SSC CGL Algebra: Variable Ratio Equality Trick

Learn How to Apply SSC CGL Algebra Variable Ratio Equality Trick: It's Easy

Master the variable ratio equality trick in SSC CGL Algebra. Learn to solve hard algebra problems in competitive tests quickly and boost your exam score.

The chosen problem with its solution highlights use of the concept for quick mental solution of the apparently hard algebra problem.

Let us showcase the problem and its solution.

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Solving the Difficult SSC CGL Algebra Question in a Few Steps by Variable Ratio Equality Concept (it is after all a concept, not a trick)

Problem example

If $ax+by+cz=20$, $a^2+b^2+c^2=16$ and $x^2+y^2+z^2=25$, then the value of $\displaystyle\frac{a+b+c}{x+y+z}$ is,

1. $\displaystyle\frac{4}{5}$
2. $\displaystyle\frac{3}{5}$
3. $\displaystyle\frac{5}{4}$
4. $\displaystyle\frac{5}{3}$

Understanding Variable Ratio Equality Pattern and Solving the Hard looking Problem in Just a Few Steps

The product of the LHSs of the second and third given equations is 400 and is equal to the square of the LHS of the first given equation. So by variable ratio equality concept, it follows that the ratios of the two sets of three variables are equal,

$\displaystyle\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=p$, say

It follows,

$a=xp$,

$b=yp$,

$c=zp$

Adding the three equations,

$a+b+c=p(x+y+z)$,

Or, $\displaystyle\frac{a+b+c}{x+y+z}=p$.

Substituting values of $a$, $b$ and $c$ into the first equation,

$a^2+b^2+c^2=16$,

Or, $p^2(x^2+y^2+z^2)=16$,

Or, $p=\displaystyle\frac{4}{5}=\frac{a+b+c}{x+y+z}$.

Answer: Option a: $\displaystyle\frac{4}{5}$.

Let us state the variable ratio equality concept we have used,

If $\displaystyle\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$, then $(a^2+b^2+c^2)(x^2+y^2+z^2)=(ax+by+cz)^2$ and vice versa.

Let us show in two ways how this happens.

Now be Convinced and Know the Proof of The Variable Ratio Equality Relation by Assuming Variable Ratios are Equal

To prove,

If $\displaystyle\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$, prove that, $(a^2+b^2+c^2)(x^2+y^2+z^2)=(ax+by+cz)^2$.

If $\displaystyle\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$, assuming each ratio to be equal to a dummy variable $p$ gives,

$a=px$,

$b=py$,

$c=pz$.

Substituting the values in RHS of the product,

$(ax+by+cz)^2$

$=p^2(x^2+y^2+z^2)^2$.

Substituting the values of $a$, $b$ and $c$ in the LHS product,

$(a^2+b^2+c^2)(x^2+y^2+z^2)$

$=p^2(x^2+y^2+z^2)^2$

$=(ax+by+cz)^2$ Proved.

But the proof the relation using other way round involves quite a bit of deductive steps. Let us show this proof also.

It never hurts to know more about these intriguing concepts.

Here is the Second Proof of Variable Ratio Equality Relation Applying Other Way Round Approach: But It's Longer

To prove,

If $(a^2+b^2+c^2)(x^2+y^2+z^2)=(ax+by+cz)^2$, prove that $\displaystyle\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$.

Expanding both sides,

$a^2x^2+b^2y^2+c^2z^2+a^2(y^2+z^2)$

$\hspace{20mm}+b^2(z^2+x^2)+c^2(x^2+y^2)$

$=a^2x^2+b^2y^2+c^2z^2+2(abxy+bcyz+cazx)$,

Or, $(a^2y^2-2abxy+b^2x^2)+(a^2z^2-2cazx+c^2x^2)$

$\hspace{20mm}+(b^2z^2-2bcyz+c^2y^2)=0$

Or, $(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0$.

By zero sum of square terms algebraic principle, each of the square terms on the LHS must be zero,

$ay-bx=0$,

Or, $\displaystyle\frac{a}{x}=\frac{b}{y}$, and

$az-cx=0$,

Or, $\displaystyle\frac{a}{x}=\frac{c}{z}$.

So,

$\displaystyle\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$.

This relation will be true for two sets of more number of variables also.

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The list of Difficult algebra problem solving in a few steps quickly is available at, Quick algebra.

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Guided help on Algebra in Suresolv

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Suresolv Algebra Reading and Practice Guide for SSC CHSL, SSC CGL, SSC CGL Tier II and Other Competitive exams.

The guide list of articles includes ALL articles on Algebra in Suresolv and is up-to-date.

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