Q&A - PSLE Math

[Vanilla Cake]

Suz855:

A,B, C and D shared a box of pens. A received 20% of the total number of pens which B, C and D received altogether. B received 50% of the total number of pens which A, C and D received altogether. C received 80% of the total number of pens which A, B and D received altogether. D received 6 pens. How many pens were in the box at first?

A : B+C+D
20% : 100%
1 : 5
Multiply by 3
3 : 15

B : A+C+D
50% : 100%
1 : 2
Multiply by 6
6 : 12

C : A+B+D
80% : 100%
4 : 5
Multiply by 2
8 : 10

LCM of 6, 3 and 9 is 18

A: 3u
B: 6u
8u
1u

D received 1u ie 6 pens
Total number of pens in the box at first = A+B+C+D = 3u+6u+8u+1u=18u = 18x6 = 108

Source: http://www.orlesson.org/orp/09Ma/2009-Math-SA2-Nanyang.pdf - 5 marks.

Pls type the original question accordingly to the source.
A -> Ali
B -> Bala
C -> Clyde
D -> Dingyi

Mod: Pls merge this question in http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=280&start=1750.

[Vanilla Cake]

pslemum:

Anybody can help with this question:

Hi pslemum
Welcome to KSP!
In future, pls post P5/P6 Maths questions in http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=280&start=0 or http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=6373&start=0. It's helpful for potential problem solvers in one way or another if you are able to quote the source or provide the given answer if available.

pslemum:

Container x had thrice as many jelly beans as Container Y. After 15% of the jelly beans in container X and 20% of the jelly beans in container Y were transferred to container W, the number of jelly beans in container W increased by 39%. COntainer W had 834 jelly beans in the end. How many jelly beans were there in container X in the end?

Before
X : Y : W
3u : u : Unknown

It's not important to find the unknown for W.

Change
15% of the jelly beans in container X and 20% of the jelly beans in container Y were transferred to container W.
15% of X = 15/100x3u = 0.45u
20% of Y = 20/100xu = 0.2u

In the end, number of jelly beans in container W increased by 39% due to 15% of X+20% of Y ie 0.45u+02u = 0.65u

Container W had 834 jelly beans in the end means 139% -> 834
39% -> 834/139x39 = 234 (increase in the number of jelly beans)

0.65u -> 234
u -> 360

X in the end -> 3u-0.45u = 2.55u = 2.55x360 = 918.

Number of jelly beans in container X in the end = 918.

Mods: Pls merge this post in http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=280&start=1750
Thanks.

[raysusan]

Vanilla Cake:

Suz855:

A,B, C and D shared a box of pens. A received 20% of the total number of pens which B, C and D received altogether. B received 50% of the total number of pens which A, C and D received altogether. C received 80% of the total number of pens which A, B and D received altogether. D received 6 pens. How many pens were in the box at first?

A : B+C+D
20% : 100%
1 : 5
Multiply by 3
3 : 15

B : A+C+D
50% : 100%
1 : 2
Multiply by 6
6 : 12

C : A+B+D
80% : 100%
4 : 5
Multiply by 2
8 : 10

LCM of 6, 3 and 9 is 18

A: 3u
B: 6u
8u
1u

D received 1u ie 6 pens
Total number of pens in the box at first = A+B+C+D = 3u+6u+8u+1u=18u = 18x6 = 108

Source: http://www.orlesson.org/orp/09Ma/2009-Math-SA2-Nanyang.pdf - 5 marks.

Pls type the original question accordingly to the source.
A -> Ali
B -> Bala
C -> Clyde
D -> Dingyi

Mod: Pls merge this question in http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=280&start=1750.

sorry i dont understand the part of x3,x6,x2
why the multiply so weird :?

[abc_parent]

To Vanilla Cake!

[abc_parent]

Thanks to Vanilla Cake and ksi!

[underthesea]

James Ang:

The solution supplied with the set of prelims paper is wrong for question 18 and it is solved using algebra, and in one of the step the solution showed 7 instead of 9. The correct answer is Vanilla = 72 and Chocolate = 138 after correcting for that wrong step. question 17 on the two candles is interesting, it is exactly the same as what

I mentioned weeks ago, source is a sec 1 textbook, but can be easily solved by drawing a model. Again the supplied solution uses algebra which p6 students may not be taught to solve using algebra.

Vanilla Cake:

Susan has some vanilla and chocolate sweets in her bag. If she adds 20 vanilla sweets into the bag, the percentage of vanilla sweets in the bag becomes 40%. If she adds 30 chocolate sweets, there will be 70% of chocolate sweets in the bag. How many vanilla and chocolate sweets does she have? [5 marks][Answer given: 64 vanilla sweets and 126 chocolate sweets]

Thank you for your time and effort. I appreciate your help as I think that the answer given is incorrect.

My answers are : Vanilla sweets: 72 and Chocolate sweets: 138
Took me about 10 min plus using 2 different methods to crack this problem sum.Pls help to clarify if I am correct or the given answer is correct.

Can you provide the solution without using algebra? Many thanks.

[pecalis]

My Humble attempt:


Add 20 Vanilla sweets, ratio of Van : Choc becomes 40 : 60 = 4 : 6
+5 +5 +5 +5
[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] [ ]
+5 +5 +5 +5 +5 +5

When add 30 Choc sweets, ratio of Van : Choc becomes 30 : 70 = 3 : 7
Remove the 20 vanilla sweets, and change the above [ ] to 3u each, so can get the correct ratio of 3 : 7
So,

+5 + 5 + 5 +5
[3u] [3u] [3u] [3u]
[3u] [3u] [3u] [3u] [3u] [3u]
+5 +5 +5 +5 +5 +5 becomes

[4u] [4u] [4u]
[4u] [4u] [4u] [4u] [4u] [4u] [4u]

=[3u] [3u] [3u] [3u]
[3u] [3u] [3u] [3u] [3u] [3u]+5 +5 +5 +5 +5 +5 +30
\t\t\t\t

Therefore, 28u = 18u + 60, 10u = 60, u = 6
No. of vanilla sweets is 12u = 12*6 = 72
No. of Choc sweets is 18u + 30 = 18*6 + 30 = 138

[tianzhu]

Hi


Hope this helps.
Best wishes

http://farm5.static.flickr.com/4127/4982279423_aee19502f7_b.jpg\">

[James Ang]

You can follow tianzhu's diagram which is correct.


To add,

The rate for candle A is 1/5 per hour, means 4/20 per hour (4u)
The rate for candle B is 1/4 per hour, means 5/20 per hour (5u)
That's why burning rate is 4:5

Candle A [][][ 4u ] =5h
Candle B [][ 5u. ] = 4h
[] = 1 u, total length is 6u, so

4/6 x 5h = 3 and 1/3 h
Or
5/6 x 4 h = 3 and 1/3 h

underthesea:

James Ang:

The solution supplied with the set of prelims paper is wrong for question 18 and it is solved using algebra, and in one of the step the solution showed 7 instead of 9. The correct answer is Vanilla = 72 and Chocolate = 138 after correcting for that wrong step. question 17 on the two candles is interesting, it is exactly the same as what
I mentioned weeks ago, source is a sec 1 textbook, but can be easily solved by drawing a model. Again the supplied solution uses algebra which p6 students may not be taught to solve using algebra.
[quote=\"Vanilla Cake\"]Susan has some vanilla and chocolate sweets in her bag. If she adds 20 vanilla sweets into the bag, the percentage of vanilla sweets in the bag becomes 40%. If she adds 30 chocolate sweets, there will be 70% of chocolate sweets in the bag. How many vanilla and chocolate sweets does she have? [5 marks][Answer given: 64 vanilla sweets and 126 chocolate sweets]

Thank you for your time and effort. I appreciate your help as I think that the answer given is incorrect.

My answers are : Vanilla sweets: 72 and Chocolate sweets: 138
Took me about 10 min plus using 2 different methods to crack this problem sum.Pls help to clarify if I am correct or the given answer is correct.

Can you provide the solution without using algebra? Many thanks.[/quote]

[BryBenMum]

Vanilla Cake:

BryBenMum:

Ailing, Bala, Chris and Dawn shared the cost of a gift for their friend. Ailing paid 1/5 of the amt paid by Bala, Chris and Dawn. Chris paid 40% of the amt paid by Ailing and Bala. The total amt paid by Ailing and Bala is equal to that paid by Chris and Dawn. Given that Dawn paid $32 more than Ailing for the gift, what was the cost of the gift?

A : B+C+D
1 : 5
5 : 25

A+B : C+D
1 : 1
15 : 15

A: 5u
B: 15u-5u = 10u
C+D: 25u-10u=15u

C paid 40% of A+B -> 40/100x15u=6u
15u-6u=9u

D-A = 9u-5u = 4u

4u-> $32
u -> $8

Cost of gift = 30u = 30x$8 = $240

Thank You !