Hi Daddy,Daddy wrote:Hi,

Need help on this question. Thank first.

Please see the video for a model/fraction approach to this question

http://youtu.be/8AuR5LrOZkc

hope it helps! thanks.

Feel free to let me know if any part is unclear!

Academic support for Primary 5

Post
by **ILLUSDUCA** » Tue Jul 08, 2014 9:36 pm

Hi Daddy,Daddy wrote:Hi,

Need help on this question. Thank first.

Please see the video for a model/fraction approach to this question

http://youtu.be/8AuR5LrOZkc

hope it helps! thanks.

Feel free to let me know if any part is unclear!

- MathIzzzFun
- KiasuGrandMaster
**Posts:**2358**Joined:**Wed Mar 09,**Total Likes:**5

Post
by **MathIzzzFun** » Tue Jul 08, 2014 11:30 pm

.

Last edited by MathIzzzFun on Wed Jul 23, 2014 3:57 pm, edited 1 time in total.

Post
by **belnanna** » Wed Jul 09, 2014 6:42 pm

Would you please teach the advance model instead? My child doesnt understand the algebra way yet. Thanks in advance.speedmaths.com wrote:Hi,Daddy wrote:Hi, i have on question to ask. Thank.

One possible methodis to use "Bags and Boxes".

A similar question / solution is given below.

Cheers

speedmaths

=

Question:

The ratio of mangoes to the ratio of watermelons in a shop was 5:6 at first.

After the shopkeeper bought another 18 mangoes, and sold 98 of the watermelons, the ratio becomes 8:5.

What is the total number of mangoes and watermelons in the shop now?

One Possible Solution:

Using the “Advanced Model Method”

or the “Boxes and Bags Method.”

At first:

Let there be[5 Boxes]of mangoes at first.

Let there be[6 Boxes]of watermelons at first.

(The ratio of mangoes to the ratio of watermelons in a shop was 5:6 at first)

Changes:

The shopkeeper bought another 18 mangoes, and sold 98 of the watermelons.

Finally:

Let there be(8 Bags)of mangoes finally.

Let there be(5 Bags)of watermelons finally.

(The ratio becomes 8:5)

(There was a total of (13 Bags) of fruits finally)

Mangoes:

[5 Boxes]+ 18 → (8 Bags)

(x 6)

[30 Boxes]+ 108 → (48 Bags)

[30 Boxes]→ (48 Bags) – 108 ………..Row 1

Watermelons:

[6 Boxes]– 98 → (5 Bags)

(x 5)

[30 Boxes]– 490 → (25 Bags)

[30 Boxes]→(25 Bags)+ 490………..Row 2

Using theEqualisation Method:

Row 1 = Row 2

[30 Boxes] = [30 Boxes]

[30 Boxes] → (48 Bags) – 108 → [30 Boxes] → (25 Bags) + 490

(48 Bags) – 108 → (25 Bags) + 490

(23 Bags) – 108 → 490

(23 Bags) → 490 + 108 = 598

(1 Bag) → 598 ÷ 23 = 26

(13 Bags) → 13 x 26 =338 (fruits)

Hope this helps.

Cheers

speedmath

.

- Kangkangteo
- BlueBelt
**Posts:**345**Joined:**Sun May 11,**Total Likes:**1

Post
by **Kangkangteo** » Wed Jul 09, 2014 8:46 pm

belnanna wrote:Would you please teach the advance model instead? My child doesnt understand the algebra way yet. Thanks in advance.speedmaths.com wrote:Hi,Daddy wrote:Hi, i have on question to ask. Thank.

One possible methodis to use "Bags and Boxes".

A similar question / solution is given below.

Cheers

speedmaths

=

Question:

The ratio of mangoes to the ratio of watermelons in a shop was 5:6 at first.

After the shopkeeper bought another 18 mangoes, and sold 98 of the watermelons, the ratio becomes 8:5.

What is the total number of mangoes and watermelons in the shop now?

One Possible Solution:

Using the “Advanced Model Method”

or the “Boxes and Bags Method.”

At first:

Let there be[5 Boxes]of mangoes at first.

Let there be[6 Boxes]of watermelons at first.

(The ratio of mangoes to the ratio of watermelons in a shop was 5:6 at first)

Changes:

The shopkeeper bought another 18 mangoes, and sold 98 of the watermelons.

Finally:

Let there be(8 Bags)of mangoes finally.

Let there be(5 Bags)of watermelons finally.

(The ratio becomes 8:5)

(There was a total of (13 Bags) of fruits finally)

Mangoes:

[5 Boxes]+ 18 → (8 Bags)

(x 6)

[30 Boxes]+ 108 → (48 Bags)

[30 Boxes]→ (48 Bags) – 108 ………..Row 1

Watermelons:

[6 Boxes]– 98 → (5 Bags)

(x 5)

[30 Boxes]– 490 → (25 Bags)

[30 Boxes]→(25 Bags)+ 490………..Row 2

Using theEqualisation Method:

Row 1 = Row 2

[30 Boxes] = [30 Boxes]

[30 Boxes] → (48 Bags) – 108 → [30 Boxes] → (25 Bags) + 490

(48 Bags) – 108 → (25 Bags) + 490

(23 Bags) – 108 → 490

(23 Bags) → 490 + 108 = 598

(1 Bag) → 598 ÷ 23 = 26

(13 Bags) → 13 x 26 =338 (fruits)

Hope this helps.

Cheers

speedmath

.

Post
by **hweelit** » Wed Jul 09, 2014 9:32 pm

hi anyone can help to solve, thanks

mrs ang bought a certain number of ceramic pots. she bought 4 smaller ones at $18 each and some bigger ones at $24 each. if the average cost of all the ceramic pots was $21, how many bigger ceramic pots did she buy?

mrs ang bought a certain number of ceramic pots. she bought 4 smaller ones at $18 each and some bigger ones at $24 each. if the average cost of all the ceramic pots was $21, how many bigger ceramic pots did she buy?

Post
by **tanxexy** » Thu Jul 10, 2014 9:13 am

Hi,hweelit wrote:hi anyone can help to solve, thanks

mrs ang bought a certain number of ceramic pots. she bought 4 smaller ones at $18 each and some bigger ones at $24 each. if the average cost of all the ceramic pots was $21, how many bigger ceramic pots did she buy?

Here is a suggested solution:

Difference between the average cost and the smaller ones = $21 - 18 = $3

Total difference = $3 x 4 = $12

This difference must be compensated by the bigger ones to bring the average up to $21.

Difference between the average cost and the bigger ones = $24 - 21 = $3

Therefore, number of bigger ones = $12 / 3 =

Cheers...

Post
by **tanxexy** » Thu Jul 10, 2014 9:27 am

Another approach is to use the concept of area.tanxexy wrote:Hi,hweelit wrote:hi anyone can help to solve, thanks

mrs ang bought a certain number of ceramic pots. she bought 4 smaller ones at $18 each and some bigger ones at $24 each. if the average cost of all the ceramic pots was $21, how many bigger ceramic pots did she buy?

Here is a suggested solution:

Difference between the average cost and the smaller ones = $21 - 18 = $3

Total difference = $3 x 4 = $12

This difference must be compensated by the bigger ones to bring the average up to $21.

Difference between the average cost and the bigger ones = $24 - 21 = $3

Therefore, number of bigger ones = $12 / 3 =4

Cheers...

Area B = The shortfall of the smaller ones from the average

Area A = The excess of the bigger ones above the average

Area A = Area B

3 x 4 = 3 x ?

Therefore, ? =

Last edited by tanxexy on Fri Jul 11, 2014 9:52 am, edited 1 time in total.

Post
by **HAPPYH** » Fri Jul 11, 2014 8:22 am

Very clearly explained. Thank you so muchtanxexy wrote:Another approach is to use the concept of area.tanxexy wrote:Hi,hweelit wrote:hi anyone can help to solve, thanks

mrs ang bought a certain number of ceramic pots. she bought 4 smaller ones at $18 each and some bigger ones at $24 each. if the average cost of all the ceramic pots was $21, how many bigger ceramic pots did she buy?

Here is a suggested solution:

Difference between the average cost and the smaller ones = $21 - 18 = $3

Total difference = $3 x 4 = $12

This difference must be compensated by the bigger ones to bring the average up to $21.

Difference between the average cost and the bigger ones = $24 - 21 = $3

Therefore, number of bigger ones = $12 / 3 =4

Cheers...

Area A = The shortfall of the smaller ones from the average

Area B = The excess of the bigger ones above the average

Area A = Area B

3 x 4 = 3 x ?

Therefore, ? =4

- MathIzzzFun
- KiasuGrandMaster
**Posts:**2358**Joined:**Wed Mar 09,**Total Likes:**5

Post
by **MathIzzzFun** » Sun Jul 13, 2014 9:33 pm

.

Last edited by MathIzzzFun on Wed Jul 23, 2014 3:58 pm, edited 1 time in total.

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