# Maths Question

- In a journey, a car travels 240 km. If the speed of the car is increased by 15 km/h, the journey takes 4/5 of an hour less. Find the original speed of the car. ANS: 60 km/h - A fraction is in the form n/n+4, where n is a positive integer. If both the numerator and the denominator are increased by3, then the fraction will be increased by 1/9. Find the original fraction. ANS: 5/9 - Mary bought a certain number of fish for \$600. After 1 year, the total number of fish has increased by 1. Mary then decides to sell all the fish at \$3 per fish less than the original cost. However, she stills makes a profit of \$ 27. How many fish did Mabel buy original? ANS: 10 - Simplify and express (i/ 1 +2i) + (1-i/ 2-i) in the form a+bi.ANS: 1 - It is given that the quadratic equation (x+2)(x+4)=5-m has no real roots.The range of possible values of m is m >6. If m is the smallest integral value, solve the given equation and express the answers in the form a+bi. ANS: -3±3i

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• 最佳解答

- 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

資料來源： 土扁
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• 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.

資料來源： My Calculation
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