Answer
451.5k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)