Q&A - P5 Math
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speedmaths.com:
thanks but i can not understand the workings
Hi,the shadowed snake:
hi could someone help me
nanyang primary school p5 2011 paper 2 sa2 Q 18
http://test-paper.info/filemgmt_data/files/P5%20Maths%202011%20SA2%20Nanyang.PDF
Q18.
When you see the words “at first” in the question (last sentence), it could be a “Work Backward” type of question. Not all the time, but in this case it is.
There were 280 tarts left.
Tarts in small boxes = 3/4 x 280 = 210 tarts
Before giving the 8 small boxes (or 8 x 7 = 56 tarts),
she would have 210 + 56 = 266 tarts
This is after selling half the small boxes
Before selling half the small boxes,
She would have 2 x 266 = 532 tarts
532 tarts were packed in 532/7 = 76 small boxes
Number of big boxes = 76 / 4 = 19 big boxes
Number of tarts in 19 big boxes = 19 x 10 = 190 tarts
Total number of tarts at first = 532 + 190 = 722 tarts
Hope this helps.
Cheers
speedmaths.com
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the shadowed snake:
hope this helps...hi could someone help me
nanyang primary school p5 2011 paper 2 sa2 Q 18
http://test-paper.info/filemgmt_data/files/P5%20Maths%202011%20SA2%20Nanyang.PDF
In the end,
number of tarts in small boxes : big boxes = 3u :1u ** total 4u
Total number of tarts left = 240 --> 4u
tarts in small boxes --> 3/4 x 280 = 210
201/7 --> 30 boxes
so, before she gave 8 small boxes to friends,
number of small boxes --> 30 + 8 = 38
Since half of the small boxes were sold,
total number of small boxes, at first = 38 x 2 = 76
At first, there were 4 times as many small boxes as big boxes
so number of big boxes at first --> 76/4 = 19
Total number of tarts baked
= 76 x 7 + 19 x 10 = 722
cheers. -
MathIzzzFun:
hi i do not understand this part
hope this helps...the shadowed snake:
hi could someone help me
nanyang primary school p5 2011 paper 2 sa2 Q 18
http://test-paper.info/filemgmt_data/files/P5%20Maths%202011%20SA2%20Nanyang.PDF
In the end,
number of tarts in small boxes : big boxes = 3u :1u ** total 4u
Total number of tarts left = 240 --> 4u
tarts in small boxes --> 3/4 x 280 = 210
201/7 --> 30 boxes
so, before she gave 8 small boxes to friends,
number of small boxes --> 30 + 8 = 38
Since half of the small boxes were sold,
total number of small boxes, at first = 38 x 2 = 76
At first, there were 4 times as many small boxes as big boxes
so number of big boxes at first --> 76/4 = 19
Total number of tarts baked
= 76 x 7 + 19 x 10 = 722
cheers.
Total number of tarts left = 240 --> 4u
tarts in small boxes --> 3/4 x 280 = 210
201/7 --> 30 boxes
Total number of tarts baked
= 76 x 7 + 19 x 10 = 722 -
the shadowed snake:
Can you solve this :
hi i do not understand this partMathIzzzFun:
[quote=\"the hi could someone help me
nanyang primary school p5 2011 paper 2 sa2 Q 18
http://test-paper.info/filemgmt_data/files/P5%20Maths%202011%20SA2%20Nanyang.PDF
hope this helps...
In the end,
number of tarts in small boxes : big boxes = 3u :1u ** total 4u
Total number of tarts left = 240 --> 4u
tarts in small boxes --> 3/4 x 280 = 210
201/7 --> 30 boxes
so, before she gave 8 small boxes to friends,
number of small boxes --> 30 + 8 = 38
Since half of the small boxes were sold,
total number of small boxes, at first = 38 x 2 = 76
At first, there were 4 times as many small boxes as big boxes
so number of big boxes at first --> 76/4 = 19
Total number of tarts baked
= 76 x 7 + 19 x 10 = 722
cheers.
Total number of tarts left = 240 --> 4u
tarts in small boxes --> 3/4 x 280 = 210
201/7 --> 30 boxes
Total number of tarts baked
= 76 x 7 + 19 x 10 = 722
\"Melvin had thrice as many 50-cent coins as 20-cent coins. If he had a total 280 coins, how many 50-cent coins did he have ?\"
cheers. -
Four children Ken, Lionel, Melvin and Nick each have some marbles.
The number of marbles that Ken has is 1/2 of the total number of marbles that Lionel, Melvin and Nick have.
The number of marbles that Melvin has is 1/4 of the total Ken, Lionel and Nick have.
The number of marbles that Lionel has is 2/3 of the total number of marbles that Ken, Melvin and Nick have.
If Nick has 44 marbles, how many marbles should Lionel give to Melvin so that Ken & Melvin have the same number of marbles?
Thanks! -
bookwormkids:
HiFour children Ken, Lionel, Melvin and Nick each have some marbles.
The number of marbles that Ken has is 1/2 of the total number of marbles that Lionel, Melvin and Nick have.1:2--> total 3
The number of marbles that Melvin has is 1/4 of the total Ken, Lionel and Nick have.1:4--> total 5
The number of marbles that Lionel has is 2/3 of the total number of marbles that Ken, Melvin and Nick have.2:3--> total 5
If Nick has 44 marbles, how many marbles should Lionel give to Melvin so that Ken & Melvin have the same number of marbles?
Thanks!
The total number of marbles remain the same.
Ken : Ken+Lionel+Melvin+Nick --> 5u : 15u
Melvin: Ken+Lionel+Melvin+Nick --> 3u : 15u
Lionel: Ken+Lionel+Melvin+Nick --> 6u: 15u
Nick --> 15u - 5u - 3u - 6u = 1u; Nick has 44 marbles
1u --> 44
Melvin --> 132 marbles
Ken --> 220 marbles
220 - 132 = 88 --> Lionel needs to give 88 marbles to Melvin
cheers. -
Hi, Can anyone help in these questions,
(1) Hanson had some one-dollar coins and some fifty-cent coins. The ratio of the number of one-dollar coins to the number of fifty-cents coins ha had was 2:5. Hanson took 140 fifty-cents coins to the bank and changed these fifty-cents coins for the same value of one-dollar coins. In the end, the number of one-dollar coins to the number of fifty-cents coins he had became 5:2. Find the value of the fifty-cent coins Hanson had at first.
(2) AB Primary School organised a 2-day Charity Funfair. Admisson tickets were sold to every visitors at a cost of $40. On the first day, 120 more children visited the funfair than adults. The number of children on the second day was 25% more than the number of children on the first day. The number of adults on the second day was 5% less than the number of adults on the first day. There were a total of 2570 people at the funfair ont he second day.
(a) Find the number of adults at the funfair on the first day.
(b) What was the total sum of money collected from the sale of the admission tickets for both days?
(3) Hotel H and Hotel B had a total of 5200 guests. After 3/4 of the guests in Hotel H and 3/5 of the guests in Hotel B checked out, Hotel B has 260 more guests than Hotel H. How many guests did Hotel H have at first?
TIA. -
MathIzzzFun:
Hibookwormkids:
Four children Ken, Lionel, Melvin and Nick each have some marbles.
The number of marbles that Ken has is 1/2 of the total number of marbles that Lionel, Melvin and Nick have.1:2--> total 3
The number of marbles that Melvin has is 1/4 of the total Ken, Lionel and Nick have.1:4--> total 5
The number of marbles that Lionel has is 2/3 of the total number of marbles that Ken, Melvin and Nick have.2:3--> total 5
If Nick has 44 marbles, how many marbles should Lionel give to Melvin so that Ken & Melvin have the same number of marbles?
Thanks!
The total number of marbles remain the same.
Ken : Ken+Lionel+Melvin+Nick --> 5u : 15u
Melvin: Ken+Lionel+Melvin+Nick --> 3u : 15u
Lionel: Ken+Lionel+Melvin+Nick --> 6u: 15u
Nick --> 15u - 5u - 3u - 6u = 1u; Nick has 44 marbles
1u --> 44
Melvin --> 132 marbles
Ken --> 220 marbles
220 - 132 = 88 --> Lionel needs to give 88 marbles to Melvin
cheers.
Hi MathIzzzFun,
Thank you so much!
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MJNB:
this question was discussed ...Hi, Can anyone help in these questions,
(1) Hanson had some one-dollar coins and some fifty-cent coins. The ratio of the number of one-dollar coins to the number of fifty-cents coins ha had was 2:5. Hanson took 140 fifty-cents coins to the bank and changed these fifty-cents coins for the same value of one-dollar coins. In the end, the number of one-dollar coins to the number of fifty-cents coins he had became 5:2. Find the value of the fifty-cent coins Hanson had at first.
TIA.
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&p=817421#p817421
cheers. -
MJNB:
Day 1Hi, Can anyone help in these questions,
(2) AB Primary School organised a 2-day Charity Funfair. Admisson tickets were sold to every visitors at a cost of $40. On the first day, 120 more children visited the funfair than adults. The number of children on the second day was 25% more than the number of children on the first day. The number of adults on the second day was 5% less than the number of adults on the first day. There were a total of 2570 people at the funfair ont he second day.
(a) Find the number of adults at the funfair on the first day.
(b) What was the total sum of money collected from the sale of the admission tickets for both days?
TIA.
Adults --> 20 units --- 5% = 1 unit
Children --> 20 units + 120 ----- 25% = 5 units + 30
Day 2
Adults --> 19 units
Children --> 25 units + 150
Total --> 44 units + 150 = 2570
1 unit --> 55
..can you continue from here ?
cheers.
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