
By: geniuskids (offline) Sunday, September 01 2013 @ 03:44 AM CDT (Read 1614 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

Active Member
Registered: 11/12/11 Posts: 169





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

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





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.
(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.
Trust this explains, Cheerio!

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





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 josarah, thanks , I do understand your method to Q18 better!
Thanks to both of you !

Active Member
Registered: 11/12/11 Posts: 169



