EquationsPYQ Jan 26Question 4200 of 155
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If 2x+3y=34\displaystyle 2x + 3y = 34 and x+yy=138\displaystyle \frac{x+y}{y} = \frac{13}{8}, find the value of 7x+5y\displaystyle 7x + 5y.

Options

A45
B53
C75
D35
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Correct Answer

Option c75

All Options:

  • A45
  • B53
  • C75
  • D35

Detailed Solution & Explanation

Given the system of equations:
2x+3y=34\displaystyle 2x + 3y = 34 ---(Equation 1)
x+yy=138\displaystyle \frac{x+y}{y} = \frac{13}{8} ---(Equation 2)

From Equation 2, we can simplify the expression:
xy+yy=138\frac{x}{y} + \frac{y}{y} = \frac{13}{8}
xy+1=138\frac{x}{y} + 1 = \frac{13}{8}
xy=1381\frac{x}{y} = \frac{13}{8} - 1
xy=58\frac{x}{y} = \frac{5}{8}
8x=5y    y=8x58x = 5y \implies y = \frac{8x}{5} ---(Equation 3)

Now, substitute the value of y\displaystyle y from Equation 3 into Equation 1:
2x+3(8x5)=342x + 3\left(\frac{8x}{5}\right) = 34
2x+24x5=342x + \frac{24x}{5} = 34
Multiply the entire equation by 5\displaystyle 5 to eliminate the denominator:
10x+24x=17010x + 24x = 170
34x=17034x = 170
x=17034=5x = \frac{170}{34} = 5

Substitute x=5\displaystyle x = 5 back into Equation 3 to find y\displaystyle y:
y=8(5)5=8y = \frac{8(5)}{5} = 8

We need to find the value of 7x+5y\displaystyle 7x + 5y:
7x+5y=7(5)+5(8)7x + 5y = 7(5) + 5(8)
7x+5y=35+40=757x + 5y = 35 + 40 = 75

Hence, **Option C** 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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