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[small]

CJM:

the tickets for a show are priced at $22 and $8. The number of $12 tickets available is 1 1/2 times the number of $8 tickets. 2/3 of the twelve-dollar tickets and the eight-dollar tickets were sold. The ticket sales amounted to $6400. How much more would have been collected if all the tickets were sold ?

Hi CJM,

Hope this helps...

At first
$12 --- 3U
$8 --- 2U

Total tickets were sold:
$12 --- 3 x 2/3 = 2U
$8 --- 2U

Convert to price unit:
$12 --- 12 x 2 = 24U
$8 --- 8 x 2 = 16U

(24U + 16U) = 6400
40U = 6400
1U = 160

160 x $12 = $1920

$1920 more would have been collected if all the tickets were sold.

[ADoc]

andante:

ADoc:

[quote=\"andante\"]...using algebra simultaneous equations but I heard that the pupils cannot use this method to solve during PSLE exam or else marks will be deducted...

Hi,
Thank you for your reply. I hope you can help to solve the problem using model approach.

Thanks and regards,[/quote]Hi there! Hope this is useful for your explanation to your kid using models. Pls note there isn't really a need for such an \"elaborate\" model as I have done. The key is able to draw the initial model representing Kylie's Units and Pauline's in terms of Kylie. The student should be able to proceed from there.

The primary concept tested here is \"equivalence of percentage and fractions (and this allows the student to draw a model with suitable number of units for subsequent calculations).

Cheers!

http://postimage.org/image/2fqei0i3o/
[/img]

[Dharma]

CJM:

To motivate her son to study mathematics, Mrs Mani agreed to reward her son 50 cents for every problem solved correctly and to fine him 35 cents for each incorrect solution. At the end of 17 problems, neither owed anything to the other. How many problems did her son solve correctly?

Amount of reward received = Amount of fine paid

Amount = Qty x Rate
For amounts to be the same , quantity and rates must be inversely proportional to each other.

Ratio of rates => Correct: Wrong= 50 : 35 = 10 : 7
Ratio of quantity => Correct : Wrong = 7 : 10

No. Of problems solved correctly and was rewarded = 7

[andante]

ADoc:

andante:

[quote=\"ADoc\"]

Hi,
Thank you for your reply. I hope you can help to solve the problem using model approach.

Thanks and regards,

Hi there! Hope this is useful for your explanation to your kid using models. Pls note there isn't really a need for such an \"elaborate\" model as I have done. The key is able to draw the initial model representing Kylie's Units and Pauline's in terms of Kylie. The student should be able to proceed from there.

The primary concept tested here is \"equivalence of percentage and fractions (and this allows the student to draw a model with suitable number of units for subsequent calculations).

Cheers!

http://postimage.org/image/2fqei0i3o/
[/img][/quote]Thank you very much.

[meimeitan]

hi


Then, may I know answer for that question:

There were 12 more girls than boys in a club. 1/3 of the girls and 1/4 of the boys took part in a competition. Among those who took part in the competition, there were 6 more girls than boys. What fraction of the club members who did not take part in the competition were boys?

Thanks.

[atutor2001]

meimeitan:

hi


Then, may I know answer for that question:

There were 12 more girls than boys in a club. 1/3 of the girls and 1/4 of the boys took part in a competition. Among those who took part in the competition, there were 6 more girls than boys. What fraction of the club members who did not take part in the competition were boys?

Thanks.

Let girls be 3U
Girls taking part in competition will be 1U

Let boys be 4V
Boys taking part in competition will be 1V

There were 12 more girls than boys in a club: 3U = 4V + 12 ----> (1)
Those who took part in the competition, there were 6 more girls than boys : 1U = 1V + 6 ----> (2)

(2)x 3 : 3U=3V+18 ----> (3)

Put (3) into (1) : 3V+18 = 4V+12 --> 1V = 6

From (2) : 1U = 6+6 = 12

No of boys who did not take part : 3V = 3x6 = 18
Total club members = 3U + 4V = 3x12 + 4x6 = 36+24 = 60

Fraction of the club members who did not take part in the competition were boys : 18/60 = 3/10

[CJM]

atutor2001:

small:

[quote=\"CJM\"]To motivate her son to study mathematics, Mrs Mani agreed to reward her son 50 cents for every problem solved correctly and to fine him 35 cents for each incorrect solution. At the end of 17 problems, neither owed anything to the other. How many problems did her son solve correctly?

Please try to use guess and check method.

From working,her son solved 7 problemd correctly.

I recommend \"replacement method\". This is a 3 mark question and guess and check might take too much time.

If the son gets all correct : 17 x 50 = 850
For every 1 wrong, the son must return 50 + fine 35 = 85

At the end, the son gets nothing means he return 85 cents 10 times. (850/85 = 10) That is, he got 10 wrong. Therefore, no. of correct is 7.[/quote]thanks so much ... my ds' answer is correct : ) i m not sure how to do the sums so consulted you all, the math expert to get the answer

[CJM]

small:

CJM:

To motivate her son to study mathematics, Mrs Mani agreed to reward her son 50 cents for every problem solved correctly and to fine him 35 cents for each incorrect solution. At the end of 17 problems, neither owed anything to the other. How many problems did her son solve correctly?

Please try to use guess and check method.

From working,her son solved 7 problemd correctly.

thanks : )
ds uses the guess n check method :

correct / incorrect / Total / Check
8 9 0.85 cross

7 10 0 tick

the answer is 7

[kwcllf]

atutor2001:

meimeitan:

hi


Then, may I know answer for that question:

There were 12 more girls than boys in a club. 1/3 of the girls and 1/4 of the boys took part in a competition. Among those who took part in the competition, there were 6 more girls than boys. What fraction of the club members who did not take part in the competition were boys?

Thanks.

Let girls be 3U
Girls taking part in competition will be 1U

Let boys be 4V
Boys taking part in competition will be 1V

There were 12 more girls than boys in a club: 3U = 4V + 12 ----> (1)
Those who took part in the competition, there were 6 more girls than boys : 1U = 1V + 6 ----> (2)

(2)x 3 : 3U=3V+18 ----> (3)

Put (3) into (1) : 3V+18 = 4V+12 --> 1V = 6

From (2) : 1U = 6+6 = 12

No of boys who did not take part : 3V = 3x6 = 18
Total club members = 3U + 4V = 3x12 + 4x6 = 36+24 = 60

Fraction of the club members who did not take part in the competition were boys : 18/60 = 3/10

G [][][][][][][][][][][][][4][4][4]
B [][][][][][][][][][][][]

Convert both fractions to a common denominator.
Girls: 1/3 change to 4/12
Boys: 1/4 change to 3/12

The 12 extra Girls, divide them into 3 parts of [4]

From the above model:

1/3 Girls = 4 [] + 4

1/4 Boys = 3 []

Since there are 6 more Girl than Boys who took part in the competition.

1 [] = 6 – 4 = 2


Total number of [] = 12 x 2 = 24

Total number of Boys and Girls = 24 x 2 + 12 = 60

Therefore,
Number of Boys that did not take part in the competition = 9 x 2 = 18
Fraction of Boys that did not take part in the competition = 18/60 = 3/10

[Muffins]

meimeitan:

hi


Then, may I know answer for that question:

There were 12 more girls than boys in a club. 1/3 of the girls and 1/4 of the boys took part in a competition. Among those who took part in the competition, there were 6 more girls than boys. What fraction of the club members who did not take part in the competition were boys?

Thanks.

Let the boys be ϰ, and the girls be ϰ + 6

So ϰ X 4 = 4ϰ (total number of boys in the club)

(ϰ+6) X 3 = 3ϰ + 18 (total number of girls)

3ϰ + 18 = 4ϰ + 12

18 = ϰ + 12

∴ ϰ = 6

Total number of members who did not take part in the competition: 5ϰ + 12

Boys who did not take part: 3ϰ

(5ϰ+12) - 3ϰ = 2ϰ + 12

6 X 3 = 18

6 X 5 = 30

30 + 12 = 42

18/42 = 3/7

Hmm.... My working seems correct. However, you guys' answers are 3/10 though... any ideas or advice?

P.S.
Total number of boys in club = 24
Total number of girls in club = 36