Question 1201163: The Extreme Rock Climbing Club planned a climbing expedition. The total cost was $1200
, which was to be divided equally among the members going. Prior to the trip, 3
members decided not to go. If the cost per person increased by $90
, how many people went on the expedition?
Found 3 solutions by greenestamps, josgarithmetic, ikleyn: Answer by greenestamps(13200) (Show Source): Answer by josgarithmetic(39617) (Show Source): Answer by ikleyn(52781) (Show Source):
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The Extreme Rock Climbing Club planned a climbing expedition. The total cost was $1200,
which was to be divided equally among the members going. Prior to the trip, 3 members
decided not to go. If the cost per person increased by $90, how many people went on the expedition?
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Let n be the number of people went on the expedition.
The number of the members originally was (n+3).
The price per a member originally was .
The price per a person finally was .
The difference of prices is $90 (given), which gives us this "price" equation
- = 90. (1)
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| At this point, the setup is complete. |
| Now our goal is to solve the equation. |
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To do it, first simplify it by dividing both sides by 30. You will get
- = 3. (2)
At this point, you just can to "guess" the answer MENTALLY: it is n= 5.
But let's do it formally.
From (2), multiplying both sides by n*(n+3), you get
40*(n+3) - 40n = 3n*(n+3),
40n + 120 - 40n = 3n*(n+3),
120 = 3n(n+3),
40 = n*(n+3),
n^2 + 3n - 40 = 0.
Factor left side
(n+8)*(n-5) = 0.
There are two roots, n= -8 and n= 5.
According to the meaning of the problem, n= 5 is a meaningful solution while n= -8 is not, so we reject it.
ANSWER. 5 persons went on the expedition.
Solved.
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