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Q&A - P5 Math

Academic support for Primary 5
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JeffLim
YellowBelt
YellowBelt
Posts: 15
Joined: Wed Aug 21,

Q&A - P5 Math - Average with Unknown Quantlity

Post by JeffLim » Mon Jul 14, 2014 7:35 pm

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
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Daddy
BlueBelt
BlueBelt
Posts: 247
Joined: Thu Apr 30,

Re: Q&A - P5 Math

Post by Daddy » Tue Jul 15, 2014 9:37 pm

Thank ILLUSDUCA and MathIzzzfun.

Another question to ask.

The ratio of Jeremiah's age to his aunt's age is 1:2. In 15 years time, the new ratio will be 8:13.
How old was Jeremiah 8 year ago?

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

Re: Q&A - P5 Math

Post by Sruthi » Tue Jul 15, 2014 10:26 pm

Image - Hi Please help with this problem
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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 » Tue Jul 15, 2014 10:51 pm

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!
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Jx2mum
BlueBelt
BlueBelt
Posts: 262
Joined: Tue Jun 03,
Total Likes:1

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

Post by Jx2mum » Tue Jul 15, 2014 10:55 pm

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
Hi, here's a possible solution.
48-45=3kg
75-48=27kg
27/3=9 (answer)
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ILLUSDUCA
GreenBelt
GreenBelt
Posts: 146
Joined: Sat Jun 21,

Re: Q&A - P5 Math

Post by ILLUSDUCA » Tue Jul 15, 2014 10:55 pm

Daddy wrote:Thank ILLUSDUCA and MathIzzzfun.

Another question to ask.

The ratio of Jeremiah's age to his aunt's age is 1:2. In 15 years time, the new ratio will be 8:13.
How old was Jeremiah 8 year ago?

Thank..
Hello Daddy,

Since both of them will increase by the same age, we need to change our first ratio accordingly to reflect this.

Currently,
8 : 13
- 1 : 2
is 7 : 11 difference accordingly

Without using algebra for P5, we can try...
1) 2 : 4 ---> 8 : 13 = 6 : 9 difference
2) 3 : 6 ---> 8 : 13 = 5 : 7 difference
3) 4 : 8 ---> 8 : 13 = 4 : 5 difference
3) 5 : 10 ---> 8 : 13 = 3 : 3 difference <-- same!
So, the actual ratio would be 5 : 10, then + 15 years each --> 8 : 13
So, 15 years old = 3 units
1 units = 5
Jeremiah's age = 5 units = 5x5 = 25 years old
8 years ago = 25 - 8 = 17 years old

hope this helps
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ILLUSDUCA
GreenBelt
GreenBelt
Posts: 146
Joined: Sat Jun 21,

Re: Q&A - P5 Math

Post by ILLUSDUCA » Tue Jul 15, 2014 10:59 pm

Sruthi wrote:Image - Hi Please help with this problem
Hello Sruthi,

Total = 300 steps
Every 2 steps = every even-numbered steps being stepped on. 300 steps = 150 even numbered steps and 150 odd number steps, so 300 - 150 even numbered steps = 150 odd steps left unstepped.

Every 3 steps = 300 / 3 = 100 times of stepping
100 times will include 50 odd numbered steps and 50 even numbered steps because multiples of 3 = 3,6,9,12 <-- 1 odd number followed by 1 even, then 1 odd then 1 even etc.
So, because all even numbered steps were already stepped on by "every 2 steps", we should leave out the 50 even numbered steps and include the odd numbered ones.

So, 150 even numbered steps + 50 odd numbered steps = 200 steps being stepped on. the rest = 300 - 200 = 100 left unstepped.

hope this helps!
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Jerico
GreenBelt
GreenBelt
Posts: 178
Joined: Tue Jul 28,

Re: Q&A - P5 Math

Post by Jerico » Tue Jul 15, 2014 11:08 pm

Hi,

Can someone please help with the following question :-

Q : The average score of a group of participants in an online quiz was 48. After 3 participants who scored an average of 56 left the group, the average score of the remaining participants became 44. How many participants were in the group at first ?

Thank you in advance for your kind assistance.
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ILLUSDUCA
GreenBelt
GreenBelt
Posts: 146
Joined: Sat Jun 21,

Re: Q&A - P5 Math

Post by ILLUSDUCA » Tue Jul 15, 2014 11:18 pm

Jerico wrote:Hi,

Can someone please help with the following question :-

Q : The average score of a group of participants in an online quiz was 48. After 3 participants who scored an average of 56 left the group, the average score of the remaining participants became 44. How many participants were in the group at first ?

Thank you in advance for your kind assistance.
Hi Jerico,

I have read the question as "the average was 44, then 1 participant with the score of 56 came in and the average became 48" so we could use our previous method (as shared by all earlier) to solve it.

So, 56 - 44 = 12
48 - 44 = 4
12 / 4 = 3
3 participants include 1 participant with score 56 and 2 with 44.

However, since there were 3 participants with score 56, there must be 6 with 44 proportionately.
So, at first = 3 + 6 = 9 participants

hope this helps!
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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 3:29 am

Jx2mum 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
Hi, here's a possible solution.
48-45=3kg
75-48=27kg
27/3=9 (answer)
Hi,

Yes, answer sh be 9.

Cheers



speedmaths
.
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