Some amount of money is distributed to a some students equally. If the number of student is decreased by 10 then they will get 20 rupees more. If student is increased by 10 then they will get 4 rupees less. Find the amount and number of student?
If each of n students gets r rupees,
nr/(n-10) = r+20
nr/(n+10) = r-4
Now just solve for r and n.
How to solve two variable in one time please.. Help me
same as always:
clear the fractions to get
nr = (r+20)(n-10)
nr = (r-4)(n+10)
nr = nr + 20n - 10r - 200
nr = nr - 4n + 10r - 40
20n-10r = 200
4n-10r = -40
or,
2n-r = 20
2n-5r = -20
So, r = 2n-20
2n-5(2n-20) = -20
2n-10n+100 = -20
-8n = -120
n = 15
so, r = 10
15 students share 150 rupees.
To find the amount and number of students, we can use algebraic equations.
Let's assume the original amount of money distributed among the students as 'M' and the original number of students as 'N'.
According to the given information:
1) "Some amount of money is distributed to a some students equally." - This means the money is equally divided among the students, so each student receives M/N rupees.
2) "If the number of students is decreased by 10, then they will get 20 rupees more." - If we decrease the number of students by 10, the new number of students becomes N-10. And according to the information given, each student would receive M/(N-10) rupees. Additionally, they would get 20 rupees more than previously, so the expression becomes M/(N-10) = M/N + 20.
3) "If the number of students is increased by 10, then they will get 4 rupees less." - If we increase the number of students by 10, the new number of students becomes N+10. And according to the information given, each student would receive M/(N+10) rupees. Additionally, they would get 4 rupees less than previously, so the expression becomes M/(N+10) = M/N - 4.
We now have two equations:
1) M/(N-10) = M/N + 20
2) M/(N+10) = M/N - 4
To solve this system of equations, we can use the method of substitution or elimination:
First, let's rearrange equation 1 to isolate M on one side:
M/(N-10) - M/N = 20
Now, let's find a common denominator for the left side:
[NM - (N-10)M] / N(N-10) = 20
Expanding and simplifying:
[NM - NM + 10M] / N(N-10) = 20
10M / N(N-10) = 20
Cross multiplying:
10M = 20N(N-10)
10M = 20N^2 - 200N
Dividing both sides by 10:
M = 2N^2 - 20N
Now, substitute M = 2N^2 - 20N into equation 2:
(2N^2 - 20N) / (N+10) = (2N^2 - 20N) / N - 4
Cross multiplying:
(2N^2 - 20N)N = (2N^2 - 20N)(N+10) - 4N(N+10)
Expanding and simplifying:
2N^3 - 20N^2 = 2N^3 + 20N^2 - 20N^2 - 200N - 4N^2 - 40N
-12N^2 = -244N
Dividing both sides by -N:
12N = 244
N = 244/12
N = 20.33
As we can't have a non-integer value for the number of students, it seems there might be an error or inconsistency in the given information or calculations. Please recheck the problem statement or calculations to ensure accuracy.