Q&A - PSLE Math

[Winx5015]

Hi, got a word problem on Speed. Please help!


At 14 30, David left Bedok and cycled towards Pasir Ris at a speed of 10km/h. At the same time, Muthu left Pasir Ris and cycled towards Bedok at 12Km/h along the same road. If the distance between Bedok and Pasir Ris was 6Km/h, how far apart were the boys 15 minutes later?

Thanks.

[small]

Winx5015:

Hi, got a word problem on Speed. Please help!


At 14 30, David left Bedok and cycled towards Pasir Ris at a speed of 10km/h. At the same time, Muthu left Pasir Ris and cycled towards Bedok at 12Km/h along the same road. If the distance between Bedok and Pasir Ris was 6Km/h, how far apart were the boys 15 minutes later?

Thanks.

Hi,

HTH..

After 15 mins (1/4 h):
David cycled ---- 10 km/h x 1/4 h = 2.5 km
Muthu cycled ---- 12 km/h x 1/4 h = 3 km

If the distance between Bedok and Pasir Ris was 6km

6 km -5.5 km = 0.5 km

15 mins later, the two boys were 0.5 km apart.

[Winx5015]

Thanks small for your help.

[Zack7]

thinkgmama:

Zack7:

[quote=\"thinkgmama\"]Hi,


Please help me with this question :

Samuel and Bob shared some stickers in the ratio 2:9.
The total number of stickers shared was between 120 and 170
After Bob had given some stickers to Samuel, the ratio became 3:4.
How many stickers did Bob have at first?

Thank you in advance.

before giving
S:B
2:9
total = 2+9 = 11 units

after giving
S:B
3:4 (* this ratio might be intended as 4:3, the question is a little vague)
total = 3+4 = 7 units

let the number of stickers given be x
let total be y

(9/11)y - (4/7)y = x ,

hence (19/77) y = x

this gives you the relationship between total stickers y against stickers given x.
that is, for every 77 total stickers, 19 stickers have to be given to maintain the ratios above

so the question says the total stickers is between 120 and 170, so assuming that stickers are whole numbers, 19/77 will be scaled by 2 to 38/154. remember that we are dealing with ratios, not absolute numbers.

this means for every 154 stickers, 38 have to be given to maintain the above ratios.

thus, 154 is the total stickers, and BOB had (9/11)y stickers in the beginning,
which means he has 126

Hi,

Thank you for the help rendered.


The part that I do not understand id why we have to scale it by 2. Is it because we have to meet the criteria of the 'total stickers is between 120 and 170'? Therefore if its not 19/77 but if it is ('40/120' - fictional) then there will be no need to scale?

This is the part which I do not understand. Can you help please? Thank you.[/quote]yes, if your fraction obtained is 40/120, 120 falls within the range given, so you 'won't have to scale', because your ratio satisfies the absolute number requirement.

but to be very strict, you have actually scaled your ratio by 1.


e.g

5units : 3 units (This is a ratio)
5 stickers : 3 stickers (This is an absolute number relationship)
you make the transition from ratio to absolute number relationship by scaling by 1 in this case.

e.g

my leg is 2 times longer than my hand

2 Leg length : 1 hand length (This is a ratio)

(below are absolute number relationships)
1metre : 0.5metre (scale of 1/2)
2m : 1m (scale of 1)
3m : 1.5 m (scale of 1.5)
... and the possibilities go on...

so in accordance to your problem , it essentially says, my arm and leg adds up to between 2.5metres to 3.5metres... this means the only possible pair is 2m:1m



another example that i find funny :roll: ,

for every dollar you earn, the govt gives you 2 dollar as cpf

so the ratio is 1:2

but you don't earn 1 dollar. you earn say 10 000 bucks

so

your absolute number relationship is

10 000 : 20 000

if you earn 100 000 bucks, then it becomes 100 000 : 200 000 and so on...



the key point here is that ratios give relationships between 2 variables.

they are not absolute number relations.

you must recognize this(or your kid), because as you progress into higher levels of study, you will be scaling your ratios by fractions, powers, exponentials etc.

[VVJJ]

Hi, need help on this question as I'm stuck now

Anyone can help?

Gopal drove from port A to port Bwhich was 750 I'm apart at an av.s of 75 km/h. At the same time,Adil drove from port B to port A at an av.s of 50 km/h.
A) how long did it take Before Gopal's car passed Adil car ?
B) how far had Gopal travelled when his car passed Adil's car?

[Alarick78]

6hrs

Gopal travel 450
Adil travel 300

[VVJJ]

Thanks alarick… How do you get the answer?

[Sun_2010]

VVJJ:

Hi, need help on this question as I'm stuck now

Anyone can help?

Gopal drove from port A to port Bwhich was 750 I'm apart at an av.s of 75 km/h. At the same time,Adil drove from port B to port A at an av.s of 50 km/h.
A) how long did it take Before Gopal's car passed Adil car ?
B) how far had Gopal travelled when his car passed Adil's car?

At the meeting point, both Adil and Gopal together would have travelled a total distance = 750 km

Their total speed = 75kmph + 50kmph = 125kmph
( since every hour each will travel a distance of the sum of their speeds)

Time taken to meet = 750/125 = 6 hrs

In 6 hours Gopal would have travelled = 6hrs x 75kmph = 450 km

[Winx5015]

Hi, can anyone help regarding questions on Speed.


(1). At 9am,Jane and Carol started travelling from Town A to Town B. Jane travelled at an average speed that was 20km/h more than Carol's average speed. At 11am, Jane reached Town Y. How far was Carol from Town Y at that time?

(2). At 9am, Jane and Carol started travelling towards each other from their homes, which were 210km apart. Jane travelled at an average speed that was 20km/h more than Carol's average speed. If they met each other at noon, what was Jane's average speed?

Thank you

[tianzhu]

Winx5015:

Hi, can anyone help regarding questions on Speed.


(1). At 9am,Jane and Carol started travelling from Town A to Town B. Jane travelled at an average speed that was 20km/h more than Carol's average speed. At 11am, Jane reached Town Y. How far was Carol from Town Y at that time?

(2). At 9am, Jane and Carol started travelling towards each other from their homes, which were 210km apart. Jane travelled at an average speed that was 20km/h more than Carol's average speed. If they met each other at noon, what was Jane's average speed?

Thank you

Hi

As in any questions on Speed, it’s useful to draw some simple pictorial representations.

Q1) Jane travelled at a speed of 20 km/h faster than Carol.

In 1h, Jane travelled 20 km more than Carol.

From 9am to 11am ------- 2h

2*20 ------40km

Carol was 40 km away when Jane reached Town B.

Q2)It may be helpful to recall the time when daddies and mummies were paktoring.

Imagine daddy at one end of a bridge and mummy standing at the other end. Daddy and mummy ran towards one another and met each other at some point along the bridge. At the moment when they met, they covered the whole length of the bridge.

For them to meet, Jane and Carol must cover a distance equal to the distance between their houses.

From 9am to noon -------- 3h

210/3 ------- 70

Jane’s speed was 20 km/h faster.

In 1h, Jane travelled 20 km more than Carol.

(70 – 20)/2 ------ 25
25+20 ------45

Jane’s speed ------ 45 km/h

Best wishes