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Friday, July 21 2017 @ 03:38 PM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 P6 Maths! 2012 Prelims Tao Nan Paper 2 Q13 & 18
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By: geniuskids (offline)  Sunday, September 01 2013 @ 03:44 AM CDT (Read 1603 times)  
geniuskids

Pls help me on these two questions. Thank you so much!

Q13: Bottle X and Bottle Y contained different amount of water. If Bottle X leaks 10ml of water each hour and Bottle Y leaks 5ml of water each
hour, Bottle X would still have 200ml of water left when Bottle Y becomes empty.
If Bottle X loses 5ml of water each hour and Bottle Y loses 10ml of water each hour, Bottle X would still have 590ml of water left when Bottle Y is empty. How much water is there in Bottle X?

Answer : 720ml

Q18 : Mr and Mrs Pratip left their house at the same time and travelled in opposite directions.
Mrs Pratip drove at a speed 16km/h slower than Mr Pratip.
30 minutes later, Mr Pratip arrived at his destination while Mrs Pratip had only completed 1/3 of her journey.
They were 100km apart.

(a) Find Mr Pratip's speed.
(b) Find their total distance apart when Mrs Pratip had reached her destination.

Answers : a) 108km/h
b) 192km

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By: echeewh (offline)  Sunday, September 01 2013 @ 01:50 PM CDT  
echeewh

Hi <geniuskids>


Following are my worked solutions:

Q13


<1st IF>

(Rate of Leakage)
X : Y
10 : 5
2 : 1

X |-----|-----|<------- 200 --------->|
Y |-----|

<2nd IF>

X : Y
5 : 10
1 : 2

Here, we need split the 1p of Y into 2 equal units(u). Likewise, the same goes for the 2p of X. Hence , the model becomes ...

*** <------------ 590 --------------->
X |--|--|--|--|<------- 200 --------->|
Y |--|--|

From this model, we have ...

3u --> 590 - 200 = 390
1u --> 130

X = 1u + 590 --> 130 + 590 = 720 ml

==========

Q18

* X *********** H *************** Y
* |-----------------|-----|-----|-----|
<-| ********* <-|
* P1 ********* P
*************** |-> |->
************** W * W1

s(W) = s(P) - 16
t(PP1) = 30 mins = (1/2)hr = t(WW1)
d(P1W1) = 100

(a)
t(PP1) = 100 ÷ [s(P) + s(W)]
(1/2) = 100 / [s(P) + s(P) - 16]
[s(P) + s(P) - 16] = 2 × 100
2s(P) = 200 + 16 = 216
s(P) = 216 ÷ 2 = 108 km/h

(b)
s(W) = s(P) - 16 = 108 - 16 = 92 km/h
d(WW1) = 92 × (1/2) = 46 km
(1/3) of W's journey = 46
d(W1Y) = (2/3) = 2 × 46 = 92 km
d(XY) = d(P1W1) + d(W1Y)
= 100 + 92 = 192 km

==========

Trust this helps.

Do let me know again if there's further clarification.

Cheers,
Edward

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By: jo sarah (offline)  Monday, September 02 2013 @ 07:54 AM CDT  
jo sarah

Q. 18
(a) Mrs is 16 km/h slower. So, in half an hour, she travels 8 km less than Mr.
And they are 100 km apart.
So, Mr's distance is (100 + 8) / 2 = 54 km, which is done in half hour.

Therefore, Mr's speed is (54 x 2) = 108 km/h. Big Grin

(b). Since Mr has reached his destination, he won't be moving.
Mrs needs to complete the remaining 2/3 of her journey, which'll take an hour. [she completes 1/3 in 1/2 hour]
Mrs's speed is (100 - 54) x 2 = 92 km/h. [or, simply, 108 - 16 = 108]

Therefore, add 92 km to the initial 100 km = 192 km; this is the distance apart when Mrs reaches her destination. Laughing Out Loud

Trust this explains, Cheerio!

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By: geniuskids (offline)  Sunday, September 08 2013 @ 07:10 AM CDT  
geniuskids

Hi Edward , thanks, managed to solve Q13 but don't understand yr method for Q18.

Hi jo-sarah, thanks , I do understand your method to Q18 better!

Thanks to both of you ! Big Grin

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