EquationsMCQMTP Sep 24 Series IIQuestion 1040 of 221
All Questions

4 tables and 3 chairs, together, cost $2,250$ and 3 tables and 4 chairs cost $1,950$. Find the cost of 2 chairs and 1 table.

Select Your Answer (Keyboard: A, B, C, D)
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option d — None of these

All Options:

  • A550
  • B1005
  • C750
  • DNone of these

Detailed Solution & Explanation

Let the cost of 1 table be t\displaystyle t and the cost of 1 chair be c\displaystyle c. From the given information: 4t+3c=2250— (i)4t + 3c = 2250 \quad — (i)\displaystyle \text{--- (i)} 3t+4c=1950— (ii)3t + 4c = 1950 \quad — (ii)\displaystyle \text{--- (ii)} Adding equations (i) and (ii): 7t+7c=4200  ⟹  t+c=600— (iii)7t + 7c = 4200 \implies t + c = 600 \quad — (iii)\displaystyle \text{--- (iii)} Subtracting equation (ii) from equation (i): t−c=300— (iv)t - c = 300 \quad — (iv)\displaystyle \text{--- (iv)} Adding (iii) and (iv): 2t=900  ⟹  t=4502t = 900 \implies t = 450 Subtracting (iv) from (iii): 2c=300  ⟹  c=1502c = 300 \implies c = 150 We need to find the cost of 2 chairs and 1 table: Cost=2c+t=2(150)+450=300+450=750Cost=2c+t=2\displaystyle \text{Cost} = 2c + t = 2(150) + 450 = 300 + 450 = 750 Since $750$ is listed as Option c, the mathematically correct answer is Option c. However, according to the provided key, the answer is marked as Option d ("None of these"). Based on the provided key: **Option d**

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 3x+y+2x−y\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = -1 and 1x+y−1x−y=43\displaystyle 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 = 15−26\displaystyle \frac{1}{5-2\sqrt{6}}

The cost of 2\displaystyle 2 oranges and 3\displaystyle 3 apples is $28$. If the cost of an apple is doubled then the cost of $3orangesand\displaystyle oranges and5$ apples is $75$. The original cost of 7\displaystyle 7 oranges and 4\displaystyle 4 apples (in Rs) is:

In a multiple choice question paper consisting of $100questionsof\displaystyle questions of1\displaystyle mark each, a candidate gets $60\% marks. If the candidate attempted all questions and there was a penalty of $0.25$ marks for wrong answers is:

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

Ready to Master Equations?

Practice all 221 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free