Sign Up!
Login
Welcome to Test-paper.info
Friday, July 21 2017 @ 09:37 PM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 Maths - ACS SA1 2012
   |  Printable Version
By: happygolucky (offline)  Sunday, July 14 2013 @ 03:27 AM CDT (Read 1265 times)  
happygolucky

Hi

Please help with the following questions. Thanks.

Q27. The ratio of the number of stickers Aaron had to the total number of stickers Benedict and Caleb had was 2:3. The ratio of the number of stickers Benedict had to the total number of stickers Aaron and Caleb had was 4:5. Find the ratio of the number of stickers Aaron had to the number of stickers Benedict had to the number of stickers Caleb had.
Ans : 18:20:7

Q30. Every Saturday, Mabel will go for her usual exercise. If she runs up 32 steps and walks up 68 steps, she will take 184 seconds. If she runs up 50 steps and walks 50 steps, she will take 175 seconds. If she runs at the same rate throughout the whole exercise, how long will she take to run the 100 steps?
Ans : 150

Q6 . Phoebe's annual expenditure is 2/5 less than Chloe's and 1/5 more than Denise. If they spent $34020 in a year altogether, find Denise's monthly expenditure.
Ans : 675

Q17. Nicole received more money than Tricia from their mother. 40% of Tricia's money is $16 less than 25% of Nicole's money. If Nicole gives 35% of her money to Tricia, Nicole will have $12 less than Tricia.
a) How much money does Tricia has?
b) How many per cent more money does Nicole has than Tricia at first? Round off your answers to the nearest 2 decimal places.
Ans: 60, 166.67%

Forum Newbie
Newbie

Registered: 12/31/06
Posts: 2

Profile Email    
   
By: echeewh (offline)  Sunday, July 14 2013 @ 07:12 AM CDT  
echeewh

Hi <happygolucky>,

Appreciate that you try to limit the number of questions per post / thread to 2. This will allow us to respond to your questions in a more efficient manner. Thank you.

Following are my worked solutions to Q27 and 30. The rest will follow in due course.

Q27.

** - for alignment purpose


A : (B + C) ** Total
2 : *** 3 ****** 5 -- (1)

B : (A + C) ** Total
4 : *** 5 ****** 9 -- (2)

As the total number of stickers A,B,C remains the same in both ratios, we can apply the <Unchanged Total> concept.

Multiply (1) 9x , (2) 5x.

A : (B + C) ** Total
18 : 27 ****** 45 -- (3)

B : (A + C) ** Total
20 : 25 ****** 45 -- (4)

Hence, C = 45u - 18u - 20u = 7u

A : B : C
18 : 20 : 7

=========

Q30.

32R + 68W = 184 -- (1)
50R + 50W = 175 -- (2)

Eliminate W by taking the Lowest common multiple (LCM) of 68, 50 is 1700.

(1)x25:
800R + 1700W = 4600 -- (3)

(2)x34:
1700R + 1700W = 5950 -- (4)

(4)-(3):
1700R - 800R = 5950 - 4600
900R = 1350
R = 3/2

To run the 100 steps,
100R = 100 × (3/2) = 150 secs

=========

Trust this helps.

Do let me know again if there's further clarifications.

Cheers,
Edward

Forum Active Member
Active Member

Registered: 04/21/11
Posts: 623

Profile Email    
   
By: echeewh (offline)  Sunday, July 14 2013 @ 11:18 PM CDT  
echeewh

//
Q6, 17 posted on Jul 15, 12 45
//


Q3.

<Annual Expenditure>

P is (2/5) less than C, i.e. C has 5 parts (p) while P has 3 parts (p).

P is (1/5) more than D, i.e. 3p of D = (6/5).
D = (5/5) = 3p × (5/6) = (5/2)p

So we have the 1p split into 2 units(u) as shown.


P |--|--|--|--|--|--| ***** --> 3p × 2 = 6u
C |--|--|--|--|--|--|--|--|--|--| ***** --> 5p × 2 = 10u
D |--|--|--|--|--| ***** --> (5/2)p × 2 = 5u


Given that total expenditure is $34020, we have ...

6u + 10u + 5u = 34020
21u = 34020
1u --> 1620

D (Annual expenditure) = 5u --> 5 × 1620 = 8100

Monthly expenditure of D:
8100 ÷ 12 = $675

========

Q17.

*** - alignment purpose


40% T = (2/5)T
25% N = (1/4)N

Given that 40% of Tricia's (T) money is $16 less than 25% of Nicole's (N) money, we have ...

(1/4)N - (2/5)T = 16

T |--|--| ******* --> (2/5)T
N |--|--|<16> * --> (1/4)N

<Before>

T |--|--|--|--|--| ***** --> 5u
N |--|--|<16>|--|--|<16>|--|--|<16>|--|--|<16> ***** --> 8u + 64

<35% N given to T>

35% N = (7/20)N

(7/20)N = [(7/20) × 8u] + [(7/20) × 64]
= (14/5)u + (112/5)

<After>

T: 5u + (14/5)u + (112/5)
= (39/5)u + (112/5)

N: 8u + 64 - (14/5)u - (112/5)
= (26/5)u + (208/5)

T - N = 12
(39/5)u + (112/5) - (26/5)u - (208/5) = 12
(13/5)u = 12 - (112/5) + (208/5) = (156/5)
13u --> 156
1u --> 12

(a)
T (at first) = 5u --> 5 × 12 = $60

(b)
N (at first) = 8u + 64
8u + 64 --> (8 × 12) + 64 = $160

N - T = 160 - 60 = 100

% more than T = (100 ÷ 60) × 100 ~ 166.67% (nearest 2 dec places)

========

Let me know again if there's further clarifications.

Cheers,
Edward

Forum Active Member
Active Member

Registered: 04/21/11
Posts: 623

Profile Email    
   


 All times are CDT. The time is now 09:37 pm.
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