Maths Question

- In a journey, .....

Let s km/h be the original speed of the car.

240/s - 240/(s + 15) = 4/5
[240/s - 240/(s + 15)] [5s(s + 15)] = (4/5) [5s(s + 15)]
1200(s + 15) - 1200s = 4s(s + 15)
1200s + 18000 - 1200s = 4s² + 60s
4s² + 60s - 18000 = 0
s² + 15s - 4500 = 0
(s - 60)(s + 75) = 0
s = 60 or s = -75 (rejected)

Hence, original speed of the car = 60 km/h


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- A fraction is in the form n/n+4, ......

(n + 3)/[(n + 4) + 3] - n/(n + 4) = 1/9
(n + 3)/(n + 7) - n/(n + 4) = 1/9
[(n + 3)/(n + 7) - n/(n + 4)] [9(n + 7)(n + 4)] = (1/9) [9(n + 7)(n + 4)]
9(n + 3)(n + 4) - 9n(n + 7) = (n + 7)(n + 4)
9n² + 63n + 108 - 9n² - 63n = n² + 11n + 28
n² + 11n - 80 = 0
(n - 5)(n + 16) = 0
n = 5 or n = -16 (rejected)
n + 4 = 9

Hence, the fraction = 5/9


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- Mary bought a certain number of fish ......

Let n be the number of fish that Mabel bought originally.

(n + 1)[(600/n) - 3] - 600 = 27
(n + 1)(600 - 3n)/n = 627
600n - 3n² + 600 - 3n = 627n
3n² + 30n - 600 = 0
n² + 10n - 200 = 0
(n - 10)(n + 20) = 0
n = 10

Number of fish that Mabel bought originally = 10


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- Simplify and express (i/ 1 +2i) + (1-i/ 2-i) ......

i/(1 + 2i) + (1 - i)/(2 - i)
= i(1 - 2i)/(1 + 2i)(1 - 2i) + (1 - i)(2 + i)/(2 - i)(2 + i)
= (i + 2)/[(1)² - (2i)²] + (2 + i - 2i + 1)/[(2)² - (i)²]
= (2 + i)/5 + (3 - i)/5
= (2 + i + 3 - i)/5
= 5/5
= 1


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- It is given that the quadratic equation (x+2)(x+4)=5-m .......

m > 6
The smallest integral value of m = 7

When m = 7:
(x + 2)(x + 4) = 5 - 7
x² + 6x + 8 = -2
x² + 6x + 10 =
x = {-6 ± √[(6)² - 4(1)(10)]}/2
x = [-6 ± √(-4)]/2
x = (-6 ± 2i)/2
x = -3 ± i

1. Let A(km/h) be the original speed, and B(h) be the original time
AxB = 240km......(i); (A+15)x 4/5B = 240......(ii)
from formula i; B = 240/A
then, substitue B = 240/A into formula ii

(A+15) x (4x240)/5A = 240
(A+15) x 4 x 240 = 240 x 5A
4A + 60 = 5A
A = 60 (km/h)

2. let f = n/(n+4); then f + 1/9 = (n+3) / (n+3+4)

then [n/(n+4)] + 1/9 = (n+3)/(n+7)
(9n+n+4)/9(n+4) = (n+3)/(n+7)
(10n+4)x(n+7) = 9(n+3)x(n+4)
10n^2 + 74n + 28 = 9n^2 + 63n + 108
n^2 + 11n - 80 = 0
so, n = 5 or -16

because n is positive integer, so n = 5
then, f = 5/5+4 = 5/9

3. Let A be the original cost of fish, B be the original number of fish
Then AB = 600......(i); (A-3)(B+1) = 627......(ii)
from formula ii, AB + A - 3B - 3 = 627
600 + (600/B) -3B = 630
600 - 3B^2 = 30B
B^2 + 10B - 200 = 0
then B = 10 or -20
so, the original number of fish is 10

4. (i/ 1 +2i) + (1-i/ 2-i) = {i(2-i) + (1-i)(1+2i)}/{(1+2i)(2-i)}
= (2i+1+1+2i-i+2)/(2-i+4i+2)
= (4+3i)/(4+3i)
= 1

5. (x+2)(x+4) = 5-m; when m is greater than 6,
x^2 + 2x + 4x + 8 = 5-m
x^2 + 6x + 8 = 5-m
when m is greater than 6 and smallest intergal value, then m = 7
x^2 + 6x + 8 = -2
x^2 + 6x + 10 = 0
x = [-6+/-(36-40)^(1/2)]/2
x = (-6+/-2i)/2
x = -3±i

2010-11-06 17:15:45 補充：
For the model answer of question 5, -3±3i, that cannot be calculated, because 45 cannot be divided by 4 with natural number.