Q&A - PSLE Math

[tianzhu]

Pei Chun P6 Prelim 2010: Q13. 450 people took part in a race. 97 of them were children and the rest were adults. 3/5 of the women and ¾ of the men were married while the rest of them were singles. There were 15 more married men than married women. (a) How many single were there? (b) What fraction of the people who took part in the race were men? [4 marks][Answer given: 116, 28/75]


Hi

You may wish to draw a MD.

450-97 ----- 353
3M-3W ------15
M is 5 more than W

4W + 20 + 5W ------353
1W ------- 37
1M -------42

a)Singles --------- (37*2)+42 ------116
b)Men -------42*4 ------168
168/450 ------ 28/75

Best wishes

[Dharma]

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

10u + 10p = 40/3 …………(1)
24u + 6p = 14 ………………(2)
6u + 6p = 8 ……………………(1) x 3/5 = (3)

18u = 6 ………… (2) – (3)
1u = 1/3

Shortest time taken = 30u = 30(1/3) = 10s

River Valley Primary School P6 Prelim 2010 P2
Q16:
A rabbit and a tortoise competed in a 5.2 km race. The rabbit ran at a speed of 20 km/h while the tortoise’s speed was 3 km/h. The tortoise ran continuously to complete the race. However the rabbit ran for 1 min and rested for 20 min. It ran the next 2 min and rested or 20 min. It then ran for the next 3 min and rested for 20 min and so forth. At this rate, how many minutes more would it take the rabbit than the tortoise to complete the race? [5 marks][Answer given: 11.6 min]

Distance rabbit ran in :
1 min => 20 x 1/60 = 1/3km
2 mins => 20 x 2/60 = 2/3 km
(1/3 + 2/3 + 3/3 + …. + N/3) is less than 5.2
1/3 (1 + 2 + 3 + …. + N ) is less than 5.2
For distance of 5km
1 + 2 + 3 + …. + N = 15
N = 5

For the last 0.2km; time taken = 0.2km/20km/hr = (1/100) hr = 3/5 min = 0.6 min
Total time taken by rabbit = (5 x 6)/2 + 20 x 5 + 0.6 = 115.6min

Total time taken by tortoise = 5.2/3 x 60 mins= 104 mins
Difference in time = 115.6 mins – 104 mins = 11.6 mins

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

(a)
The question is a bit ambiguous. It can be interpreted in 3 different ways. I chose the 3rd way to interpete.
Difference in time when Tom alights at the 2 bus stops is the additional time taken by the bus to travel from Bus Stop A to Bus Stop B = 1/30 X 60 = 2 mins = 120 s

(b)
Earliest time taken (stop at Bus Stop A)
= ( 20/30 x 60 + 800/40 ) mins = 60 mins = 1 hr
Tom reaches home at 1405 hrs

Nanyang Primary School P6 Prelim 2010 Maths paper 2 Q18 [5 marks]-
http://www.postimage.org/image.php?v=aVyqS29

[tianzhu]

Rosyth P6 Prelim 2010 P2: Q18: Xavier, Yati and Zul each had a certain number of stamps. At first, Yati had 200 stamps more than Xavier and Zul had ¾ the number of stamps Yati had. After Yati gave away 1/8 of her stamps to Xavier, she had 60 fewer stamps than Xavier. What was the ratio of the number of stamps Xavier had to the number of stamps Yati had to the number of stamps Zul had at first? Give your answer in the simplest form. [5 marks][Answer given: 42:52:39]


Hi

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

At First
X ------8units
Y ------- 8 units+200
Z -------3/4 of Y

In the End
X -------9 units + 25
Y ------7 units + 175
Z remains the same

7units+175+60 ----- 9 units +25
1 unit ------105

X ------- 105*8 ------ 840
Y -------1040
Z -------780 (3/4 of Y)


Therefore X:Y:Z ---------- 42:52:39

Best wishes

[tianzhu]

Maha Bodhi P6 Prelim 2010: Paper 2: Q17. A group of 328 P5 children were divided into two teams at the P5 camp this year. There were 8 more children in Team B than in Team A. There were also 50% more boys in Team B than in Team A. If 3/8 of the girls were in Team B, (a) what percentage of the pupils in Team B are girls? (b) what percentage of the children are boys? Leave your answers as mixed numbers in the simplest form. [5 marks][Answers given: 28 and 4/7%, 60 and 40/41%]


Hi
Hope this helps.

For those students who are comfortable with Simultaneous Equations, you may use them. Or you may prefer to work with MD or letters(alphabets)

Team A -------160
Team B ----168

Team A
BBGGGGG --------160

Team B
BBBGGG ---------168
BG -----------56

BBGGGGG --------160
Therefore GGG --------160 – (56*2) -------- 48
G ----------16

a)Boys ------120
Girls ----48
% girls ------- 48/168*100 ---------28 and 4/7%

b)Boys -------5*40 -------200
Girls -------128
% Boys ---------200/328 *100 ------ 60 and 40/41%

Best wishes

[Vanilla Cake]

Hi Uncle Tianzhu / Uncle Dharma,


Sorry to bother both of you. Pls refer to this http://psle2010a.blogspot.com/2010/09/volume-p6-2010-sa2-hhk-p2-q-18.html for the unusual volume question from SHHK P6 Prelim 2010 Paper 2 Q18 [5 marks]. Could you pls paste and post this question c/w diagram in KSP forum to share among readers and offer your views in addition to Uncle Observer's solution. I had tried to post this diagram but the size was too big. How do you control the size of the image/diagram that you want to post here?
Thanks.

[atutor2001]

Vanilla Cake:

Hi Uncle Tianzhu / Uncle Dharma,


Sorry to bother both of you. Pls refer to this http://psle2010a.blogspot.com/2010/09/volume-p6-2010-sa2-hhk-p2-q-18.html for the unusual volume question from SHHK P6 Prelim 2010 Paper 2 Q18 [5 marks]. Could you pls paste and post this question c/w diagram in KSP forum to share among readers and offer your views in addition to Uncle Observer's solution. I had tried to post this diagram but the size was too big. How do you control the size of the image/diagram that you want to post here?
Thanks.

Hi Vanilla cake

During my kids' time, they were taught to remember this as the \"candle question\" because the question was given as 2 candles of different lengths being burnt at different time.

The solution given was based on algebra - kind of difficult for P6 to digest.

Remembered teaching my kids to use model which is easier. The height of each tank is first divided into parts, equal to the \"number of half-hour\" Then by comparing the portion after 11 pm (being equal in height) we can sub-divide the units into equal size. We will then get 4 units = 5cm. The height of tank A will be 36 units, which will then work out to be 45cm.

Sorry don't know how to draw and post the proper model.

[moomoo7368]

Hi vanilla cake, thanks for your links regarding 2010 prelim papers!

[kancheongmum]

Vanilla Cake:

Hi Uncle Tianzhu / Uncle Dharma,


Sorry to bother both of you. Pls refer to this http://psle2010a.blogspot.com/2010/09/volume-p6-2010-sa2-hhk-p2-q-18.html for the unusual volume question from SHHK P6 Prelim 2010 Paper 2 Q18 [5 marks]. Could you pls paste and post this question c/w diagram in KSP forum to share among readers and offer your views in addition to Uncle Observer's solution. I had tried to post this diagram but the size was too big. How do you control the size of the image/diagram that you want to post here?
Thanks.

Hi Vanilla Cake

Uncle Observer's solution is very difficult for my ds to understand and I find it the same too. So I used the 'candle method' which Uncle Observer had posted before. If I am not wrong is the equal fraction method. Let me try and I hope is not too complicated to understand. Tianzhu and Dharma pls give your view and correct me if I am wrong. Tks

Tank A emptied in 6 hrs --- 1 hour emptied 1/6 tank
Tank B emptied in 4 hrs --- 1 hour emptied 1/4 tank
Tank A in 4 hrs and Tank B in 2.5 hrs will reach the same height.
That means remaining tank to emptied
Tank A === 1 - (1/6 x 4) --- 1/3 tank
Tank B === 1 - (1/4 x 2.5) --- 3/8 tank

1/3 of height of Tank A is equal to 3/8 of height of Tank B
1/3 ---- 3/9

Height of Tank A 9u
Height of Tank B 8u

9u - 8u = 1u --- 5cm
Height of Tank A is 9 x 5 = 45cm

[Vanilla Cake]

kancheongmum:

Hi Vanilla Cake


Uncle Observer's solution is very difficult for my ds to understand and I find it the same too. So I used the 'candle method' which Uncle Observer had posted before. If I am not wrong is the equal fraction method. Let me try and I hope is not too complicated to understand. Tianzhu and Dharma pls give your view and correct me if I am wrong. Tks

Tank A emptied in 6 hrs --- 1 hour emptied 1/6 tank
Tank B emptied in 4 hrs --- 1 hour emptied 1/4 tank
Tank A in 4 hrs and Tank B in 2.5 hrs will reach the same height.
That means remaining tank to emptied
Tank A === 1 - (1/6 x 4) --- 1/3 tank
Tank B === 1 - (1/4 x 2.5) --- 3/8 tank

1/3 of height of Tank A is equal to 3/8 of height of Tank B
1/3 ---- 3/9

Height of Tank A 9u
Height of Tank B 8u

9u - 8u = 1u --- 5cm
Height of Tank A is 9 x 5 = 45cm

Hi kancheongmum,

Yes, you are right as it took me about 15 minutes plus to understand Uncle Observer's solution considering the fact that I am a year 2 (Sec 2) student. :oops:
Refer to http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=6373&postdays=0&postorder=asc&start=2420.
Another similar question also came out in the River Valley Primary School P6 Prelim Maths 2010 Paper 2 Q17 [5 marks] :

Candle A and candle B are of the same length. Candle A, which is broader, can burn for 5 h while Candle B, the thinner candle, can burn for 4 h. If both candles are lighted at the same time, how long does it take for Candle A to be twice as long as Candle B?
Answer : 3⅓ hours

Off-topic. The newly released package of 10 schools per subject for 2010 P6 Prelim top papers comes at a price tag of $40 for 4 subjects.
4 subjects x 10 schools = 40 papers
$40 ÷ 40 = $1 per paper

Interesting number and pls don't get me wrong. I am not trying to promote these top papers. Just a casual remark to ease the PSLE 2010's tension and pressure.

[Dharma]

kancheongmum:

Vanilla Cake:

Hi Uncle Tianzhu / Uncle Dharma,


Sorry to bother both of you. Pls refer to this http://psle2010a.blogspot.com/2010/09/volume-p6-2010-sa2-hhk-p2-q-18.html for the unusual volume question from SHHK P6 Prelim 2010 Paper 2 Q18 [5 marks]. Could you pls paste and post this question c/w diagram in KSP forum to share among readers and offer your views in addition to Uncle Observer's solution. I had tried to post this diagram but the size was too big. How do you control the size of the image/diagram that you want to post here?
Thanks.

Hi Vanilla Cake

Uncle Observer's solution is very difficult for my ds to understand and I find it the same too. So I used the 'candle method' which Uncle Observer had posted before. If I am not wrong is the equal fraction method. Let me try and I hope is not too complicated to understand. Tianzhu and Dharma pls give your view and correct me if I am wrong. Tks

Tank A emptied in 6 hrs --- 1 hour emptied 1/6 tank
Tank B emptied in 4 hrs --- 1 hour emptied 1/4 tank
Tank A in 4 hrs and Tank B in 2.5 hrs will reach the same height.
That means remaining tank to emptied
Tank A === 1 - (1/6 x 4) --- 1/3 tank
Tank B === 1 - (1/4 x 2.5) --- 3/8 tank

1/3 of height of Tank A is equal to 3/8 of height of Tank B
1/3 ---- 3/9

Height of Tank A 9u
Height of Tank B 8u

9u - 8u = 1u --- 5cm
Height of Tank A is 9 x 5 = 45cm

Hi kancheongmum,

Your solution is clear and correct.