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 By: 2000123 (offline)  Thursday, June 21 2012 @ 08:05 AM CDT (Read 881 times)
2000123

Jolene accidentally spilled ink onto her result slip.Her results were as follow. A question mark means that there were ink spilled on that number
English 84
Chinese 7?
Mathematics 9?
Science 91
Average 86
What could the smallest possible difference between her Chinese and Mathematics marks be?

10/11kg of sugar is packed into little bags.If each bag contain only 1/7kg of sugar,how much sugar is left unpacked?

In a survey,some pupils were asked if they preferred volleyball or table-tennis.13/20 of the pupils chose volleyball,75% of the pupils chose table-tennis and 5% of the pupils did not choose any of the 2 sports.
Given that 90 pupils chose both volleyball and table-tennis,how many pupils took part in the survey?

Amy savings was 40% less than Baoyu's savings at first.After Amy donated \$52 and Baoyu donated \$60, Baoyu's savings became 5 times as much as Amy's.What was their total savings at first?

A coin box contained only 20-cents and 50-cents in the ratio of 4:5.When 16 fifty-cent coins were taken out and replaced by some twenty-cent coins,the number of fifty-cent coins left in the box was 7/8 of the twenty-cent coins.The total value of all the coins remained the same.Find the sum of money in the coin box.

The numbers 1 to 500 are written in columns as shown.A - represents that the box is blank.

A B C D E
1 2 3 4 5
9 8 7 6 -
- 10 11 12 13
17 16 15 14 -
- 18 19 20 21
25 24 23 22 -
- 26 ... ... ...
If this pattern continues, in which column will the number 500 be written?

Junior

Registered: 03/07/12
Posts: 21

 By: echeewh (offline)  Thursday, June 21 2012 @ 08:54 PM CDT
echeewh

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

Active Member

Registered: 04/21/11
Posts: 627

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