Learning OutcomesCompute length from a situation expressed in fraction.Solve problems in real context involving computation oflength.
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
Solve problem in real context involving computation oflengthExample 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.
Solve problem in real context involving computation oflengthExample 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
Solve problem in real context involving computationof 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