Q&A - PSLE Math

[MathIzzzFun]

Neat:

MathIzzzFun:

Hi


Please help with these 2 questions:


2. Mr Samad made some toys to sell. He sold 20% of the toys on Monday. He sold 27 toys on Tuesday. On Wednesday he sold 10% of the remaining toys. After that he found that the toys he had left was 1 more than 1/2 the number of toys he had made. How many toys did he made?

Thank You.

Hi

Could you check the source of the question, while the answer is a whole number (115), it does not make sense because (1/2 of 115 + 1) is not a whole number.

cheers.

Hi MathIzzzFun

The answer is 115 and it is from River Valley Primary, Preliminary 2010. Paper 2, question 18.

Cheers

o..in that case, I think the original question is likely about money instead of toys...I have changed the toys to $$ below....

Mr Samad had some money. He spent 20% of the money on Monday. He spent $27 on Tuesday. On Wednesday he spent 10% of the remaining money. After that he found that the money he had left was $1 more than 1/2 the original amount he had. How much money did he have at first ?

Original amount --> 10u
Spent on Monday --> 2u, Left --> 8u
Spent on Tuesday --> $27, Left --> 8u - $27
Spent on Wednesday --> 0.8u - $2.70, Left --> 7.2u - 24.30

So, 7.2u - $24.3 = 5u + $1
1u --> 11.5
Original amount --> 10 x $11.5 = $115

cheers.

[tianzhu]

Michaelia0816:

Tianzhu, pls help,

1) Starting from the same position on a straight road,Da Wei jogging at an average speed of 1.4km/h slower than Cheng Kiat, the two boys were 6.6km apart after jogging for 45 minutes.
(a) Find the average speed of Da Wei.
(b) If Cheng Kiat continued jogging for another 30 minutes and completed his jog, find the total distance covered by him. Give your answer correct to 1 decimal place.

Hi girl

It helps to start with a simple diagram.

You may use MD or work in term of units.

1h ------ 1.4 km

45 minutes ------- 1.05 km

Cheng Kiat -------- 1 unit + 1.05

Da Wei ------- 1 unit

1 unit + 1unit + 1.05 -------- 6.6

1 unit ------- 2.775

Average speed of Da Wei ------ 2.775/0.75 -------- 3.7 km/h

Average speed of Cheng Kiat ------- 3.7 + 1.4 ------- 5.1 km/h

Total distance covered by Cheng Kiat --------- 5.1*1.25 ------- 6.4 km

Alternatively you may also use the concept of combined speed if you’ve taught in school.

Combined speed of the two boys -------- 6.6/0.75 ------- 8.8km/h

Average speed of Da Wei ------ (8.8 – 1.4)/2 -------- 3.7 km/h

Average speed of Cheng Kiat ------- 3.7 + 1.4 ------- 5.1 km/h

Total distance covered by Cheng Kiat --------- 5.1*1.25 ------- 6.4 km

Best wishes

[speedmaths.012624com]

kwcllf:

Hi please help


Three groups of pupils A,B,C worked on a project. B worked on the project during the first 4 days. A and C then took over and worked together for the next 6 days. After which, A was left to complete the remaining work in 9 days. The ratio of the amount of work done by A,B,c is 3:1:2. How long will each group take to complete the project without involvement of other groups?

Thanks

Hi,

Another possible solution can be found on page 568.
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&hilit=Beaker+A&start=5670

Cheers


speedmaths.com


.

[ozora]

Th breadth of rectangle is reduced by 20%. To maintain the same area, its length is increased . A) what is the % increase in its length?

B) what is the % increase in its perimeter given that the ratio of its original length to its original breadth is 9:5? Answer round off to hundredths.
Need some help.

[speedmaths.012624com]

ozora:

Th breadth of rectangle is reduced by 20%. To maintain the same area, its length is increased . A) what is the % increase in its length?

B) what is the % increase in its perimeter given that the ratio of its original length to its original breadth is 9:5? Answer round off to hundredths.
Need some help.

Hi,

One possible solution:

Area = Length x Breadth
Area = L x B

Let the L and B be 10 each.
(Okay, it is a square, but all squares are rectangles.)

Area = 10 x 10 = 100

Breadth reduced by 20% from 10 to 8.

(new L) x 8 = 100
(new L) = 100 / 8 = 12.5

Can you work from here?

Cheers



speedmaths.com


.

[ozora]

speedmaths.com:

ozora:

Th breadth of rectangle is reduced by 20%. To maintain the same area, its length is increased . A) what is the % increase in its length?

B) what is the % increase in its perimeter given that the ratio of its original length to its original breadth is 9:5? Answer round off to hundredths.
Need some help.

Hi,

One possible solution:

Area = Length x Breadth
Area = L x B

Let the L and B be 10 each.
(Okay, it is a square, but all squares are rectangles.)

Area = 10 x 10 = 100

Breadth reduced by 20% from 10 to 8.

(new L) x 8 = 100
(new L) = 100 / 8 = 12.5

Can you work from here?

Cheers



speedmaths.com


.

Thanks but does the 2ndpart suppose to assist in answering the first part of answer.

[YumYum]

Hi, can anyone pls help with this Qn: http://i49.tinypic.com/23uoib.jpg\">

[charsen]

http://i50.tinypic.com/1z3a4bm.jpg\">

YumYum:

Hi, can anyone pls help with this Qn: http://i49.tinypic.com/23uoib.jpg\">

Here is the solution that I have attempted.

http://i50.tinypic.com/1z3a4bm.jpg\">

The final answer for (a) is 1078m2

[speedmaths.012624com]

ozora:

speedmaths.com:

[quote=\"ozora\"]Th breadth of rectangle is reduced by 20%. To maintain the same area, its length is increased . A) what is the % increase in its length?

B) what is the % increase in its perimeter given that the ratio of its original length to its original breadth is 9:5? Answer round off to hundredths.
Need some help.

Hi,

One possible solution:

Area = Length x Breadth
Area = L x B

Let the L and B be 10 each.
(Okay, it is a square, but all squares are rectangles.)

Area = 10 x 10 = 100

Breadth reduced by 20% from 10 to 8.

(new L) x 8 = 100
(new L) = 100 / 8 = 12.5

Can you work from here?

Cheers



speedmaths.com


.

Thanks but does the 2ndpart suppose to assist in answering the first part of answer.[/quote]Hi,

Yes, students usually use the 2nd part to help answer the first part.

For the first part, when you deal with percentage change, and area (or volume), you can use the original numbers (9 and 5), or you can use your own numbers (10 and 10, which is slightly easier)

Whichever set of numbers you use, you should get the answer as 25%, for the first part.

We normally encourage students to use 2 methods, when they have the time:
One method to get the answer
Another method to check the answer.

For the second part, when you deal with perimeter, you must use the numbers (9 and 5) given in the question.
Breadth drop from 5 to 4 (drop of 20%)
Length increase from 9 to 11.25 (increase of 25%)

Old perimeter is 9 + 9 + 5 + 5 = 28
New perimeter is 11.25 + 11.25 + 4 + 4 = 30.5

Can you work on the second part from here?

Hope this helps.

Cheers


speedmaths.com


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[charsen]

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