Tutor MathsGuru: Ask me for your burning Maths questions!

[firebird]

Dear Maths guru


Many thanks

Very sorry I could not view your model working due to the software at my end.

Could you kindly help.

With best regards
firebird

[starlight1968sg]

firebird:

Dear mathsguru


Please help me on the following P6 maths:

1) Sophie and Rachel share blue ribbons in the ratio of 5 : 2.
They want to buy some red ribbons so that the total number of ribbons they have will be tripled. In what ratio must they divide the red ribbons so that Rachel will have twice the number of ribbons she has now?

I got 17:4. But the given answer is 6:1.

2) Faizal has a rod. He needs 6 minutes to saw it into 4 pieces. How many minutes does he need to saw the rod into 28 pieces?

I got 42 minutes. But the given answer is 54 minutes

4) There were some horses and people on a farm. There were 62 more legs than heads. There were 26 more heads than tails.

a) How many people were there?
b) How many horses were there?

I really do not know where to start for question 4 a) & b)

5) Sugar and water are mixed in the ratio 1:5 to make syrup. The volume of the syrup is 800 cm3.

a) Find the volume of sugar in the syrup.
b) Find the volume of water in the syrup.
c) How much more sugar must be added to the surup so that the ratio of the sugar to water in the syrup becomes 1:3?

Many thanks
firebird

Hi Firebird and Maths Guru,
I tried Q1 but still unsure. The confusion part is there are both blue and red ribbons involved.

For Q2, the trick is \" only 3 cuts are needed to get 4 pieces\".

For Q4, I can get the number of people but still struggling with part b.

Thanks for sharing.

[Dharma]

starlight1968sg:

Hi Firebird and Maths Guru,

I tried Q1 but still unsure. The confusion part is there are both blue and red ribbons involved.

For Q2, the trick is \" only 3 cuts are needed to get 4 pieces\".

For Q4, I can get the number of people but still struggling with part b.

Thanks for sharing.

Hi starlight1968sg,

a)\tThere are 26 more heads than tails. This means that the 26 extra heads do not have tails and therefore the 26 are the number of people.

b)\tTails : 1u (No. of horses)
Heads : 1u + 26
Legs : 1u + 88

No. of legs = 26(2) + 1u(4) = 4u + 52

4u + 52 = 1u + 88
3u = 36
1u = 12
No. of horses = 12

[CoffeeCat]

firebird:

Dear Maths guru


Many thanks

Very sorry I could not view your model working due to the software at my end.

Could you kindly help.

With best regards
firebird

try right-clicking and saving as a picture file? i mean surely most computers in his age can see a picture file?

[Millicent]

Dear Mathsguru,


Thanks once again for your great help

[firebird]

Dear Mathsguru and coffeeCat


Good afternoon.

Now I could clearly view mathsguru’s answers.

Many thanks
firebird

[firebird]

Dear Starlight1968sg


Please refer to your request for the following maths.

1) Sophie and Rachel share blue ribbons in the ratio of 5 : 2.
They want to buy some red ribbons so that the total number of ribbons they have will be tripled. In what ratio must they divide the red ribbons so that Rachel will have twice the number of ribbons she has now?

My suggestion:

a) Note, the colour of the ribbon is not an issue for this question.
b) Addition of the ribbons is the real issue here.
c) Let Sophie be S and Rachel be R. Ok so far?
So, S:R = 5:2 Total 7 units (given)
Now S and R want to triple their quantity = 7 x 3 = 21
Therefore, we get S + R = 21
Now R also want to double, but within the quantiy 21.
Therefore R = 2 x 2 = 4
As R is now 4, S = 21- 4 =17

S : R
(Now) 17 : 4
(Originally) 5 : 2 (minus "original" from "now")
-------
12 : 2 (17-5=12 and 4-2=2)
====
To simplyfy further divide 12:2 by 2
We should get 6 : 1

So 6:1 is the ratio of distribution of the addition of ribbons.

2) Faizal has a rod. He needs 6 minutes to saw it into 4 pieces. How many minutes does he need to saw the rod into 28 pieces?

Just take a 10cm string. Imagine it to be a rod. Count and cut 3 times only.
I you will get 4 pieces.

4) There were some horses and people on a farm. There were 62 more legs than heads. There were 26 more heads than tails.

a) How many people were there?
b) How many horses were there?

First of all I assume you know answer to a) which is 26.

Now let total Horses be H, each horse has 4 legs and 1 head
Now let total People be P=26 (answer from a) 26 x 2 = 52 legs and 26 heads

I suggest we Less total heads from total legs as follows:

{52 people legs} + (4 x H)} MINUS {H+ 26 (people head)} = 62 (given)
52 + 4H - H - 26 = 62
3H + 26 = 62
3H = 36
H = 12

I hope the above helps

With best regards
firebird

[/list][/quote]

[P5G]

Pl help.


Paper 1 Q13.
http://www.orlesson.org/orp/09Ma/2009-Math-SA1-MGS.pdf


Worksheet answer is (4) 135 [not likely].


Please advise how to get (3) 125 as Benny's score?


TIA.

[mmmouse]

Anyone knows how to solve this Pr. 3 Maths question?


5 erasers + 3 rulers = S$5.05
3 erasers + 6 rulers = S$3.95
2 erasers + 2 rulers = ???

Thanks!

[CoffeeCat]

mmmouse:

Anyone knows how to solve this Pr. 3 Maths question?


5 erasers + 3 rulers = S$5.05
3 erasers + 6 rulers = S$3.95
2 erasers + 2 rulers = ???

Thanks!

oh wellz its a simultaneous linear equation question.
if 5 erasers + 3 rulers = $5.05
then 10 erasers + 6 rulers = $10.10
and since 3 erasers + 6 rulers = $3.95
then 7 erasers = $6.15
however $6.15 is not divisible by 7 unless there's a mistake somewhere, so its time for another method.

notice that the question is asking for 2 erasers and 2 rulers.
So a quicker method is to combine both information given to derive something like ...

Notice that
5 erasers + 3 rulers = S$5.05 --> (1)
(2 more eraser than rulers)
3 erasers + 6 rulers = S$3.95 --> (2)
(3 less erasers than rulers)

so from (1) we get
15 erasers + 9 rulers = $15.15
(notice 15 is 6 more than 9)
from (2) we get
6 erasers + 12 rulers = $7.90
(notice 6 is 6 less than 12)

(because lcm of 2 and 3 is 6)

so therefore
21 erasers + 21 rulers = $23.05
2 erasers + 2 rulers = ($23.05/21) * 2 = ...

Therefore I have reason to suspect your question is wrong =.=.[/b]