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Q&A - P5 Math

Academic support for Primary 5
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speedmaths.com
BrownBelt
BrownBelt
Posts: 559
Joined: Mon Oct 12,
Total Likes:3

Re: Q&A - P5 Math - Average with Unknown Quantlity

Post by speedmaths.com » Wed Jul 16, 2014 3:32 am

ILLUSDUCA wrote:
JeffLim wrote:Hi there,
Need help on this.

The average mass of a group of boys is 45 kg. If one more boy who weighs 75 kg joins the group, the new average mass of the group will become 48 kg. Find the total number of boys in the group.
Thank you in advance.
Jeff
Hello Jeff!

Here's one way to do this:
75kg - 45kg = 30kg difference
48kg - 45kg = 3kg difference
30/3 = 10 students (9 of them weighing average of 45kg and 1 of them weighing 75kg)

Hope this helps!

Hi,

The answer sh be 9 boys,
as the boy who weighs 75 kg did NOT actually join the group.

Cheers



speedmaths
.
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Sruthi
BlueBelt
BlueBelt
Posts: 243
Joined: Thu Nov 12,

Re: Q&A - P5 Math

Post by Sruthi » Wed Jul 16, 2014 6:38 am

Image _ Please help with this question
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ILLUSDUCA
GreenBelt
GreenBelt
Posts: 146
Joined: Sat Jun 21,

Re: Q&A - P5 Math - Average with Unknown Quantlity

Post by ILLUSDUCA » Wed Jul 16, 2014 10:26 am

speedmaths.com wrote:
ILLUSDUCA wrote:
JeffLim wrote:Hi there,
Need help on this.

The average mass of a group of boys is 45 kg. If one more boy who weighs 75 kg joins the group, the new average mass of the group will become 48 kg. Find the total number of boys in the group.
Thank you in advance.
Jeff
Hello Jeff!

Here's one way to do this:
75kg - 45kg = 30kg difference
48kg - 45kg = 3kg difference
30/3 = 10 students (9 of them weighing average of 45kg and 1 of them weighing 75kg)

Hope this helps!

Hi,

The answer sh be 9 boys,
as the boy who weighs 75 kg did NOT actually join the group.

Cheers



speedmaths
.
thanks for clarification speedmaths.com! I kind of just left it open typing the bracket and the words in it.
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ILLUSDUCA
GreenBelt
GreenBelt
Posts: 146
Joined: Sat Jun 21,

Re: Q&A - P5 Math

Post by ILLUSDUCA » Wed Jul 16, 2014 10:30 am

Sruthi wrote:Image _ Please help with this question
Hello Sruthi,

In figure 2, shaded areas are 6 small triangles.
So, 6 small triangles = 108cm^2
1 small triangle = 108 / 6 = 18 cm^2

And putting any 2 small triangles together forms a square, so, 18 x 2 = 36cm^2.
So, the area of any square formed by 2 small triangles = 36cm^2
Given that sides of any square will have the same length, and area of square = L x B, 36 = 6 x 6. The length of 1 side of the square is 6cm.

Figure 1 is made up of 16 (not 24, miscalculation) sides of a square, so it's perimeter is 16 x 6 = 96cm.
(thanks mathizzzfun for correction)

hope this helps!
Last edited by ILLUSDUCA on Wed Jul 16, 2014 11:23 am, edited 1 time in total.
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MathIzzzFun
KiasuGrandMaster
KiasuGrandMaster
Posts: 2358
Joined: Wed Mar 09,
Total Likes:5

Re: Q&A - P5 Math

Post by MathIzzzFun » Wed Jul 16, 2014 10:55 am

..
Last edited by MathIzzzFun on Wed Jul 23, 2014 3:52 pm, edited 1 time in total.
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tracywham
OrangeBelt
OrangeBelt
Posts: 84
Joined: Thu Jun 23,

Re: Q&A - P5 Math

Post by tracywham » Wed Jul 16, 2014 11:57 am

Hi, I have doubts on the answer for this question:

A machine takes 60 seconds to cut an iron rod into 6 equal parts, At this rate, how long will it take the machine to cut the rod into 9 equal parts?

The book answer is 96s. Is it correct? How? I keep getting 90s.. Please advise. Thank you
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speedmaths.com
BrownBelt
BrownBelt
Posts: 559
Joined: Mon Oct 12,
Total Likes:3

Re: Q&A - P5 Math

Post by speedmaths.com » Wed Jul 16, 2014 12:12 pm

tracywham wrote:Hi, I have doubts on the answer for this question:

A machine takes 60 seconds to cut an iron rod into 6 equal parts, At this rate, how long will it take the machine to cut the rod into 9 equal parts?

The book answer is 96s. Is it correct? How? I keep getting 90s.. Please advise. Thank you
Hi,

"6 equal parts" means 5 cuts
"9 equal parts" means 8 cuts

5 cuts → 60 secs
1 cut → ??
8 cuts → ??


Can you solve from here?

Hope this helps.

speedmaths
.
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gingerbelle
KiasuNewbie
KiasuNewbie
Posts: 2
Joined: Thu Aug 14,

Re: Q&A - P5 Math

Post by gingerbelle » Thu Jul 17, 2014 5:02 pm

Image
Please help me with these 2 questions. Thank you.
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tanxexy
OrangeBelt
OrangeBelt
Posts: 40
Joined: Tue Nov 12,

Re: Q&A - P5 Math

Post by tanxexy » Fri Jul 18, 2014 10:05 am

gingerbelle wrote: Please help me with these 2 questions. Thank you.
Hi gingerbelle,

1) A group of pupils calculated their average score for a Maths test. If one of them scored 9 marks more, their average score would be 81. If one of them scored 3 marks less, their average score would be 78. How many pupils were there in the group?

Not sure if it helps if you look at this problem as an “Excess/Shortage” scenario:
If everyone is to be given the score of 81, there is a shortage of 9 marks.
If everyone is only given the score of 78, there is an excess of 3 marks.

Excess/Shortage = 3 – (–9) = 3 + 9 = 12
Gap = 81 – 78 = 3
Number of pupils = 12/3 = 4


2) A group of 100 pupils participated in a Maths competition. Their average score was 76 marks. If the average score for the boys was 80 marks and for the girls was 70 marks, how many girls participated in the Maths competition?

Each boy contributed above the average: 80 – 76 = 4 marks
Each girl required 76 – 70 = 6 marks to meet the average.

The boys' marks above the average will compensate for the shortfall of the girls' marks:
1 girl (6 marks) would need 1.5 boy (4 marks x 1.5)
Therefore, 3 boys (3 x 4 = 12 marks) would help to bring up the score of 2 girls (2 x 6 = 12 marks) to the average.

So, there were 2 girls for every 3 boys => girls made up of 2/5 of the group.
Number of girls = 2/5 x 100 = 40


Hope the above is clear. Please feel free to clarify.

cheers...
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gingerbelle
KiasuNewbie
KiasuNewbie
Posts: 2
Joined: Thu Aug 14,

Re: Q&A - P5 Math

Post by gingerbelle » Fri Jul 18, 2014 3:19 pm

Thank you so much for your clear explanation. I got the answers through trial and error but could not explain it to my son. That's why he wrote "dunno why" on his worksheet :thankyou:
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