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 By: Kuro (offline)  Sunday, June 22 2014 @ 09:00 PM CDT (Read 3690 times)
Kuro

I need help on 5 questions, 3 of the questions are in the attachment. Thanks

After using only some yellow and blue beads to make a necklace, she had 5/7 of yellow beads and 3/5 of blue beads left.
Given that the number of beads left was 2820, hw many blue beads did she have at first? Ans 500

Q2. 20% of Jack's stamps is equal to 30% of Vindra's stamps. 25% of Jack's stamps is equal to 35% of Vindra's and Kumar's stamps. Given that Jack and Kumar had a total of 242, how many stamps did Vindra have? Ans 154

Junior

Registered: 12/31/06
Posts: 27

 By: echeewh (offline)  Sunday, June 22 2014 @ 11:38 PM CDT
echeewh

Hey <Kuro>

Following are my worked solutions for Q1, 2: ( the remaining 3 Q in attachment shall be done in due course.)

Q1

Given that theres an equal number of R and Y beads,
R + Y = 2 parts (2 wholes)

2RY + 1B = 3440 -- (1)
(12/7)RY + (3/5)B = 2820 -- (2)

(2)x35:
60RY + 21B = 98700 -- (3)

Compare (1),(3) to eliminate RY.

(1)x30:
60RY + 30B = 103200 -- (4)

(4)-(3):
30B - 21B = 103200 - 98700
9B --> 4500
B --> 4500 ÷ 9 = 500

=========

Q2

*** - for alignment purpose

Apply <Equal Concept> method, we have ..

(1/5)J = (3/10)V
(3/15)J = (3/10)V

J : V
15 : 10
3 : 2

(1/4)J = (7/20)VK
(7/28)J = (7/20)VK

J : VK
28 : 20
7 : 5

J : V : VK
3 : 2 **** -- (1)
7 *** : 5 -- (2)
___________

(1)x7:

J : V : VK
21 : 14 ***** -- (3)

(2)x3:

21 **** : 15 -- (4)
______________
21 : 14 : 15

K = 15u - 14u = 1u

Given that Jack(J) and Kumar(K) had a total of 242,

J + K = 242
21u + 1u --> 242
22u --> 242
1u --> 242 ÷ 22 = 11

V = 14u --> 14 × 11
= 154

========

Trust these help. Do let me know again if there' re clarifications.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: echeewh (offline)  Monday, June 23 2014 @ 01:38 AM CDT
echeewh

Hey <kuro>

Following is my worked solution for Q4.

Q4.

Base Area of B = (1/2) × 80 × 20 = 800 cm2

After equal/same vol of water is drained from A and filled in B (B at first is empty),

Total vol in (A + B ) (at first) = Total vol in (A + B ) (after) = 80 × 20 × 30 = 48000 cm3

Given that height of water level in A, B (after) are the same (h),

Vol in A / vol in B = (80 × 20 × h) / (800 × h)
= 2

Vol(A) : Vol(B )
2 : 1

Hence, 3u --> 48000
1u --> 48000 ÷ 3 = 16000 cm3

This infers that 16000 cm3 will be filled in B while the same amount is drained from A.

Given the rate of filling up B is 800 cm3/min, time taken to fill B is ...

16000 ÷ 800 = 20 mins

======

Trust this helps. Do let me know again if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: jo sarah (offline)  Monday, June 23 2014 @ 08:03 AM CDT
jo sarah

hi kuro,

Question 3:

You swing the shaded region CDE into FAE,
so that you get a quadrant BCA of radius 10 cm minus the unshaded semi-circle radius 2.5 cm -- all these in the top right quarter of the big square.
Then you add the semi-circle radius 5 cm in the top left quarter of the big circle.

So the required total shaded area = [ (1/4) x pi x 10 x 10 - (1/2) x pi x 2.5 x 2.5 ] + (1/2) x pi x 5 x 5
= pi x [ 100/4 - 6.25/2 + 25/2 ]
= 107.99 cm^2

hope you can follow the reasoning (sorry, can't reproduce the diagram)

Regular Member

Registered: 03/20/12
Posts: 111

 By: jo sarah (offline)  Monday, June 23 2014 @ 08:44 AM CDT
jo sarah

solution to question 5 attached, please.
[drawing isn't too ideal, but trust it is helpful enough]

Regular Member

Registered: 03/20/12
Posts: 111

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