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 By: Jane Lim (offline)  Friday, July 08 2011 @ 10:35 PM CDT (Read 3768 times)
Jane Lim

Pls help to solve qn 13b and qn 18.

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Registered: 10/01/10
Posts: 11

 By: w13158 (offline)  Tuesday, August 02 2011 @ 04:25 AM CDT
w13158

Quote by: Jane Lim

Pls help to solve qn 13b and qn 18.

Q13.
a. What is the length of the base of Container X?

Since the water flow rate of Tap A is 3 times that of Tap B and after 20 minutes of water flows, both containers have the same height, therefore the base Area of container Y is 3 times the base area of container X.
First need to find the base area of container Y.
50x45=2250cm2
2250/3=750 (This is the base are for Container X)
750/25=30 (Take the base area divide by the width to get the length)

b. How many litres of water can Container X hold?

Can’t solve this, unless the height of the water in either container is given. Lets says if the height is 10cm, then the solution will be:

2/5--> 25 x 30 x 10 = 7500cm3
5/5--> 5/2 x 7500 = 18750
18750 = 18.75 l

Q18.
a. The difference in time when he alights at the two different by stops.

Since bus stop B is 1 km away from bus stop A and the speed of the bus is 30km/h, therefore you needs to calculate the time taken by the bus for 1km distance.

Time=Distance/Speed = 1/30 x 60 mins
=2 mins

b. The earliest time he can reach home if he boards the bus from the interchange 20km away at 1305.

To solve this question, you need to calculate the time taken for both routes A and B. His walking speed given is 40m/min.
800/40=20mins (time taken for route A)
730/40=18 ¼ mins (time taken for route

Next, calculate the time taken for the bus to travel 20km at a speed of 30km/h.
20/30 x 60min = 40 min

40+20=60 (Total time taken for route A)
40+2+18 ¼ = 60 ¼ (Total time taken for route
Therefore the time taken for route A is shorter, so choose route A.

1305 + 60 min = 1405.