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An amount of money was divided between some people in such a way that

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An amount of money was divided between some people in such a way that  [#permalink]

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New post 17 Aug 2017, 01:29
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Question Stats:

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An amount of money was divided between some people in such a way that if there had been 4 more people, everyone would have $16 less. but if there had been 4 less people, everyone would have got $24 more. How many people were there in group? ?

A. 12
B. 16
C. 20
D. 24
E. 32

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Re: An amount of money was divided between some people in such a way that  [#permalink]

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New post 17 Aug 2017, 01:49
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Bunuel wrote:
An amount of money was divided between some people in such a way that if there had been 4 people, everyone would have $16 less. but if there had been 4 less people, everyone would have got $24 more. How many people were there in group? ?

A. 12
B. 16
C. 20
D. 24
E. 32


Let n be the number of people with p dollars
total = pn

(n+4)(p-16)=pn
np+4p-16n-64 = pn
4p-16n = 64.......(i)

(n-4)(p+24)=pn
-4p+24n = 96.......(ii)

on solving we get
8n = 160
n=20
C
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Re: An amount of money was divided between some people in such a way that  [#permalink]

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New post 17 Aug 2017, 02:25
Ans : C

Solution - Let the no of people be x and the amount that each one gets be y . Thus the total amount - xy

Given Condition 1 - (x+4)(y-16) = xy (The total amount remains the same because there is a change in the no of people and the amount that each one gets according to the given condition).

SOlving - y-4x=16

GIven condition 2 - (x-4)(y+24) = xy. Solving , -y+6x=24.

Solving both the equations - x=20
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Re: An amount of money was divided between some people in such a way that  [#permalink]

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New post 17 Aug 2017, 02:41
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@ Brunel - I agree with dave13 .

I too assumed if there had been 4 more people in the first part of the problem.
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An amount of money was divided between some people in such a way that  [#permalink]

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New post 17 Aug 2017, 03:34
I dont understand why i always write wrong equations when it comes to such problems. here is what i did, --- logically my approach should be correct ...

Let X be a total sum
N = number of persons
y = sum each owns after division

x / n+4 = y-16 from here follows---> x= n+4y-64

x / n-4= y+24 from here follows ---> x = n-4y-96

after this i simply got stuck :) what next ?
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Re: An amount of money was divided between some people in such a way that  [#permalink]

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New post 17 Aug 2017, 08:42
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Answer is clearly C


As seen from pic. It is easily understandable.
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Re: An amount of money was divided between some people in such a way that  [#permalink]

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New post 21 Aug 2017, 15:44
Bunuel wrote:
An amount of money was divided between some people in such a way that if there had been 4 more people, everyone would have $16 less. but if there had been 4 less people, everyone would have got $24 more. How many people were there in group? ?

A. 12
B. 16
C. 20
D. 24
E. 32


We can let the number of people = n and the amount of money each originally received = m; thus, the total original amount of money is mn.

If there had been 4 more people, everyone would have $16 less:

(n + 4)(m - 16) = mn

nm + 4m - 16n - 64 = mn

4m - 16n - 64 = 0

4m = 16n + 64

m = 4n + 16

And, had there been 4 fewer people, everyone would have received $24 more:

(n - 4)(m + 24) = mn

nm - 4m + 24n - 96 = mn

-4m + 24n - 96 = 0

-4m = -24n + 96

m = 6n - 24

Since m = 4n + 16 and m = 6n - 24, we have:

4n + 16 = 6n - 24

40 = 2n

20 = n

Answer: C
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An amount of money was divided between some people in such a way that  [#permalink]

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New post 22 Aug 2017, 09:09
1
dave13 wrote:
Bunuel wrote:
An amount of money was divided between some people in such a way that if there had been 4 people, everyone would have $16 less. but if there had been 4 less people, everyone would have got $24 more. How many people were there in group? ?

A. 12
B. 16
C. 20
D. 24
E. 32


Bunuel you know I dont understand why i always write wrong equations when it comes to such problems. here is what i did, --- logically my approach should be correct ...

Let X be a total sum
N = number of persons
y = sum each owns after division

x / n+4 = y-16 from here follows---> x= n+4y-64

x / n-4= y+24 from here follows ---> x = n-4y-96

after this i simply got stuck :) what next ?

dave13 : You're so close! :-)

I can't tell where your calculations went wrong, but your original equations are fine.

"N" value in both of your calculations, after you set up the equations, is incorrect. See below.

Just to make it clearer, I have substituted, for X, that which equals your X: Ny (# of people * amt per person = total amt).

X = Ny. Substitute Ny for X on LHS.

First:
\(\frac{Ny}{(n+4)}\) = (y - 16)
Ny = (N + 4)(y - 16)
Ny = Ny - 16N + 4y - 64
0 = -16N + 4y - 64 (i)

Second:
\(\frac{Ny}{(n-4)}\) = y + 24
Ny = (N - 4)(y + 24)
Ny = Ny + 24N - 4y - 96
0 = 24N - 4y - 96 (ii)

Add (ii) and (i)

24N - 4y - 96
-16N + 4y - 64
-----------------
8N - 160
8N = 160
N = 20

Hope that helps.
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Re: An amount of money was divided between some people in such a way that  [#permalink]

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New post 22 Aug 2017, 09:22
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Total amount is same in each case :
(X+4)*16 = (x-4)*24
solve for X ..
It will come 20
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Re: An amount of money was divided between some people in such a way that  [#permalink]

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