1. Learning Outcomes
Compute length from a situation expressed in fraction.
Solve problems in real context involving computation of
length.

2. Compute length from a situation expressed in fraction
Example 1:
Points to note
of 240 cm
* of = ×
=
= × 240 cm
=
=
= 96cmcm
Example 2:
1
2 of 64 km *Convert the mixed number into
4 improper fraction
9
= 64 km
4
= ____ km
4
= km
NOTES:
Cancellation Method should be emphasized to find the answer
involving multiplication of fraction by the whole number.
25
Example: 3 × 125 cm = 3 × 25 cm
5 1
= 75 cm

3. Solve problem in real context involving computation of
length
Example 1:
5
Jafri cycles of a kilometre to reach his school. Sivam cycles 450 m to
8
reach the same school.
Between Jafri and Sivam , who cycles further to school?
List the 5
* km of
clues given 8
*450 m
Convert
fraction of
km to m 5× 1000
8
5000
= m
8
= 625 m
Comparing
distances to get the
Jafri cycles 625 m
Sivam cycles 450 m
Answer: Jafri cycles further than Sivam to reach their
school.

4. Solve problem in real context involving computation of
length
Example 2:
Safiq took part in the cross-country run which is 15 km long. He took a
2
rest after running of the way. How far was he from the finishing line?
3
point?
List the Total distance = 15 km
clues given
2
Running distance = of 15 km
3
2
× 15 km
Carry out the plan
3
by multiplying the
denominators
= ________ km
3
= km
Find the unfinish
distance by
1 5 km
Total distance = 15 km
- 1 0 km
Running distance = 10 km
Unfinished distance = 5 km 5 km

5. Solve problem in real context involving computation
of length
Example 3:
Diagram 1 shows the road map of four places , K , L , M and N.
7500 m
L
N
K
M
The distance from K to L is the same as the distance from L to M. The
1
distance from M to N is of the distance from K to M .
4
What is the distance , in km , from K to N?
List clues
given Distance given
KL =LM = 7500 m
1 1
MN = of KM = × 7500 m = 1875 m
4 4
Distance from K to N = 7500 m + 7500 m + 1875 m
= 16875 m km m
1 6 8 7 5
= 16.875 km
or
16 875 = 16.875 km
1 000