中四數學題(English)

Let x and x+1 be two consecutive positive integers :

1/x + 1/(x+1) = 11/30

(x+1) + x = 11/30 * x(x+1)

30(2x+1) = 11x(x+1)

11x^2 - 49x -30 = 0

(11x+6)(x-5) = 0

x=5 or x=-6/11(rejected)

So 5 and 6 are two consecutive positive integers that the sum of their reciprocals is 11/30.

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Note that 12 fewer bottles of wine = 1 dozen fewer bottles of wine.
Let x be the original price per dozen bottles of wine.
We have :

900/x - 900/(x+150) = 1

900(x+150) - 900x = x(x+150)

135000 = x^2 + 150x

x^2 + 150x - 135000 = 0

(x - 300)(x + 450) = 0

x= 300 or x=-450(rejected)

So the original price per dozen bottles of wine is 300.

1. Let the smaller number be x, the larger number will be (x +1).
So 1/x + 1/(x + 1) = 11/30
30(x + 1) + 30x = 11x(x + 1)
30x + 30 + 30x = 11x^2 + 11x
11x^2 - 49x - 30 = 0
(11x + 6)(x -5) = 0
x = -6/11 (rej.) or 5,
so the 2 numbers are 5 and 6.
2.
Let original price per dozen bottles of wine be x.
Let number of bottles of wine = y dozens.
So original value = xy = 900.............(1)
New price per dozen = x + 150.
New quantity = y - 1.
So (x + 150)(y -1) = 900.............(2)
From (1) y = 900/x, sub. into (2) we get
(x + 150)(900/x - 1) = 900
(x + 150)(900 - x) = 900x
900x - x^2 + 135000 - 150x - 900x = 0
x^2 + 150x - 135000 = 0
(x - 300)(x + 450) = 0
x = 300 or -450(rej.)
So price is $300/dozen.