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[blitz]

Belle2011:

blitz:

Help please.

Tracy counted the number of 10cents coins, 20cent coins and 50cents coins in her piggy bank. The number of 10c coins was 24 more than the number of 50c coins. After spending 3/5 of the 10c coins, 1/4 of the 20c coins and 18 50c coins, she had an equal number of 10c and 20c coins. The number of 50c coins now formed 40% of the remaining number of coins.

(a) How many 10c coins did Tracy have at first?
(b) How much did she spend in all?

I use model to solve:
So you would see that 2 portions of 10c coins the same as 3 parts of 20c coins. To make them equal, you will get
10c coins will have 15u,
20c coins will have 8u and
50c coins will have 15u-24.

After spending, the leftover
10c coins will be 6u,
20c coins will be 6u and
50c coins will be 15u-24-18 = 15u-42.

Since (15u-42)/(6u + 6u + (15u-42)) = 40% = 2/5,
u =6

So there are 15u = 15*6 = 90 10c coins.

You will be able to solve part b too.

I am sorry, I still dont understand. How did 2 units of 10c become 3 parts of 20c?

[Belle2011]

Blitz,

Sorry I dont know how to draw model here.
After spending…, she has equal number of 10c coins and 20c coins left.
How many 10c coins left? Since she spent 3/5, she would have 2/5 left.
So there are 2 units of 10c.
How many 20c coins left? Since she spent 1/4, she would have 3/4 left.
So there are 3 parts of 20c.

[blitz]

Thank you for trying to explain, but I am really dumb…I still dont understand.

[Superbugs]

Pls help on this question:


There are some marbles and some boxes. If one marble is put in each box, one marble will be left without a box. If two marbles are put in each box, one box will remain empty. How many marbles and how many boxes are there?

From the question, I deduce that when 2 marbles are put in a box, there will be shortage of 2 marbles. Hence using multiples

1. Multiples of 1 - 1,2,3,4 etc
2. Excess of 1 - 2,3,4,5 etc
3. Multiples of 2 - 2,4,6,8 etc
4. Shortage of 2 - 0,2,4,6 etc
Hence my answer is 2 marbles and 1 box.
If I use excess and shortage mtd, I also got the same answers. However, there seems to be more than 1 answer. How come? Which is the correct mtd to use and what is the answers?



Kindly advise if

[chrisho]

Can u pls help me eith this.


A group if children collected some seashells. If each child receives 8 seashells, they would need 6 more seashells. If each child receives 6 seashells, they would have extra 4 seashells. How many children were there?

[MathIzzzFun]

Superbugs:

Pls help on this question:


There are some marbles and some boxes. If one marble is put in each box, one marble will be left without a box. If two marbles are put in each box, one box will remain empty. How many marbles and how many boxes are there?

From the question, I deduce that when 2 marbles are put in a box, there will be shortage of 2 marbles. Hence using multiples

1. Multiples of 1 - 1,2,3,4 etc
2. Excess of 1 - 2,3,4,5 etc
3. Multiples of 2 - 2,4,6,8 etc
4. Shortage of 2 - 0,2,4,6 etc
Hence my answer is 2 marbles and 1 box.
If I use excess and shortage mtd, I also got the same answers. However, there seems to be more than 1 answer. How come? Which is the correct mtd to use and what is the answers?



Kindly advise if

Hi

when using the listing method for this question, need to note that the number of boxes is the same. So, in looking for a \"match\" - both the number of boxes and total must match - in this case, answer is 3 boxes & 4 marbles.

cheers.

[MathIzzzFun]

chrisho:

Can u pls help me eith this.


A group if children collected some seashells. If each child receives 8 seashells, they would need 6 more seashells. If each child receives 6 seashells, they would have extra 4 seashells. How many children were there?

Hi

after each child receives 6 seashells, 4 is left.
In order to give 2 more to each child (ie so that each child has with the remaining 4 seashells, we need 6 more seashells ie total 10 is needed so that 2 more can be given to each child -> number of children = 10/2 = 5

number of seashells = 5 x 6 + 4 = 34

5 children and 34 seashells.

alternatively, the listing method can also be used.

cheers.

[blitz]

Belle2011:

blitz:

Help please.

Tracy counted the number of 10cents coins, 20cent coins and 50cents coins in her piggy bank. The number of 10c coins was 24 more than the number of 50c coins. After spending 3/5 of the 10c coins, 1/4 of the 20c coins and 18 50c coins, she had an equal number of 10c and 20c coins. The number of 50c coins now formed 40% of the remaining number of coins.

(a) How many 10c coins did Tracy have at first?
(b) How much did she spend in all?

I use model to solve:
So you would see that 2 portions of 10c coins the same as 3 parts of 20c coins. To make them equal, you will get
10c coins will have 15u,
20c coins will have 8u and
50c coins will have 15u-24.

After spending, the leftover
10c coins will be 6u,
20c coins will be 6u and
50c coins will be 15u-24-18 = 15u-42.

Since (15u-42)/(6u + 6u + (15u-42)) = 40% = 2/5,
u =6

So there are 15u = 15*6 = 90 10c coins.

You will be able to solve part b too.

I don't understand the part in blue. How to change the 5u of 10c into 15u? And also the 20c, how to change it into 8u?

[MathIzzzFun]

blitz:

Belle2011:

[quote=\"blitz\"]Help please.

Tracy counted the number of 10cents coins, 20cent coins and 50cents coins in her piggy bank. The number of 10c coins was 24 more than the number of 50c coins. After spending 3/5 of the 10c coins, 1/4 of the 20c coins and 18 50c coins, she had an equal number of 10c and 20c coins. The number of 50c coins now formed 40% of the remaining number of coins.

(a) How many 10c coins did Tracy have at first?
(b) How much did she spend in all?

I use model to solve:
So you would see that 2 portions of 10c coins the same as 3 parts of 20c coins. To make them equal, you will get
10c coins will have 15u,
20c coins will have 8u and
50c coins will have 15u-24.

After spending, the leftover
10c coins will be 6u,
20c coins will be 6u and
50c coins will be 15u-24-18 = 15u-42.

Since (15u-42)/(6u + 6u + (15u-42)) = 40% = 2/5,
u =6

So there are 15u = 15*6 = 90 10c coins.

You will be able to solve part b too.

I don't understand the part in blue. How to change the 5u of 10c into 15u? And also the 20c, how to change it into 8u?[/quote]Hi

hope this helps - http://www.flickr.com/photos/62167097@N02/6027487907/in/photostream

cheers.

[blitz]

THank you for the model drawing. Will forward it to my child.