Q&A - PSLE Math

[Drdj]

Hi, could you explain this Andrew Er Assessment Book question to me?


THe question is $1,120 was spent on pots and pans. 1 Pan is $20, 1 Pot is $60. 8 more pans than pots were bought. How many pots and pans were bought?

Using algebra, and solving 2 simultaneous equtions - 20x + 60 y = 1,120
and x -y = 8 (where x = no. of pots), I get y = 12, and x = 20 (Which is the book's answer)

Using this method,
Difference in cost of pans = $20 x 8 pans = $160
1 unit = cost of pots
Hence, 1 unit + $160 = cost of pans
Hence 1 unit = cost of pots = ($1,120 - $160) / 2 = $480

No. of pots = $480/$60 = 8, No. of pans = $640/$20=32
But this is wrong (since difference is 24, not 8). Why?

Thanks in advance.

tianzhu:

Denise:

A seamstress used up 64m of lace in 10 days. Everyday the length of lace used is increased by two times, compared to the before. How many days did she take to use a length of 1/4m

Hi, you may wish to check on this sentence again.

Everyday the length of lace used is increased by two times, compared to the before.

Is it more appropriate to replace by with to?

Everyday the length of lace used is increased to two times, compared to the before.

[mrswongtuition]

Drdj:

Hi, could you explain this Andrew Er Assessment Book question to me?


THe question is $1,120 was spent on pots and pans. 1 Pan is $20, 1 Pot is $60. 8 more pans than pots were bought. How many pots and pans were bought?

Using algebra, and solving 2 simultaneous equtions - 20x + 60 y = 1,120
and x -y = 8 (where x = no. of pots), I get y = 12, and x = 20 (Which is the book's answer)

Using this method,
Difference in cost of pans = $20 x 8 pans = $160
1 unit = cost of pots
Hence, 1 unit + $160 = cost of pans
Hence 1 unit = cost of pots = ($1,120 - $160) / 2 = $480

No. of pots = $480/$60 = 8, No. of pans = $640/$20=32
But this is wrong (since difference is 24, not 8). Why?

Thanks in advance.

This is how I would teach my students (Sorry if it is confusing as I do not know how to explain by typing):

Information from question:
Total cost of Pots & Pans = $1120
1 Pan = $20
1 Pot = $60
There are 8 more pans than pots.

Therefore, by number of pots and pans:
Pot !________?_________!
Pan !________?_________!_____8_____!

If we take away the cost of 8 pans, we will get the cost of equal number of pots and pans.

By cost, $20 x 8 = $160, $160 was spent on the 8 more pans bought.
$1120 - $160 = $960 (amount spent on equal number of pots and pans)

By grouping of costs,
1 pot + 1 pan = $20 + $60 = $80

$960/$80 = 12
There are 12 sets.

Therefore,
Pots = 12
Pans = 12 + 8 = 20

To double check that the answer is correct:
12 x $60 = 720 (Cost of pots)
20 x $20 = 400 (Cost of pans)
$720 + $400 = $1120 (Total cost of pots and pans)

[suiyuan]

Hi, Mrs Wong


How would you teach your students to solve the following problems?(if feasible by Model Drawing)

1) There are 32 rooms for 71 pupils at a student hostel. At the boys’ side, each room is shared by 2 boys. At the girls’ side, each room is shared by 3 girls. If each room is fully occupied by either 2 boys or by 3 girls, find the number of boys and girls respectively.

2) Alan has 1.5 times as many marbles as Bernard while Carl has 2.5 times as many marbles as Bernard. During a game, both Alan and Carl lost to Bernard and for every 1 marble that Alan lost to Bernard; Carl lost 2 marbles to Bernard. At the end, Alan has 21 marbles left, while Carl has 27 marbles left. How many marbles did Alan lose to Bernard?

3) When Jenny was 40 years old, her son was twice her daughter’s age. Jenny will be twice her son’s age when her daughter is 28 years old. How old will Jenny be when her daughter is 20 years old?

Thank you for your help.

[mrswongtuition]

Hi Suiyuan,


For Question 1, I would use Guess & Check:
Information from question:
32 rooms (all full)
71 pupils total
Each room either 2 boys or 3 girls

Rooms for Girls/\tNo. of Girls/\tRoom for Boys/\tNo. of Boys/\tTotal Rooms/\tTotal Pupils/\tCheck
16\t48\t16\t32\t32\t80\tNO
10\t30\t22\t44\t32\t74\tNO

Based on the above,
For every decrease in 6 rooms for girls, there is a decrease in 6 pupils, therefore, for every decrease in 1 room for girls, there is a decrease in 1 pupil in total.

We need 71 pupils in total, therefore total pupils need to decrease by 3 pupils, which means that we need to decrease the number of rooms for girls by 3 rooms.

Then you will get:
Rooms for Girls/\tNo. of Girls/\tRoom for Boys/\tNo. of Boys/\tTotal Rooms/\tTotal Pupils/\tCheck
7\t21\t25\t50\t32\t72\tYES

The answer is:
21 girls
50 boys

*This is how I would teach Guess & Check method so that students do not waste time trying all the possibilities but just try 2 and zoom in directly on the answer.

(Sorry I do not know how to post table here)

[suiyuan]

Hi, Mrs Wong.


Thank you for your quick response.

For Q1,besides G and C, is there any alternative way?

You may post images by following the steps in these links.
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=2699&start=15
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=3767&start=30

[suiyuan]

[quote][IMG]http://i697.photobucket.com/albums/vv338/mrsw[/quote]
Hi Mrs Wong

You may wish to repost the image again.The link doesn't seem to work.
Alternatively, you may click on URL.

[Tang]

suiyuan:

Hi, Mrs Wong


How would you teach your students to solve the following problems?(if feasible by Model Drawing)

1) There are 32 rooms for 71 pupils at a student hostel. At the boys’ side, each room is shared by 2 boys. At the girls’ side, each room is shared by 3 girls. If each room is fully occupied by either 2 boys or by 3 girls, find the number of boys and girls respectively.

2) Alan has 1.5 times as many marbles as Bernard while Carl has 2.5 times as many marbles as Bernard. During a game, both Alan and Carl lost to Bernard and for every 1 marble that Alan lost to Bernard; Carl lost 2 marbles to Bernard. At the end, Alan has 21 marbles left, while Carl has 27 marbles left. How many marbles did Alan lose to Bernard?

3) When Jenny was 40 years old, her son was twice her daughter’s age. Jenny will be twice her son’s age when her daughter is 28 years old. How old will Jenny be when her daughter is 20 years old?

Thank you for your help.

1.
G + B = 71 [x 2]
2G + 2 B = 2 x 71 = 142

G/3 + B/2 = 32 [x 6]
2 G + 3 B = 6 x 32 = 192

B = 192 - 142 = 50 boys

G = 71 - 50 = 21 girls

Or
Suppose all boys, 32 x 2 = 64
71 - 64 = 7 [short fall]
3 - 2 = 1 [Difference in number]
7 / 1 = 7 Rooms for girls

32 - 7 = 25 Rooms for boys

7 x 3 = 21 girls
25 x 2 = 50 boys


2.

A : B : C = 3 : 2 : 5
Change ratio -1 : +3 : -2
AN = 21 and CN = 27

3 Units - Part = 21 [x 2]
6 Units - 2 Parts = 2 x 21 = 42

5 Units - 2 Parts = 27

Unit = 42 - 27 = 15

Part = 3 x 15 - 21 = 24 marbles.


3. Refer to Uncle Observer's Blog - Maths-Whole Number for Model Drawing and answer.

[suiyuan]

suiyuan:

[quote][IMG]http://i697.photobucket.com/albums/vv338/mrsw

Hi Mrs Wong

You may wish to repost the image again.The link doesn't seem to work.
Alternatively, you may click on URL.[/quote]Hi Mrs Wong

I am sorry to inconvenience you again.Could you help to repost the image as that particular link doesn't seem to work.

[Drdj]

Dear all,


Please help me to solve this sum - from CHIJ Prelim 2008
Thanks in advance

http://www.postimage.org/

[AnnT2009]

Please help,


Q1 Mary and Jason each have some money. If Mary spent$80 per day and Jason spent $40 per day. Mary will have $500 left when Jason has spent all his money. If Mary spent $40 per day and Jason spent $80 per day. Mary will have $1100 left when Jason spent all his money. What is the amount of money Jason has?
Ans:$400

Q2 Mr Tan bought three times as many badges as toy cars and spent $144 in total. He spent $84 more on toy cars than on badges. Given that a toy car cost $10.40 more than a badge. What is the cost of a badge?
Ans:$1

TIA.