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# Form 4 and Form 5 Quadratic KBAT Questions

The questions (No. 8, 9 & 10) are in picture (in Malay). Question 7 is not KBAT.

*KBAT = Kemahiran Berfikir Aras Tinggi (Higher Order Thinking Skills)

Possible solution are as follows:

8) let the initial average speed & time be v and t respectively.

Same distance of 15 km is travelled one way.

Since 15 minutes = 15/60 = 1/4 hours, we have:

15 = vt

Or

t = 15/v… (i)

And

15 = (v + 5)(t – 1/4)…(ii)

Put (i) into (ii)

15 = (v + 5)(15/v – 1/4)
15v = (v + 5) (15 – v/4)
60v = (v + 5)(60 – v)
60v = 60v – v^2 + 300 – 5v
v^2 + 5v – 300 = 0
(v + 20)(v – 15) = 0

Since v > 0,

v = 15 km/h (Answer)

9) Let the initial cost shared per student & the initial number of students be c & n respectively.

Then,

cn = 600

Or,

c = 600/n… (i)

And,

(c – 5)(n + 10) = 600… (ii)

Put (i) into (ii)

(600/n – 5)(n + 10) = 600
(600 – 5n)(n + 10) = 600n

600n + 6000 – 5n^2 – 50n = 600n

5n^2 + 50n – 6000 = 0
n^2 + 10n – 1200 = 0
(n + 40)(n – 30) = 0

Since n > 0,

n = 30 ahli kelab siber (Answer)

10) Let the speed of train A be v.

6 minutes
= 6/60 hours = 1/10 hours

In 6 minutes,

Let the Distance travelled by train A = x
x = (v) x (1/10)
x = (v/10)…(i)

Let the Distance travelled by train B = y.
y = (v + 20) x (1/10)
y = (v + 20)/10… (ii)

Since these distances (x and y) are perpendicular to each other forming a right triangle with hypotenuse = 10

We have by Pythagoras Theorem,

x^2 + y^2 = 10^2… (iii)

Put (i) & (ii) into (iii):

(v/10)^2 + ((v + 20)/10)^2 = 10^2

v^2 + (v + 20)^2 = 10000
2v^2 + 40v + 400 = 10000
2v^2 + 40v – 9600 = 0
v^2 + 20v – 4800 = 0
(v + 80)(v – 60) = 0

Since v > 0,

v = 60 km/h (for train A, Answer)

And

20 + 60 = 80 km/h (for train B, Answer)

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Anonymous Christian

Jonathan Ramachandran is founder at AnonymousChristian.org which serves as a Christian Blog discussing the Greater Hope of Biblical Christian in Christ Alone. He is also an IMO (International Math Coach) and a featuring Guitarist for many bands.

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