Tutor MathsGuru: Ask me for your burning Maths questions!

[josephgiri]

ongmt:

1) At a party, there were some ballons, 25 ballon burst & 10% of remaining ballons fly away. If only 60% of the ballons were left, how many ballons were there at first?

9 units -> 60%
1 unit -> 6 2/3%
10 units -> 66 2/3%

100% - 66 2/3% = 33 1/3%
33 1/3% -> 25 balloons
100% -> 75 balloons

There were 25 balloons at first.

[MathIzzzFun]

ongmt:

1) At a party, there were some ballons, 25 ballon burst & 10% of remaining ballons fly away. If only 60% of the ballons were left, how many ballons were there at first?



3) A fan club had 150 member last yr. This yr, the no of male member reduced by 20% and number of female member increas by 20%. As a result, there are now as man male member as female members. How many memebers does the club have this year?

Plse help :udaman:

Hi

Q1.
9/10 remaining balloons (after 25 burst)= 3/5 Total = 9/15 Total
Total --> 15units
Remaining balloons --> 10 units
5 units --> 25
15 units --> 75
There were 75 balloons at first.

Q3.
4/5 male = 12/10 female
12/15 male = 12/10 female
Last year
Male --> 15 units
Female --> 10 units
total 25 units --> 150
Male --> 90
Female --> 60
This year, Male --> 72, Female --> 72, total 144 members.

cheers.

[imacsg]

Hi,

Can you help to solve the below question

find x : 10, 16, 26, 40, 58, x

Thanks

[MathIzzzFun]

imacsg:

Hi,

Can you help to solve the below question

find x : 10, 16, 26, 40, 58, x

Thanks

Hi

x = 80

One way to solve number pattern is to find the difference of two consecutive terms... and repeat for pattern of \"difference\" until the difference is a constant.

For this case, you can work out the 1st level difference pattern : 6,10,14,18
and then down one more level and get the 2nd level difference pattern :
4,4,4
So, the 5th number in the 1st level difference pattern is 18+4 = 22
and x = 58 + 22 = 80

cheers.

[imacsg]

Hi

Thanks.

I nearly fainted when I see, the teacher asked them to use

this equation an2 + bn + c to solve above question:

Using Fx 95Sg Plus calculator, set Mode 3 (Equation Mode), Then choose 1.
The calculator will have 3 columns (a), (b) and (c). Arrange your 2 equations like this:

(a) (b) (c)
b + c = 8
2b + c = 8


Hit, 1=,1=,8=, 2=, 1=, 8=, Then, hit “=” you will see X=0, hit “=” again, you will see Y=8.
It means b=0 and c=8 !!!

Therefore, the general formula is 2n2 + (0Xn) + 8
or 2n2 + 8



My son sec2 also do not know how to use the calculator , equation mode and this is only P5 question
.

[MathIzzzFun]

imacsg:

Hi

Thanks.

I nearly fainted when I see, the teacher asked them to use

this equation an2 + bn + c to solve above question:

Using Fx 95Sg Plus calculator, set Mode 3 (Equation Mode), Then choose 1.
The calculator will have 3 columns (a), (b) and (c). Arrange your 2 equations like this:

(a) (b) (c)
b + c = 8
2b + c = 8


Hit, 1=,1=,8=, 2=, 1=, 8=, Then, hit “=” you will see X=0, hit “=” again, you will see Y=8.
It means b=0 and c=8 !!!

Therefore, the general formula is 2n2 + (0Xn) + 8
or 2n2 + 8



My son sec2 also do not know how to use the calculator , equation mode and this is only P5 question
.

Hi
I believe your child in the GEP.
Yes, the \"correct way\" is to use the quadratic equation to find the nth term when the 2nd level difference is a constant, for eg if the question ask for the 100th number. But for this question, there is no need to work out the nth term because it is quite easy to find x.

cheers.

[imacsg]

Hi MathIzzzFun,

Thanks for your reply. My son is not in the GEP, but in the best class. I find if the child is not so strong in maths, if you teach them this method, they will not understand.

Cheer.

[ongmt]

Jug X and Jug Y contain different amounts of water at first. 50% of water in Jug X was poured into Jug Y, Then 40% of water in Jug Y was poured into Jug X. The final amunt of water in Jug X to Jug Y was 7:6

a) what wa the ratio of the amount of water in Jug X to Jug Y at first
b) if there were 10 litres of water in Jug X at first, how much water was there in Jug X at the end.

plse help:salute:

[MathIzzzFun]

ongmt:

Jug X and Jug Y contain different amounts of water at first. 50% of water in Jug X was poured into Jug Y, Then 40% of water in Jug Y was poured into Jug X. The final amunt of water in Jug X to Jug Y was 7:6

a) what wa the ratio of the amount of water in Jug X to Jug Y at first
b) if there were 10 litres of water in Jug X at first, how much water was there in Jug X at the end.

plse help:salute:

Hi
a)
in the end,
X :Y --> 7u: 6u -- 6u represents 60% of Y after transfer of 40%.
Before 2nd transfer, X:Y --> 7u - 4u : 6u + 4u = 3u : 10u
-- 3u = 50% of X after transfer,
at first,
X: Y --> 6u: 7u

b) X, at first : in the end --> 6u : 7u = 10 litres : 7/6 x 10 litres
In the end, X --> 11 2/3 litres

cheers.

[yapperz2000]

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Hi!
I need help with this question.
Couldn't find the solution as it was catholic high prelims maths 2008
:udaman: