Q&A - PSLE Math

[jieheng]

Q1


a)

The difference between the sweets = 380 - 76 = 304

Extra sweets for each pupil = 15 - 7 = 8

No. of pupils = 304 / 8 = 38 (Ans)

b)


No. of sweets that Miss Lam has = 38 * 7 + 380 = 646 (Ans)

OR

No. of sweets that Miss Lam has = 38 * 15 + 76 = 646 (Ans)

[jieheng]

Q2


One set of equal no of oranges and pears = 7 * 5 = 35

Cost of 35 oranges = 35 / 7 * 2 = $ 10

Cost of 35 pears = 35 / 5 * 3 = $ 21

The difference between one set of fruits = 21 - 10 = $ 11

Given she paid $ 33 more for pears than oranges

No. of sets of fruits = 33 / 11 = 3

a) She paid = 3 * ( 10 + 21) = $ 93 (Ans)

b) No. of oranges and pears she bought = 3 * (35 + 35) = 210 (Ans)

[shurley197323]

Thanks jieheng. Don’t mind 1 more question.

2/5 of the counters in a box were red and tge rest were blue.
After putting 48 blue counters in the box, 3/4 of the counters were blue. How many counters were in the box at first?

[jieheng]

At First

Red Blue
2u 3u

+ 48counters

New Ratio 1u 3u
Change to 2u 6u
(There is no change in the no. of Red counters)


Different in the ratio of after and before for blue counters = 6u - 3u =3u
3u----->48

1u----->16

No. of counters at first = 5 u = 5 * 16 = 80 (Ans)

[shurley197323]

jieheng:

At First

Red Blue
2u 3u

+ 48counters

New Ratio 1u 3u
Change to 2u 6u
(There is no change in the no. of Red counters)


Different in the ratio of after and before for blue counters = 6u - 3u =3u
3u----->48

1u----->16

No. of counters at first = 5 u = 5 * 16 = 80 (Ans)

Hi.sorry .My kid had not learn ratio yet. Got other solution?

[jieheng]

At First ,


Red [][]

Blue [][][]

After ,

Red [][]

Blue [][][] + 48

Red is 1/4 , Blue is 3/4

Red * 3 = Blue

[][] [][] [][] = [][][] + 48

[][][] = 48

[] = 16

At First ,

There are [][][]][] = 5 * 16 = 80 counters (Ans)





[/img]

[shurley197323]

Thanks, thnaks jieheng, hope you won’t feel frustrated to answer so many questions.

[jieheng]

shurley197323:

Thanks, thnaks jieheng, hope you won't feel frustrated to answer so many questions.

Hi ,

You are welcome.

No problem.

Regards,

[tianzhu]

Hi Lynn2


Ok, now the solutions to one of your earlier “coins questions”.

Usually, I solve it making some assumption.Pretend that the number of $1 coins is equal to the number of 50 cents coins, you'll have (17*1) + 15.50 more of value in Box B.

An alternative solution is MD. It hinges on “number and value” of the different coins.
Another way is to use the “Number* Value Method by working in tabulated form. This is a tweaked version of the usual “Number*Value’ table.

Hope this helps.

Best wishes

http://farm6.static.flickr.com/5014/5557734869_1bdc3c4e23_z.jpg\">

http://farm6.static.flickr.com/5062/5557734951_9232b0d90a_z.jpg\">

[Lynn2]

Hi Tianzhu and all,


Thanks for all the helps.I will walk through with all the various way that you have provided and sure hope can clear that block in her mind.

Will post again should I failed to accomplish the mission...

Cheers
Lynn2

tianzhu:

Hi Lynn2

Ok, now the solutions to one of your earlier “coins questions”.

Usually, I solve it making some assumption.

An alternative solution is MD. It hinges on “number and value” of the different coins.
Another way is to use the “Number* Value Method by working in tabulated form. This is a tweaked version of the usual “Number*Value’ table.

Hope this helps.

Best wishes

http://farm6.static.flickr.com/5014/5557734869_1bdc3c4e23_z.jpg\">