
By: sundarshoba (offline) Tuesday, July 03 2012 @ 09:23 AM CDT (Read 1277 times)



sundarshoba 
Please assist to work out the steps for the attached questions. Thks

Chatty
Registered: 12/31/06 Posts: 53





By: jo sarah (offline) Wednesday, July 04 2012 @ 10:12 AM CDT



jo sarah 
Hi,
let me take no. 14 first.
class 6A class 6B
girls: 4 u 6 p ['u' means units; 'p' means parts]
boys: 3 u 1 p
Now, total girls = 2 x total boys
thus, 4 u + 6 p = 2 (3 u + 1 p)
= 6 u + 2 p
so, 2 u = 4 p
i.e., 1 u = 2 p
that being so, the table becomes:
class 6A class 6B
girls: 8 p 6 p ['u' means units; 'p' means parts]
boys: 6 p 1 p
and so, 2 p > 8 pupils
14 p > 7 x 8 = 56 pupils in class 6A
Cheerio!

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





By: jo sarah (offline) Wednesday, July 04 2012 @ 10:21 AM CDT



jo sarah 
Hi,
let me take no. 14 first.
class 6A class 6B
girls: 4 u 6 p ['u' means units; 'p' means parts]
boys: 3 u 1 p
Now, total girls = 2 x total boys
thus, 4 u + 6 p = 2 (3 u + 1 p)
= 6 u + 2 p
so, 2 u = 4 p
i.e., 1 u = 2 p
that being so, the table becomes:
class 6A class 6B
girls: 8 p 6 p ['u' means units; 'p' means parts]
boys: 6 p 1 p
and so, 2 p > 8 pupils
14 p > 7 x 8 = 56 pupils in class 6A
Cheerio!

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





By: echeewh (offline) Wednesday, July 04 2012 @ 06:12 PM CDT



echeewh 
Q15
Hi Sundar,
Pls find my worked solution as follows:
M = Midpoint
X = Meeting point
**** = Buffer for alignment purpose
A*******************M*********X********* B

>******************<300m>>******<
J****************************<**********M
*****************************J1,M1
S(J) = 4 km/h more than S(M)
(a)
Extra (Additional) distance travelled by J to meet or pass M
= 2 x (distance between M and X)
= 2 x 300 m = 600 m = 0.6 m
Time taken for J to reach X
= Extra (Additional) distance ÷ Speed Difference
= 0.6 ÷ 4
= 0.15 hr = 9 min
Hence, time when J passed M = 9.09 am
(b)
Time taken by J from X to B = 3 min
Time taken by M from B to X = 9 min (since this is the time taken by both M and J to reach X from their respective starting points B and A)
As Distance is constant, we can use the Time and Speed ratios method, where Speed ratio is the reverse / inverse of the Time ratio.
T(J) : T(M)
3 : 9
S(J) : S(M)
9 : 3
Given that Speed Difference, S(J)  S(M) = 4 km/h, we have ...
6u > 4
1u > 4 ÷ 6 = 2/3 hr
S(J) = 9u > 9 x (2/3) = 6 km/h
S(M) = 3u > 3 x (2/3) = 2 km/h
Hence, Distance AB = S(J) x Time taken by J to complete AB
= 6 x (12/60) = 1.2 km
Time taken by M to complete AB = 1.2 / 2 = 0.6 hr = 36 min
Time when M reached B = 9.36 am
=========================================
Trust the above helps.
Do let me know again if theres any clarification.
Cheers,
Edward

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





By: sundarshoba (offline) Wednesday, July 04 2012 @ 10:53 PM CDT



sundarshoba 
Thank you very much Mr.Edward & Sarah!!!

Chatty
Registered: 12/31/06 Posts: 53



