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Divide 243 into three parts such that half of the first part, one-third of the second part and one-fourth of the third part are all equal.

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Hint- Here, we will be assuming the first three parts of 243 as three different variables and then will be using a substitution method to solve the obtained three equations in three variables.

Here, we have to divide 243 into three parts. Let us assume that the first part be $x$, the second part be $y$ and the third part be $z$.
Clearly, $x + y + z = 243{\text{ }} \to {\text{(1)}}$
Given, half of the first part, one-third part of the second part and one-fourth of the third part are all equal to each other i.e., $\dfrac{x}{2} = \dfrac{1}{3}\left( y \right) = \dfrac{1}{4}\left( z \right) \Rightarrow \dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}$
Here, we will represent all the other variables in terms of variable x.
Now, considering $\dfrac{x}{2} = \dfrac{y}{3}$, we can write variable y in terms of variable x as under
$ \Rightarrow y = \dfrac{{3x}}{2}{\text{ }} \to (2{\text{)}}$
Now, considering $\dfrac{x}{2} = \dfrac{z}{4}$, we can write variable z in terms of variable x as under
$ \Rightarrow z = \dfrac{{4x}}{2} \Rightarrow z = 2x{\text{ }} \to {\text{(3)}}$
Substitute equations (2) and (3) in equation (1), we gwt
$x + \dfrac{{3x}}{2} + 2x = 243 \Rightarrow \dfrac{{2x + 3x + 4x}}{2} = 243 \Rightarrow 9x = 486 \Rightarrow x = 54$
Put the above obtained value of x in equation (2), we get
$ \Rightarrow y = \dfrac{{3 \times 54}}{2} = 81$
Put the above obtained value of x in equation (3), we get
$ \Rightarrow z = 2 \times 54 = 108$
Hence, 243 is divided into 54, 81 and 108 as the three parts respectively.
Therefore, 243 is divided into three parts such that the half of the first part which is 54, one-third of the second part which is 81 and one-fourth of the third part which is 108 are all equal.

Note- In these type of problems, according to the problem statement we will obtain three equations in three variables and then from two of the above equations, we will be representing the two different variables (y and z) in terms of the other variable (x) and then substituting these values in the third equation to solve for the value of x.

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