Q&A - P5 Math

[sunset_dae]

Thanks alot!


Can help to solve the following also:

In a school, the children are divided into two camps, A and B.
The ratio of the number of children in Camp A to Camp B is 2: 3. All the children in Camp A are boys. In camp B, the ratio of the number of boys to girls is 1:3.
(i) What fraction of the children in the school are boys?
(ii) if there are 60 more boys in camp A than B, how many girls are there in the school?



In a fruit stall, there were 175 apples and 25 oranges.
Later, an equal number of apples and oranges were brought into the stall. The ratio of the number of apples to the number of oranges became 4:1. Find the total number of apples and oranges brought into the stall


thanks alot!

[MathIzzzFun]

sunset_dae:

Thanks alot!


Can help to solve the following also:

In a school, the children are divided into two camps, A and B.
The ratio of the number of children in Camp A to Camp B is 2: 3. All the children in Camp A are boys. In camp B, the ratio of the number of boys to girls is 1:3.
(i) What fraction of the children in the school are boys?
(ii) if there are 60 more boys in camp A than B, how many girls are there in the school?



In a fruit stall, there were 175 apples and 25 oranges.
Later, an equal number of apples and oranges were brought into the stall. The ratio of the number of apples to the number of oranges became 4:1. Find the total number of apples and oranges brought into the stall


thanks alot!

Hi

For q2, it is similar to the first two questions. Use \"same gap/differnce\" to solve. Number of apples more than oranges = 175 - 25 = 150 --> this remains the same since same number of apples and oranges are added.

You may want to try it on your own.

q1.
Camp A : Camp B = 2 : 3
Camp B, boys : Camp B, girls = 1:3 (total 4)

make the number for Camp B the same .. lowest common multiple for 3 & 4 = 12

Camp A: Camp B = 2 : 3 = 8u : 12u, Total = 20u
Camp B, boys : Camp B, girls = 1 : 3 = 3u : 9u (total 12u = Camp B children)

Total number of boys = 8u (camp A) + 3u ( camp B) = 11u
Boys as fraction of total number of children in school = 11/20

If there are 60 more boys in Camp A than Camp B, then
8u - 3u = 60, 5u = 60
1u = 12

Number of girls = 9u = 9 x 12 = 108

cheers.

[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