hey there,

You have posted 6 Questions in one post and without any proper labelling of ur questions. I suggest that in future , pls do limit or keep the number of Questions to 2 or 3 in one post so that it wont take us (solution providers) so much time to do all at one go.

-----------------------------------------------------------

Q1.

Total = 86 x 4 = 344

344 - 84 - 91 = 169

Smallest possible difference betw C and M implies the one with the lower marks (C) must be maximum while the one with higher marks (M) must be minimum.

Hence, C = 79; M = 169 - 79 = 90

Smallest possible difference = 90 - 79 = 11

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

Q2.

(10/11) ÷ (1/7) = (10/11) x 7 = 70/11 = 6 4/11 ? 6 bags

6 x (1/7) = 6/7 kg

Fraction of sugar left = (10/11) - (6/7) = (70/77) - (66/77) = (4/77) kg

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

Q3.

V: 13/20

T: 75% = 3/4 = 15/20

None: 5% = 1/20

Total: 29/20

The extra 9/20 (or 9u) represents the number of pupils in the survey who preferred both V and T. In other words, this 9u is counted twice in the Total.

Given that 90 pupils chose both V and T, we have ...

9u --> 90

1u --> 10

Number of pupils who took part in survey:

= 20u --> 20 x 10

= 200

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

Q4.

<Before>

A : B

6 : 10

3 : 5

<Process>

Amy (A) donated $52 and Baoyu ( B ) donated $60

<After>

A : B

1 : 5

There are several methods/techniques in solving this: Model / Cross-Multiply / Simultaneous to name a few.

The concept here is known as <Changed Qtys> and my preferred method is <Cross-Multiply>

<Cross-Multiply>

(3u - 52) / 1 = (5u - 60) / 5

5 x (3u - 52) = 1 x (5u - 60)

15u - 260 = 5u - 60

15u - 5u --> 260 - 60

10u --> 200

1u --> 20

Total savings (at first) = 8u --> 8 x 20 = $160

Alternative method:

<Model>

(***) - for alignment purpose

<before>

A |------|------|------|

B |------|------|------|------|------|

<process>

As B has 5p in the beginning and B donated $60, we can rewrite this as...

1u ( B ) (at end) = 1p ( B ) - 12

So the <after> model for B will be as follows (5 units), while A will have 1 of such unit.

<after>

A |---|--| --> (1p - 12)

B |---|--|---|--|---|--|---|--|---|--| --> (5p - 60)

******* |

****** 12

Now, comparing A's model <before> and <after>, we have ...

3p - (1p - 12) --> 52

2p + 12 --> 52

2p --> 52 - 12 = 40

1p --> 20

Total savings (at first) = 8p --> 8 x 20 = $160

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

Q5.

(***) - buffer for alignment purpose

<Before>

20cts : 50cts

**4 : 5

<Process>

16 x 50cts = 800cts

800 ÷ 20 = 40 (20cts) coins

Hence, 40 (20cts) coins added, 16 (50cts) coins removed.

<After>

**8 : 7

Applying <Changed Qtys> concept and using <Cross-Multiply> method as in Q4, we have ...

(4u + 40) / 8 = (5u - 16) / 7

7 x (4u + 40) = 8 x (5u - 16)

28u + 280 = 40u - 128

40u - 28u --> 280 + 128

12u --> 408

1u --> 34

<Before>

20cts: 4 x 34 = 136

Value: 136 x 20 = 2720

50cts: 5 x 34 = 170

Value: 170 x 50 = 8500

Hence, total sum of money in coin box

= 2720 + 8500

= $112.20

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

Q6.

By going thru' the table, you will gather the following patterns shown:

A: multiple of 8 + 1

B: multiple of 8 + 2; multiple of 8 + 6 (after subtracting 2)

C: multiple of 4 + 3

D: multiple of 2 + 4; multiple of 8 + 4

E: multiple of 8 + 5

Work 500 through each of these patterns in the respective columns, you will have in Column D as shown:

500 - 4 = 496

496 is a multiple of 2 and 8.

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

Trust the above helps.

Do let me know if any of these is different from the answerkey.

Cheers,

Edward