Hey Shirley,

Following pls find my worked solution:

Q1.

Using <Work Backwards> to draw model from <After> to <Before>

<After>

B |----|<-260->|

H |----|

<Process>

3/4 of the guests in H and 3/5 of the guests in B checked out.

B: (2/5) --> 1p + 260

(5/5) --> [(1p + 260) x 5] / 2 = (5p/2) + 650

H: (1/4) --> 1p

(4/4) --> 4p

<Before>

B |-----|-----|--|<-----650---->|

H |-----|-----|-----|-----|

We can now divide 1p into 2 equal units (u) as follows for both B and H models.

B |--|--|--|--|--|<-----650---->|

H |--|--|--|--|--|--|--|--|

Given that there were 5200 guests in both B and H at first, we have ...

13u + 650 --> 5200

13u --> 5200 - 650 = 4550

1u --> 350

H (at first) = 8u --> 8 x 350 = **2800**

Alternative method to use (without drawing model) is the Simultaneous method as shown below:

H + B = 5200 -- (1)

(2/5)B - (1/4)H = 260 -- (2)

Transforming (2) to whole numbers ...

(2)x20:

8B - 5H = 5200 -- (3)

Since question is to find number of guests in H at first, we eliminate (B ) as follows:

(1)x8:

8H + 8B = 41600 -- (4)

Comparing (3),(4) ...

(4)-(3):

8H + 5H --> 41600 - 5200

13H --> 36400

H --> **2800**

=======================================

Q2

Using ratio and <Restating the Problem> technique, we have ..

A : C

5 : 1 --> (1)

A --> M : W

****** 1 : 3 --> (2)

C --> B : G

****** 5 : 3 --> (3)

Since C = 1p while B+G = 8u, we have ...

(1)x8:

A : C

40 : 8 --> (4)

Now that A = 40p while M+W = 4u, we have ...

(2)x10:

A --> M : W

***** 10 : 30 --> (5)

We use ratios in (3), (4), (5) to calculate the following:

Given that there were 450 more female than male audience in the concert, we have ...

Females = W + G = 30u + 3u = 33u

Males = M + B = 10u + 5u = 15u

Females - Males = 450

33u - 15u --> 450

18u --> 450

1u --> 25

W = 30u --> 30 x 25 = **750**

===========================

Trust the above helps.

Do let me know again if these do not agree with your Answerkey or if you have further clarification.

Cheers,

Edward