Tutor MathsGuru: Ask me for your burning Maths questions!

[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

[firebird]

Dear Muffins


Your reply is very clear.

Many thanks for your help.

Regards
firebird

[pineapple tarts]

Dear Mathsguru,


wow.... many thanks for your solutions. It's very clear but I still need some time to digest it :oops: I wonder how these P5/6 handle such questions. Anyway, I just want to say thank you!

And thanks to Dharma too

[Vanilla Cake]

firebird:

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).

Is this question (4 marks) from Pei Chun P6 CA1 2009 paper 2 Q11? http://www.orlesson.org/orp/09Ma/2009-P6-Math-CA1-PeiChun.pdf.

[trytry]

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

Before
ALI.................A...A....A....A....A....|
Bala................B...B....B....B.........|.....1134
......................{...4groups..}....A=1134-948=186
.........................=.948...............

After
ALI.......................A....A....A....A........|
Bala......................B....B....B..............|....897


1 group of A and B = 1134-897=237

4 group of A and B=237 x4=948

therefore 1A=1134-948=186

ALi has 5A=186*5=930
then BALA has 1134-930=204 at first



Hope you understand

[Dharma]

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

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?

A & B – 12 days – 1 job
A & B – 1 day – 1/12 job
A & B – 3 days – ¼ job

B alone – 5 days – (5/12 – ¼) job = 1/6 job
B alone – 1/6job – 5 days
B alone – 6/6 job – 5 x 6 = 30 days
B alone – 1 day – 1/30 job

A alone – 1 day – (1/12 – 1/30)job = (5/60 – 2/60)job = 3/60 job = 1/20 job
A alone – 20/20 job – 20 days

If they work alone; A will take 20 days and B will take 30 days to complete the job.

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?

A : 1u
B & C : 4u

B : 1p
A & C : 5p

A : B : C = 1u : 1p : 4u – 1p or = 1u : 1p : 5p – 1u

4u – 1p = 5p – 1u
5u = 6p

A : B : C = 1u : 5/6u : 25/6u – 1u = 1u : 5/6u : 19/6u = 6u : 5u : 19u
19u – 5u = $56
14u = $56
1u = $4

Cost of watch = 30u = 30 x $4 = $120