Tutor MathsGuru: Ask me for your burning Maths questions!

[CoffeeCat]

Vanilla Cake:

Pls help to provide worked solutions and thanks in advance.

Answer given: m=1

http://postimage.org/image/2uvidl31g/

This kinda puzzle, the concept is similar to sudoku, if you or your sister has played before.
Simply put, in solving such puzzles, need to look for an opening.
Here it is at row2 column 3 (the left of m), it is \"under the influence of\" or affected by numbers 1, 2, 3 and 4, thus leaving 5 as the only viable option left.
After this, i think there might be other ways to proceed. One is again look for another square that is affected by 4 other numbers, example row 2 column1, which is 3.
If without any previous sudoku experience, can link this concept to chess, like \"under attack\" from rook or bishop.

[CoffeeCat]

Vanilla Cake:

Five classes A,B,C,D and E took part in an international chess competition.
Each class sent in 2 participants.
The rules of the competition were:
(a) Participants do not compete with each other for more than 1 game,
(b) Participants from the same class cannot compete with each other.
After a few rounds of competitions, it turned out that all had completed different number of games, except for a participant from class A.
How many games had the 2 participants from class A completed?

I think you or your sister might have done \"similar\" but easier questions.

This question is like a puzzle, requires some drawing. First Draw 10 dots to represent the 10 person (called 1, 2, .., 10, where 1,2 from same class, ..., 9 & 10 from same class). You can draw them in the form of 10 vertices of a 10-gon, where 2 dots (eg 1 and 2) are horizontally on the same level, and there are 5 levels (5 classes).
Notice that the pairs are not identified with anyclass yet.



The clue is : all had completed a different number of games.
So the maximum is 8, minimun 0, 9 possible different numbers, indeed there must exist one who has the same number as some other person.
Now we construct the players who completed 8 and 0 games.
Take person 1 and 2. 1 must play with the other 8 but can't play with 2. So you connect dot 1 with dot 3 4 5 6 7 8 9 10.
Now out of the 10 dots, everyone has played 1 game except for 2, so it will be the player with 0 game.
Now we shall construct players who completed 7 and 1.
take person 3 and 4. 4 has played only 1 game so far, we will let it be the player with only 1 game. So connect 3 with 5, 6, 7, 8 , 9, 10.
So you continue this process, while taking the next pair from the same class, leaving 1 untouched and the other playing more games.
Eventually you should reach 1 pair (the last pair) who both completed 4 games.

Notice that the question doesn't specify which of the 2 participants from class A, because it doesn't and shouldn't made any difference in this case.

[CoffeeCat]

Chan09:

Need help


2) a figure is made up of 2 identical squares. 1/6 of each square is shaded. What fraction of the whole figure is shaded?
Ans: 1/11

Thank you :?

This is a funny question. I suspect there is a diagram involved. The fact that the ans is not 1/6 shows that there is overlapping between the 2 identical squares. In fact, it might be the case that the overlapping part is shaded, and if it represents 1 unit, then the total area of the figure will be 5+5+1 =11 units.

[Jcong]

In 2008, there were 883 pupils in Greenwood Primary School and 1/3 of them were girls. After merging with Blackforest Girls’ School in 2009, 3/4 of the pupils in the merged school are now girls.


How many girls were from Blackforest Girls School?

[MathIzzzFun]

Jcong:

In 2008, there were 883 pupils in Greenwood Primary School and 1/3 of them were girls. After merging with Blackforest Girls' School in 2009, 3/4 of the pupils in the merged school are now girls.


How many girls were from Blackforest Girls School?

Hi Jcong,

883 is not divisible by 3, is it 882?

Find the number of boys in Greenwood Primary. Since Blackforest is a girl school, the number of boys will not change, after the merger. So number of boys = 1/4 total number of students in merged school

After that, find the number of girls and then subtract the number of girls originally from Greenwood to get the original number of girsl in Blackforest Girls

cheers.

[Bunny27]

Need help for this question:


1. Machine A can make 240 cookies in 5 minutes and Machine B can make 128 cookies in 4 minutes. How long will it take to make 2000 cookies if both machines are operating at the same time ?

Thanks

mathsguru:

Original Title: Let MathsGuru Answer Your Burning


Maths Questions!

:welcome:

Dear Parents,

Are you frustrated/stuck when helping your child solve his/her Maths questions? Are you inclined to use Algebra most of the time? Do you have difficulty trying to use diagrams or other heuristic methods (that Primary School students learn) to solve?

:idea: Post your questions here and see how MathsGuru solve them to the best of her ability. Detailed solutions will be posted back in this thread.

So start asking and watch this space!!

Cheers :celebrate: ,
MathsGuru

P/S (Disclaimer, in case you're wondering...):
Although MathsGuru is a full-time Maths tutor, this thread is meant to be an absolutely free resource for parents (or even children) with no strings attached. Just someone who's passionate about Maths and wanna spread the fun in learning Maths with others.

[CoffeeCat]

Bunny27:

Need help for this question:


1. Machine A can make 240 cookies in 5 minutes and Machine B can make 128 cookies in 4 minutes. How long will it take to make 2000 cookies if both machines are operating at the same time ?

Thanks

For such rate questions involving 2 or more workers, 1 generic method is just to find the combined rate.
Rate of Machine A : 48 cookies in 1 min
Rate of Machine B : 32 cookies in 1 min
Combined rate of A & B : 80 cookies in 1 min
Hence time taken : 2000 / 80 = 25 mins

[xuanwilliam]

avell:

Dear all


Pls help me for the following questions:-

Qn (1)
4/5 of John's money is equal to 3/8 of Sarah's money. Express Sarah's money as a ratio of John's money in it simplest form (draw a model to help you to solve the sum.

Qn (2)
The ratio of boys and girls on a school bus was 2 : 3. When 16 girls alighted from the bus, the ratio of the remaining boys to girls became 6:5. Find the number of children who remained on the bus.

Qn (1)
Make the Numerator of the fractions the same : 4/5*3=12/15
3/8*4=12/32

So the denominator is the total amount for each person.
The ratio would be
J:S
15:32
Qn(2)
Before
B:G
2:3

After
B:G
6:5

The number of boys did not change so....
Before
B:G
2:3
*6 *6
12:18

After
B:G
6:5
*2 *2
12:10

18-10=8
8u = 16
1u = 2
12+10=22
22u = 44

Number of Children remained : 44

[Bunny27]

Thank you!


uote=\"CoffeeCat\"]

Bunny27:

Need help for this question:

1. Machine A can make 240 cookies in 5 minutes and Machine B can make 128 cookies in 4 minutes. How long will it take to make 2000 cookies if both machines are operating at the same time ?

Thanks

For such rate questions involving 2 or more workers, 1 generic method is just to find the combined rate.
Rate of Machine A : 48 cookies in 1 min
Rate of Machine B : 32 cookies in 1 min
Combined rate of A & B : 80 cookies in 1 min
Hence time taken : 2000 / 80 = 25 mins[/quote]

[Musicstar]

Dear all,


Please help me these question. Thank you.

1) In May, the ratio of Karen’s weight to Huiyu’s weight was 6:5. In June , Huiyu put on 5kg and Karen lost 3kg. As a result, both of them had the same weight. What was Huiyu’s weight in May?

2)The ratio of the number of Sally’s stickers to the number of Willy’s stickers was 4:1. After Sally bought another 13 stickers and Willy received 7 more stickers from his teacher, the ratio of Sally’s stickers to Willy’s stickers became 3:1 . How many stickers did Sally have at first?

3) Peter, Ali and Samy shared some marbles in the ratio of 2:4:7. They decided to play some games with their marbles in the playground. In the first game, Samy lost 8 marbles to Peter. Peter then had the same number of marbles as Ali. In the second game, Samy won some marbles from Ali but Ali lost some of his marbles in the tall grass and could not find them. The new ratio of marbles among the three boys was
Peter : Ali : Samy
2 : 1 : 3

(a) How many marbles did Peter have after the 1st game?
(b) How many more marbles does Samy have then Ali after 2nd game?