Tutor MathsGuru: Ask me for your burning Maths questions!

[Brenda10]

Hifive:

Brenda10:

[quote=\"Hifive\"]Hi everyone


Need help on the following questions:

1) A class was fixing a puzzle of their photos. In the first hour, the ratio of the number of fixed pieces to unfixed pieces is 1:5. In the next hour, they fixed another 15 pieces of the puzzle. Finally, the number of fixed pieces became 60% of the unfixed pieces. How many pieces were there in the puzzle?


Source : Catholic High SA2 2009 P5

Thanks in advance.

Working:
60/100 = 3/5


At First
Fixed unfixed
1 5
x8 x8
8 40

At End
Fixed Unfixed
3 5
x6 x6
18 30

18 - 8 = 10 units and 40 - 30 = 10 units


10 units = 15
1 unit = 1.5
48 units = 48 x 1.5 = 72

Thank you.

Hi Brenda 10

The method you use is different from VC's mum but both got the same answer. I am trying to study the 2 methods and explain both ways to my DS. Could you pls explain why you multiply \"at first\" by 8 and \"at end\" by 6?

Thanks in advance.[/quote]Hi Hifive

Since the total number of puzzles remains the same, we must make the total numbers of puzzle the same by using common multiple. As such the \"At First\" and \"At End\" is 48 units.

Note: 6 is taken from At First : 1 + 5
8 is taken from At End : 3 + 5

BTW, I find VC's mum method easier, you may want to consider using this after her explanation.

Thank you.

[Vanilla Cake]

Brenda10:

Hi Hifive


Since the total number of puzzles remains the same, we must make the total numbers of puzzle the same by using common multiple. As such the \"At First\" and \"At End\" is 48 units.

Note: 6 is taken from At First : 1 + 5
8 is taken from At End : 3 + 5

BTW, I find VC's mum method easier, you may want to consider using this after her explanation.

Thank you.

Hi Brenda10,

Thks for helping to explain to Hifive, my mum and I also have difficulties in understanding my younger sister's P5 Maths workings. Too many shortcuts and will eventually lead to her downfall in 2011 PSLE Maths exams .

[Brenda10]

Vanilla Cake:

Brenda10:

Hi Hifive


Since the total number of puzzles remains the same, we must make the total numbers of puzzle the same by using common multiple. As such the \"At First\" and \"At End\" is 48 units.

Note: 6 is taken from At First : 1 + 5
8 is taken from At End : 3 + 5

BTW, I find VC's mum method easier, you may want to consider using this after her explanation.

Thank you.

Hi Brenda10,

Thks for helping to explain to Hifive, my mum and I also have difficulties in understanding my younger sister's P5 Maths workings. Too many shortcuts and will eventually lead to her downfall in 2011 PSLE Maths exams .

Hi VC

You're welcome.

Same here, my dd also likes to solve the maths in her own method therefore I also have same concern. Sometime I will put up her working and seek for advice if I have doubt.

Let's hope that by next year they will be more mature and everything is in order and they do well in the PSLE .

Cheers :celebrate:

[Hifive]

Hi Brenda 10


Thanks for your explanation. Now can understand better.

Cheers.

[Brenda10]

Hifive:

Hi Brenda 10


Thanks for your explanation. Now can understand better.

Cheers.

Hi Hifive

You're welcome and I wish your DS good luck and all the best in the coming PSLE.
:celebrate:

[Hifive]

Hi Brenda 10


My DS is in P5 this year. Will chat with you more cos’ I read from the forum that your DD is also P5 this year.

Cheers

[Brenda10]

Hifive:

Hi Brenda 10


My DS is in P5 this year. Will chat with you more cos' I read from the forum that your DD is also P5 this year.

Cheers

Hi Hifive

Sorry as I'm a little bit blur this morning. :oops:

I'm Sure we will have more to chat especially next year.

Cheers

[trytry]

I need help in 2 average questions below:


Q1. 20 students took a test. The teacher ranked them from the highest score (1st) to the lowest score (20th). The top 10 students had an average of 92 marks and the next 10 students had an average of 80 marks. Also, the top 5 students had an average score that was 12 marks higher than the average of next 15 students. Find the lowest possible score of the student in the 20th rank. \t\t\t\t\t\t\t
\t\t
2. There are 5 members in John's family : Father, Mother, elder sister Helen, John and younger sister Lucy. Just before Lucy was borned, the average age of the family was 19 yrs old and the age of John's father is same as the sum of ages of the rest of family. 2 years ago, the average age of the family was 35.2 yrs and the age of John's mother was 8 years more than the sum of ages of John and Lucy. The current average age of the family, excluding Helen, is 39.5 years. Find the present age of John.


Thanks.

[Vanilla Cake]

trytry:

I need help in 2 average questions below:

Your 2 questions are from National Mathematical Olympiad of Singapore - Special Round set on 31 July 2010.
Q1 is from Q20 [5 marks]
Q2 is from Q16 [5 marks]

trytry:

Q1. 20 students took a test. The teacher ranked them from the highest score (1st) to the lowest score (20th). The top 10 students had an average of 92 marks and the next 10 students had an average of 80 marks. Also, the top 5 students had an average score that was 12 marks higher than the average of next 15 students. Find the lowest possible score of the student in the 20th rank

Refer to http://psle2010a.blogspot.com/2010/08/nmos-q20.html.
\t\t\t\t\t\t\t

trytry:

2. There are 5 members in John's family : Father, Mother, elder sister Helen, John and younger sister Lucy. Just before Lucy was borned, the average age of the family was 19 yrs old and the age of John's father is same as the sum of ages of the rest of family. 2 years ago, the average age of the family was 35.2 yrs and the age of John's mother was 8 years more than the sum of ages of John and Lucy. The current average age of the family, excluding Helen, is 39.5 years. Find the present age of John.

Refer to http://psle2010a.blogspot.com/2010/08/nmos-2010-q16.html

while waiting for KSP members to respond.

VC's mum

[atutor2001]

trytry:

I need help in 2 average questions below:


Q1. 20 students took a test. The teacher ranked them from the highest score (1st) to the lowest score (20th). The top 10 students had an average of 92 marks and the next 10 students had an average of 80 marks. Also, the top 5 students had an average score that was 12 marks higher than the average of next 15 students. Find the lowest possible score of the student in the 20th rank. \t\t\t\t\t\t\t
.

The solution can be broken down into 2 parts :

Part 1

Through normal calculation (model preferred) we can find the average score of the following bands :

1st to 5th : Average = 95
6th to 10th : Average = 89
11th to 20th : Average = 80

Part 2

Concept :

a) If there are odd number of consecutive numbers, the average will be the middle score.
b) If we want the last score to be smallest we must make the front score as big as possible.

Applying concept (a) to band (1st - 5th position) starting with the middle position i.e. 3rd position being equal to the average i.e. 95, we can build the score of the other positions

3rd(95)
2nd(96) 3rd(95) 4th(94)
1st(97) 2nd(96) 3rd(95) 4th(94) 5th(95)

From concept (b), we would like the 5th score to be as high as possible but that is not possible because the question stated \"no 2 scores are the same\". So the highest possible score for the 5th is 95.

Similarly for band (6th - 10th position), we get

8th(89)
7th(90) 8th(89) 9th(88.)
6th(91) 7th(90) 8th(89) 9th(88.) 10th(87)

Similarly, the biggest possible score for the 10th position is 87

For the last band (11th - 20th position), we want the front position to be as big as possible. But the biggest for position 11th must be less than 87 i.e. 11th position is 86. We then built downwards until position 19th

11th(86) 12th(85) ........ 18th(79) 19th(78.)

The total score of 11th to 20th = 10 x 80 = 800

The total score of 11th to 19th position = 86+85+...+79+78= 738

The smallest 20th score = 800 - 738 =62