4. Solutions for RatiosΒΆ

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

  1. For 13:24 and 34:23 we have their ratio as

  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:


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 \(f=\frac{be}{a}\).


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:


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,


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

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

\[\Rightarrow~x=ak, y=bk,~\text{and}~z=ck\]

Now we will consider right side.

\[R.H.S. = \frac{(ak+bk+ck)^3+(a+b+c)^3}{(ak+bk+ck)^2+(a+b+c)^2}\]

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


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

\[\frac{x}{2b+2c-a} = \frac{xk}{4z+4x-2y+4x+4y-2z-2y-2z+x}\]

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

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


Squaring we have following,


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


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


Hence, the required result is proven.

  1. Clearly,

\[\left(\frac{2a}{a+b}\right)\left(\frac{2b}{b+c}\right)\left(\frac{2c}{c+a}\right)= \left(\frac{2b}{a+b}\right)\left(\frac{2c}{b+c}\right)\left(\frac{2a}{c+a}\right)=\]
  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:

  1. As shown in previous problem the eliminant is:


All remaining questions are left as exercises to the reader.