# Form 5 Maths Problems

1) Kenny buys a massage chair for \$11000 and a hi-fi set for \$20000. The massage chair and the hi-fi set will depreciate by 20% and 36% each year respectively. It is known that after t years, the value of the massage chair will be \$3000 less than that of the hi-fi set. Find the value of t.

2) A cruise ship set of from town A at 6am. It travelled 840 km downstream to town B and then travelled upstream back to town A. It took 37 hours 30 minutes for the whole journey. Given that the speed of the current was 3km/h, find the time when the cruise ship arrived at town B.

3) Two taps A and B are used to fill up a tank with water. If tap A is used alone to fill up the tank, it takes 25 minutes longer than using tap B alone. If each of tap A and tap B is turned on alone for 20 minutes one by one and then they are turned on together for 10 minutes, the tank will be full.

i) Find the time taken for tap A alone to fill up the tank.

ii) Find the time taken for tap B alone to fill up the tank.

4) Jack cycles from place A to place B at a constant speed. If he increases his speed by 3km/h, he will arrive at B half an hour earlier. Given that the distance between A and B is 30km, find Jack's original cycling speed

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1.

after t years,

massage chair: 11000 (1 - 20%)^t

hi-fi set: 20000 (1 - 36%)^t

20000 (1 - 36%)^t - 11000 (1 - 20%)^t = 3000

20 (0.64)^t - 11 (0.8)^t = 3

20 (0.8)^2t - 11 (0.8)^t = 3

let x = (0.8)^t

20x^2 - 11x = 3

20x^2 - 11x - 3 = 0

(4x - 3)(5x + 1) = 0

x = 4/3 or -1/5 (rejected)

4/3 = (0.8)^t

log (4/3) = t log (0.8)

t = log (4/3) / [log 0.8]

2.

downstream speed: x + 3 km/hr

time required: 840/(x+3)

upstream speed: x - 3 km/hr

time required: 840/(x-3)

total time: [840/(x+3)] + [840/(x-3)] = 37.5

840(x-3) + 840(x+3) = 37.5(x+3)(x-3)

112(x-3) + 112(x+3) = 5(x+3)(x-3)

112x + 112x = 5x^2 - 45

5x^2 - 224x - 45 = 0

(5x + 1)(x - 45) = 0

x = -1/5 (rejected) or 45

3i.

let a, b be the flowrate of A and B respectively

(1/a) - (1/b) = 25 => b - a = 25ab

20a + 20b + 10(a+b) = 1 => a + b = 1/30

(1/30 - a) - a = 25a(1/30 - a)

1/30 - 2a = 25a/30 - 25a^2

1 - 60a = 25a - 750a^2

750a^2 - 85a + 1 = 0

(10a-1)(75a-1) = 0

a = 1/10 (rejected) or 1/75

A required 75mins to fill the tank

3ii.

b = 1/30 - 1/75 = 45/2250 = 1/50

B required 50mins to fill the tank

4.

let x be the original speed

time required for original speed

30/x

time required for new speed

30/(x+3)

thus,

[30/x] - [30/(x+3)] = 0.5

30(x+3) - 30x = 0.5x(x+3)

60x + 180 - 60x = x^2 + 3x

x^2 + 3x - 180 = 0

(x-12)(x+15) = 0

x = 12 or -15(rejected)