SOLUTIONS OF NETWORK IN GRAPH THEORY

NETWORK IN
GRAPH THEORY

CHAPTER 5
5.1 NETWORK

NETWORK IN FORM 4
GRAPH THEORY
MATHEMATICS
SOLUTIONS KSSM

COMPREHENSIVE PRACTICE COMPREHENSIVE
PRACTICE
QUESTIONS 1 – 10
TEXT BOOK
This booklet aims to help students in solving Pg 148 – Pg 150
NETWORK IN GRAPH THEORY problems from the
textbook. By Wong SK
01 July 2021 ( PDPR )

No. 1

No. 2(a)

No. 2(b)

No. 3

No. 3

No. 5a(i)

No. 5a(ii)

No. 5b

No. 6
OR

No. 6

No. 7

No. 7

Directed and Weighted graph, so the graph must have arrow to show the direction
and value for the distance.

Shortest distance = 3080m
shortest distance in km = 3080/1000 ( 1 km = 1000m )
= 3.08 km

No. 7

No. 9(b)

No. 9(b)

Commission :
Agent = 25% x RM100 = 25/100 x RM100 = RM25
Manager = 2% x RM100 = 2/100 x RM100 = RM2
b(i) 18 x RM2 x 30 = RM1080
b(ii) Let a = number of policies

25 x a ≥ 1000
25a ≥ 1000
a ≥ 1000 /25
a ≥ 40 ( at least )

No. 10
OR

No. 10

(a) 1st day = 4
2nd day = 4 + 3(4) = 4 + 12 = 16

3rd day = 4 + 3(4) + 3(12) = 4 + 12 + 36 = 52

(b) 4th day = 4 + 3(4) + 3(12) + 3(36)

5th day = 4 + 3(4) + 3(12) + 3(36) + 3(108) = 484

(c) 8th day = 4 + 3(4) + 3(12) + 3(36) + 3(108) + 3(324) + 3(972) + 3(2916)
- [ 4 + 3(4) + 3(12) ]

= 4 + 12 + 36 + 108 + 324 + 912 + 2916 + 8748 - [ 52 ]
= 13120 – 52 ( because the life span of a cell is 5 days )

= 13068

OR

No. 10