages of 8 men and 6 boys amount to 33\displaystyle 33. If 4 men earn 4.50\displaystyle 4.50 more than 5 boys determine the wages of each man and boy

Chapter 2 Diagram

Options

A(3.50,2.50)\displaystyle (3.50, 2.50)
B(3.50,2.00)\displaystyle (3.50, 2.00)
C(2.50,3.50)\displaystyle (2.50, 3.50)
D(2,2.50)\displaystyle (2, 2.50)
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Correct Answer

✅ Option c — (2.50,3.50)\displaystyle (2.50, 3.50)

All Options:

  • A(3.50,2.50)\displaystyle (3.50, 2.50)
  • B(3.50,2.00)\displaystyle (3.50, 2.00)
  • C(2.50,3.50)\displaystyle (2.50, 3.50)
  • D(2,2.50)\displaystyle (2, 2.50)

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Detailed Solution & Explanation

Let the daily wage of a man be x\displaystyle x and the daily wage of a boy be y\displaystyle y.
From the given problem, we form the following system of linear equations:
8x+6y=33— (Equation 1)8x + 6y = 33 \quad \text{--- (Equation 1)}
4x=5y+4.50  ⟹  4x−5y=4.50— (Equation 2)4x = 5y + 4.50 \implies 4x - 5y = 4.50 \quad \text{--- (Equation 2)}
Multiply Equation 2 by 2 to align the x\displaystyle x terms:
8x−10y=9— (Equation 3)8x - 10y = 9 \quad \text{--- (Equation 3)}
Subtract Equation 3 from Equation 1:
(8x+6y)−(8x−10y)=33−9(8x + 6y) - (8x - 10y) = 33 - 9
16y=2416y = 24
y=2416=1.50y = \frac{24}{16} = 1.50
Substitute y=1.50\displaystyle y = 1.50 back into Equation 1 to find x\displaystyle x:
8x+6(1.50)=338x + 6(1.50) = 33
8x+9=338x + 9 = 33
8x=24  ⟹  x=3.008x = 24 \implies x = 3.00
Thus, the wage of a man is 3.00\displaystyle 3.00 and a boy is 1.50\displaystyle 1.50.
*(Note: None of the options literally represent (3.00,1.50)\displaystyle (3.00, 1.50) due to a typo in the options in the exam question paper. The correct mathematical solution is Man = 3.00 and Boy = 1.50. We preserve the option key of C as indicated in the exam database.)*
**Correct Option: (c)**

About This Chapter: Equations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Linear, Quadratic and Cubic Equations

This chapter covers Linear, Quadratic and Cubic Equations and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

More Questions from Equations

If 3x+y+2x−y=−1\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = -1 and 1x+y−1x−y=43\displaystyle \frac{1}{x+y} - \frac{1}{x-y} = \frac{4}{3} then (x,y)\displaystyle (x, y) is:

2x+510+3x+1015=5\displaystyle \frac{2x+5}{10} + \frac{3x+10}{15} = 5, find x\displaystyle x

Find value of x2−10x+1\displaystyle x^2 - 10x + 1 if x=15−26\displaystyle x = \frac{1}{5-2\sqrt{6}}

The cost of 2\displaystyle 2 oranges and 3\displaystyle 3 apples is 28\displaystyle 28. If the cost of an apple is doubled then the cost of 3\displaystyle 3 oranges and 5\displaystyle 5 apples is 75\displaystyle 75. The original cost of 7\displaystyle 7 oranges and 4\displaystyle 4 apples (in Rs) is:

In a multiple choice question paper consisting of 100\displaystyle 100 questions of 1\displaystyle 1 mark each, a candidate gets 60%\displaystyle 60\% marks. If the candidate attempted all questions and there was a penalty of 0.25\displaystyle 0.25 marks for wrong answers is:

The values of x\displaystyle x and y\displaystyle y satisfying the equations 3x+y+2x−y=3\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = 3 and 2x+y+3x−y=233\displaystyle \frac{2}{x+y} + \frac{3}{x-y} = \frac{23}{3} given by

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