How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $\frac{a}{c} = \frac{2}{5}$ but that is not a correct answer.
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $\frac{a}{c} = \frac{2}{5}$ but that is not a correct answer.
Divide a:b with c:d i.e. $\frac{(\frac{a}{b})}{(\frac{c}{d})} = \frac{(\frac{2}{5})}{\frac{5}{2}}$
$\implies$ $\frac{a}{c}*\frac{d}{b} = \frac{4}{25}$
$\implies$ $\frac{a}{c}*\frac{3}{2} = \frac{4}{25}$
$\implies$ $\frac{a}{c}= \frac{8}{75}$
Now talking about the approach, So you must try to figure the ratio whose value is required from all the ratios given to you. Like in this question, we just divided the first 2 ratios and after putting the values we got the answer.
Hint $\rm\,\ \dfrac{a}{c}\, =\, \dfrac{a}{\color{#C00}b} \dfrac{\color{#C00}b}{\color{#0A0}d} \dfrac{\color{#0A0}d}c\, =\, \dfrac{2}5 \dfrac{2}3 \dfrac{2}5$
Remark $\ $ This is a special case of ubiquitous multiplicative telescopy
$\rm \frac{a_1}{a_n}\, =\, \frac{a_1}{\color{#C00}{a_2}}\frac{\color{#C00}{a_2}}{\color{#0A0}{a_3}}\frac{\color{#0A0}{a_3}}{\cdots}\cdots\dfrac{\cdots}{\color{brown}{a_{n-1}}}\frac{\color{brown}{a_{n-1}}}{a_n}$
These ratios are just simple equations. For example $a:b=2:5$ is $a= \frac{2}{5}b$ No need for confusing tricks here. Just substitutions : $ a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$ So that $ a:c = 8:75 $