If x+y, y+z, z+x are in the ratio 6:7:8 and x+y+z = 14 then the value of x is.

Option a: 6. Solution: Given the continuous ratio (x+y) : (y+z) : (z+x) = 6 : 7 : 8 , let us write: x + y = 6k, y + z = 7k, z + x = 8k Summing these three equations together: (x+y) + (y+z) + (z+x) = 6k + 7k + 8k 2(x + y + z) = 21k We are given x + y + z = 14 . Substituting this in: 2(14) = 21k 28 = 21k k = 28/21 = 4/3 Now we find x : From y + z = 7k , we substitute k = 4/3 : y + z = 7(4/3) = 28/3 Since x + (y + z) = 14 , we solve for x : x = 14 - (y+z) = 14 - 28/3 = 42 - 28/3 = 14/3

If $x+y, y+z, z+x$ are in the ratio $6:7:8$ and $x+y+z = 14$ then the value of $x$ is.

The correct answer is Option a: $6$. Given the continuous ratio $(x+y) : (y+z) : (z+x) = 6 : 7 : 8$, let us write: $$x + y = 6k, \quad y + z = 7k, \quad z + x = 8k$$ Summing these three equations together: $$(x+y) + (y+z) + (z+x) = 6k + 7k + 8k$$ $$2(x + y + z) = 21k$$ We are given $x + y + z = 14$. Substituting this in: $$2(14) = 21k \implies 28 = 21k \implies k = \frac{28}{21} = \frac{4}{3}$$ Now we find $x$: From $y + z = 7k$, we substitute $k = \frac{4}{3}$: $$y + z = 7\left(\frac{4}{3}\right) = \frac{28}{3}$$ Since $x + (y + z) = 14$, we solve for $x$: $$x = 14 - (y+z) = 14 - \frac{28}{3} = \frac{42 - 28}{3} = \frac{14}{3}$$

Ratio, Proportion, Indices, LogarithmMCQMTP Nov 19Question 820 of 305

All Questions

If $\displaystyle x+y, y+z, z+x$ are in the ratio $\displaystyle 6:7:8$ and $\displaystyle x+y+z = 14$ then the value of $\displaystyle x$ is.

Options

\displaystyle 6

\displaystyle 7

\displaystyle 8

\displaystyle 10

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Correct Answer

✅ Option a — $\displaystyle 6$

All Options:

A $\displaystyle 6$
B $\displaystyle 7$
C $\displaystyle 8$
D $\displaystyle 10$

Detailed Solution & Explanation

Given the continuous ratio

\displaystyle (x+y) : (y+z) : (z+x) = 6 : 7 : 8

, let us write:

x + y = 6k, \quad y + z = 7k, \quad z + x = 8k

Summing these three equations together:

(x+y) + (y+z) + (z+x) = 6k + 7k + 8k

2(x + y + z) = 21k

We are given

\displaystyle x + y + z = 14

. Substituting this in:

2(14) = 21k \implies 28 = 21k \implies k = \frac{28}{21} = \frac{4}{3}

Now we find

\displaystyle x

:
From

\displaystyle y + z = 7k

, we substitute

\displaystyle k = \frac{4}{3}

y + z = 7\left(\frac{4}{3}\right) = \frac{28}{3}

Since

\displaystyle x + (y + z) = 14

, we solve for

\displaystyle x

x = 14 - (y+z) = 14 - \frac{28}{3} = \frac{42 - 28}{3} = \frac{14}{3}

About This Chapter: Ratio, Proportion, Indices, Logarithm

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

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Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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If x+y,y+z,z+x\displaystyle x+y, y+z, z+xx+y,y+z,z+x are in the ratio 6:7:8\displaystyle 6:7:86:7:8 and x+y+z=14\displaystyle x+y+z = 14x+y+z=14 then the value of x\displaystyle xx is.