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## Tuesday, July 29, 2008

### Henry Park Pri Sch 2007 PSLE Math Prelim Q48

A lorry, a van and a car set off at the same time travelling at a constant speed of 60 km/h, 80 km/h and 120 km/h respectively. The lorry and the van were travelling from Town G to Town H while the car was travelling from Town H to Town G. The car passed the lorry 2 minutes after passing the van.
a) Find the ratio of the distances travelled by the lorry to the van to the car at the moment when the car passed the van.
b) Find the distance between Town G and H.

Solution

a)
Since all of them started at the same time, the distance covered by each vehicle is proportionate to their respective speeds.

Lorry : Van : Car
60 : 80 : 120
3 : 4 : 6 (Answer)

b)
Ratio of the distance covered by the lorry to the distance covered by the car is 3:6 or 1:2. Hence, for every 2 units of distance the car travelled, the lorry travelled 1 unit.

Ratio of distance covered by the van to the distance covered by the car is 4:6 or 2:3. Hence, for every 3 units of distance the car travelled, the van travelled 2 units.

Van covered 2 units for every 3 units car covered.
Hence if the van covered 6 units, the car would have covered 9 units.

Lorry covered 1 unit for every 2 units car covered.
Hence, if the lorry covered 5 units, the car would have covered 10 units.

Distance = speed x time

The speed of the car was 120 km/h.
When the car passed the lorry, it was 2 min after it had passed the van.

Distance covered by the car within those 2 minutes ----
120 km/h x 1/30 h = 4 km

1 unit ----- 4 km

(Total distance)
15 units ----- 4 km x 15 = 60 km

Answer: The distance between Towns G and H is 60 km.

Anonymous said...

you are only confusing the students.

Excel Eduservice said...

The way to do it is to read and refer to the graph while you read. I know it is a little difficult scrolling up and down, but that is the inherent problem learning from the computer screen.

I have highlighted this point before, that computers can never replace the board as far as teaching is concerned.

The ideal is still traditional face to face learning. If the computer can replace teachers, MOE would have done that a long time ago. Training teachers is not cheap. It costs the government (and your tax money) quite a bomb.

Anonymous said...

it doesn't shows how you obtained 15 units, or how you arrive at the model diagram (of total 15 units) - like why the van covers 6 units. The best method would be to use algebra. That is, 1/2 (x + 4) + 4 = 2/3x, where x is the distance the car meets the van.

Excel Eduservice said...

P6 students are not trained to use algebra to work out speed sums.

The 15 units is based on trial and error, which is actually guess and check - an accepted method in PSLE Math.

Observer said...

I am afraid although the answer to part b is correct, the workings do not seem quite right.

Based on your drawing, for the 2 minutes, the distance is covered by both the lorry and the car and not the car alone. So it should have been

(60 + 120) km/h x (2/60 h) = 6 km.

6 km is 1 unit
From part a), this is the difference in distance between the van and the lorry (4 units - 3 units)

Total distance is 10 units (the combined distance covered by the van and car)
1 unit --> 6 km
Total distance (10 units) --> 60 km

Observer said...

My sincere apology, the workings for part b is correct. But we could have made use of the answer to part a to work out part b instead of having to figure out a different 15 units.

DV : DL : DC = 3 : 4 : 6

1 unit (DL - DV) --> (60 + 120) x (2/60) = 6 km (Distance covered by Van and Car in 2 minutes)

Total distance: 10 units (DL + DC) --> 10 X 6 = 60 km.

Observer said...

Let me try to explain how 15 units are derived.

DV DL DC
----------
3 : 4 : 6

As the distance is the same, make them the same 3 x 5 --> 15 and 5 x 3 --> 15,
DV DL DC
----------
1 : 2 = 5 : 10
2 : 3 = 6 : 9
----------

In this case, 1 unit is the distance covered by the car in 2 minutes and total distance is 10 units.

Observer said...

In P5, pupils are already using algebra to solve their problem sums (although they were not told that it is algebra). Eg

3 Apples + 2 Oranges --> \$2.00
2 Apples + 3 Oranges --> \$2.50

6 Apples + 4 Oranges --> \$4.00
6 Apples + 9 Oranges --> \$7.50

5 Oranges --> \$3.50
1 Orange --> \$0.70

3 Oranges --> \$2.10

2 Apples --> 2.50 - 2.10 = \$0.40
1 Appple --> \$0.20

Excel Eduservice said...

It is not quite right to say that is algebra. Algebra has very specific rules. Example, a x b = ab. On the other hand, there is no such thing as apples x oranges = applesoranges in models.

Many people who have not taught PSLE Maths get the wrong concept that algebra is used in models. While some of the principles used are similar, it is wrong to say that algebra has been used. Algebra is a specific topic, with its own rules, which models cannot follow in totality.