Rs 9,000 were divided equally among a certain number of persons. Had there been 20 persons more, each would have got Rs 160 less. Find the original number of persons.
Last updated date: 17th Mar 2023
•
Total views: 211.7k
•
Views today: 2.90k
Answer
211.7k+ views
Hint- Here, we will be proceeding by making an equation in one variable by using the formula i.e., Amount divided among each person = $\dfrac{{{\text{Total amount to be divided}}}}{{{\text{Total number of persons}}}}$ and then solving for the unknown.
Let the original number of persons be x among which Rs 9000 is to be divided.
For the amount that each person will be getting, we will divide the total amount Rs 9000 by the original number of persons among which Rs 9000 is to be divided.
i.e., Amount that each person will be getting = Rs $\dfrac{{9000}}{x}$.
Now, when the total number of persons among which Rs 9000 is to be divided are 20 persons more than the original number of persons (x).
i.e., Number of persons amount which Rs 9000 is to be divided = (x+20) persons
Hence, amount that each person will be getting = Rs $\dfrac{{9000}}{{x + 20}}$
According to problem statement, $\dfrac{{9000}}{x} - \dfrac{{9000}}{{x + 20}} = 160$
Now, we will solve the above equation by taking x(x+20) as LCM to get the value of x.
$
\Rightarrow \dfrac{{9000}}{x} - \dfrac{{9000}}{{x + 20}} = 160
$
$
\Rightarrow \dfrac{{9000\left( {x + 20} \right) - 9000x}}{{x\left( {x + 20} \right)}} = 160
$
$
\Rightarrow \dfrac{{9000x + 180000 - 9000x}}{{x\left( {x + 20} \right)}} = 160 \\
$
$
\Rightarrow \dfrac{{180000}}{{x\left( {x + 20} \right)}} = 160 \\
$
By applying cross multiplication in the above equation, we get
$
\Rightarrow 180000 = 160x\left( {x + 20} \right)
$
$
\Rightarrow 1125 = x\left( {x + 20} \right)
$
$
\Rightarrow 1125 = {x^2} + 20x
$
$
\Rightarrow {x^2} + 20x - 1125 = 0 \\
$
$
\Rightarrow {x^2} + 45x - 25x - 1125 = 0
$
$
\Rightarrow x\left( {x + 45} \right) - 25\left( {x + 45} \right) = 0 \\
$
$
\Rightarrow \left( {x + 45} \right)\left( {x - 25} \right) = 0 \\
$
Either $\left( {x + 45} \right) = 0 \Rightarrow x = - 45$ or $\left( {x - 25} \right) = 0 \Rightarrow x = 25$
As we know that here x is representing the original number of persons among which Rs 9000 is getting divided and the number of persons will always be positive so, only positive values of x should be considered.
i.e., $x = - 45$ is neglected.
So, $x = 25$ is the correct value.
Therefore, the original number of persons among will Rs 9000 is to be divided are 25 persons.
Note- In this particular problem, we can easily see that if the number of persons (among which Rs 9000 is divided) is increased, then the amount that each person will get will automatically decrease because both of these are inversely proportional to each other.
Let the original number of persons be x among which Rs 9000 is to be divided.
For the amount that each person will be getting, we will divide the total amount Rs 9000 by the original number of persons among which Rs 9000 is to be divided.
i.e., Amount that each person will be getting = Rs $\dfrac{{9000}}{x}$.
Now, when the total number of persons among which Rs 9000 is to be divided are 20 persons more than the original number of persons (x).
i.e., Number of persons amount which Rs 9000 is to be divided = (x+20) persons
Hence, amount that each person will be getting = Rs $\dfrac{{9000}}{{x + 20}}$
According to problem statement, $\dfrac{{9000}}{x} - \dfrac{{9000}}{{x + 20}} = 160$
Now, we will solve the above equation by taking x(x+20) as LCM to get the value of x.
$
\Rightarrow \dfrac{{9000}}{x} - \dfrac{{9000}}{{x + 20}} = 160
$
$
\Rightarrow \dfrac{{9000\left( {x + 20} \right) - 9000x}}{{x\left( {x + 20} \right)}} = 160
$
$
\Rightarrow \dfrac{{9000x + 180000 - 9000x}}{{x\left( {x + 20} \right)}} = 160 \\
$
$
\Rightarrow \dfrac{{180000}}{{x\left( {x + 20} \right)}} = 160 \\
$
By applying cross multiplication in the above equation, we get
$
\Rightarrow 180000 = 160x\left( {x + 20} \right)
$
$
\Rightarrow 1125 = x\left( {x + 20} \right)
$
$
\Rightarrow 1125 = {x^2} + 20x
$
$
\Rightarrow {x^2} + 20x - 1125 = 0 \\
$
$
\Rightarrow {x^2} + 45x - 25x - 1125 = 0
$
$
\Rightarrow x\left( {x + 45} \right) - 25\left( {x + 45} \right) = 0 \\
$
$
\Rightarrow \left( {x + 45} \right)\left( {x - 25} \right) = 0 \\
$
Either $\left( {x + 45} \right) = 0 \Rightarrow x = - 45$ or $\left( {x - 25} \right) = 0 \Rightarrow x = 25$
As we know that here x is representing the original number of persons among which Rs 9000 is getting divided and the number of persons will always be positive so, only positive values of x should be considered.
i.e., $x = - 45$ is neglected.
So, $x = 25$ is the correct value.
Therefore, the original number of persons among will Rs 9000 is to be divided are 25 persons.
Note- In this particular problem, we can easily see that if the number of persons (among which Rs 9000 is divided) is increased, then the amount that each person will get will automatically decrease because both of these are inversely proportional to each other.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
