QOTD | Quantitative Aptitude | Algebra

Question of The Day23-03-2020

If a + b = cx, b + c = ay and c + a = bz then what will be the approximate value of \(\frac{7}{8}(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1})\) ?

Answer

Correct Answer : d ) 0.875

Explanation :

Consider the given equations

\(a+b=cx\)

Or, \(a+b+c=cx+c\)

\(\frac{1}{x+1}=\frac{c}{a+b+c}......(1)\)

\(b+c=ay\)

Or, \(a+b+c=ay+a\)

\(\frac{1}{y+1}=\frac{a}{a+b+c}......(2)\)

\(c+a=bz\)

Or,\(a+b+c=bz+b\)

\(\frac{1}{z+1}=\frac{b}{a+b+c}......(3)\)

Adding equation (1), (2), and (3) we get

\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=\frac{a+b+c}{a+b+c}\)

\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1\)

Therefore,

\(\frac{7}{8}(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1})=\frac{7}{8}=0.875\)

Hence, (d) is the correct answer.

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS etc.

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