Q&A - P3 Math

[lizawa]

Jenlimtz:

Hi


Pls help:

There were 420 more marbles in box A than box B. When 150 marbles was transferred from box B to box A. There were 4 times as many marbles in box A as box B.
a) How many marbles are there in box B now?
b) How many marbles were there in box A at first?

TIA.

Solution :
http://i52.tinypic.com/21j9890.jpg\">

[chloecube]

lizawa:

chloecube:

3 similar rings and 2 similar chains cost $1200
2 similar rings and 1 similar chain cost $750
Find out the cost of 1 ring.

2 rings + 1 chain = $750
4 rings + 2 chains = $1500
3 rings + 2 chains = $1200 (given)
1 ring = $1500 - $1200 = $300

:thankyou:

[chloecube]

lizawa:

Jenlimtz:

Hi


Pls help:

There were 420 more marbles in box A than box B. When 150 marbles was transferred from box B to box A. There were 4 times as many marbles in box A as box B.
a) How many marbles are there in box B now?
b) How many marbles were there in box A at first?

TIA.

Solution :
http://i52.tinypic.com/21j9890.jpg\">

sorry, i am not good in Math, correct me if i am wrong.

lizawa method is right, but why must add 150 two times in the 2nd model?

3units = 420 + 150 = 570
1 unit = 190
Box B has 190 marble

Box A has 190+150+420 = 760 marbles at first.

[imacsg]

Can any one help?

Tin A contains some 20 cents and Tin B contains some 50 cents coins. There are 9 more coins in Tin A than in Tin B. The amount of money in Tin A is $2.70 less than that in Tin B.
How many 50 cents coins are there in Tin B?
Solution is below:
9 x 0.20 = 1.80
2.70 + 1.80 = 4.50
4.50 / 0.30 = 15 coins - 50 cents

But I do not understand, can anyone explain?

[MathIzzzFun]

imacsg:

Can any one help?

Tin A contains some 20 cents and Tin B contains some 50 cents coins. There are 9 more coins in Tin A than in Tin B. The amount of money in Tin A is $2.70 less than that in Tin B.
How many 50 cents coins are there in Tin B?
Solution is below:
9 x 0.20 = 1.80
2.70 + 1.80 = 4.50
4.50 / 0.30 = 15 coins - 50 cents

But I do not understand, can anyone explain?

Hi

If there were equal number of coins, then it would be easy to know the number of coins because each 50 cent coin is 30 cents more than a 20 cents coin and then by dividing the \"extra\" by 30 cents, one would get the number of coins for each.

So, to make the number of coins the same, we could either ADD 9 x 50 cents coins, or take away the 9 x 20 cents coins. Either way, we will add to the difference in amount - $4.50 for adding 9x50 cents coins and $1.80 for removing 9 x 20cents coins.

So, we remove the 9 x 20 cents coins, then the difference will increase to
$ 2.70 + $1.80 = $4.50, now dividing by 30cents, will give the number of coins each for 50 cents and 20 cents (remaining, need to add 9 to get original number of 20 cent coins).

If we add 9 x 50 cent coins, then the difference will increase to
$2.70 + $4.50 = $ 7.20, now dividing by 30 cents ie 7.20 /0.30 = 24, will give number of number of coins each for 20 cents and 50 cents -need to subtract 9 to get original number of 50 cent coins, since we have added 9 x 50 cents coins -> 24 - 9 = 15 x 50 cents coins.

cheers.

[chloecube]

MathIzzzFun:

imacsg:

Can any one help?

Tin A contains some 20 cents and Tin B contains some 50 cents coins. There are 9 more coins in Tin A than in Tin B. The amount of money in Tin A is $2.70 less than that in Tin B.
How many 50 cents coins are there in Tin B?
Solution is below:
9 x 0.20 = 1.80
2.70 + 1.80 = 4.50
4.50 / 0.30 = 15 coins - 50 cents

But I do not understand, can anyone explain?

Hi

If there were equal number of coins, then it would be easy to know the number of coins because each 50 cent coin is 30 cents more than a 20 cents coin and then by dividing the \"extra\" by 30 cents, one would get the number of coins for each.

So, to make the number of coins the same, we could either ADD 9 x 50 cents coins, or take away the 9 x 20 cents coins. Either way, we will add to the difference in amount - $4.50 for adding 9x50 cents coins and $1.80 for removing 9 x 20cents coins.

So, we remove the 9 x 20 cents coins, then the difference will increase to
$ 2.70 + $1.80 = $4.50, now dividing by 30cents, will give the number of coins each for 50 cents and 20 cents (remaining, need to add 9 to get original number of 20 cent coins).

If we add 9 x 50 cent coins, then the difference will increase to
$2.70 + $4.50 = $ 7.20, now dividing by 30 cents ie 7.20 /0.30 = 24, will give number of number of coins each for 20 cents and 50 cents -need to subtract 9 to get original number of 50 cent coins, since we have added 9 x 50 cents coins -> 24 - 9 = 15 x 50 cents coins.

cheers.

MathIzzzFun, u are fabulous! :salute:

i go :shock: after reading the solution

[imacsg]

Hi MathIzzzFun,


I think you are fabulous also. Thanks a lot.

[Aberc]

Regarding the question about the marbles, 150 has to be added twice because once the marbles are removed from box b, the difference between box a and b will increase by 150. When these are added to box a, the difference is increased by another 150.


Take for example if A has 5 sweets and B has 3 sweets, the diff between them is 2 sweets. If B ate 1 sweet, The diff between them will increase by one, that is a diff of 3. If B didn’t eat the sweet but gave it to A instead, the diff between them will increase by a further one, that is a diff of 4. A would have 6 sweets now and B has 2.

This type of question is sometimes known as internal transfer. I find the book ‘thinking maths’ by onsponge to be very good.

[chloecube]

Aberc:

Regarding the question about the marbles, 150 has to be added twice because once the marbles are removed from box b, the difference between box a and b will increase by 150. When these are added to box a, the difference is increased by another 150.


Take for example if A has 5 sweets and B has 3 sweets, the diff between them is 2 sweets. If B ate 1 sweet, The diff between them will increase by one, that is a diff of 3. If B didn't eat the sweet but gave it to A instead, the diff between them will increase by a further one, that is a diff of 4. A would have 6 sweets now and B has 2.

This type of question is sometimes known as internal transfer. I find the book 'thinking maths' by onsponge to be very good.

i see...i get what you mean after some thinking, its so tricky. what level problem sum is this?

[MathIzzzFun]

chloecube:

MathIzzzFun:

[quote=\"imacsg\"]Can any one help?

Tin A contains some 20 cents and Tin B contains some 50 cents coins. There are 9 more coins in Tin A than in Tin B. The amount of money in Tin A is $2.70 less than that in Tin B.
How many 50 cents coins are there in Tin B?
Solution is below:
9 x 0.20 = 1.80
2.70 + 1.80 = 4.50
4.50 / 0.30 = 15 coins - 50 cents

But I do not understand, can anyone explain?

Hi

If there were equal number of coins, then it would be easy to know the number of coins because each 50 cent coin is 30 cents more than a 20 cents coin and then by dividing the \"extra\" by 30 cents, one would get the number of coins for each.

So, to make the number of coins the same, we could either ADD 9 x 50 cents coins, or take away the 9 x 20 cents coins. Either way, we will add to the difference in amount - $4.50 for adding 9x50 cents coins and $1.80 for removing 9 x 20cents coins.

So, we remove the 9 x 20 cents coins, then the difference will increase to
$ 2.70 + $1.80 = $4.50, now dividing by 30cents, will give the number of coins each for 50 cents and 20 cents (remaining, need to add 9 to get original number of 20 cent coins).

If we add 9 x 50 cent coins, then the difference will increase to
$2.70 + $4.50 = $ 7.20, now dividing by 30 cents ie 7.20 /0.30 = 24, will give number of number of coins each for 20 cents and 50 cents -need to subtract 9 to get original number of 50 cent coins, since we have added 9 x 50 cents coins -> 24 - 9 = 15 x 50 cents coins.

cheers.

MathIzzzFun, u are fabulous! :salute:

i go :shock: after reading the solution [/quote]

imacsg:

Hi MathIzzzFun,

I think you are fabulous also. Thanks a lot.

hi imacsg, chloecube:

thkq, u r welcome

cheers.