
By: chino (offline) Tuesday, August 09 2011 @ 01:49 AM CDT (Read 1914 times)



chino 
Tin A contains some 20cents coins and Tin B contains some 50cents 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 50cents coins are there in Tin B ?
i need alternatives solving instead of guess and check. Please help, Thank u.

Newbie
Registered: 03/14/10 Posts: 14





By: anniechew12 (offline) Tuesday, August 09 2011 @ 05:10 AM CDT



anniechew12 
Quote by: chinoTin A contains some 20cents coins and Tin B contains some 50cents 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 50cents coins are there in Tin B ?
i need alternatives solving instead of guess and check. Please help, Thank u.
Tin A : 20 cents coins
Tin B : 50 cents coins
First, make Tin A & Tin B have same contains of coins
9 more in Tin A than B =9* .20 = $1.80
Amount in Tin A will be $2.70$1.80 = $4.50 less than Tin B
Different amount between Tin A & Tin B = 50 cents  20cents = 30 cents
$4.50/$0.30 = 15
Tin A = 15*0.20 + 9*0.20 = $4.80
Tin B = 15*0.50 = $7.50

Newbie
Registered: 06/23/11 Posts: 5





By: chino (offline) Tuesday, August 09 2011 @ 08:14 PM CDT



chino 
Thank u very much for your clear explanation.
Can this method be used for the following question? ive seen some answers but stil not quite understand.
There are a total of 28 cows and ducks in a farm. If the total number of cows' legs is 40 more than the ducks' legs,
find the no. of cows and ducks in the farm.

Newbie
Registered: 03/14/10 Posts: 14





By: anniechew12 (offline) Wednesday, August 10 2011 @ 07:32 AM CDT



anniechew12 
Quote by: chinoThank u very much for your clear explanation.
Can this method be used for the following question? ive seen some answers but stil not quite understand.
There are a total of 28 cows and ducks in a farm. If the total number of cows' legs is 40 more than the ducks' legs,
find the no. of cows and ducks in the farm.
Cow's legs is 40 more than duck's legs
40/4 = 10 cows
2810 = 18 cows & ducks
Since 18 cows & ducks have the same legs
1 cow'legs = 2 duck'legs
18/3 = 6
No of cows = 6 + 10 = 16
no of ducks = 6*2 = 12
I hope this solution can help you. It looks simple but difficult for me to solve it.

Newbie
Registered: 06/23/11 Posts: 5





By: suz855 (offline) Wednesday, August 10 2011 @ 10:10 PM CDT



suz855 
There are a total of 28 cows and ducks in a farm. If the total number of cows' legs is 40 more than the ducks' legs,
find the no. of cows and ducks in the farm.
Alternatively, you can solve this problem using the assumption method
If all were cows, 28 x 4 = 112 legs
If there were 27 cows, 108legs
1duck 2legs
difference in legs 106
If there were 26 cows 104legs
2 ducks 4 legs
diff in legs 100
base on the above, observe that an increase in duck cause a reduction of 6 legs
(112106=6, 106100=6 .... etc)
Given that there are 40 more cow legs than duck legs
the gap to close will be 112  40 = 72
72/6 = 12 ducks
28 12 = 16 cows.

Junior
Registered: 12/31/06 Posts: 16





By: chino (offline) Sunday, August 14 2011 @ 10:39 PM CDT



chino 
Thank you all for explaining the steps for solving this cow duckie questions.
Yes it looks simple but how come i didnt think of solving it in this method???
how do our kids have the time to think in the examination to solve such tricky questions.
Time factor is so crucial.

Newbie
Registered: 03/14/10 Posts: 14





By: cimman (offline) Monday, August 22 2011 @ 12:10 PM CDT



cimman 
Quote by: chinoThank you all for explaining the steps for solving this cow duckie questions.
Yes it looks simple but how come i didnt think of solving it in this method???
how do our kids have the time to think in the examination to solve such tricky questions.
Time factor is so crucial.
not really. This method can be used for all problems of this nature, if the total quantity of the 2 animals are known, if not then, Guess and Check is needed.
This method can be memorized and can basically be used "as is" for all problems of this nature. No need to think. Have your child do 10 problems of this nature using the same method. I'm sure your child will be able to handle this category of problems after that. Again, the caveat is that the total quantity must be known.

Newbie
Registered: 05/01/11 Posts: 5



