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

\[\frac{4}{5}*\frac{6}{6} = \frac{24}{30} \]


\[\frac{5}{6}*\frac{5}{5} = \frac{25}{30} \]

Therefore, 5:6 is greater.

  1. Similarly, as in previous case

\[\frac{34}{33}*\frac{67}{67} = \frac{2278}{2211} \]


\[\frac{68}{67}*\frac{33}{33} = \frac{2244}{2211} \]

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:


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

  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

\[\frac{x}{q+r-p}=\frac{y}{r+p-q}=\frac{z}{p+q-r}=k \Rightarrow~x=(q + r - p)k, y = (p + r - q)k~\text{and}~z = (p + q - r)k\]

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

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

\[my+nz-lx=\frac{k}{l}\\ nz+lx-my=\frac{k}{m}\\ lx+my-nz=\frac{k}{n} \]

Pairing two at a time gives us:

\[x=\frac{k}{2l}\left(\frac{1}{m}+\frac{1}{n}\right)\\ y=\frac{k}{2m}\left(\frac{1}{l}+\frac{1}{n}\right)\\ z=\frac{k}{2n}\left(\frac{1}{m}+\frac{1}{l}\right) \]

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.