Tutor MathsGuru: Ask me for your burning Maths questions!

[mathsguru]

ttyh:

Pls help me in this...Thanks..

Complete the addition square by finding the missing numbers. (imagine all the numbers inside a 9 grid square)

+ 200 300
50 250 A
60 B 360

A = ?
B = ?

Hi TTYH,

Here's my solution.

Cheers,
MathsGuru

http://www.postimage.org/image.php?v=gxMxFOi

[mathsguru]

YLH88:

In a lecture hall, there were 2 classes of students seated inside, and the ratio of the number of boys to the number of girls was 5 : 6. 40% of the boys and 37.5% of the girls in the hall belonged to Class A.

Given that Class A had 21 more girls than boys, how many more students did Class B have than Class A ?

Hi YLH88,

Here's my solution. You can encourage your child to explore using fraction of a unit in the diagram. In this way, can save time by not re-drawing the diagram in terms of LCM.


MathsGuru

http://www.postimage.org/image.php?v=gxMy6eS

[Dharma]

YLH88:

Hi Mathsguru,


Need your help with the below question :

In a lecture hall, there were 2 classes of students seated inside, and the ratio of the number of boys to the number of girls was 5 : 6
40% of the boys and 37.5% of the girls in the hall belonged to Class A.
Given that Class A had 21 more girls than boys, how many more students did Class B have than Class A ?


Thank you so much!

Boys : 5u = 40u
Girls : 6u = 48u

Boys (Class A) : 4/10 x 40u = 16u
Girls (Class A) : 3/8 X 48u = 18u ... [ 37.5% = 3/8]

2u = 21
1u = 10.5

Boys (Class B) : 40u – 16u = 24u
Girls (Class B) : 48u – 18u = 30u

Additional No. of students in Class B = 54u – 34u = 20u = 20 x 10.5 = 210

[Dharma]

pineapple tarts:

Dear Mathsguru,


Aiyoh, I thot I am very clever with such questions already. Sikali I got stuck with this one. Pls help me.

There are 50 and pen and pencil. Each pen cost $6 and each pencil cost $5. The total cost is $47 MORE THAN the total cost of the pencil. How many pen and how many pencil are there?

There are 50 and pen and pencil. Each pen cost $6 and each pencil cost $5. The total cost of pens is $47 MORE THAN the total cost of the pencil. How many pen and how many pencil are there?

Pens : 1u
Pencil : 1p

Total cost of pens : 1u x $6 = $(6u)
Total cost of pencils : 1p x $5 = $(5p)

Given
1u + 1p = 50 ………….. (1)
$(6u) – $(5p) = $47
6u – 5p = 47 …………….(2)
5u + 5p = 250 …………(3) = (1) x 5
11u = 297 …….[(2) + (3)]
1u = 27 (No. of pens)
1p = 50-27 = 23 (No. of pencils)

[firebird]

Dear Mathsguru


Good evening.

Please help me on the following P6 maths:

1) If Ahmad and Ben work together, they can compete a job in 12 days. If Ahmad and Ben worked together for 3 days, followed by Ben who worked alone for the next 5 days, they can finish 5/12(fraction) of the job. How many days will each of them need to complete the job if they work alone?

2) Allyson, Betty and Charlene shared to buy a watch for their father. Allyson paid 1/4(fraction) of the total share of Betty and Charlene. Betty paid 1/5 (fracion) the total share of Allyson and Charlene. If Charlene paid $56 more than Betty, how much did the watch cost?

Many thanks
firebird

[persistent]

There are 50 and pen and pencil. Each pen cost $6 and each pencil cost $5. The total cost of pens is $47 MORE THAN the total cost of the pencil. How many pen and how many pencil are there?


Pens : 1u
Pencil : 1p

Total cost of pens : 1u x $6 = $(6u)
Total cost of pencils : 1p x $5 = $(5p)

Given
1u + 1p = 50 …………… (1)
$(6u) – $(5p) = $47
6u – 5p = 47 …………….(2)
5u + 5p = 250 …………(3) = (1) x 5
11u = 297 …….[(2) + (3)]
1u = 27 (No. of pens)
1p = 50-27 = 23 (No. of pencils)

Hi Dharma,

I like you solution to this question! Its elegant!
Thanks for teaching me!

[YLH88]

Hi Mathsguru, Dharma, Coffeecat,


Thank you so much for the solution !!

[david59]

persistent:

There are 50 and pen and pencil. Each pen cost $6 and each pencil cost $5. The total cost of pens is $47 MORE THAN the total cost of the pencil. How many pen and how many pencil are there?


Pens : 1u
Pencil : 1p

Total cost of pens : 1u x $6 = $(6u)
Total cost of pencils : 1p x $5 = $(5p)

Given
1u + 1p = 50 ………….. (1)
$(6u) – $(5p) = $47
6u – 5p = 47 …………….(2)
5u + 5p = 250 …………(3) = (1) x 5
11u = 297 …….[(2) + (3)]
1u = 27 (No. of pens)
1p = 50-27 = 23 (No. of pencils)

Hi Dharma,

I like you solution to this question! Its elegant!
Thanks for teaching me!

I agree. In my humble opinion, using Aglebra is easier to solve many problem sums. I would prefer to teach using Algebra too. However most maths teachers would not want to do it for one reason or other.
In this forum, I do not use Algebra as some parents and students may not understand how to use it. Emm... maybe we should make Algebra the norm in this forum. :?

[firebird]

Dear Mathsguru


Good morning.

Please help me on the following P6 math:

1) Ali and Bala had a total of 1134 marbels. After each of them had given away some marbles, the number of marbles Ali has was 4 times the number of marbles he had given away and the number of marbles Bala had was 3 times the number of marbles he had given away. They had a total of 897 marbles left. How many marbles did Bala had first?

My working via simultaneous equation a) 5 Ali + 4 Bala = 1,134
b) 4 Ali + 3 Bala = 897 and c) Subtracting b from a is 1 Ali + 1 Bala =237
MY ANSWER IS 204 FOR BALA (which was the given answer).

Please let me know:
(i) Is simultaneous equation method to find solution allowed for PSLE?
(ii) How to find the solution based on model mehtod for the question 1.


Thank you.
firebird

[Muffins]

firebird:

Dear Mathsguru


Good morning.

Please help me on the following P6 math:

1) Ali and Bala had a total of 1134 marbels. After each of them had given away some marbles, the number of marbles Ali has was 4 times the number of marbles he had given away and the number of marbles Bala had was 3 times the number of marbles he had given away. They had a total of 897 marbles left. How many marbles did Bala had first?

My working via simultaneous equation a) 5 Ali + 4 Bala = 1,134
b) 4 Ali + 3 Bala = 897 and c) Subtracting b from a is 1 Ali + 1 Bala =237
MY ANSWER IS 204 FOR BALA (which was the given answer).

Please let me know:
(i) Is simultaneous equation method to find solution allowed for PSLE?
(ii) How to find the solution based on model mehtod for the question 1.


Thank you.
firebird

Hi firebird, simultaneous equations are allowed in PSLE. And there is another way you can solve this: Ali has 5 units of marbles while Bala has 4 parts of marbles, so 5u + 4p = 1134, and in the end, they had minused 1 unit and part respectively, 4u + 3p = 897.

Because we are trying to find out Bala's (4 parts) amount, we have to make the units equal and minus the numbers. So, we change the 5 units and 4 units to their lowest common multiple, 20 units. So, (5u + 4p) X 4 = 20u + 16p = 1134 X 4 = 4536, and (4u+3p) X 5 = 20u + 15p, and 897 X 5 = 4485.

Simply, 20u + 16p = 4536, and 20u + 15p = 4485.
So 1p (20u+16p - 20u+15p) = 4536 - 4485 =51. Bala had 4p in the beginning, so 51X4 = 204