Thanks chia-chan. But the algebraic eqn is too complex for psle to solve..

Since this is a GEP paper Math Olympiad Q, they should be able to solve. Not for mainstream PSLE.

Is this ok?

Firstly, please allow me to explain the following : for any given fixed duration of time, the ratio of the distances traveled by two objects is equal to the ratio of their respective speeds. So if two cars travel at 10km/h and 15km/h, the ratio of their speeds is 2:3. In an hour the ratio of the distances traveled by the cars will be 2:3 as well.

If in a given period of time, and we need not know exactly what the amount of time is, the boy travels 10 units. If he had increased his speed by 20%, he would had traveled 12 units. If he had increased his speed by 30%, he would had traveled 13 units.

In the two situations described in the question, the two times taken for the boy to reach school are the same, which is 5 minutes earlier than if he had walked at his normal pace.

So this is how both situations pan out: Situation one, when he walks at 20% faster: 12 units + 12 units + 12 units Situation two, when he walks 500m at his usual speed and then the rest of the way at 30% faster: 10 units + 13 units + 13 units Since 10 units is equivalent to 500m, (12+12+12) or (10+13+13) = 36 units must therefore represent 1800m

Now why is it that we can be sure that it is 12+12+12 units and 10+13+13 units? Since 13 units are 1 unit more than 12 units, two 13 units can each contribute 1 unit to bump the 10 units to 12 units. Please do not mistaken this for guess-and-check. The concept illustrated is basically that speed is proportional to the distance traveled.

Suggested solution: In a given period of time, the boy walks 10 units so that if he walks 20% and 30% faster he has walked 12 units and 13 units respectively. In case one, 12u + 12u + 12u In case two, 10u + 13u + 13u where 10u is equivalent to 500m. 36 units is thus equivalent to 1800m (ans)

If worked out, we will find out that the boy walks at 60m/min and he takes 30 minutes to reach school. If anyone wishes to know how that can be calculated, please kindly flag me. Thank you.

Tks Xavier! Can u explain in situation one why there r 3 sets of 12uits?

Good morning, Caroline. Sure of course. Let us narrow down to the period of time at which the boy walks 500m at his normal speed. I assigned the distance at 10 units. In that same amount of time, the boy will cover 12 units if his walking speed is increased by 20%. Do you understand this part?

Yup.

Excellent. So after this first period of time, the boy walks 13 units in the second situation while still maintaining 12 units in the first situation, correct?

That would be in the second period of time, which is the same length of time as the first period.

Ok

And..?

Now in two periods of time, the boy has walked 10 units + 13 units for a total of 23 unit in the second situation. In the same two periods of time, the boy has walked 12 units + 12 units for a total of 24 units in the first situation.

That puts the boy 1 unit behind in the second situation as compared to the first situation over two periods of time.

So by the third period, the boy will cover the same distance in both the first and second situation as he has a 1 unit speed advantage per period in the second situation over the first.

Thus the 12u + 12u + 12u over three periods of time in the first situation and the 10u + 13u + 13u over the same three periods of time in the second situation. Remember that the distance from his home to school is the same for both situations and he uses the same amount of time for both situations as well.

Thanks Adrian! Is very clear

Hi Adrain why is one complete turn 2 x 3.14 x 2.9?

Is it finding circumference?

Yes...

https://www.facebook.com/photo.php?fbid=10153085696153004&set=p.10153085696153004&type=1&theater

Izam Marwasi Raymond Ng thank you :)

Y muz plus 20 twice?

Referring to the model I drew, the 4 green units is made up of the 52 and twice of the 20. It is difficult to see this just by reading the question because it is not obvious. That's why we draw models as it shows the relationship between the quantities.

Whenever one transfers a quantity to another party, the change in the difference is always double of the transferred quantity. Let me give an illustration. Let us say that Ali and Bob each has $10. If Ali gives $1 to Bob, Ali will be left with $9 and Bob ends up with $11. The difference is $2 ($11 - $9) though only $1 has changed hands.

Their total age 5 years ago was 66 - 10 = 56. So, 5 years ago, 4u = 56 1u = 14 Lenny was 14 years old, 5 years ago. Lenny is 19 years old now.

Thanks both

For questions that deals with age, do apply the thought of "difference remains the same". 5 years ago, total age = 66 - 10 = 56 As mentioned by Edlynn Rose and Vei Li Soo, 4u = 56 1u = 14 Now, Lenny's age = 14 + 5 = 19 years old

Why did you subtract 10 ?

5years for each person... Paul Booth

To return to the scenario 5 years ago, you have to subtract 5 years from Mag's present age and 5 years from Lenny's present age. That is why 10 years are subtracted from the sum of Mag's and Lenny's present ages.

Looking at the models, we know that we can't divide 66 by 4 because of the two boxes of 5. So to create equal units, we take away 2 groups of 5 from the bars and from the 66. Then we can say that the 4 black units equals to 56.

Again from reading the question, it may not be obvious especially if it is an unfamiliar problem. Drawing models will help us and younger students understand the problem situation better.