Hey there,

Pls find the worked solution:

Q15.

Packets A, B, C contained equal amt of cement in the end. Using model method, you start drawing from the end ( where there are equal amts).

Skill employed = Work Backwards

By Working Backwards, we have ...

(i) 1/3 transferred from C to A --> C would have 2 parts left. Hence C, A --> 2p each.

Working backwards, we add 1 part back to C while subtract 1 part from A.

Hence, C --> 3p; A --> 1p.

(ii) 1/3 transferred from B to C --> B would have 2 parts left. Hence, B --> 2p.

Working backwards, we add 1 part back to B while subtract 1 part from C.

Hence, B --> 3p; C --> 2p

(iii) 1/3 transferred from A to B --> A would have 2 parts left. From (i), A was left with 1p, Using model technique, we need to split this 1p into 2 smaller units (u). Hence, A --> 2u. Likewise, from (ii), the 3p in B would be splitted into 6u (3p x 2) while the 2p in C splitted into 4u (2p x 2).

Working backwards, we add 1 unit back to A while subtract 1 unit from B.

Hence, A --> 3u; B --> 5u

Therefore, at the start, A --> 3u; B --> 5u; C --> 4u. Total of these --> 36kg ( Total Unchanged concept - Total Before = Total After ).

12u --> 36 kg

1u --> 3kg

Packet B at first = 5u --> 5 x 3 = 15 kg.

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Q16.

Beside the skills used below, Model method ( using Work Backwards technique ) can also be used to derive the same answer.

Male = Female members at the end

Given that Males (M) reduces by 20% ( 1/5 ), M will be left with 4/5.

Given that Females (F) increases by 20% ( 1/5 ), F will be 6/5.

Using Equal concept, we have

4/5 M = 6/5 F

12/15 M = 12/10 F ( make numerator same )

Hence, M = 15u , F = 10u ( at the start = last year )

Now we have

25u --> 150 members

1u --> 6 members

Total members this year = 12u (M) + 12u (F) = 24u --> 24 x 6 = 144

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Trust the above helps.

Cheers,

Edward