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 By: sundarshoba (offline)  Tuesday, July 03 2012 @ 09:23 AM CDT (Read 1316 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: 627

 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

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