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# Divide 240 into three parts so that $\dfrac{1}{3}$ of the first, $\dfrac{1}{4}$ of the second and $\dfrac{1}{5}$ of the third part are equal.

Last updated date: 13th Jun 2024
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Hint: Here, we will first write the given statement mathematically by assuming the three parts to be some variable. Then we will add all the parts and equate it to 240. We will use the mathematical equation formed by us to substitute the values of the second and third parts in terms of the first part. Hence, we will find the value of one variable. Substituting further, we will get the value of each part and hence, we will be able to find the required answer.

Let the three parts be $a,b$ and $c$.
Now, according to the question,
The number 240 is divided into three parts so that $\dfrac{1}{3}$ of the first, $\dfrac{1}{4}$ of the second and $\dfrac{1}{5}$ of the third part are equal.
Hence, if the first part is $a$ , second part is $b$ and the third part is $c$, then, we can say that,
$\dfrac{1}{3}$ of $a$ is equal to $\dfrac{1}{4}$ of $b$ is equal to $\dfrac{1}{5}$ of $c$
Now, writing this mathematically, we get,
$\Rightarrow \dfrac{1}{3}a = \dfrac{1}{4}b = \dfrac{1}{5}c$…………………………….. $\left( 1 \right)$
Now, we will write every part in terms of $a$.
Hence, the first part will remain the same, i.e. $a$
From equation $\left( 1 \right)$ , we can say that,
$\dfrac{a}{3} = \dfrac{b}{4}$
Hence, by cross multiplying,
$\Rightarrow b = \dfrac{{4a}}{3}$
Therefore, the second part $b = \dfrac{{4a}}{3}$………………………… $\left( 2 \right)$
Again, from equation $\left( 1 \right)$, we get
$\dfrac{a}{3} = \dfrac{c}{5}$
Hence, by cross multiplying,
$\Rightarrow c = \dfrac{{5a}}{3}$
Therefore, the third part $c = \dfrac{{5a}}{3}$………………………………… $\left( 3 \right)$
Now, since we have divided the number 240 into three parts i.e. $a,b$ and $c$.
Hence, the summation of these parts will give us the number 240
$a + b + c = 240$
Here, substituting $b = \dfrac{{4a}}{3}$ and $c = \dfrac{{5a}}{3}$ from $\left( 2 \right)$ and $\left( 3 \right)$, we can write this as:
$\Rightarrow a + \dfrac{{4a}}{3} + \dfrac{{5a}}{3} = 240$
Now, taking the LCM as 3 in the LHS,
$\Rightarrow \dfrac{{3a + 4a + 5a}}{3} = 240$
$\Rightarrow \dfrac{{12a}}{3} = 240$
Now, solving further, we get,
$\Rightarrow 4a = 240$
Dividing both sides by 4,
$\Rightarrow a = 60$
Therefore, the first part $a = 60$
Hence, the second part $b = \dfrac{{4a}}{3} = \dfrac{{4 \times 60}}{3} = 4 \times 20 = 80$
And, the third part $c = \dfrac{{5a}}{3} = \dfrac{{5 \times 60}}{3} = 5 \times 20 = 100$
Therefore, we should divide 240 into 60, 80 and 100 if we want that $\dfrac{1}{3}$ of the first, $\dfrac{1}{4}$ of the second and $\dfrac{1}{5}$ of the third part are equal.
Hence, this is the required answer.

Note: In order to check whether we have divided the given number correctly or not, we can substitute the values in $\left( 1 \right)$ i.e.
$\dfrac{1}{3}a = \dfrac{1}{4}b = \dfrac{1}{5}c$
This is the statement which is required to be proved.
Hence, if the three parts, in which we have divided 240, satisfies this equality, then our answer is correct.
Therefore, substituting $a = 60$ , $b = 80$ and $c = 100$ , we get,
$\dfrac{1}{3} \times 60 = \dfrac{1}{4} \times 80 = \dfrac{1}{5} \times 100$
$\Rightarrow 20 = 20 = 20$
Clearly, they are equal.
Hence, it is proved that 240 should be divided as 60, 80 and 100 if we want that $\dfrac{1}{3}$ of the first, $\dfrac{1}{4}$ of the second, and $\dfrac{1}{5}$ of the third part are equal.
Hence, this is the required answer.