Q&A - P5 Math

[jieheng]

Oracle:

Help ... please ...


Everyday, Krishnan received 50 cents more than Edward for his allowance. However, Edward spent 80 cents more than Krishnan during their recess everyday. Over a period of time, Krishnan managed to save $100 and Edward managed to save $35.

(i) How many days would Krishnan take to save $100?
(ii) How much was Krishnan's savings every day?

Thank you very much.

The amount of money that Krishnan save more than Edward every day = 0.50 + 0.80 = 1.30

Over a period of time , the amount of money that Krishnan save more than Edward = 100 - 35 = 65

i) No of days that Krishnan would take to save $100 = 65 / 1.30 = 50

ii) Krishnan's savings every day was = $100 / 50 = $2

[pupilview]

ponyo:

pupilview:

The enrolment of school A is twice of the enrolmentof school B.If 1260 pupils from school A transfer to school B,school B's enrolment would become thrice that of school A.How many pupils are there in school A?


Please help me to solve the above question using the model method.

Total remains constant.
Before : A:B= 2:1(total parts =3)
After: A:B=1:3 (total parts=4)

Common multiple of 3&4=12, ie
Before: A:B=8:4 (total=12 units)
After: A:B=3:9 (total=12 units)
8-3 units=5u were transferred from A to B
5units-->1260
1u=252
At first A has 8 units,=252*8=2016

Model drawing:
Before
A: [u][u][u][u] [u][u][u][u]
B: [u][u][u][u]

After
A: [u][u][u]
B: [u][u][u][u] [u] [u][u][u][u]

Diff=5u=1260

hope that helps...

My DD's reasoning:

A: 2u
B: 6u

6u - 2u = 1260
4u = 1260
u = 315

So School A's enrollment is (315x2)+1260= 1890.
:roll: :? Somebody please help.Which is the correct answer??

[Oracle]

Bingo JieHeng! I got it! Thanks.

[Oracle]

More questions:


(1) A bag contained 12 more 50-cent coins than 20-cent coins. The total amount of 50-cent coins was $15 more than the total amount of 20-cent coins. How many 20-cent coins were there in the bag?

(2) There were 12 more motorbikes than cars parked at a carpark. The total number of car wheels was 80 more than the total number of motorbike wheels. How many cars were there at the carpark?

Thank you.

[MathIzzzFun]

Oracle:

More questions:


(1) A bag contained 12 more 50-cent coins than 20-cent coins. The total amount of 50-cent coins was $15 more than the total amount of 20-cent coins. How many 20-cent coins were there in the bag?

(2) There were 12 more motorbikes than cars parked at a carpark. The total number of car wheels was 80 more than the total number of motorbike wheels. How many cars were there at the carpark?

Thank you.

Make quantity of both items (20-cent & 50cent) same.

Basic concept --> number of each item = total difference/unit difference

Q1.
Suppose the extra 12 x 50-cent coins are removed so that there are now equal number of 20-cent and 50-cent coins

total value removed from 50-cent coins --> 12 x $0.50= $6

$15- $6 = $9 --> now, the value of all the 50-cent coins is $9 more than value of all 20-cent coins.

Number of each coin type --> $9/($0.50-$0.20) = 30

Number of 20-cent coins = 30
Number of 50-cent coins = 30 + 12 = 42

cheers.

[MathIzzzFun]

Oracle:

More questions:


(1) A bag contained 12 more 50-cent coins than 20-cent coins. The total amount of 50-cent coins was $15 more than the total amount of 20-cent coins. How many 20-cent coins were there in the bag?

(2) There were 12 more motorbikes than cars parked at a carpark. The total number of car wheels was 80 more than the total number of motorbike wheels. How many cars were there at the carpark?

Thank you.

Make quantity of both items (motorbikes & Cars) same.

Basic concept --> number of each item = total difference/unit difference

Suppose we removed extra 12 motorbikes.
Number of motorbike wheels removed = 12 x 2 = 24
80 + 24 = 104 --> there are now 104 more car wheels than motorbike wheels

Number of each vehicle type = 104 /(4-2) = 52

Number of cars = 52
Number of motorbikes = 52 + 12 = 64

by the way, these are similar to the question you posted at
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&p=955333#p955333

cheers.

[MathIzzzFun]

pupilview:

ponyo:

[quote=\"pupilview\"]The enrolment of school A is twice of the enrolmentof school B.If 1260 pupils from school A transfer to school B,school B's enrolment would become thrice that of school A.How many pupils are there in school A?


Please help me to solve the above question using the model method.

Total remains constant.
Before : A:B= 2:1(total parts =3)
After: A:B=1:3 (total parts=4)

Common multiple of 3&4=12, ie
Before: A:B=8:4 (total=12 units)
After: A:B=3:9 (total=12 units)
8-3 units=5u were transferred from A to B
5units-->1260
1u=252
At first A has 8 units,=252*8=2016

Model drawing:
Before
A: [u][u][u][u] [u][u][u][u]
B: [u][u][u][u]

After
A: [u][u][u]
B: [u][u][u][u] [u] [u][u][u][u]

Diff=5u=1260

hope that helps...

My DD's reasoning:

A: 2u
B: 6u

6u - 2u = 1260
4u = 1260
u = 315

So School A's enrollment is (315x2)+1260= 1890.
:roll: :? Somebody please help.Which is the correct answer??[/quote]


School A has 2016 pupils.

http://i48.tinypic.com/347flzk.png\">

cheers.

[pupilview]

MathIzzzFun , your pictorial explanation on the school enrollment was excellent. Thanks.

[Oracle]

Thanks MathIzzzFun,


Obviously have not fully understood the concept … sigh … thanks very much. Will try harder.

[Xiao Hu]

Hi there,

Couldn’t figure out the solution to this question, so would like to seek maths gurus’ help here.

Sumi had 3 times as much money as June. Every day, Sumi spent $18 and June spent $12. When June had spent all her money, Sumi had $270 left.
(a) How many days did June take to spend all her money?
(b) How much money did Sumi have at first?

Sol given:
Diff between Sumi and June ->$270-$0=$270
Groups of diff ->$270 / $6 = 45
(a) Days ->45 /3 = 15
(b) At first June -> $12 * 15 = $180
Sumi had -> $1803=$540

What I couldn’t figure out is why divide 45 by 3 to arrive at the number of days.

What if the question is now changed to Sumi had 4 times as much money?
Sumi had 4 times as much money as June. Every day, Sumi spent $18 and June spent $12. When June had spent all her money, Sumi had $270 left.
(a) How many days did June take to spend all her money?
(b) How much money did Sumi have at first?

Since June no change, then she would have spent all her $180 in 15 days by spending $12 a day. Sumi would have 4$180=$720 (bec Sumi had 4 times as much). So after 15 days, Sumi would have spent $18 *15 =$270, left with $720-$270=$450.

Then diff = $450 - $ 0 = $450
Groups of diff = $450 / $6 =75
Now what do I use to divide 75 with to arrive at number of days?
If 75 / 4 = 18.75, the 4 is from the 4 times as much, which might be the same as the solution given for the 1st question. This is obviously not correct.
I must have missed something …

Thanks in advance,
Xiaohu.