Tutor MathsGuru: Ask me for your burning Maths questions!

[singmathstutor]

We should be looking at the question in perspective from the time Andrew alighted at Woodlands.


In Andrew’s case, to reach his destination (home) after alighting at Woodlands,
Total time he needs is 20min for resting and
30min for walking home at a speed of 6km/h in a distance of 3km.
T= 3km/6km/h
=1/2h
=30min

Therefore, the 20min+30min=50min is as such.

In Ben and Edmund’s case, we should start comparing the time just when Edmund has started his return trip to JP to pick Ben. Since 15min is used to make that trip, and both the boys would have reached Orchard at the same as Andrew reaching home, therefore time left for their journey to Orchard is 50min-15min=35min

That 15min is only applicable to Ben and Edmund’s journey, so you should not have added 15min to Andrew’s return trip.

Hope this would have helped to clear your doubts

[Dharma]

Herbie:

I have one speed qn which need help.


A lorry travelled at an average speed of 72km/h from Pink Town to Happy Town. 50 minutes later, a van started out from the same point towards the same destination at an average speed of 90km/h, reaching Happy Town at the same time.

a. How long did the van take to travel from Pink Town to Happy Town?
b. If both the lorry and the van arrived at Happy Town at 4.05pm at what time did the lorry start out??

Tx.

For constant distance;
Speed ratio=> Lorry : Van = 4 : 5
Time ratio => Lorry : Van = 5 : 4
5u – 4u = 1u = 5/6hrs

a)\tTime taken by van = 4u = 4 x 5/6 hrs = 3 1/3hrs
b)\tTime taken by lorry = 5u = 25/6hrs = 4 1/6hrs
Time lorry left Happy Town => 4 1/6 hrs before 4.05pm = 11.55am

[Herbie]

Thanks Dharma.


I have another speed qn.

Town S and Town T are 600km apart.
At 9am, a van travelling at an uniform spped left Town S for Town T
At the same time, a car set off from Town T to Town S at an uniform speed which was 12km/h faster than the van’s

Find the average speed of the car if the two vehicles met at 3pm.

TX

[Herbie]

Sorry, i have another speed qn.


Amy and Billy walked from their home to school.

A-----B ----- School

Amy started walking from Point A at a speed of 120m/min.
At 9.05am, Amy reached Point B, and at the same time, Billy started walking from Point B towards the school.

At 9.15am, Billy was 100m ahead of Amy.

Find Billy’s walking speed.
If Billy reached the school 2 min ahead of Amy, what was the distance Billy covered from Point B to the school?

Tx

[Dharma]

Herbie:

Thanks Dharma.


I have another speed qn.

Town S and Town T are 600km apart.
At 9am, a van travelling at an uniform spped left Town S for Town T
At the same time, a car set off from Town T to Town S at an uniform speed which was 12km/h faster than the van's

Find the average speed of the car if the two vehicles met at 3pm.

TX

When the van and car met, they had already travelled for 6 hrs.
Sum of speeds of car & van = 600km/6hrs = 100km/hr
Difference in speed of car & van = 12km/hr

Speed of car = (100 + 12)km/hr divided by 2 = 56km/hr

[Dharma]

Herbie:

Sorry, i have another speed qn.


Amy and Billy walked from their home to school.

A-----B ----- School

Amy started walking from Point A at a speed of 120m/min.
At 9.05am, Amy reached Point B, and at the same time, Billy started walking from Point B towards the school.

At 9.15am, Billy was 100m ahead of Amy.

Find Billy's walking speed.
If Billy reached the school 2 min ahead of Amy, what was the distance Billy covered from Point B to the school?

Tx

From 9.05am to 9.15am; Amy walked 1200m [120m/min x 10min]
Billy walked 1300m [1200m + 100m]
Billy’s walking speed = 1300m/10min = 130m/min

Speed ratio => Amy : Billy = 12 : 13

For fixed distance (From B to school)
Time ratio => Amy : Billy = 13 : 12
1u = 2min
12u = 24min

Distance (B to school) = 130km/hr x 24min = 3120m

[Almighty]

singmathstutor:

We should be looking at the question in perspective from the time Andrew alighted at Woodlands.


In Andrew's case, to reach his destination (home) after alighting at Woodlands,
Total time he needs is 20min for resting and
30min for walking home at a speed of 6km/h in a distance of 3km.
T= 3km/6km/h
=1/2h
=30min

Therefore, the 20min+30min=50min is as such.

In Ben and Edmund's case, we should start comparing the time just when Edmund has started his return trip to JP to pick Ben. Since 15min is used to make that trip, and both the boys would have reached Orchard at the same as Andrew reaching home, therefore time left for their journey to Orchard is 50min-15min=35min

That 15min is only applicable to Ben and Edmund's journey, so you should not have added 15min to Andrew's return trip.

Hope this would have helped to clear your doubts

Thankyou Singmathstutor
Still trying to understand. Will come back to u if i dont get it...
:roll:

[Almighty]

mathsguru:

Almighty:

[quote=\"mathsguru\"]
Hi math monster,
I used yr algebra method.I am stuck..Pl correct me.. This is my solution:
Berfore:
A : 3m + 55
B : 5m - 55
After A gave B $55/-
A : B = 3m : 5m
B gives 20 % = (53 - 55) X 20% + 10
= m-1

So, A gets = (3m +55) + m-1
= 4m + 54
B left with = (5m -55) - m -1
= 4m -56
HOw can 4m +54 = 4m -56 n solve for m??Where did i go wrong??

If MM not active,Can MAth guru help me out? Pl.refer to P :165 for the method taught by MM.
I am not comfortable using Model...So, Pl.advise!!!

Hi Almighty,

Think u misread the question. It says \"If Brian gave Andrew 20% of THIS money\" not \"his money\". I think it's referring to 20% of $55 and not his entire sum of money.


MathsGuru

[/quote]Thanks MAthguru.I got it now!!!!

Tang, Here is the algebra solution ...
\tAt First :\t\t\t
\t\t\t\t
\tA = 3m+ 55 \t\t\t
\tB = 5m - 55\t\t\t
\t\t\t\t
\tAfter \t\t\t
\t 20% of 55 = 11\t\t\t
\t\t\t\t
\t5m - 55 -11 -10 = 5m -76\t\t\t
\t3m + 55 + 11 +10 = 3m + 76\t\t\t
\t\t\t\t
\t5m - 76 + 108 = 3m + 76\t\t\t
\t5m + 32 = 3m + 76\t\t\t
\t5m = 3m + 76 - 32\t\t\t
\t2m = 44\t\t\t
\tm = 22\t\t\t
So,\tA + B = 3m +55 + 5m -55 \t\t\t
\t = 8M\t\t\t
\t8m = 22 X 8 = 176

[Almighty]

Edmund drove Andrew from Jurong Point to Woodlands at a constant

speed of 75 Km/h. Andrew rested for 20 mins before walking 3km home at a speed of
6km/h.Edmund took 15 minutes to return to Jurong Point to pick Ben up. HE then drove
to Orchard at 75Km/h.The three people reached their destinations at the same time.
What was the distance Edmund drove from Jurong Point to Orchard?

Almighty:

singmathstutor:

We should be looking at the question in perspective from the time Andrew alighted at Woodlands.

In Andrew's case, to reach his destination (home) after alighting at Woodlands,
Total time he needs is 20min for resting and
30min for walking home at a speed of 6km/h in a distance of 3km.
T= 3km/6km/h
=1/2h
=30min

Therefore, the 20min+30min=50min is as such.

In Ben and Edmund's case, we should start comparing the time just when Edmund has started his return trip to JP to pick Ben. Since 15min is used to make that trip, and both the boys would have reached Orchard at the same as Andrew reaching home, therefore time left for their journey to Orchard is 50min-15min=35min

That 15min is only applicable to Ben and Edmund's journey, so you should not have added 15min to Andrew's return trip.

Hope this would have helped to clear your doubts

Thankyou Singmathstutor
Still trying to understand. Will come back to u if i dont get it...
:roll:

We should not take 15 mins as the time taken by Edmund to drop Andrew to Woodlands because:
Edmund's speed is only given and not the time taken .On his return from Woodlands to Jurong, his time is only given which means the speed need not be the same as 75km/h.
Am i right?pl.clarify :?

[singmathstutor]

Hi Almighty,


Edmund did not use 15min to alight Andrew, he used that time to get back to JP to fetch Ben. I think you have interpreted wrongly here.