
By: Kuro (offline) Sunday, July 07 2013 @ 09:30 PM CDT (Read 1674 times)



Kuro 
Q1. There are two bags of jelly beans labelled A and B. In bag A, there are 470 red jelly beans and 480 green jelly beans.
In Bag B, there are 350 red jelly beans and 200 green jelly beans. Some red and green jelly beans from bag B are transferred to bag A.
As a result, 60% of the jelly beans in Bag A and 20% of those in Bag B are red.
How many red and how many green jelly beans are transferred from Bag B to Bag A?
Ans: Red=310, Green=40
Q2. Rui Wen had 42 more beads than Shamila. Each of them gave away some of their beads to their friends.
The number of beads Shamila gave away was 4/7 of the number of beads Rui Wen had at first.
The number of beads Rui Wen gave away was 2/3 of the number of beads Shamila had at first.
Both had an equal number of beads left. How many beads did Rui Wen have at first?
Ans: 735
Q3, Q4 and Q5 are in attachment Q456.jpg.
Thanks for help rendered.

Junior
Registered: 12/31/06 Posts: 27





By: echeewh (offline) Sunday, July 07 2013 @ 10:55 PM CDT



echeewh 
Hey <Kuro>,
Appreciate that you post 2  3 questions at any one time in future as it will take some time to provide you with the worked solutions.
Following is my worked solution for Q2, as well as a reference to the concept and method of how to solve your Q1. As for the remaining questions in your attachment, will work on them and provide the solutions in due course.
// Q5 posted Jul 8, 2013, 14 30 //
// Q3 posted Jul 8, 2013, 16 00 //
// Q4 posted Jul 9, 2013, 07 15 //
Q1.
This question is very similar to several ones done recently. You may refer to the concept and method of how to solve this problem by referring to this thread/post listed "P6 Percentage" (date last posted: Apr 8, found in pg 13 as at Jul 8, 2013).
Apply <Unchanged Total > concept , i.e. total number of Red (R) jellybeans in Bags A + B <after> = total number of Red (R) jellybeans in Bags A + B <before>. Likewise, for Green (G). Given the % of R + G left in Bags A, B , the <After> ratios in the two bags can be established. Let the ratio in bag A be the parts (p) qty , while the ratio in bag B be the units (u) qty. We can then form 2 equations from here by applying the <Unchanged Total > concept; hence, use the <Simultaneous method> to solve for either the (p) or (u) qty, to answer the question.
Let me know again if you are not able to solve it. Thank you.
========
Q2.
<before>
R: <42>
S: 
<process>
S gave away (4/7)R at first
R gave away (2/3)S at first
Common multiple of 7,3 is 21.
So, can split the equal portions of R,S <Before> into 21 equal units (u). Hence, the <Before> model is transformed as shown:
<Before>
R: 21u + 42
S: 21u
(4/7)R:
(4/7) × (21u + 42) = 12u + 24
(2/3)S:
(2/3) × 21u = 14u
<After>
R:
21u + 42  14u = 7u + 42
S:
21u  12u  24 = 9u  24
Given that both had equal number of beads left,
7u + 42 = 9u  24
9u  7u = 42 + 24
2u > 66
1u > 33
R (at first):
21u + 42 > (21 × 33) + 42 = 735
=======
// posted Jul 8, 2013, 16 00 //
Q3.
(a)
Capacity of tank = 95 × 32 × 48 = 145920 cm3
(b)
Refer to tilted tank in Figure B.
The crosssectional area of the tilted tank in B is in the form of a triangle.
Vol of water in tank = Crosssectional Area × height
The height, in this case, is the 32 cm breadth (width) of the tank on which the tilted tank rested.
Given that water level reached the halfway mark of the length of the base, we have ...
95 ÷ 2 = 47.5
Vol of water in tank = [(48 × 47.5) ÷ 2] × 32 = 36480
Vol of water needed to fill up tank
= 145920  36480
= 109440 cm3
(c)
Height of water in tank (Figure A)
= 36480 ÷ (95 × 32)
= 12 cm
========
// posted Jul 8, 2013, 14 30 //
Q5.
See attachment for worked solution.
========
Trust this helps.
Do let me know again if there's further clarifications.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 627





By: echeewh (offline) Monday, July 08 2013 @ 03:00 AM CDT



echeewh 
hey <kuro>,
Q3 and 5 posted Jul 8, 1600 and 1430 respectively.
========
Do let me know again if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 627





By: echeewh (offline) Monday, July 08 2013 @ 06:15 PM CDT



echeewh 
// posted Jul 9, 2013, 07 15 //
Q4.
Number of Squares (SQ)
= Square of Pattern Number (n) = n × n
(a)
Pattern 5 = 5 × 5 = 25
Number of Sticks (ST) in Pattern 5
= 4 + 9 + 9 + 9 + 9 + (12 × 2) = 40 + 24 = 64
(b)
Pattern 105 = 105 × 105 = 11025
(c)
Number of Squares = 625
This is in Pattern 25. {Square Root of 625}
In every Pattern, there is always 1 Square with 4 Sticks.
n = Pattern Number
Number of Squares with 3 Sticks (SQ3)
= (n  1) × 3
Number of Squares with 2 Sticks (SQ2)
= [(n × n)  SQ3  1]
SQ3 = (25  1) × 3 = 72
SQ2 = (625  72  1) = 552
Number of Sticks (ST) in Pattern 25
= (72 × 3) + (552 × 2) + 4
= 216 + 1104 + 4
= 1324
=========
Trust this helps.
Do let me know again if there's further clarifications.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 627



