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

Tin A contains some 20-cents coins 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 ?

Newbie

Registered: 03/14/10
Posts: 14

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

Quote by: chino

Tin A contains some 20-cents coins 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 ?

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: 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.

Cow's legs is 40 more than duck's legs
40/4 = 10 cows
28-10 = 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
(112-106=6, 106-100=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: 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.

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

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