Q&A - P5 Math
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pupilview:
Total remains constant.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.
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\")
http://i48.tinypic.com/347flzk.png\">
cheers. -
MathIzzzFun , your pictorial explanation on the school enrollment was excellent. Thanks.
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Thanks MathIzzzFun,
Obviously have not fully understood the concept … sigh … thanks very much. Will try harder. -
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. -
Xiao Hu:
Hi
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 -> $180*3=$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.
You may use MD
Sumi:June ------ 18:12 ------ 3:2
270 + 3 units ------ 6 units (Sumi had 3 times as much money as June)
3 units ------- 270
1 unit ------ 90
June@first ------- 180
Number of days taken by June to spend her money ------ 180/12 ------- 15
Sumi@first ------- 540 (270 + 3 units)
For the second part, the answer is 9 and 432.
Best wishes -
Xiao Hu:
total = spent + leftHi 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 -> $180*3=$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.
Sumi's spending : June's spending = $ 18 : $12 = 3u : 2u
Sumi's total --> Spent + left
Sumi's total --> 2u x 3 = 6u = 3u + $270
3u --> $270, 1u --> $90
Sumi's money --> 6 x $90 = $540
June's money --> 2 x $90 = $180
with these, you can work out the number of days needed for June to spend all her money.
cheers. -
tianzhu:
HiXiao Hu:
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 -> $180*3=$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.
You may use MD
Sumi:June ------ 18:12 ------ 3:2
270 + 3 units ------ 6 units (Sumi had 3 times as much money as June)
3 units ------- 270
1 unit ------ 90
June@first ------- 180
Number of days taken by June to spend her money ------ 180/12 ------- 15
Sumi@first ------- 540 (270 + 3 units)
For the second part, the answer is 9 and 432.
Best wishes
Hi tianzhu,
Thanks very much!! Really appreciate your help here.
Wishing you a good evening,
Xiaohu -
MathIzzzFun:
Hi MathIzzzFun ,
total = spent + leftXiao 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 -> $180*3=$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.
Sumi's spending : June's spending = $ 18 : $12 = 3u : 2u
Sumi's total --> Spent + left
Sumi's total --> 2u x 3 = 6u = 3u + $270
3u --> $270, 1u --> $90
Sumi's money --> 6 x $90 = $540
June's money --> 2 x $90 = $180
with these, you can work out the number of days needed for June to spend all her money.
cheers.
Yes!! Thanks so much for showing your workings so clearly. Same Unitary approach as Tianzhu's.
Awesome gurus!!
You both save my day, thanks again,
Xiaohu -
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.
Here's a step-by-step breakdown:![](\"http://i45.tinypic.com/6ekfnr.jpg\")
![](\"http://i45.tinypic.com/zmmet0.jpg\")
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Oracle:
Thanks MathIzzzFun,
Obviously have not fully understood the concept ... sigh ... thanks very much. Will try harder.![](\"http://i47.tinypic.com/344pmpv.jpg\")
The total amount of 50-cent coins was $15 more than the total amount of 20-cent coins.
Total Amount of 50-cents = Total Amount of 20-cents + $15
(Number of 20-cents * $0.50) + (12 * $0.50) = (Number of 20-cents * $0.20) + $15
Number of 20-cents * $0.30 = $9
Number of 20-cents = 30
Is this allowed?
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