
By: Kuro (offline) Sunday, June 22 2014 @ 09:00 PM CDT (Read 3640 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

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: 625





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: 625





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 semicircle radius 2.5 cm  all these in the top right quarter of the big square.
Then you add the semicircle 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



