
By: pyeo (offline) Monday, September 08 2014 @ 12:43 PM CDT (Read 1600 times)



pyeo 
1. Simon and Paul set out on their bicycles at the same time and from the same place to ride to a nearby swimming pool. Simon rode three times as long as Paul rested on the trip. Paul rode four times as long as Simon rested. The two reached the pool at the same time. Who rode faster?
2. A shop sells sweets in bags of 7 and 20. What is the largest number of sweets that cannot be purchased exactly? Justify your answer.
if you can provide help to the above, it would be very very much appreciated.
From a stressed mum!!
;(

Newbie
Registered: 10/09/11 Posts: 8





By: echeewh (offline) Monday, September 08 2014 @ 11:37 PM CDT



echeewh 
Hey <pyeo >
//updated 9/9/14 11:05pm
My sincere apologies to you and all viewers for producing a wrong solution earlier.
The 'corrected' solution is updated below
//
Following is my worked solution for Q1:
***  for alignment purpose
Q1
< Time taken / Duration >
(C)=cycled; (R)=rested
S(C) : P(R)
3 : 1
S(R) : P(C)
1 : 4
Given both S, P reached the pool at the same time, total time taken (duration) by each of them is the same.
S(C) + S(R) = P(C) + P(R)
3p + 1u = 4u + 1p
3p  1p = 4u  1u
2p = 3u
p = (3/2)u
Hence,
S(C) = 3p = 3 × (3/2)u = (9/2)u
P(C) = 4u
S(C) took (9/2)u time, and P(C) took 4u time to reach the pool.
Since distance travelled by both is the same and P(C) cycled for a shorter time (duration),
Paul rode faster.
===========
Trust this helps.
Do let me know again if this is different from your <Answerkey> or if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 625





By: pyeo (offline) Tuesday, September 09 2014 @ 12:47 AM CDT



pyeo 
Thank you so much for your help.
Your step by step explanation makes it easy to understand.
Can I be cheeky and ask for help with question 2.
Once again thank you.
From a stressed mum!

Newbie
Registered: 10/09/11 Posts: 8





By: echeewh (offline) Tuesday, September 09 2014 @ 10:38 PM CDT



echeewh 
Hey <pyeo>
Let me answer Q2 as follows:
Following is my opinion of a possible reasonably good solution ( and if I can understand what the question / setter needs ):
This looks like an openended question with no one single definite number. But importantly that number one gives must be reasonably good to be justified.
Any reasonably good answer can be one so long as the number given is not a multiple of 7, 20, 27 or any of these combinations. Smallest number is 29.
Based on my reasoning, the answer can be any of these following and so on ...
1002, 603, 50, ...
( so i think one can give an answer in the tens, hundreds, thousands, ten thousands, hundred thousands, or millions, etc )
========
Trust this helps.
Do let me know again if this is different from your <Answerkey> or if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 625





By: jo sarah (offline) Wednesday, September 10 2014 @ 03:03 AM CDT



jo sarah 
2. A shop sells sweets in bags of 7 and 20. What is the largest number of sweets that cannot be purchased exactly? Justify your answer.
(the question seems to invite an infinite no. of solutions; or it is rather vague, since the "largest' can't be established... Just my thought.)
well, let's look at what is the no. of sweets that can be purchased exactly.
It means a no. that is a multiple of both 7 and 20.
Since 7 and 20 have no common factors, the smallest no. that is a multiple of both is 7 x 20 = 140.
Thus, any no. that is a multiple of 140 will be a no. of sweets that can be purchased exactly.
It therefore implies any no. that isn't a multiple of 140 is a no. that can't be purchased exactly.
As such, how to tell the "largest" no.?

Regular Member
Registered: 03/20/12 Posts: 111



