Q&A - P5 Math

[esj152104]

A trapezium is divided into 4 parts. A and B form half the trapezium. C and D form the other half. The areas of A and B are in the ratio 2:3.

The areas of C and D are in the ratio 1:2.
(i) What fraction of the trapezium is part D?
(ii) If part B is bigger than part A by 12cm square , find the area of the trapezium.

Pls help to solve this. Thanks

[MathIzzzFun]

esj152104:

A trapezium is divided into 4 parts. A and B form half the trapezium. C and D form the other half. The areas of A and B are in the ratio 2:3.

The areas of C and D are in the ratio 1:2.
(i) What fraction of the trapezium is part D?
(ii) If part B is bigger than part A by 12cm square , find the area of the trapezium.

Pls help to solve this. Thanks

Hi

pls post diagram.

I will assume that Area of (A+B) = 1/2 area of trapezium = Area of (C+D)

Then
A: B = 2:3 = 6u : 9u
D= 1:2 = 5u : 10u

a)D as fraction of area of trapezium = 10u/30u = 1/3

b) B-A = 3u = 12 cm2, 1u = 4cm2
Area of trapezium = 30u = 30 x 4 cm2 = 120 cm2


cheers.

[esj152104]

Hi,


Thanks for your help in solving the question, as for the diagram the question was not given any diagram.

[esj152104]

Hi,


how do you derive the ratio to 6u:9u and 5u:10u
A: B = 2:3 = 6u : 9u
D= 1:2 = 5u : 10u

[wahwah]

Hi,

Need help to solve the following Math questions. Many thanks in advance.

There were a total of 2440 TV sets in Warehouse A and Warehouse B. After 2/3 of the stock in Warhouse A and 2/5 of the stock in Warehouse B were sold, there were 120 more TV sets in Warehouse B than in Warehouse A. How many TV sets were there in each warehouse at first?

The cost of a camera and a watch was $2200. The usual price of the camera was 20% more than its cost and the usual price of the watch was 10% more than its cost. During a promotion, both items were given a 10% discount of their usual price. As a result, there was a profit of $131. What was the cost of the camera?

An aquarium owner sold 98 rainbow fish on the first day and earned a profit of $44.10 for each of the rainbow fish sold. On the second day, 133 rainbow fish were sold and the aquarium owner gained a profit of 40% of the cost. If the money collected on both days were the same, how much dud each rainbow fish cost?

[MathIzzzFun]

[quote="esj152104"]Hi,


how do you derive the ratio to 6u:9u and 5u:10u
A: B = 2:3 = 6u : 9u
D= 1:2 = 5u : 10u

Hi

Total of A+B = 2+3 = 5 parts
Total of C+D = 1+2 = 3 shares
Since A+B=C+D, you need to "equalize" - common multiple is 15, so multiply A+B by 3, and multiply C+D by 5 to get total of 15 units each.

cheers.

[tianzhu]

Hi


I'll take Q2 and Q3.

Where are these questions from? Quite interesting ones.

Camera

Cost Price ------ 100 units
Usual Price ----- 120 units
Promotion price ------108 units

Watch

Cost Price ------ 100 parts
Usual Price ----- 110 parts
Promotion price ------ 99 parts

100 units + 100 parts ------ 2200
108 units + 99 parts ------ 2331

1 unit + 1 part ------ 22

9 units + 99 units + 99 parts -----2331
9 units + (99*22) ------2331

1 unit ----- 17

Hence 1 part -----5

Cost of camera ------17*100 -----1700


First day

Profit earned ------ 98*44.1 ------ 4321.8

Second day

CP per fish ----- 10 units
SP per fish ----- 14 units

Total selling price -----133 * 14 ------ 1862 units

1862 units ----- 980 units + 4321.8
1 unit ----- 4.9

CP per fish ------4.9*10 -------49.

Best wishes

[tianzhu]

Hi


I think MD is a neat way to solve Q1.

I’ll give you some pointers.

The units showing TV sets in warehouse A and B are different. They are not of identical sizes.

We shall use units for WA and parts for WB,

WA
Sold ----- 2 units -----6 units
Left ----- 1 unit ----- 3 units

WB
Sold ----- 2 parts
Left ----- 3 parts

3 parts ------ 120 more than 3 units
Hence, 1 part ------1 unit + 40

WB ------ 5 parts ----- 5 units + 200

14 units + 200 ------ 2440
1 unit ------ 160

WA ------ 1440
WB ------ 1000

Best wishes

[tianzhu]

teddy7171:

Hi tianzhu,


Thank you.

Hi

You're welcome.

Best wishes

[esj152104]

Hi,


Thks for solving the question for me.