Hey Tping,

Following pls find my worked solutions:

Q1.

(1/5) x (12/5) = (12/25) km

(12/25) + (3/25) = (15/25) km = (3/5) km

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Q2.

*** - buffer space for alignment purpose

<after>

B |-----|<--260-->| --> (1p + 260)

H |-----| ********** --> (1p)

<process>

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

1/4 of guests in H, 2/5 of guests in B remained.

As (2/5) guests in B remained behind, split this (1p) into 2 equal units (u) and (260) into 130 each. Likewise, the (1p) in H is divided into (2u). So the (1p + 260) in B becomes 2 x (1u + 130), while the (1p) in H becomes (2u).

At start <before>, there should be 5 x (1u + 130) in B, while there should be 4 x (2u) in H.

Hence the <Before> model becomes as shown ..

<before>

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

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

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

B + H --> 5200

5u + 650 + 8u --> 5200

13u --> 5200 - 650 = 4550

1u --> 350

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

Alternative method ( Simultaneous method):

H + B = 5200 -- (1)

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

Transforming (2) into whole numbers...

(2)x20:

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

As question is interested in knowing number of guests in H at first, we shall eliminate (B ).

(1)x8:

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

(4)-(3):

8H - (-5H) --> 41600 - 5200

13H --> 36400

H --> 2800

Number of guests in H at first = **2800**

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Trust the above helps.

Do let me know if theres any clarification.

Cheers,

Edward