Q&A - PSLE Math

[tianzhu]

brastilava:

Hi, please help in these 2 questions:


1) 7 rows of soldiers with 20 soldiers in each row were assembled on a parade square. The soldiers were spaced out equally to form the perimeter of the parade square. The number of soldiers on the length of the parade square was twice the number of soldiers on its breadth. How many soldiers were on each side of the rectangle?

2) I am a 2-digit number. I am a multiple of 4. When I am divided by 9, there will be a remainder of 6. When I am divided by 7, there will be a remainder of 5. Who am I?

Hi

Good Morning.

For Q1, please refer to this similar question.

The answers to your question are 48 and 24.

7*20 ------ 140

140 + 4 ------144

6 units ------ 144

You may carry on from here.

For Q2, use “Systematic Listing”

It may be a bit tedious, but this is the sort of durian bao chiak type of questions if students are careful in their calculations. All one needs to do is to list down the multiples and make appropriate adjustment.

The answer is 96.

Best wishes

http://farm6.staticflickr.com/5272/6897367566_8d8cd99d93_z.jpg\">

[ksmama]

[quote][/quote]

MathIzzzFun:

match3frog:

Please help with the following question:

Hi mathsizzfun,
Regardg yr ans to the foll questn, my kid calculated a diff way n the ans is different. Please comment on why his method is not acceptable. Is it because when d ques asks for average sp, we have to calculate based on whole dist n time taken n not assume aver sp same throughout? Thanks.

Diff in speed-> 65/3 =21.67km/h
Speed of car 70+21.67 = 91.67km/h.





A car was travelling from Town P to Town Q.

After completing 2/7 of the journey in 1/2 hour, the car passed a van travelling at a constant speed of 70km/h in the same direction.
3 hours later, the car reached Town Q but the van was still 65 km away from Town Q.

(a) Find the average speed of the car.

(b) If the van left Town P at 0810, at what time would it arrive at Town Q?

Thank you

5/7of distance PQ --> 3 x 70 + 65 = 275 km
distance PQ --> 275/5 x 7 --> 385 km
total time taken by car --> 3. 5 h
average speed of car --> 385/3. 5 --> 110 km/h

cheers.

[ksmama]

Sori my post somewh in the middle of incomplete quote. Pls bear with my mess n read thru

[jonaandr]

please help to solve the problem.


Mrs Hodge bought three times as many key chains as stuffed toys and spent $124 in total. She spent $64 more on the stuffed toys than the key chains. Given that a stuffed toy cost $8.40 more than a key chain, find the cost of a key chain.

Thanks a lot!

[tianzhu]

jonaandr:

please help to solve the problem.


Mrs Hodge bought three times as many key chains as stuffed toys and spent $124 in total. She spent $64 more on the stuffed toys than the key chains. Given that a stuffed toy cost $8.40 more than a key chain, find the cost of a key chain.

Thanks a lot!

Hi

A neat way is to use MD.Due to time constraints, I shall give you only some pointers.

Amount@stuffed toys ------- 94

Amount@key chains ------- 30

Draw 3 boxes to show number of key chains.

Draw 1 box to show number of stuffed toys.

Cost of 1 unit of key chains -------30/3 ------- 10

Cost of 1 unit of key chains -------94

94 – 10 ------ 84

This extra $84 is due to the number of stuffed toys costing $8.40 more than a key chain.

Number of stuffed toys ------ 84/8.4 ------- 10

Number of key chains ------- 30

Cost per key chain ------- 30/30 ------- 1

Best wishes

[cimman]

brastilava:

Hi, please help in these 2 questions:


2) I am a 2-digit number. I am a multiple of 4. When I am divided by 9, there will be a remainder of 6. When I am divided by 7, there will be a remainder of 5. Who am I?

some terminology foundation first:
http://i45.tinypic.com/21oas2d.png\">

coming up with the equation:
http://i49.tinypic.com/dev04m.png\">

plugging in the numbers:
http://i48.tinypic.com/1zlsm88.png\">

[rajamalar]

hi thank you for your method it is fambous

[cimman]

rajamalar:

hi thank you for your method it is fambous

you're welcome.

[imacsg]

At 9am Jerome left P Town for Y Town which was 540 km away. 2 hours later, Michael left P town for Y Town. He travelled at an average speed of 40 km/h faster than Jerome and overtook him at 3 pm.

a) What was Jerome’s average speed?
b) How far was Jerome from Y town when Michael reached his destination?

Can anyone help the above question? Thanks.

[MathIzzzFun]

imacsg:

At 9am Jerome left P Town for Y Town which was 540 km away. 2 hours later, Michael left P town for Y Town. He travelled at an average speed of 40 km/h faster than Jerome and overtook him at 3 pm.

a) What was Jerome's average speed?
b) How far was Jerome from Y town when Michael reached his destination?

Can anyone help the above question? Thanks.

Michael left Town P at 11 am and overtook Jerome at 3 pm ie 4 h later
40 km/h x 4h = 160 km --> Michael travelled 160km more than Jerome in 4h -- this is the \"catch up\" distance which is the distance travelled by Jerome from 9am to 11 am (2h)
Jerome's speed = 160 km/2 h = 80 km/h

alternatively, we can also use time ratio/speed ratio
When they meet,
time taken by Michael --> 11 am to 3 pm --> 4h
time taken by Jerome --> 9 am to 3pm --> 6h
Speed ratio of Michael : Jerome = 6 : 4 --> 3u : 2u
1u --> 40km/h, Jerome's speed --> 2 x 40 km/h = 80 km/h

with this, I think you can work out part (b)


cheers.