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On a z day outdoor adventure expedition, 4 children consumed food cost

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On a z day outdoor adventure expedition, 4 children consumed food cost [#permalink]

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New post 03 Jan 2018, 21:18
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Question Stats:

85% (01:10) correct 15% (01:11) wrong based on 13 sessions

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On a z day outdoor adventure expedition, 4 children consumed food costing x dollars and supplies costing t dollars. For the same food and supply costs per child per day, what would be the total cost of food and supplies consumed by 8 children during a 10 day outdoor adventure?

A. 20(x + t)/z

B. (5x + 5t)/z

C. 32(x + t)/z

D. 20z/(x + t)

E. 20(x + z)/t

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Re: On a z day outdoor adventure expedition, 4 children consumed food cost [#permalink]

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New post 03 Jan 2018, 23:06
Cost of z days for 4 children = x+t $
Cost per day per children = \(\frac{(x+t)}{(4z)}\) $
cost of 8 children for 10 days = \(\frac{(x+t)}{(4Z)}\)* 8 *10= 20(x+t)/z $

Ans A
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Re: On a z day outdoor adventure expedition, 4 children consumed food cost [#permalink]

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New post 03 Jan 2018, 23:21
can be solved directly by using unitary method but giving detail method as below

Cost of z days for 4 children = x+t
Cost per day per children = (x+t)/(4z) (we will divide as per day, per child cost will decrease)
cost of 8 children for 10 days =( (x+t)/(4Z))* 8 *10= 20(x+t)/z (we will multiply as 10 day, 8 child cost will increase)
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Re: On a z day outdoor adventure expedition, 4 children consumed food cost [#permalink]

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New post 03 Jan 2018, 23:38
A is the answer as per my understanding. This is how I solved
Z days, 4 Children, Total Food Cost(FC) X, Supply cost(SC) T = X+T
For 8 children total cost will be 2(X+T)
Now total cost for 8 Children per day will be 2(X+T)/Z
So for 10 days will be 10*2(X+T)/Z = 20(X+t)/Z

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Re: On a z day outdoor adventure expedition, 4 children consumed food cost   [#permalink] 03 Jan 2018, 23:38
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