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The worst explanation I have ever seen
Shaunak3004
(1) x/y = x/(y+10) + 1
(2) x/y = x/(y-15) - 3

Taking (1)
x/y = (x+y+10)/(y+10)
xy + 10x = xy + y^2 + 10y
10x = y^2 + 10y ...(1)

Taking (2)
x/y = (x - 3y + 45)/(y-15)
xy - 15x = xy - 3y^2 + 45y
15x = 3y^2 - 45y
5x = y^2 - 15 ...(2)

Multiplying (2) by 2 and substituting 10x from (1) in (2)
2y^2 - 30y = y^2 + 10y
y = 40

Substituting y = 40 in (2)
5x = 1000
x = 200
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Hi Bunuel KarishmaB

I assumed X as total money and N as total number of initial boys, and formed equations to get N and X.

Wanted to check if I missed any faster way or logic to solve this question?
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This one is most efficient way by forming the equation.

If you are good with manipulation you can insert the options and play along but it might be complicated sometimes. So good to go with algebra in my view


agrasan
Hi Bunuel KarishmaB

I assumed X as total money and N as total number of initial boys, and formed equations to get N and X.

Wanted to check if I missed any faster way or logic to solve this question?
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Rickooreoisb
A fixed amount of money is shared equally among some boys.

  • If 10 more boys had shared the same amount, each would receive $1 less.
  • If 15 fewer boys had shared the same amount, each would receive $3 more.

What is the total amount of money?

A. $120
B. $140
C. $160
D. $180
E. $200

When one variable increases by a fixed amount and the other subtracted, it is best to use equations. If n is the number of boys and A is the avg amount each has,

If 10 more boys had shared the same amount, each would receive $1 less.

\(\frac{n}{10} = A - 1\)

If 15 fewer boys had shared the same amount, each would receive $3 more.

\(\frac{15A}{3} = n - 15\)

So n = $40 and A = 5

Total money = $200

Answer (E)
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