How to solve a difficult SSC CGL level algebra question by variable ratio equality in a few steps 18 | SureSolv

How to solve a difficult SSC CGL level algebra question by variable ratio equality in a few steps 18

Variable ratio equality falls under advanced algebraic concepts, but it is easy to grasp and use

how to solve difficult algebra question variable ratio equality

Problems based on variable ratio equality may seem to be quite tough in competitive test environment unless you know the underlying concept. 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.


Solving difficult algebra question in a few steps by Variable ratio equality concept

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. $\displaystyle\frac{5}{3}$

Solution: Using variable ratio equality: Solving in mind

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.

Proof of 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 other way round involves quite a bit of deductive steps. Let us show this proof also.

Second proof of Variable ratio equality relation from other way round

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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