Hi, could someone help me with these questions? Thanks.
Q1. There were 159 elephants in the zoo. 5/7 of the female elephants and 5/8 of the male elephants were of normal weight. The number of underweight male elephants was twice the number of underweight female elephants. Find the total number of female elephants in the zoo.
Q2. Janna has some 50 cents coins. If she spends $2.50 each day, she will have 3 coins left. If she spends $4 each day, she'll have 7 coins left. What is the minimum amount of money that she has?
Q3. At a party, the number of people who put on mask was 768 more than those who did not. If 6 more people were given a mask each, the number of people wearing mask would be 11 times as many as those who did not put on a mask. When 2/3 of these party goers were female, how many male were at the party?
Q4. Grace had 40 red and 48 green beads. SHe used all the beads to make into flower bookmarks. The number of red and green beads used for each bookmark was the same. What is the maximum number of bookmarks that she could make?
Q1. 2/7 female and 3/8 male were underweight.
There were twice as many underweight male as female.
= > 6/21 female and 12/32 male were underweight.
total amount of elephants = 53 units = 159 elephants
= > 1 unit of elephant = 3 elephants
= > total female elephants = 21 units = 63 elephants
See the 2 underline numbers? The amount of males is directly twice the number of females, hence a direct comparison could be made.
Q2. I assume that you can handle some algebra, since this is P6 Maths.
3 coins = $1.50 and 7 coins = $3.50
If she spend $4, she will take x days to have $3.50 left, if she spend $2.50, she will take y days to have $1.50
= > $4x + $3.50 = $2.50y + $1.50
2.5y - 4x = 2
5y - 4x = 4
= > By trial and error, we let y be 2 and x be 1,
5(2) - 4(1) = 2 ( rejected )
Now, we let y be 4 and x be 2,
5(4) - 4(2) = 4 ( correct answer )
= > Hence minimum amount of money she has = $ ( 4(2) + 3.5 ) OR $ ( 2.5(4) + 1.5 )
= $ 11. 50
A model diagram will be better for this question, though I do not have the sources to draw it.
Q3. Since I cannot draw a model here, i will refer to the question in terms of units.
If 6 more ppl were given masks, ppl with masks will be 780 more than ppl without masks.
Also, ppl with masks will be 11 times ppl without masks.
= > 780 = 10 units of ppl
total amount of ppl = 12 units
= > total ppl = 780/10 X 12 = 936 ppl
2/3 were female
= > 1/3 were male, total male = 1/3 X 936 = 312 ppl
Q4.
factors of 40 = 1,2,4,5,8,10,20,40
factors of 48 = 1,2,3,4,6,8,12,16,24,48
Looking at the highest common factor of both numbers, it is deduce that 8 is the HCF.
= > Maximum of 8 bookmarks could be made, with 5 red beads and 6 green beads on each bookmark.
Bar Models can be drawn by using the rectangle tools in Microsoft Word. Next, upload it to websites eg photobucket and then link it to homework.sgforums.com ie post it as a picture file in the forum.
To really help the PSLE maths students, please use model drawing method (not using secondary school algebra) to solve PSLE problem sums.
OH NO....
crap sia... i forget already..... LOL.....
Originally posted by Lee012lee:Bar Models can be drawn by using the rectangle tools in Microsoft Word. Next, upload it to websites eg photobucket and then link it to homework.sgforums.com ie post it as a picture file in the forum.
To really help the PSLE maths students, please use model drawing method (not using secondary school algebra) to solve PSLE problem sums.
They are now 'playing' algebra for P6 in quite a number of Pri schools. It is still not a must thing though under normal curriculum, but I expect algebra to be brought down within the next 5 years or so.
And eeee.........you talk so much why don't you help me to load up XD
Btw, only Q2. REALLY needs a model to better express it, the rest can be work out without algebra or models. It will be good if someone is kind enough to help out the TS by loading up a model diagram for Q2.
Originally posted by Forbiddensinner:They are now 'playing' algebra for P6 in quite a number of Pri schools. It is still not a must thing though under normal curriculum, but I expect algebra to be brought down within the next 5 years or so.
And eeee.........you talk so much why don't you help me to load up XD
Btw, only Q2. REALLY needs a model to better express it, the rest can be work out without algebra or models. It will be good if someone is kind enough to help out the TS by loading up a model diagram for Q2.
It is good that you help to solve the problem sums for PSLE students.
But you need to solve the problem sums in a way that PSLE students can understand (model drawing method), not to solve the problem sums using the method that you are familiar with (secondary school algebra).
You have told dkcx "in the thread Primary 6 Maths" that you NEVER teach your students to try and guess the correct answer for problem sums but you have just used it for your solution for question two.
" By trial and error, we let y be 2 and x be 1,
5(2) - 4(1) = 2 ( rejected )
Now, we let y be 4 and x be 2,
5(4) - 4(2) = 4 ( correct answer )"
You are contradicting yourself. It is also NOT true that guess and check method cannot be used in PSLE exam as you have claimed it. The guess and check method can be used in PSLE exam but it is usually used only as a last resort.
For your information, question 2 is a variation of the "Money not Enough" type of question (ie one of the many PSLE questions classified under the different patterns and topics) in PSLE exam.
Steps to solve question 2 using the model drawing method.
1. $2.50 = Five 50-cent coins $1.50 = Three 50-cent coins
$4 = Eight 50-cent coins $3.50 = Seven 50-cent coins
2. Since the amount of the money must be the same for both cases, we will draw two equal length bars.
3. First Bar : 5 : 3 : :
Second Bar : 8 : 7 : :
4. Next, we will add 8 into the empty space in the second bar
First Bar : 5 : 3 : :
Second Bar : 8 : 7 : 8 :
5 7 + 8 = 15
since 15 can be divided by 5, we will not add more 8s into the empty space in the second bar.
6. Hence,
First Bar : 5 : 3 : 5 : 5 : 5 :
Second Bar : 8 : 7 : 8 :
7. So, the minimum amount of money she has = (8 + 7 + 8)coins x 50 cents
= $11.50
Alternatively, the minimum amount of money she has
= (5 + 3 + 5 + 5 + 5) coins x 50 cents = $11.50
Originally posted by Lee012lee:It is good that you help to solve the problem sums for PSLE students.
But you need to solve the problem sums in a way that PSLE students can understand (model drawing method), not to solve the problem sums using the method that you are familiar with (secondary school algebra).
You have told dkcx "in the thread Primary 6 Maths" that you NEVER teach your students to try and guess the correct answer for problem sums but you have just used it for your solution for question two.
" By trial and error, we let y be 2 and x be 1,
5(2) - 4(1) = 2 ( rejected )
Now, we let y be 4 and x be 2,
5(4) - 4(2) = 4 ( correct answer )"
You are contradicting yourself. It is also NOT true that guess and check method cannot be used in PSLE exam as you have claimed it. The guess and check method can be used in PSLE exam but it is usually used only as a last resort.
For your information, question 2 is a variation of the "Money not Enough" type of question (ie one of the many PSLE questions classified under the different patterns and topics) in PSLE exam.
Steps to solve question 2 using the model drawing method.
1. $2.50 = Five 50-cent coins $1.50 = Three 50-cent coins
$4 = Eight 50-cent coins $3.50 = Seven 50-cent coins
2. Since the amount of the money must be the same for both cases, we will draw two equal length bars.
3. First Bar : 5 : 3 : :
Second Bar : 8 : 7 : :
4. Next, we will add 8 into the empty space in the second bar
First Bar : 5 : 3 : :
Second Bar : 8 : 7 : 8 :
5 7 + 8 = 15
since 15 can be divided by 5, we will not add more 8s into the empty space in the second bar.
6. Hence,
First Bar : 5 : 3 : 5 : 5 : 5 :
Second Bar : 8 : 7 : 8 :
7. So, the minimum amount of money she has = (8 + 7 + 8)coins x 50 cents
= $11.50
Alternatively, the minimum amount of money she has
= (5 + 3 + 5 + 5 + 5) coins x 50 cents = $11.50
My method here is a set way of trial and error with proper workings , what I am against is random guessing of answers ( tikum-tikum ).
Even for guess and check questions, there is usually a standard set of base workings, without which you cannot solve the problem.
Once again, I will have to argue that only Q2. is worked out in a way not suitable for Pri school students, though I believe that I stated I have no other choice since I cannot draw out models.
On the otherhand, I will have to thank you for showing me a method of 'drawing' without using actual models. I hope you will not mind if I borrow your method to 'draw' models for others to see.
Hi,
This is a good site to learn more about model-drawing:
http://thesingaporemaths.com/index.html
Here is an interesting article of how US sees our maths teaching:
http://www.nychold.com/art-hoven-el-0711.pdf
Thanks!
Cheers,
Wen Shih
sigh... i miss the good old model drawing days!
Originally posted by Forbiddensinner:
My method here is a set way of trial and error with proper workings , what I am against is random guessing of answers ( tikum-tikum ).
Even for guess and check questions, there is usually a standard set of base workings, without which you cannot solve the problem.
Once again, I will have to argue that only Q2. is worked out in a way not suitable for Pri school students, though I believe that I stated I have no other choice since I cannot draw out models.
On the otherhand, I will have to thank you for showing me a method of 'drawing' without using actual models. I hope you will not mind if I borrow your method to 'draw' models for others to see.
Glad to hear that you like the method used to draw the bar model.
Baiyun's bar models are much better than mine.
I really like Baiyun's bar models. They are so beautiful.
I do not mind.
It is good of you to help to solve the problem sums for PSLE students.
Please come more often to the forum to help to solve problem sums for PSLE
students.