Sign Up!
Login
Welcome to Test-paper.info
Saturday, September 23 2017 @ 11:28 AM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 2 Speed questions.
   |  Printable Version
By: drstevewu (offline)  Monday, July 15 2013 @ 03:23 AM CDT (Read 1221 times)  
drstevewu

Hi, I am trying to solve these 2 speed questions:

1. Mary walked from home to the Church. For the first 2 minutes, she was walking at an average speed of 50 m/min. When she realized that she was going to be late for 5 minutes, she quickly increased her speed by 10m/min. As a result, she was late by 2 minutes. What is the distance between her home and the Church?

2. A delivery man has to deliver a parcel from Town A to Town B by a certain time. If he travels at an average speed of 96km/h, he will arrived at Town B 1/3 hour late. If he travels at an average speed of 90km/h, he will arrived at Town B 1/2 hour late. What is the distance between the two towns?

Thank you for your help..



Cheers,
Steve..Smile

Forum Junior
Junior


Registered: 04/20/10
Posts: 31

Profile Email    
   
By: echeewh (offline)  Monday, July 15 2013 @ 04:51 PM CDT  
echeewh

Hey there,

In view that both questions are similar, you can use the method shown in Q1 to try and work out Q2 on your own (Ans: 240 km). Should there be any problem, please do let me know. Big Grin

Following is my worked solution to Q1.

Q1.

*** - alignment purpose

H **************************** C
|---------------------------------------|
|-> ** |-> ********** |-> ** |->
M ** M1 ********* M2 ** M3

s(MM1) = 50 m/min
t(MM1) = 2 min
d(MM1) = 50 × 2 = 100 m

From M1, if M travelled at the original speed of 50 m/min, she would be late by 5 mins. I.e. M would be at M2 and it would take M another 5 mins at that original speed to reach C.

s(M1M2) = 50 m/min
s(M2C) = 50 m/min
t(M2C) = 5 min
d(M2C) = 50 × 5 = 250 m

From M1, if M were to increase her speed to 60 m/min, she would be late by 2 mins. I.e. M would be at M3 and it would take M another 2 mins at that speed to reach C.

s(M1M3) = 60 m/min
s(M3C) = 60 m/min
t(M3C) = 2 min
d(M3C) = 60 × 2 = 120 m

At M1, if M travelled at 50 m/min, M would be 250 m away from C; while M travelled at 60 m/min, M would be 120 m away from C. At this juncture, the time taken for both instances are the same. I.e.

t(M1M2) = t(M1M3)

When time is constant, the distance ratio is the same as the speed ratio. I.e.

s(M1M2) : s(M1M3)
50 : 60
5 : 6

d(M1M2) : d(M1M3)
5 : 6

d(M1M3) - d(M1M2) = d(M2M3) = d(M2C) - d(M3C)
6u - 5u = 250 - 120
1u --> 130

d(M1M3) = 6u --> 6 × 130 = 780 m

Distance HC = d(MM1) + d(M1M3) + d(M3C)
= 100 + 780 + 120
= 1000 m = 1 km

========

Trust this helps.

Do let me know again if this is different from your <Answerkey> or if there's further clarification

Cheers,
Edward

Forum Active Member
Active Member

Registered: 04/21/11
Posts: 627

Profile Email    
   
By: drstevewu (offline)  Friday, July 26 2013 @ 10:49 AM CDT  
drstevewu

Thank you Edward... Smile

Forum Junior
Junior

Registered: 04/20/10
Posts: 31

Profile Email    
   


 All times are CDT. The time is now 11:28 am.
Normal Topic Normal Topic
Locked Topic Locked Topic
Sticky Topic Sticky Topic
New Post New Post
Sticky Topic w/ New Post Sticky Topic w/ New Post
Locked Topic w/ New Post Locked Topic w/ New Post
View Anonymous Posts 
Able to Post 
HTML Allowed 
Censored Content