If\[bc:ac:ab = 1:2:3\] then find\[\dfrac{a}{{bc}}:\dfrac{b}{{ca}}\]
Answer
645k+ views
Hint: Ratio is the relation between two numbers which shows how much bigger one quantity is than another.
In a ratio between three numbers the value of each part is found by dividing the given amount by the sum of the parts in the ratio. We then multiply each number in the ratio by the value of each part in ratio \[a:b = \dfrac{a}{b}\]
Complete step-by- step solution:
Given \[bc:ac:ab = 1:3:5.......(1)\]
i.e.\[bc:ac = 1:3\] {from (1)}
\[ \Rightarrow \dfrac{{bc}}{{ac}} = \dfrac{1}{3}\]
On cancelling c from the equation, we get:
\[ \Rightarrow \dfrac{b}{a} = \dfrac{1}{3}.........(2)\]
Again from (1) \[ac:ab = 3:5\]
\[ \Rightarrow \dfrac{{ac}}{{ab}} = \dfrac{3}{5}\]
On cancelling ‘a’ from the equation, we get:
\[ \Rightarrow \dfrac{c}{b} = \dfrac{3}{5}.........(3)\]
From equations (2) and (3)
\[ \Rightarrow \dfrac{b}{a} = \dfrac{1}{3}\] and \[\dfrac{c}{b} = \dfrac{3}{5}\]
As the value of b is not same in both cases, we will be making it equal by multiplying and dividing (2) by 5
We have:
\[\dfrac{b}{a} = \dfrac{1}{3} \times \dfrac{5}{5} = \dfrac{5}{{15}}\]
\[\dfrac{b}{a} = \dfrac{5}{{15}}..........(4)\]
Compare eqn. (3) by eqn. (4)
We get \[a = 15,b = 5\] and \[c - 13\] because the value of b in both equations is 5.
i.e. \[a:b:c = 15:5:3\]
To find \[\dfrac{a}{{bc}} = \dfrac{b}{{ca}}\]
Put \[a = 15,b = 5,c = 3\]
\[\dfrac{{15}}{{5 \times 3}}:\dfrac{5}{{3 \times 15}}\]
\[ \Rightarrow \dfrac{{15}}{{15}}:\dfrac{5}{{15}}\]
\[ \Rightarrow 1:\dfrac{1}{3}\]
Multiply the above whole term with 3, we get:
\[ \Rightarrow 3 \times 1:3 \times \dfrac{1}{3}\]
\[ \Rightarrow 3:1\]
Hence, \[\dfrac{a}{{bc}}:\dfrac{b}{{ca}} = 3:1\]
Note: Consider two ratios to be \[a:b\] and \[c:d\]
Then in order to find the continued proportion for the two given ratio terms, we convert the means to a single term/number. This would, in general, be the LCM of means.
For the given ratio, the LCM of \[b\]&\[c\] will be\[bc\].
Thus, multiplying the first ratio by \[c\] and second ratio by \[b\], we have
First ratio- \[ca:bc\]
Second ratio- \[bc:bd\]
Thus, the continued proportion can be written in the form of \[ca:bc:bd\].
In ratio if \[a:b:c = x:y:z\]then we can compare \[a:b = x:y\] and \[b:c = y:z\]in ratio we can divide and multiply throughout by any number as it will not affect the ratio.
In a ratio between three numbers the value of each part is found by dividing the given amount by the sum of the parts in the ratio. We then multiply each number in the ratio by the value of each part in ratio \[a:b = \dfrac{a}{b}\]
Complete step-by- step solution:
Given \[bc:ac:ab = 1:3:5.......(1)\]
i.e.\[bc:ac = 1:3\] {from (1)}
\[ \Rightarrow \dfrac{{bc}}{{ac}} = \dfrac{1}{3}\]
On cancelling c from the equation, we get:
\[ \Rightarrow \dfrac{b}{a} = \dfrac{1}{3}.........(2)\]
Again from (1) \[ac:ab = 3:5\]
\[ \Rightarrow \dfrac{{ac}}{{ab}} = \dfrac{3}{5}\]
On cancelling ‘a’ from the equation, we get:
\[ \Rightarrow \dfrac{c}{b} = \dfrac{3}{5}.........(3)\]
From equations (2) and (3)
\[ \Rightarrow \dfrac{b}{a} = \dfrac{1}{3}\] and \[\dfrac{c}{b} = \dfrac{3}{5}\]
As the value of b is not same in both cases, we will be making it equal by multiplying and dividing (2) by 5
We have:
\[\dfrac{b}{a} = \dfrac{1}{3} \times \dfrac{5}{5} = \dfrac{5}{{15}}\]
\[\dfrac{b}{a} = \dfrac{5}{{15}}..........(4)\]
Compare eqn. (3) by eqn. (4)
We get \[a = 15,b = 5\] and \[c - 13\] because the value of b in both equations is 5.
i.e. \[a:b:c = 15:5:3\]
To find \[\dfrac{a}{{bc}} = \dfrac{b}{{ca}}\]
Put \[a = 15,b = 5,c = 3\]
\[\dfrac{{15}}{{5 \times 3}}:\dfrac{5}{{3 \times 15}}\]
\[ \Rightarrow \dfrac{{15}}{{15}}:\dfrac{5}{{15}}\]
\[ \Rightarrow 1:\dfrac{1}{3}\]
Multiply the above whole term with 3, we get:
\[ \Rightarrow 3 \times 1:3 \times \dfrac{1}{3}\]
\[ \Rightarrow 3:1\]
Hence, \[\dfrac{a}{{bc}}:\dfrac{b}{{ca}} = 3:1\]
Note: Consider two ratios to be \[a:b\] and \[c:d\]
Then in order to find the continued proportion for the two given ratio terms, we convert the means to a single term/number. This would, in general, be the LCM of means.
For the given ratio, the LCM of \[b\]&\[c\] will be\[bc\].
Thus, multiplying the first ratio by \[c\] and second ratio by \[b\], we have
First ratio- \[ca:bc\]
Second ratio- \[bc:bd\]
Thus, the continued proportion can be written in the form of \[ca:bc:bd\].
In ratio if \[a:b:c = x:y:z\]then we can compare \[a:b = x:y\] and \[b:c = y:z\]in ratio we can divide and multiply throughout by any number as it will not affect the ratio.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Give 10 examples for herbs , shrubs , climbers , creepers

What is the opposite gender of Gander class 6 english CBSE

What is the Total Duration of Football Match?

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE


