Q&A - PSLE Math

[Vanilla Cake]

Hi All,


One of my younger cousins is taking PSLE this year. I got stuck at this speed qn from Nanyang Prelim 2008 Q47 while helping him. Here it goes :

Lily is meeting a friend at a certain time. If she drives at 80 km/h, she will be 1/3 hour late. If she drives at 60 km/h, she will be 3/4 hour late. How long will the journey take if she drives at 90 km/h ?

Answer given is 1 1/9 h

TIA for your time and help.

[kiasiparent]

Vanilla Cake:

Hi All,


One of my younger cousins is taking PSLE this year. I got stuck at this speed qn from Nanyang Prelim 2008 Q47 while helping him. Here it goes :

Lily is meeting a friend at a certain time. If she drives at 80 km/h, she will be 1/3 hour late. If she drives at 60 km/h, she will be 3/4 hour late. How long will the journey take if she drives at 90 km/h ?

Answer given is 1 1/9 h

TIA for your time and help.

Speed 80:60 = 8:6
Time: 6units:8units
2 units of difference in time = 3/4 - 1/3 = 9/12 - 4/12 = 5/12 hrs
1 unit of time = 5/24 hrs

Time taken if she drive at 80km/h = 6 x 5/24 = 5/4hrs
Total distance = 5/4 x 80 = 100 km
Time taken if she drives at 90km/h = 100/90 = 1 1/9 hr

[tianzhu]

kiasiparent:

Since they return 50mins earlier, it means the father (constant speed) saved 25 mins driving to the school and 25 mins driving back.
So David met his father at 6pm - 25mins = 5.35 pm
5.35 - 5pm = 35mins
David had walked for 35 mins.

Hi kiasiparent
Thank You.

[tianzhu]

Thank you for your help.


RGPS SA1 2008
http://farm4.static.flickr.com/3639/3465444648_9b5f70ab2f_o.jpg\">

[kiasiparent]

tianzhu:

Thank you for your help.


RGPS SA1 2008
http://farm4.static.flickr.com/3639/3465444648_9b5f70ab2f_o.jpg\">

In 1 day, Jane can complete 1/10 of the strings whereas sharon can complete 1/18 of the string.

I believe the fastest way to do this question is to do guess and check.

Since they took 14 days together, try jane 1 day and sharon 13 days etc.
try jane 2 days and sharon 12 days.

Then you will realise that jane 5 days and sharon 9 days will be the correct answer as 5/10 + 9/18 = 1 whole string.

You can also do simultaneous equation.
Assume the whole string is 180 parts long.

Then Jane can do 18 parts in 1 day and sharon can do 10 parts in 1 day.
Let the no of Janes day be J and the no of Sharon days be S.
Doing simultaneous equation:
J + S = 14
18J + 10S = 180

Therefore, S = 14 - J
18J+ 10(14 - J)= 180
8J = 40
J = 5 days.

[Dharmaratnam]

tianzhu:

Thank you for your help.


RGPS SA1 2008
http://farm4.static.flickr.com/3639/3465444648_9b5f70ab2f_o.jpg\">

Hi Tianzhu,

U can also solve it this way

Jane can complete 1/10 of the strings in 1 day
Sharon can complete 1/18 of the strings in 1 day

If Jan takes N days, Sharon will take (14-N)days to complete the work.

Therefore,

N/10 + (14-N)/18 = 1

N= 5

Jane took 5 days

[faith2u]

[Moderator's note: Topics merged.]


At 1030am, a van left Town X, travelling at an average speed of 64km/h.
At 1115am, a car left Town X, travelling on the same road at an average speed of 80km/h.
(a). What time did the car catch up with the van?
(b). How far from Town X did each vehicle travel when they passed each other?

:?

[chengsmummy]

At 1115am, the first car has already travelled 64km/h * 45/60h=48km

the second car is 80-64=16km/h faster then the first car, so it takes
48/16 = 3 hours to catchup the first car.

by the time, both of the cars have travelled
80*3 = 240km

[faith2u]

Hi chengsmummy....thanks alot for your help.

:lol: :celebrate:

[mathsparks]

Section C - Questions.

Any idea if it’s compulsory to solve section c questions using models? Is it ok to solve using algebra?