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Hint: Assume that there were ‘x’ number of children. Calculate the amount of money each child would have received if there were ‘x’ children using a unitary method. Write equations in ‘x’ relating the number of children to the money. Solve the equation to get the value of parameter ‘x’.

Complete step-by-step answer:

We have data regarding dividing \[Rs.1200\] among a certain number of children. We have to find the number of children among which the money was divided.

Let’s assume that there were ‘x’ children among whom the money \[Rs.1200\] was divided.

So, the amount of money each child received \[=\dfrac{Rs.1200}{x}\].

If 5 more children were added, new number of children \[=x+5\].

The amount of money each child received \[=\dfrac{Rs.1200}{x+5}\]. This amount is \[Rs.8\] less than the previous amount which each child received.

Thus, we have \[\dfrac{1200}{x+5}=\dfrac{1200}{x}-8\].

Rearranging the term of the equation, we have \[\dfrac{1200}{x}-\dfrac{1200}{x+5}=8\].

Taking LCM of terms on the left hand side of the above equation, we have \[\dfrac{1200\left( x+5 \right)-1200x}{x\left( x+5 \right)}=8\].

Further simplifying the above equation, we have \[\dfrac{1200x+6000-1200x}{x\left( x+5 \right)}=8\].

Thus, we have \[\dfrac{6000}{x\left( x+5 \right)}=8\].

Cross multiplying the terms on both sides, we have \[8x\left( x+5 \right)=6000\].

Thus, we have \[x\left( x+5 \right)=750\].

Rearranging the terms, we have \[{{x}^{2}}+5x-750=0\]

Factoring the above equation, we have \[{{x}^{2}}+30x-25x-750=0\].

Taking out the common terms, we have \[x\left( x+30 \right)-25\left( x+30 \right)=0\].

So, we have \[\left( x+30 \right)\left( x-25 \right)=0\].

Thus, we have \[x=25,-30\].

We will only consider \[x=25\] as the number of children can’t be negative.

Hence, there were initially 25 children, which is option (d).

Note: To check if the calculated answer is correct or not, substitute the calculated value in the equation and check if it satisfies the given equation or not. Also, one must be careful about units which perform calculations. Each child is distributed the money in Rupees.

Complete step-by-step answer:

We have data regarding dividing \[Rs.1200\] among a certain number of children. We have to find the number of children among which the money was divided.

Let’s assume that there were ‘x’ children among whom the money \[Rs.1200\] was divided.

So, the amount of money each child received \[=\dfrac{Rs.1200}{x}\].

If 5 more children were added, new number of children \[=x+5\].

The amount of money each child received \[=\dfrac{Rs.1200}{x+5}\]. This amount is \[Rs.8\] less than the previous amount which each child received.

Thus, we have \[\dfrac{1200}{x+5}=\dfrac{1200}{x}-8\].

Rearranging the term of the equation, we have \[\dfrac{1200}{x}-\dfrac{1200}{x+5}=8\].

Taking LCM of terms on the left hand side of the above equation, we have \[\dfrac{1200\left( x+5 \right)-1200x}{x\left( x+5 \right)}=8\].

Further simplifying the above equation, we have \[\dfrac{1200x+6000-1200x}{x\left( x+5 \right)}=8\].

Thus, we have \[\dfrac{6000}{x\left( x+5 \right)}=8\].

Cross multiplying the terms on both sides, we have \[8x\left( x+5 \right)=6000\].

Thus, we have \[x\left( x+5 \right)=750\].

Rearranging the terms, we have \[{{x}^{2}}+5x-750=0\]

Factoring the above equation, we have \[{{x}^{2}}+30x-25x-750=0\].

Taking out the common terms, we have \[x\left( x+30 \right)-25\left( x+30 \right)=0\].

So, we have \[\left( x+30 \right)\left( x-25 \right)=0\].

Thus, we have \[x=25,-30\].

We will only consider \[x=25\] as the number of children can’t be negative.

Hence, there were initially 25 children, which is option (d).

Note: To check if the calculated answer is correct or not, substitute the calculated value in the equation and check if it satisfies the given equation or not. Also, one must be careful about units which perform calculations. Each child is distributed the money in Rupees.

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