3. Ratio#

In his book Elements of Algebra, Euler says that anything we can increase or decrease is magnitude or quantity.

We will study ratios of quantities in this chapter. Many of you might already be knowing about ratios from earlier classes. A ratio is nothing but a comparison between two quantities of same kind. For example, 4 balls vs 7 balls or 12 shoes vs 8 shoes.

A ratio between two quantities \(Q1\) and \(Q2\) is written as \(Q1:Q2\) most of the time. \(Q1\) or the first term is called the antecedent and the \(Q2\) or the second term is called the consequent.

To measure the ratio between two quantities typically we use fractions. A reduced fraction is tells us about one quantity being multiples of another or part of that. For example, 2 dozen bananas compared to 6 bananas can be written as \(\frac{2*12}{6}\) or 4.

Ratios are abstract quantities because they are sheer numbers. In practice, considering our previous example we can say 2 dozen bananas are four times 6 bananas. This four is an abstract quantity because it does not have a physical form.

Clearly, you can see that

\[\frac{Q1}{Q1} = \frac{mQ1}{mQ2}\]

so the ratio does not change if numerator or antecedent and denominator or consequent are multiplied by the same quantity.

3.1. Comparing Two Ratios#

To compare two ratios we need to make their denominator same the we can deduce if they are equal or unequal. For example consider two ratios \(r1:r2\) and \(r3:r4\). Then we can have

\[\frac{r1}{r2} = \frac{r1r4}{r2r4} ~~\&~~ \frac{r3}{r4} = \frac{r3r2}{r2r4}\]

Now since consequent is same for both the ratios we can determine which ratio is bigger or equal or smaller by determining if \(r1r4\) is bigger or equal or smaller as compared to \(r2r3\).

3.2. Ratio of Ratios#

Clearly, a ratio can be represented by a fraction of two integers. Now let us try to see what happens to ratio of two ratios:

Consider two ratios: \(a:b\) and math:c:d. Thus we can write ratio of these two ratios as fraction

\[\frac{\frac{a}{b}}{\frac{c}{d}} ~~or~~ \frac{ad}{bc} \]

Therefore we can conclude that it is equivalent to \(ad:bc\).

3.3. Surds and Ratios#

When I said a ratio can be applied as a fraction of two integers then that applied to finite numbers. A surd like \(\sqrt{x}\) is not a finite number because certain portion of its fractional part if recurring. Therefore, when a surd if involved a ratio cannot be represented as a fraction of two integers.

When we can represent a ratio exactly by two integers the quantities are called commensurable else they are called incommensurable.

3.4. Compounding Ratios#

Ratios are compounded by multiplying ratios. The multiplication is done in simple fashion. Antecedents with antecedents and consequents with consequents.

For example, consider some ratios: \(a:b\), \(c:d\) and \(e:f\). The compounded ratio is \(abc:def\).

3.5. Duplicating Ratios#

When a ratio is compounded with itself once it is called duplicate ratio. If it is compounded twice then it is called triplicate ratio. If a square root is taken then it is called subduplicate ratio.

3.6. Antecedent vs Consequent#

A ratio is of greater inequality, or less inequality or of equality depending on whether antecedent is greater or smaller or equal to the consequent.

3.7. Common Properties of Ratios#

A ratio of greater inequality is diminished and a ratio of less inequality is increased if we add same quantity to both its terms.

Let \(\frac{a}{b}\) be a ratio and we add \(x\) to both its terms so we have \(\frac{a+x}{b+x}\) as our new ratio.

Now,

\[\frac{a}{b}~-~\frac{a+x}{b+x}~=~\frac{ax-bx}{b(b+x)} \\ =\frac{x(a-b)}{b(b+x)}.\]

and \(a-b\) is positive or negative depending upon the fact whether \(a\) is greater or smaller than \(b\). Hence,

\[\text{if}~a> b,~~ \frac{a}{b}>\frac{a+x}{b+x} \text{and if}~ a< b,~~ \frac{a}{b}<\frac{a+x}{b+x}\]

Similarly, it can be proved that a ratio of greater inequality is increased and a ratio of less inequality is decreased if we subtract same quantity to both its terms.

When two or more ratios are equal we can denote that by a single symbol which helps us in many ways.

Consider,

\[\text{If}~\frac{a}{b}=\frac{c}{d}=\frac{e}{f}= ... , \text{each of these ratios}~=\left(\frac{pa^n+qc^n+re^n+ ...}{{pb^n+qd^n+rf^n}+ ...}\right)^\frac{1}{n} \text{Let}~\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=...=k; \text{the}~a=bk, c=dk, e=fk, ...; \text{Hence},~pa^n=pb^nk^n,~qc^n=qd^nk^n,~re^n=rf^nk^n,...; \implies \left(\frac{pa^n+qb^n+re^n+ ...}{{pb^n+qd^n+rf^n}+ ...}\right) = \left(\frac{pb^nk^n+qd^nk^n+rf^nk^n+ ...}{{pb^n+qd^n+rf^n}+ ...}\right) = k^n \implies \left(\frac{pa^n+qb^n+re^n+ ...}{{pb^n+qd^n+rf^n}+ ...}\right)^\frac{1}{n} = k\\\]

Similarly, we can say that when a a number of fractions are equal, each of them is equal to the sum of all the numerators divided by the sum of denominators,

\[\text{If}~\frac{a}{b}=\frac{c}{d}=\frac{e}{f}= ... , \implies \frac{a+c+e}{b+d+f} = \frac{bk+dk+fk}{b+d+f} = k.\]

if \(\frac{a_1}{b_1}, \frac{a_2}{b_2}, ..., \frac{a_n}{b_n}\) be unequal fractions, of which denominators are all having the same sign, then the fraction

\[\frac{a_1 + a_2 + ... + a_n}{b_1 + b_2+ .. +b_n} \]

lies in magnitude between the greatest and least of them.

Let all the denominators are positive. Let \(\frac{a_r}{b_r}\) be the least fraction, and denote it by \(k\); then

\[\frac{a_r}{b_r} = k; \therefore a_r = kb_r; \frac{a_1}{b_1} = k; \therefore a_1 = kb_1; \frac{a_2}{b_2} = k; \therefore a_2 = kb_2; \text{and so on;} \therefore \text{by addition,} a_1 + a_2 + ... + a_n > (b_1 + b_2 + ... + b_n)k; \therefore\frac{a_1 + a_2 + ... + a_n}{b_1 + b_2 + ... + b_n}> k; \text{that is,}>\frac{a_r}{b_r}\]

Similarly, we may prove that

\[\frac{a_1 + a_2 + ... + a_n}{b_1 + b_2 + ... + b_n}> k; \text{that is,}<\frac{a_s}{b_s} \]

where \(\frac{a_s}{b_s}\) is greatest of the given functions.

If we have two equations containing three unknown variables in the first degree such as

(1)#\[a_1x + b_1y + c_1z = 0\]
(2)#\[a_2x + b_2y + c_2z = 0\]

Rewriting these we can have following;

\[a_1\left(\frac{x}{z}\right) + b_1\left(\frac{y}{z}\right) + c_1 =0, a_2\left(\frac{x}{z}\right) + b_2\left(\frac{y}{z}\right) + c_2 =0,\]

Solving these we obtain,

\[\frac{x}{z}=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},~~~\frac{y}{z}=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1} \]

or,

(3)#\[\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1}=\frac{z}{a_1b_2-a_2b_1}\]

So we can see that for \(x\) coefficients of \(y\) and \(z\) are used and for \(y\), that of \(x\) and \(z\) are used and so on in a cyclic manner as given in image below:

Figure made with TikZ

Cross multiplication of co-efficients

Multiply the coefficients as indicated by the arrows. The ascending arrows are positive and descending ones are negative. This is called the Rules of Cross Multiplication.

Now expanding on previous discussion consider following set of three equations:

(4)#\[a_1x + b_1y + c_1z=0\]
(5)#\[a_2x + b_2y + c_2z=0\]
(6)#\[a_3x + b_3y + c_3z=0\]

So by our previous rule of cross-multiplication between eq. (4) and (5) we have

\[\frac{x}{z}=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},~~~\frac{y}{z}=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\]

Now if we take these ratios to be equal to \(k\) then multiplying and substituting in equation (6) we have

\[a_3(b_1c_2-b_1c_2)+b_3(c_1a_2-c_2a_1)+c_3(a_1b_2-a_2b_1) = 0 \]

This particular relation is known as the eliminant of the given equations.

3.8. Problems#

  1. Find the ratio of ratios \(4:7\) and \(11:5\).

  2. Find the ratio of ratios \(13:24\) and \(34:23\).

  3. Find out that among the following pair of ratios which one is greater in the pair:

    1. \(4:5\) and \(5:6\),

    2. \(34:33\) and \(68:67\) and

    3. \(123:63\) and \(233:120\).

  4. Find the duplicate, triplicate and subduplicate of following ratios:

    1. \(2(x+3):y-5\)

    2. \(a^2+ab+b^2:a^3-b^3\)

  5. If \(x-2:3(x+7)\) has a ratio of \(7:9\) then find \(x\).

  6. Find two numbers in the ratio of \(11:17\) where one exceeds other by 126.

  7. What number must be added to each of the terms of \(5:13\) to make it equal to \(5:7\)?

  8. What number must be subtracted from each of the terms of \(33:37\) to make it equal to \(7:9\)?

  9. If \(x-2:y-3=3:5\), find the ratio of \(9x-2y:4x+2y\).

  10. If

\[\frac{a}{b}=\frac{c}{d}=\frac{e}{f}, \]

prove that

\[\frac{7a^4b^2+8a^2c^2-5e^4f}{7b^6+8b^2f^2-5e^5} = \frac{a^4}{b^4}. \]

  1. If

\[\frac{a}{b}=\frac{b}{c}=\frac{c}{d}, \]

prove that \(\frac{a}{d}\) is equal to

\[\sqrt{\frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}}.. \]

  1. If

\[\frac{x}{q+r-p}=\frac{y}{r+p-q}=\frac{z}{p=q-r} \]

prove that

\[(q-r)x + (r-p)y + (p-q)z = 0. \]

  1. If

\[\frac{y+z}{pb+qc}=\frac{z+x}{pc+qa}=\frac{x+y}{pa+qb}, \]

prove that

\[\frac{2(x+y+z)}{a+b+c} = \frac{(b+c)x+(c+a)y+(a+b)z}{ab+bc+ca}. \]

  1. If

\[\frac{x}{x}=\frac{y}{b}=\frac{z}{c}, \]

prove that

\[\frac{x^3+a^3}{x^2+a^2}+\frac{y^3+b^3}{y^2+b^2}+\frac{z^3+c^3}{z^2+c^2}=\frac{(x+y+z)^3+(a+b+c)3}{(x+y+z)^2+(a+b+c)^2}.\]

  1. If

\[\frac{2y+2z-x}{a}=\frac{2z+2x-y}{b}=\frac{2x+2y-z}, \]

prove that

\[\frac{x}{2b+2c-a}=\frac{y}{2c+2a-b}=\frac{z}{qa+2b-c}. \]

16. If \((a^2+b^2+c^2)(x^2+y^2+z^2)=(ax+by+cz)^2\) show that \(x:a=y:b=z:c\).

  1. If

\[x=\frac{a-b}{a+b},~y=\frac{b-c}{b+c},~z=\frac{c-a}{c+a} \]

prove that

\[(1+x)(1+y)(1+z) = (1-x)(1-y)(1-z). \]

  1. If \(l(my+nz-lx)=m(nz+lx-my)=n(lx+my-nz),\) prove that

\[\frac{y+z-x}{l}=\frac{z+x-y}{m}=\frac{x+y-z}{n} \]

19. Show that the eliminant of \(ax+by+cz=0, cx+by+az=0, bx+ay+cz=0\), is \(a^3+b^3+c^3-3abc\).

20. Eliminate \(x, y, z\) from the equations \(ax+hy+gz=0, hx+by+fz=0, gx+fy+cz=0\).

  1. If \(x=cy+bz, y=az+cx, z=bx+ay\), prove that

\[\frac{x^2}{1-a^2}=\frac{y^2}{1-b^2}=\frac{z^2}{1-c^2} \]

  1. If \(a(y+z)=x, b(z+x)=y, c(x+y)=z\) prove that \(bc+ca+ab+2abc=1\).

  2. Solve the following equations:

    1. \(3x-4y+7z=0, 2x-y-2z=0, 3x^3-y^3+z^3=18\).

    2. \(x+y-z=0, 3x-2y+17z=0, x^3+3y^3+2z^3=167\).

    3. \(4xy-7yz-3zx=0, 4xy-21yz+3zx=0, x+2y+3z=19\).

    4. \(3x^2-2y^2+5z^2=0, 7x^2-3y^2-15z^2=0, 5x-4y+7z=0\).

  3. If

\[\frac{l}{\sqrt{a}-\sqrt{b}}+\frac{m}{\sqrt{b}-\sqrt{c}}+\frac{n}{\sqrt{c}-\sqrt{a}}=0, \frac{l}{\sqrt{a}+\sqrt{b}}+\frac{m}{\sqrt{b}+\sqrt{c}}+\frac{n}{\sqrt{c}+\sqrt{a}}=0,\]

show that

\[\frac{l}{(a-b)(c-\sqrt{ab})}=\frac{m}{(b-c)(a-\sqrt{bc})}=\frac{n}{(c-a)(b-\sqrt{ac})} \]

  1. Solve the following equations:

    1. \(ax+by+cz=0, bcx+cay+abz=0,xyz+abc(a^3x+b^3y+c^3z)=0\).

    2. \(ax+by+cz=a^2x+b^2y+c^2z=0,x+y+z+(a-b)(b-c)(c-a)=0\).

  2. If \(a(y+z)=x, b(z+x)=y, c(x+y)=z\) prove that

\[\frac{x^2}{a(1-bc)}=\frac{y^2}{b(1-ca)}=\frac{z^2}{c(1-ab)}. \]

  1. If \(ax+hy+gz=0, hx+by+fz=0, gx+fy+cz=0\), prove that

\[\frac{x^2}{bc-f^2}=\frac{y^2}{ca-g^2}=\frac{z^2}{ab-h^2}~~\text{and} \]
\[(bc-f^2)(ca-g^2)(ab-h^2)=(gh-af)(fg-ch)(hf-bg) \]