1. ## WOrd problems

Q: A party of boys purchased a boat and upon payment for the same discovered that if thay had numbered 3 more, they would have paid a dollar apiece less; but if they had numbered 2 less,they would have paid apiece more. How many boys were there, and what did the boat cost?

let x= number of boys and y = number of dollars each paid. let xy = represents the number of dollars the boat cost

i created these two sim. equations but i am wrong. This is due to my interpretation of the question. How do i interpret in probably to get the right equations

$x + 3 + y = xy$
$x-2 = x(y+1)$

2. ## Re: WOrd problems

Originally Posted by bigmansouf
Q: A party of boys purchased a boat and upon payment for the same discovered that if thay had numbered 3 more, they would have paid a dollar apiece less; but if they had numbered 2 less,they would have paid apiece more. How many boys were there, and what did the boat cost?

let x= number of boys and y = number of dollars each paid originally

let $P$ be the price paid

$xy = P$

$(x+3)(y-1) = P$

$xy+3y-x-3 = P$

$P + 3y-x-3 = P$

$x = 3(y-1) \implies y > 1 \text{ and } y \in \mathbb{Z}$ ... also, $x$ is a multiple of $3$

so, what value of $y$ satisfies $xy=P$ and $(x+3)(y-1) = P$ ? (guess & check)

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method 2 ...

$P = \text{ price paid}$

$x = \text{ number of boys}$

$\dfrac{P}{x}$ is the price each paid

3 more in the party, each boy pays 1 dollar less

$\dfrac{P}{x} - 1 = \dfrac{P}{x+3}$

$\dfrac{P(x+3)}{x(x+3)} - \dfrac{x(x+3)}{x(x+3)} = \dfrac{Px}{x(x+3)}$

$P(x+3) - x(x+3) = Px$

$Px + 3P - x^2 - 3x = Px \implies x^2 + 3x - 3P = 0$

$x = \dfrac{-3 \pm \sqrt{9 + 12P}}{2}$

since $x$ is a positive integer, the value $(9+12P)$ must be a perfect square number ...

$x = \dfrac{-3 + \sqrt{9 + 12P}}{2} \implies P = 6 \implies x = 3$

3. ## Re: WOrd problems

Originally Posted by bigmansouf
A party of boys purchased a boat and upon payment for the same
discovered that if thay had numbered 3 more, they would have
paid a dollar apiece less; but if they had numbered 2 less,they
would have paid apiece more.
How many boys were there, and what did the boat cost?
Unclear: you omitted "how much more" apiece.
Looks like you used 1, therefore do did Skeeter.
But that doesn't seem to work (unless I need another coffee!).
Using Skeeter's results (p=6, x=3) has this result:

p x y
6 3 2 (original)
6 6 1 (3 more boys)
6 1 6 (2 less boys) : only 1 boy left, who pays 4 more...

4. ## Re: WOrd problems

Here is how it can work out:

Originally Posted by bigmansouf
Q: A party of boys purchased a boat and upon payment for the same discovered that if thay had numbered 3 more, they would have paid
a dollar apiece less; but if they had numbered 2 less,they would have paid [a dollar] apiece more. How many boys were there,
and what did the boat cost?

let x= number of boys and y = number of dollars each paid. let xy = represents the number of dollars the boat cost
From the first hypothetical:

xy = (x + 3)(y - 1)

xy = xy - x + 3y - 3

0 = -x + 3y - 3

x = 3y - 3

and

From the second hypothetical:

xy = (x - 2)(y + 1)

xy = xy + x - 2y - 2

0 = x - 2y - 2

x = 2y + 2

____________________________

Solve simultaneously:

x = 3y - 3
x = 2y + 2
-------------

3y - 3 = 2y + 2

y - 3 = 2

y = 5

Substitute the y-value back into one of those equations to find x:

x = 2(5) + 2

x = 10 + 2

x = 12

$There \ \ were \ \ 12 \ \ boys, \ \ and \ \ the \ \ boat \ \ cost \ \ 60 \ \ dollars.$

And you can check what happens with the cost apiece per boy when there are three more boys or two fewer boys.

5. ## Re: WOrd problems

Originally Posted by bigmansouf
A party of boys purchased a boat and upon payment for the same
discovered that if thay had numbered 3 more, they would have
paid a dollar apiece less; but if they had numbered 2 less,they
would have paid apiece more.
How many boys were there, and what did the boat cost?
Geezzzzz......SO:
x boys bought a boat: each paid y dollars.
If instead (x+3) boys bought the boat, each would pay (y-1) dollars.
If instead (x-2) boys bought the boat, each would pay (y+1) dollars.