# 4. Solutions for Ratios¶

1. We have 4:7 and 11:5. Therefore, ratio of these two ratios is

$\frac{4}{7}*\frac{5}{11}=\frac{20}{77}.$
1. For 13:24 and 34:23 we have their ratio as

$\frac{13}{24}*\frac{23}{34}=\frac{299}{816}.$
1. a. For these two we make their denominators equal. We multiply and divide first by 6 and second by 5 then we have

$\frac{4}{5}*\frac{6}{6} = \frac{24}{30}$

and

$\frac{5}{6}*\frac{5}{5} = \frac{25}{30}$

Therefore, 5:6 is greater.

1. Similarly, as in previous case

$\frac{34}{33}*\frac{67}{67} = \frac{2278}{2211}$

and

$\frac{68}{67}*\frac{33}{33} = \frac{2244}{2211}$

Clearly, 34:33 is greater.

c. Following the principal from last two problems we have: $$\frac{123}{63} = \frac{123*120}{63*120} = \frac{14760}{7860}$$ and $$\frac{233}{120} = \frac{233*63}{120*63} = \frac{14679}{7860}$$. Clearly, 123:63 is greater than 233:63.

1. a. Given ratio is 2(x+3):y-5. Therefore, duplicate, triplicate and subduplicate are given as follows:

$\left(\frac{2(x+3)}{y-5}\right)^2, \left(\frac{2(x+3)}{y-5}\right)^3 \text{and} \sqrt{\frac{2(x+3)}{y-5}}.$

Expanding the first two will give:

$\frac{2x^2+12x+18}{y^2-10y+25}~\text{and}\frac{2x^3+6x^2+18x+54}{y^3-5y^2+25y-125}.$

b. The given ratio is $$a^2+ab+b^2:a^3-b^3$$ which can be simplified to $$1:a-b$$ and then our required ratios are $$1:(a-b)^2, 1:(a-b)^3~\text{and}~1:\sqrt{a-b}$$. Expansion gives following

$\frac{1}{a^2-2ab+b^2}~\text{and}~\frac{1}{a^3-3a^2b+3ab^2-b^3}.$
1. Given,

$\frac{x-2}{x(x+7)} = \frac{7}{9} \Rightarrow~9x-18=21x+47 \Rightarrow~12x=-165 \Rightarrow~x=-\frac{55}{4}.$

6. Let us say one no. is x the clearly second is x+126. Now the ratio is x:x+126 which should be equal to 11:17. Therefore, we have

$\frac{x}{x+126}=\frac{11}{17} \Rightarrow~17x=11x+1386 \Rightarrow~x=231.$
1. Let us say the no. to be added is x then we have from problem

$\frac{5+x}{13+x}=\frac{5}{7} \Rightarrow~35+7x=65+5x \Rightarrow~x=15.$
1. It is similar as last problem. Let us say number is x then we have

$\frac{33-x}{37-x}=\frac{7}{9} \Rightarrow~197-9x=259-7x \Rightarrow~x=19$

Since x=19 out ratio is 14:18 i.e. 7:9 is our desired ratio.

9. Clearly, x is 3+2 i.e 5 and y is 5+3 i.e. 8 which satisfies the first pair of ratios. Substituting these in third ratio yields 29:26.

10. Let, $$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$$ then if we multiply and divide numerator with $$a^4$$ and denominator with $$b^4$$ then we will have following:

$\frac{a^4\left(7b^2+\frac{8c^2}{a^2}-\frac{5e^4f}{a^4}\right)}{b^4\left(7b^2+\frac{8f^2}{b^2}-\frac{5e^5}{b^4}\right)}$

Now, let us look at second fractions carefully inside parentheses in both numerator and denominator. From first first and last ratios we can say that $$f=\frac{be}{a}$$.

$\Rightarrow~\frac{f^2}{b^2}=\frac{b^2e^2}{a^2b^2}=\frac{e^2}{a^2}$

Now, similarly e can be written as $$\frac{cf}{d}$$,

$\Rightarrow~\frac{e^2}{a^2} = \frac{c^2f^2}{d^2a^2}$

Clearly, from first two ratios $$\frac{c}{a}=\frac{d}{b}$$. Therefore, second term in consideration comes to be equal. Similarly, third can be proven as well and is left to reader to do that.

11. Given that, $$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k$$ let us rewrite all in terms of a and k. Clearly, $$c=dk,~b=dk^2,~\text{and}~a=dk^3$$. Now, assuming all in terms of d and k and rewriting without square root we have following:

$\frac{d^5k^{15}+d^2k^4d^2k^2+d^3k^9d2k^2}{d4k^8dk+d^4+d^2k^4dkd^2} \Rightarrow~\frac{d^5k^{15}+d^4k^6+d^5k^{11}}{d^5k^9+d^4+d4k^5}$

Now, multiplying and dividing numerator by $$d^4k^6$$ and denominator by $$d^4$$ i.e. in other words we are taking out these common factors. This will reduce the fraction to following:

$\frac{d^4k^6(dk^9+1+dk^5)}{d^4(dk^9+1+dk^5)}$

Now, clearly this has reduced to $$\frac{d^4k^6}{d^4}$$. Taking square and eliminating one power of d gives us $$\frac{dk^3}{d}$$ which is nothing but ratio of a and d.

1. Let us equate given ratios to a constant k. Then, we have

\begin{align}\begin{aligned}\frac{x}{q+r-p}=\frac{y}{r+p-q}=\frac{z}{p+q-r}=k\\\Rightarrow~x=(q + r - p)k, y = (p + r - q)k~\text{and}~z = (p + q - r)k\end{aligned}\end{align}

Now substituting these in the given equation we get,

$(q - r)(q + r - p)k + (r - p)(p + r - q)k + (p - q)(p + q - r)k = 0$

Now it is as simple as multiplying and cancelling terms which are equal which will make left hand side 0.

1. Let,

$\frac{y+z}{pb+qc}=\frac{z+x}{pc+qa}=\frac{x+y}{pa+qb}=k$
$\Rightarrow~y+z=(pb+qc)k,~z+x=(pc+qa)k,~\text{and}~x+y=(pa+qb)k$

Adding all left hand sides and respective right hand sides we have

$2(x+y+z) = (pa+pb+pc+qa+qb+qc)k$
$\Rightarrow~\frac{2(x+y+z)}{a+b+c} = \frac{(pa+pb+pc+qa+qb+qc)k}{a+b+c}$
$\Rightarrow~\frac{(pa+pb+pc+qa+qb+qc)(ab+bc+ca)k}{(a+b+c)(ab+bc+ca)}$
$\begin{split}\Rightarrow~\frac{\begin{array}{l}(pa b^2+p b^2 c+pabc+qabc+qb c^2+q c^2 a+\\ pabc+pb c^2+p c^2 a+q a^2 b+qabc+qc a^2+\\ p a^2 b+pabc+pc a^2+qa b^2+q b^2 c+qabc)k\end{array}}{(a + b + c)(ab + bc + ca)}\end{split}$
$\Rightarrow~\frac{(a+b+c)(pab+pbc+pca+qab+qbc+qca)k}{(a+b+c)(ab+bc+ca)}$
$\Rightarrow~(p+q)k$
$\Rightarrow~\frac{(a+b+c)(p+q)k}{(a+b+c)}$
$\Rightarrow~\frac{(pab+pbc+pca+qab+qbc+qca)k}{a+b+c}$
$\Rightarrow~\frac{2(x+y+z)}{a+b+c}.$
1. Let,

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$
$\Rightarrow~x=ak, y=bk,~\text{and}~z=ck$
$\therefore~L.H.S.=\frac{a^3k^3+a^3}{a^2k^2+a^2}+\frac{b^3k^3+b^3}{b^2k^2+b^2}+\frac{c^3k^3+c^3}{c^2k^2+c^2}$
$\Rightarrow~\frac{(a+b+c)(k^3+1)}{k^2+1}$

Now we will consider right side.

$R.H.S. = \frac{(ak+bk+ck)^3+(a+b+c)^3}{(ak+bk+ck)^2+(a+b+c)^2}$
$\Rightarrow~\frac{(a+b+c)^3(k^3+1)}{(a+b+c)^2(k^2+1)}$
$\therefore~L.H.S.=R.H.S.$

15. As we have done so far we can equate given ratios to k and then we can write

$2y+2z-x=ak,2z+2x-y=bk,~\text{and}~2x+2x-z=ck$

Now considering first ratio among to be proven for equality we have

$\frac{x}{2b+2c-a} = \frac{xk}{4z+4x-2y+4x+4y-2z-2y-2z+x}$
$\Rightarrow~\frac{xk}{9x}=\frac{k}{9}$

Similarly, 2nd and 3rd ratios can be proven equal to same.

1. Multiplying we have following on L.H.S.

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

Squaring we have following,

$a^2x^2+b^2y^2+c^2z^2+2abxy+2bcyz+2cazx$

Now, subtracting R.H.S from L.H.S we have following

$a^2y^2+b^2x^2-2abxy+a^2z^2+c^2x^2-2cazx+b^2z^2+c^2y^2-2bcyz=0$
$(ay-bx)^2+(za-cx)^2+(bz-cy)^2=0$

Now since squares are not negative hence all there expressions in parentheses must be zero.

$\therefore~ay=cx,~za=cx,~\text{and}~bz=cy$
$\therefore~\frac{x}{a}=\frac{y}{b},~\frac{x}{a}=\frac{c}{z},~\text{and}~\frac{z}{c}=\frac{y}{b}$

Hence, the required result is proven.

1. Clearly,

$\left(\frac{2a}{a+b}\right)\left(\frac{2b}{b+c}\right)\left(\frac{2c}{c+a}\right)= \left(\frac{2b}{a+b}\right)\left(\frac{2c}{b+c}\right)\left(\frac{2a}{c+a}\right)=$
$\frac{8abc}{(a+b)(b+c)(c+a)}=\frac{8abc}{(a+b)(b+c)(c+a)}$
1. We can rewrite given ratios as:

$\begin{split}my+nz-lx=\frac{k}{l}\\ nz+lx-my=\frac{k}{m}\\ lx+my-nz=\frac{k}{n}\end{split}$

Pairing two at a time gives us:

$\begin{split}x=\frac{k}{2l}\left(\frac{1}{m}+\frac{1}{n}\right)\\ y=\frac{k}{2m}\left(\frac{1}{l}+\frac{1}{n}\right)\\ z=\frac{k}{2n}\left(\frac{1}{m}+\frac{1}{l}\right)\end{split}$

Now let us compute the first of the ratios which are to be proven equal.

$\frac{y+z-x}{l} = \frac{k}{2l}\left\{\frac{1}{lm}+\frac{1}{mn}+\frac{1}{ln}+\frac{1}{mn}-\frac{1}{lm}-\frac{1}{ln}\right\}$
$\Rightarrow~\frac{y+z-x}{l} =\frac{k}{lmn}$

Similarly, others can be also proven equal.

1. Applying the formula for eliminant we can write it as:

$a(bc-a^2)+c(ab-c^2)+b(ca-b^2)=0$
$\Rightarrow~a^3+b^3+c^3-3abc=0.$
1. As shown in previous problem the eliminant is:

$ah^2+bg^2+ch^2-3abc=0$

All remaining questions are left as exercises to the reader.