EquationsMCQPYQ Nov 19Question 986 of 221
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2x+510+3x+1015=5\displaystyle \frac{2x+5}{10} + \frac{3x+10}{15} = 5, find x\displaystyle x

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option b — $9.58$

All Options:

  • A$10.58$
  • B$9.58$
  • C$9.5$
  • DNone of these

Detailed Solution & Explanation

The given equation is:
2x+510+3x+1015=52x+510+3x+1015=5\displaystyle \frac{2x+5}{10} + \frac{3x+10}{15} = 5
Taking the Least Common Multiple (LCM) of $10\displaystyle and &#36;15, which is $30\displaystyle , we can rewrite the equation as:<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mrow><mn>3</mn><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><mo>+</mo><mn>5</mn><mo stretchy="false">)</mo><mo>+</mo><mn>2</mn><mo stretchy="false">(</mo><mn>3</mn><mi>x</mi><mo>+</mo><mn>10</mn><mo stretchy="false">)</mo></mrow><mn>30</mn></mfrac><mo>=</mo><mn>5</mn></mrow><annotation encoding="application/x-tex">3(2x+5)+2(3x+10)30=5\displaystyle \frac{3(2x+5) + 2(3x+10)}{30} = 5</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:2.113em;vertical-align:-0.686em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.427em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">30</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">3</span><span class="mopen">(</span><span class="mord">2</span><span class="mord mathnormal">x</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">5</span><span class="mclose">)</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">2</span><span class="mopen">(</span><span class="mord">3</span><span class="mord mathnormal">x</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">10</span><span class="mclose">)</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.686em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">5</span></span></span></span></span><br>Multiplying both sides by &#36;30:
3(2x+5)+2(3x+10)=1503(2x+5) + 2(3x+10) = 150
Expanding the terms:
6x+15+6x+20=1506x + 15 + 6x + 20 = 150
12x+35=15012x + 35 = 150
12x=150−3512x = 150 - 35
12x=11512x = 115
x=11512≈9.58x = 11512\displaystyle \frac{115}{12} ≈\displaystyle \approx 9.58
Hence, **Option B** is the correct answer.

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.

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