Question

If 391 is divided into three parts proportional to \[\dfrac{1}{2}:\dfrac{2}{3}:3\], then the first part is?

Answer 1

Hint: This type of problem can be solved by proportionate method. First, we have to consider the total number 391 and we have to find the first part. We have to divide \[\dfrac{1}{2}\] by the sum of all ratios, that is \[\dfrac{1}{2}+\dfrac{2}{3}+3\]. Do necessary calculations and simplify the obtained expression. Then, multiply the obtained expression with the total number 391. Do necessary calculation in the division and find the final answer.

Complete step by step answer:
According to the question, we are asked to find the first part of 391 in the ratio \[\dfrac{1}{2}:\dfrac{2}{3}:3\].
 We have been given the ratio is \[\dfrac{1}{2}:\dfrac{2}{3}:3\].
Let us first assume the ratio \[\dfrac{1}{2}:\dfrac{2}{3}:3\] to be A:B:C.
Here, A is the ratio \[\dfrac{1}{2}\], B is the ratio \[\dfrac{2}{3}\] and c is the ratio 3.
We have to find the first part A.
To find the first part A, we have to divide \[\dfrac{1}{2}\] by the sum of the ratios A,B and C.
Let us find A+B+C.
A+B+C=\[\dfrac{1}{2}+\dfrac{2}{3}+3\]
Let us take the LCM 6 to simplify.
\[\Rightarrow A+B+C=\dfrac{1\times 3+2\times 2+3\times 6}{6}\]
On simplifying further, we get
\[A+B+C=\dfrac{3+4+18}{6}\]
Let us now add the numerator of the RHS.
\[\Rightarrow A+B+C=\dfrac{25}{6}\]
Now, we need to divide the first part A by the sum of A, B and C.
\[\Rightarrow \dfrac{A}{A+B+C}=\dfrac{\dfrac{1}{2}}{\dfrac{25}{6}}\]
We know that \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\]. Using this property of division, we get
\[\dfrac{A}{A+B+C}=\dfrac{1}{2}\times \dfrac{6}{25}\]
We know that 6 is the product of 3 and 2.
\[\Rightarrow \dfrac{A}{A+B+C}=\dfrac{1}{2}\times \dfrac{2\times 3}{25}\]
Here, 2 are common in the numerator and denominator. Let us cancel 2 from the numerator and denominator.
\[\Rightarrow \dfrac{A}{A+B+C}=\dfrac{3}{25}\] ------------(1)
To find the first part A of 391, we have to multiply 391 with the expression (1).
\[\Rightarrow \dfrac{A}{A+B+C}\times 391=\dfrac{3}{25}\times 391\]
We know that 391 multiplied by 3 is 1173, that is \[3\times 391=1173\].
Substituting in the above equation, we get
\[\Rightarrow \dfrac{A}{A+B+C}\times 391=\dfrac{1173}{25}\]
On dividing 1173 by 25, we get
\[\dfrac{A}{A+B+C}\times 391=46.92\]
Hence, the first part of 391 dividing in the ratio \[\dfrac{1}{2}:\dfrac{2}{3}:3\] is 46.92.

Note:
Whenever we get such types of problems, it is always advisable to assume the ratios to be A, B and C to avoid confusion. We never get negative values on multiplying with the sum of ratios. It is not necessary to find the second and third parts as it is not mentioned in the question. We can similarly find the second and third parts of 391 dividing in the ratio \[\dfrac{1}{2}:\dfrac{2}{3}:3\].