Q&A - PSLE Math

[ozora]

Th breadth of rectangle is reduced by 20%. To maintain the same area, its length is increased . A) what is the % increase in its length?

B) what is the % increase in its perimeter given that the ratio of its original length to its original breadth is 9:5? Answer round off to hundredths.
Need some help.

[speedmaths.012624com]

ozora:

Th breadth of rectangle is reduced by 20%. To maintain the same area, its length is increased . A) what is the % increase in its length?

B) what is the % increase in its perimeter given that the ratio of its original length to its original breadth is 9:5? Answer round off to hundredths.
Need some help.

Hi,

One possible solution:

Area = Length x Breadth
Area = L x B

Let the L and B be 10 each.
(Okay, it is a square, but all squares are rectangles.)

Area = 10 x 10 = 100

Breadth reduced by 20% from 10 to 8.

(new L) x 8 = 100
(new L) = 100 / 8 = 12.5

Can you work from here?

Cheers



speedmaths.com


.

[ozora]

speedmaths.com:

ozora:

Th breadth of rectangle is reduced by 20%. To maintain the same area, its length is increased . A) what is the % increase in its length?

B) what is the % increase in its perimeter given that the ratio of its original length to its original breadth is 9:5? Answer round off to hundredths.
Need some help.

Hi,

One possible solution:

Area = Length x Breadth
Area = L x B

Let the L and B be 10 each.
(Okay, it is a square, but all squares are rectangles.)

Area = 10 x 10 = 100

Breadth reduced by 20% from 10 to 8.

(new L) x 8 = 100
(new L) = 100 / 8 = 12.5

Can you work from here?

Cheers



speedmaths.com


.

Thanks but does the 2ndpart suppose to assist in answering the first part of answer.

[YumYum]

Hi, can anyone pls help with this Qn: http://i49.tinypic.com/23uoib.jpg\">

[charsen]

http://i50.tinypic.com/1z3a4bm.jpg\">

YumYum:

Hi, can anyone pls help with this Qn: http://i49.tinypic.com/23uoib.jpg\">

Here is the solution that I have attempted.

http://i50.tinypic.com/1z3a4bm.jpg\">

The final answer for (a) is 1078m2

[speedmaths.012624com]

ozora:

speedmaths.com:

[quote=\"ozora\"]Th breadth of rectangle is reduced by 20%. To maintain the same area, its length is increased . A) what is the % increase in its length?

B) what is the % increase in its perimeter given that the ratio of its original length to its original breadth is 9:5? Answer round off to hundredths.
Need some help.

Hi,

One possible solution:

Area = Length x Breadth
Area = L x B

Let the L and B be 10 each.
(Okay, it is a square, but all squares are rectangles.)

Area = 10 x 10 = 100

Breadth reduced by 20% from 10 to 8.

(new L) x 8 = 100
(new L) = 100 / 8 = 12.5

Can you work from here?

Cheers



speedmaths.com


.

Thanks but does the 2ndpart suppose to assist in answering the first part of answer.[/quote]Hi,

Yes, students usually use the 2nd part to help answer the first part.

For the first part, when you deal with percentage change, and area (or volume), you can use the original numbers (9 and 5), or you can use your own numbers (10 and 10, which is slightly easier)

Whichever set of numbers you use, you should get the answer as 25%, for the first part.

We normally encourage students to use 2 methods, when they have the time:
One method to get the answer
Another method to check the answer.

For the second part, when you deal with perimeter, you must use the numbers (9 and 5) given in the question.
Breadth drop from 5 to 4 (drop of 20%)
Length increase from 9 to 11.25 (increase of 25%)

Old perimeter is 9 + 9 + 5 + 5 = 28
New perimeter is 11.25 + 11.25 + 4 + 4 = 30.5

Can you work on the second part from here?

Hope this helps.

Cheers


speedmaths.com


.

[charsen]

.

[YumYum]

Hi Charsen, thanks.

[charsen]

YumYum:

Hi Charsen, thanks.

This question needs a bit of visualisation as in how the goat graze.I hope you can understand my working.

[Michaelia0816]

Thx Tianzhu for the yesterday question and the pervious day questions!