Q&A - PSLE Math

[tianzhu]

Maha Bodhi School P6 Prelim 2010 Maths paper 2 Q15 [4 marks]-Speed question.

There are 30 steps between Levels 5 and 6. A toy robot takes 13⅓ seconds to move up 20 steps if it moves up the first 10 steps at its highest speed and the next 10 steps in its lowest speed. It will take 14 seconds to reach Level 6 from Level 5 if it moves up the first 24 steps at its highest speed and the next 6 steps at its lowest speed. Find the shortest time it will take to reach Level 6 from Level 5.
Answer Given : 10 seconds

Hi

Hope this helps.
A quick one.

If students are comfortable with Simultaneous Equations, then this question should not bother them too much.

For those who are not familiar with SE, we can work around this problem with MD or using letters.

To draw a meaningful comparison, the number of steps must be the same.

10HS + 10LS -------13 1/3
24HS +6 LS ------- 14

Consider 60 steps.

30HS+30LS ------40
48HS+12LS ------28

Therefore 18LS takes 12 seconds longer than 18 HS
1LS takes 12/18 or 2/3 seconds longer than HS

If 6LS steps are replaced by HS, you’ll save 6*2/3 ------4s

Therefore shortest time taken is (14-4) ------10s

Best wishes

[tianzhu]

1. Nanyang Primary School P6 Prelim 2010 Maths paper 2 Q18 [5 marks]- To find area of the shaded parts. Answer given is 9.12 cm²

The figure below is created using the curved lines (arcs) of quadrants with radius 1 cm, 2cm and 3 cm. Find the area of the shaded parts. (Take pie as 3.14)

Hi VC

Please load the image, if possible.

Please try to follow the calculations without the my usual colourful picture.
This question looks intimidating at first glance, but it’s not really so.

Rearrange the shaded areas.

Consider 3cm quadrant.
Area of 3cm square ------3*3 -----9
Area of radius 3cm quadrant ------1/4*3.14*9 -------7.065
9-7.065 ----- 1.935
Area of shaded area -------- 9 - 2(1.935) ----- 5.13 ------------(A)

Consider 1cm square
Area of 1cm square -------1
Area of 1cm quadrant ------ 1/4 *3.14*1 ------- 0.785
1-0.785 ----- 0.215
Area of shaded area ------- 1-2(0.215 ) ----- 0.57 --------------(B)

Area of shaded area ---------5.13 - 0.57 -------- 4.56 ----------(A) minus(B)

Total shaded area -------2(4.56) ------ 9.12.

Best wishes

[tianzhu]

Maha Bodhi School P6 Prelim 2010 Maths paper 2 Q18 [5 marks]-Speed question.

Tom is travelling on a bus that is moving at a uniform speed of 30 km/h. If he alights at Bus Stop A, he will walk home by Route A which is 800m away from his flat. He can also alight 1 km down the road at Bus Stop B and take the 730 m Route B home. If Tom walks at a uniform speed of 40 m/min, find:
(a) the difference in time when he alights at the two different bus stops.
(b) the earliest time he can reach home if he boards the bus from the interchange 20 km away at 1305.

Answers given: (a) 40 min (b) 1405

Hi VC

I am not too sure about part (a)
Does it mean the time taken to walk from the bus stops to his home for the different routes?

Route A is the fastest route.

40m -----1 min
800m ---- 20 mins

Route B
730/40 ---- 18.25mins
But you need to add the bus’s travelling time.
30km -------60mins
1km -------2 mins
Therefore total time taken --------20.25mins

If the question is referring to the difference in time between routes A and B, the answer is 20.25mins - 20 mins which is 15 seconds.

Part (b) is more straightforward

Time taken by bus to travel 20km ------ 20/30*60 ------ 40mins
Time taken to reach home ---------20+40 -------60mins or 1h
Therefore, he reaches home at 1405

Best wishes

[tianzhu]

Red Swastika P6 Prelim: 2010 paper 2: Q16. Kylie and Leonard have $4576 altogether. 1/8 of Leonard’s money is $38 more than 25% of Kylie’s money. (a) How much does Kylie have? (b) What fraction of Leonard’s money is Kylie’s money? (Express your answer in its simplest form.) [5 marks][Answer given: $1424, 89/197]


Hi
A quick one.
Draw a MD.

8*38 -------304

8 units + 304 + 4 units ------- 4576
12 units ------ 4576-304
1 unit ----- 356

a)Kylie -------4*356 -----1424

b)Leonard ------ 4576 – 1424 --------3152
Fraction --------- 1424/3152 --------89/197

Best wishes

[tianzhu]

Pei Chun P6 Prelim 2010: Q17. In 2007, there were 90 more girls than boys in ABC School. In 2008, there were 1/3 fewer girls and the number of boys increased by 5/9, resulting in a total of 660 pupils. (a) What fraction of the pupils in 2007 were boys? Express your answer in its simplest form. (b) In 2009, the number of girls increased while the number of boys remained the same as in 2008. The ratio of the number of girls to the number of boys in 2009 was the same as the ratio in 2007. How many pupils were there in 2009? [5 marks][Answer given: 3/7, 980]


Hi
Hope this helps.
You may wish to draw a MD.

2007
Girls ------9units+90
Boys ------9units

2008
Girls ------6units+60
Boys ------14units (increased by 5/9)

20 units +60 ------660
1 unit -------30

a)In 2007, boys --------9*30 -----270 and girls --------360
Therefore 270/630 --------3/7

In 2008, boys ------ 14*30 ------ 420

In 2007 Girls: Boys ------- 360:270 ------4:3

In 2009 Boys ------420, Girls -----560

Therefore total number ------- (420+560) -------980

Best wishes

[Vanilla Cake]

tianzhu:

Hi VC


Please load the image, if possible.
Please try to follow the calculations without the my usual colourful picture.
This question looks intimidating at first glance, but it’s not really so.

Rearrange the shaded areas.

Consider 3cm quadrant.
Area of 3cm square ------3*3 -----9
Area of radius 3cm quadrant ------1/4*3.14*9 -------7.065
9-7.065 ----- 1.935
Area of shaded area -------- 9 - 2(1.935) ----- 5.13 ------------(A)

Consider 1cm square
Area of 1cm square -------1
Area of 1cm quadrant ------ 1/4 *3.14*1 ------- 0.785
1-0.785 ----- 0.215
Area of shaded area ------- 1-2(0.215 ) ----- 0.57 --------------(B)

Area of shaded area ---------5.13 - 0.57 -------- 4.56 ----------(A) minus(B)

Total shaded area -------2(4.56) ------ 9.12.

Dear Uncle Tianzhu,
I still have problems in loading up pictures/diagrams via Flickr/Photobucket. So, pls refer to http://psle2010a.blogspot.com/2010/09/circle-ny-p6-2010-sa2-p2-q18.html and thanks for your explanation.

[Vanilla Cake]

tianzhu:

Maha Bodhi School P6 Prelim 2010 Maths paper 2 Q18 [5 marks]-Speed question.

Tom is travelling on a bus that is moving at a uniform speed of 30 km/h. If he alights at Bus Stop A, he will walk home by Route A which is 800m away from his flat. He can also alight 1 km down the road at Bus Stop B and take the 730 m Route B home. If Tom walks at a uniform speed of 40 m/min, find:
(a) the difference in time when he alights at the two different bus stops.
(b) the earliest time he can reach home if he boards the bus from the interchange 20 km away at 1305.

Answers given: (a) 40 min (b) 1405

Hi VC

I am not too sure about part (a)
Does it mean the time taken to walk from the bus stops to his home for the different routes?

Route A is the fastest route.

40m -----1 min
800m ---- 20 mins

Route B
730/40 ---- 18.25mins
But you need to add the bus’s travelling time.
30km -------60mins
1km -------2 mins
Therefore total time taken --------20.25mins

Part (b) is more straightforward

Time taken by bus to travel 20km ------ 20/30*60 ------ 40mins
Time taken to reach home ---------20+40 -------60mins or 1h
Therefore, he reaches home at 1405

Best wishes

Dear Uncle Tianzhu,
Pls refer to http://psle2010a.blogspot.com/2010/09/maha-modhi-p6-speed-q18.html. Both of you have different answers for (a) and different from the answer given.

[tianzhu]

Hi VC


Is 9.12 the correct answer?

Best wishes

[tianzhu]

Hi VC


If the question is referring to the difference in time between routes A and B, the answer is 20.25mins - 20 mins which is 15 seconds.

My answer is different from the WS.

Best wishes

[Vanilla Cake]

tianzhu:

Hi VC


Is 9.12 the correct answer?

Best wishes

9.12 is the given answer but I have not gone through yet. Uncle Observer's answer is the same as yours.