Q&A - PSLE Math
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janice099:
For a fixed distanceHello there, this is my first reply. Thanks for helping
John can walk to school in 40 minutes while his brother, Jack, takes 20 minutes. If they both start walking to school at the same time, how long will it take for John to be 2/5 as far as Jack.
Time ratio of John to Jack is 2:1
Speed ratio of John to Jack will be 1:2
For a fixed time;
Distance ratio of John to Jack will be 1 : 2 same as the speed ratio.
Canβt understand how the distance ratio be 2:5? -
[Moderator's note: Topics merged.]
A car and a coach travelled from Town A to Town B. The coach left Town A at 0648 and it took 5 hours to reach Town B. The car started 30 minutes later than the coach and it took 4 hours to reach Town B. At what time did the car catch up with the coach? :?: -
A car and a coach travelled from Town A to Town B. The coach left Town A at 0648 and it took 5 hours to reach Town B. The car started 30 minutes later than the coach and it took 4 hours to reach Town B. At what time did the car catch up with the coach?
Information from question
(1) Both car and coach travelled the same distance from Town A to Town B meaning that distance in this case was constant.
(2) The car started 1/2 hour later and took less time than the coach to reach Town B meaning that the car travelled faster than the coach.
(3) Duration of journey for the car was 4 hours and for the coach was 5 hours.
Solution
Find the 1st common multiple of 4 and 5 and you will get 20.
This was the distance between both Towns A and B which was 20 units.
Coach took 5 hours to travel 20 units
Coach took 1 hour to travel 4 units
Car took 4 hours to travel 20 units
Car took 1 hour to travel 5 units
But the coach started 1/2 hour earlier and had travelled:
(1/2 x 4) units = 2 units.
So, the car had to catch up for this \"2 units\" of the distance.
From above, car could travel faster than the coach :
Difference in 1 hour = (5-4)units = 1 unit.
This means that for 1 hour, the car could catch up by 1 unit.
For 2 units, the car would require 2 hours.
0648.......30 min..0718..........2 hours .........0918
Hence, the car caught up with the coach at 0918 h -
Vanilla Cake:
A car and a coach travelled from Town A to Town B. The coach left Town A at 0648 and it took 5 hours to reach Town B. The car started 30 minutes later than the coach and it took 4 hours to reach Town B. At what time did the car catch up with the coach?
Information from question
(1) Both car and coach travelled the same distance from Town A to Town B meaning that distance in this case was constant.
(2) The car started 1/2 hour later and took less time than the coach to reach Town B meaning that the car travelled faster than the coach.
(3) Duration of journey for the car was 4 hours and for the coach was 5 hours.
Solution
Find the 1st common multiple of 4 and 5 and you will get 20.
This was the distance between both Towns A and B which was 20 units.
Coach took 5 hours to travel 20 units
Coach took 1 hour to travel 4 units
Car took 4 hours to travel 20 units
Car took 1 hour to travel 5 units
But the coach started 1/2 hour earlier and had travelled:
(1/2 x 4) units = 2 units.
So, the car had to catch up for this \"2 units\" of the distance.
From above, car could travel faster than the coach :
Difference in 1 hour = (5-4)units = 1 unit.
This means that for 1 hour, the car could catch up by 1 unit.
For 2 units, the car would require 2 hours.
0648.......30 min..0718..........2 hours .........0918
Hence, the car caught up with the coach at 0918 h
5 - 4 = 1 h
1 - 1/2 = 1/2 h -> so car reached Town B 1/2 h earlier.
Hence car caught up at the mid-point, i.e after 4/2 = 2 h.
0648....30 min.......0718..........2 h .........0918
The car caught up with the coach at 0918. -
Hi alamak, let me (try to) solve your question:
Dist = SAME
Every 1 hour, the coach travelled 1/5 of the distance, so in 1/2 an hour, the coach would travel 1/10 of the distance. When the coach had 9/10 of the distance left, the car started from Town A. In every 1 hour, the car would travel 1/4 of the distance. So in 4 hours, the coach would have travelled 4/5 more, and 1/10 + 4/5 is 9/10. o the car reached when it had 1/10 of its distance left, which it would be able to travel in 1/2 an hour, so the car's total time, 4 hours, minus the 1/2 hour, equals 7/2 hours or 3 and a half hours. So, with a timeline, 6.48 + 3 1/2 hours equals 9.18 am, or in 24 hour time 0918.
Hope my working and answer is right!
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Many ways to solve the same problem. Use the method u r comfortable with. Cheers
http://www.postimage.org/ -
tianzhu:
Once again, many many thanks..my DD and I misunderstood the question.Hi Dharma
Thanks for the correction.I made a mistake.(use 20% more instead of 120% more)
Price of venision per kg should be $6.55.
Price of beef per kg is $7.20.
Best wishes.
Hi small
Please consider my earlier post as null and void.
The answer is $47.80.
Best wishes
[quote]The cost of 2kg of beef is 120% more than the cost of 1kg of venison. 1kg of mutton costs 20% more than the cost of 1kg of beef.
1 kg of beef ( 160%) vs 1kg of vensison (100%)
120%/2 +100% =160% :([/quote]
:oops: -
Hi all, thanks for all your solutions!
I see the picture now..Haha
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Thank you for your help.
How to show that the sum of all angles is equal to 360 degrees?
Q35 RGPS SA2 2007
http://www.wendykoh.com/07/primary6-rgs-maths.pdf -
tianzhu:
Partial c + d + f = 180 degree (triangle)Thank you for your help.
How to show that the sum of all angles is equal to 360 degrees?
Q35 RGPS SA2 2007
http://www.wendykoh.com/07/primary6-rgs-maths.pdf
The other portion of c + b + a + e = 180 degree (triangle, External angle of a triangle = 2 internal opposite angles = a + e)
So all angles add up to 360 degree.
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