Sign Up!
Login
Welcome to Test-paper.info
Wednesday, November 22 2017 @ 01:16 AM CST
 Forum Index >  Test Paper Related >  Primary 6 Matters
 Help on 5 maths questions
   |  Printable Version
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

Q1. Betty had 3440 red, yellow and blue beads. The number of red and yellow beads that Betty had was equal.
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

Forum Junior
Junior

Registered: 12/31/06
Posts: 27

Profile Email    
   
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


Forum Active Member
Active Member

Registered: 04/21/11
Posts: 627

Profile Email    
   
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

Forum Active Member
Active Member

Registered: 04/21/11
Posts: 627

Profile Email    
   
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)

Forum Regular Member
Regular Member

Registered: 03/20/12
Posts: 111

Profile Email    
   
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]

Forum Regular Member
Regular Member

Registered: 03/20/12
Posts: 111

Profile Email    
   


 All times are CST. The time is now 01:16 am.
Normal Topic Normal Topic
Locked Topic Locked Topic
Sticky Topic Sticky Topic
New Post New Post
Sticky Topic w/ New Post Sticky Topic w/ New Post
Locked Topic w/ New Post Locked Topic w/ New Post
View Anonymous Posts 
Able to Post 
HTML Allowed 
Censored Content