Hi there,

Following pls find my worked solutions:

Q1 (solution posted 6 Apr)

There are different ways in which this can be organised. To organise this easily and optimising the usage of these 40 tables (T), it can be done in the following manner using multiple of 6 *(note: my answer is not exhaustive as there can be various combination)*

Why multiple of 6?

Multiple of 8 and 10 cannot be used for the Sat party of 120pax. Such grouping will exceed the max number of tables (40) it has.

Multiple of 6 - such grouping requires minimal reorganising of the tables. As the party number increases from Thu to Sat, additional grouping of 2T (plus other grouping) will be added. Existing grouping of 2T (from previous day's party) does not require reorganisation. Hence, minimal work with maximum efficiency.

Table seating:

1T = 4p

2T = 6p

3T = 8p

4T = 10p

(i) Thursday party of 64:

(10 × 6) + 4 = **(10 × 2T) + (1 × 1T)**

(ii) Friday party of 98:

(15 × 6) + 8 = **(15 × 2T) + (1 × 3T)**

{Add in 12T - 5 groups of 2T; remaining 2T joined with 1T from Thu party to cater for 1 group of 8pax}

(iii) Saturday party of 120:

(20 × 6) = **(20 × 2T)**

{Add in remaining 7T - 3 groups of 2T; remaining 1T joined with 1T from Fri party to cater for 1 group of 6pax}

Q2

In 1 day --> (1/12)total of food consumed

In 3 days --> (3/12)total of food consumed

Remaining food --> (9/12)total

Average food consumption per person per day --> (1/12) ÷ 40 = (1/480)

In 1 day, 48 people would consume [48 × (1/480)]total of food --> (1/10)total of food

Since remaining food is (9/12)total, this quantity of food would last [(9/12) ÷ (1/10)] days.

(9/12) ÷ (1/10) = **7.5 days**

Q3

Using <List Out> method , we have ...

5 runners (A,B,C,D,E):

{with A,B positions fixed}

A B C

A B D

A B E

{with A,C positions fixed}

A C B

A C D

A C E

{with A,D positions fixed}

A D B

A D C

A D E

{with A,E positions fixed}

A E B

A E C

A E D

If A comes 1st, there are 12 combinations in which other runners can finish in 2nd and 3rd positions.

Given there are 5 runners, the above combinations can be repeated for each of B,C,D,E if anyone of them were to finish in 1st position.

Hence, total combinations = 5 × 12 = **60**

(b)

10 runners (A,B,C,D,E,F,G,H,I,J)

{with A,B positions fixed}

there are 8 combinations in which the other 8 runners will finish in 3rd.

{with A position fixed}

there are 9 combinations in which the other 9 runners will finish in 2nd.

So with A position fixed (1st), there are (9 × 8 = 72) combinations.

Given there are 10 runners, the above combinations can be repeated for each of B,C,D,E,F,G,H,I,J if anyone of them were to finish in 1st position.

Total combinations = 10 × 72 = **720**

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The above combination is about Permutation (as order is important).

For 5 runners, its 5P3.

Using a calculator, this gives **60**

For 10 runners, its 10P3.

Using a calculator, this gives **720**

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Trust these help.

Do let me know again if this is different from your Answerkey or if there's further clarification.

Cheers,

Edward