Monday, August 30, 2010

Percentage P5 2009 SA2 P2 Q17 Ai Tong

A seamstress had some white and some black buttons.60% of the buttons were white.He bought another 120 buttons.Now the percentage of white buttons became only 40% of the total.How many white buttons did he have at first?


W      B
---------
3   :   2 ---- 5 units
+ 120
---------
2   :   3 ----- 5 parts
---------

5 p --> 5 u + 120
1 p --> 1 u + 120/5 = 1 u + 24

2 p --> 2 u + 2 x 24 = 2 u + 48

3p --> 3u + 3 x 24 = 3 u + 72


3 u + ? --> 2 u + 48
2 u + (120 - ?) --> 3 u + 72

Both equations will end up with 1 u --> 48 -


Solution 1
-----------
W [][][][][][]
B  [][][][][][][][][]

5 u --> 120
1u --> 120 /5 = 24

W: 6 u --> 6 x 24 = 144 white buttons.


Solution 2
------------
W [][][][][][][][]
B  [][][][][][][][][][][][]

10 u --> 120
1u --> 120/10 = 12

W: 6 u --> 12 x  6 = 72 white buttons.


Solution 3
-----------
W [][][][][][][][][][][][][][]
B  [][][][][][][][][][][][][][][][][][][][][]

15 u --> 120
1 u --> 120/15 = 8

W: 12 u --> 12 x 8 = 96 white buttons.

etc.......

Friday, August 27, 2010

Speed P6

Jack cycles to his friend's house and then returns home by the same route. He always cycles at 4 km/h when going uphill, 12 km/h when going downhill and 6 km/h when on level ground. If his total cycling time is 2h 20mins, what is the distance he cycles in km?


D1 - uphill, D2 - downhill, D3 - level ground
Time from Home to Friend's house --> D1/4 + D2/12 + D3/6


D3 - level ground, D2 - uphill, D1 - downhill
Time from Friend's house to Home --> D3/6 + D2/4 + D1/12


Total time --> D1 (1/4 + 1/12) + D2 (1/12 + 1/4) + D3 (1/6 + 1/6)
                 = 1/3 D1 + 1/3 D2 + 1/3 D3
                 = 1/3 D

1/3 D --> 2 20/60 = 7/3

D --> 3 x 7/3 = 7 km

7 x 2 = 14 km


The distance he cycles is 14 km.

Challenging P6

A pair of hamsters can give birth to a pair of baby hamsters every month. Each pair of baby hamsters needs 1 month to mature and then this pair of hamsters will in turn give birth to a pair of baby hamsters every month. How many pairs of hamsters are there after 6 months? (4 marks)


The last digit represents the number of pairs of hamsters born. The front digits/alphabet represent the parents.  Remember to add 1 (S).

Wednesday, August 25, 2010

Challenging P5 Chong Fu

Nicholas has two 6-sided dice. One dice has numbers 1 ,3, 5, 7, 11 and 13 while the other dice has numbers 4, 8, 18, 24, 30 and 36. The dice are tossed together and their values added.
a)What fraction of the outcomes adds up to a sum divisible by 7?
b)What fraction of the outcomes adds up to a sum divisible by 3?


There are 36 outcomes (6 x 6 = 36)

Multiples of 7 --> 6 outcomes
(3 + 4, 3 + 18, 5 + 30, 11 + 24, 13 + 8, 13 + 36)

6 ÷ 36 = 1/6

a) Fraction of outcomes of sum divisble by 7 is 1/6.


Multiples of 3 --> 9 outcomes
(1 + 8, 3+ 18, 3 + 24, 3 + 30, 3 + 36, 5 + 4, 7 + 8, 11 + 4, 13 + 8)

9 ÷ 36 = 1/4

b)Fraction of outcomes of sum divisible by 3 is 1/4.

Sunday, August 1, 2010

NMOS 2010 Q1











A1 = A2

C1 = C2

So B1 = B2, which is

4 x 5 = 20 cm²

NMOS 2010 Q2

Q2. Adrian, Bryan, Charles, David, Elvin and Fedric played against one another in a singles tournament such that each one of them played every other player only once, and none of the games ended with a draw. If Adrian won 4 games, Bryan won 3 games, Charles won 2 games, David won 2 games and Elvin won 2 games, how many games did Fedric win?


6 x 5 / 2 = 15


15 - 4 - 3 - 2 - 2 - 2 = 2 games – won by F

NMOS 2010 Q3

Q3. At 6 a.m. in the morning, Andrew (at Town A) and Benjamin (at Town B) were travelling in opposite directions towards each other. They met each other along the way at 12 noon. Andrew reached Town B at 4 p.m. If they travelled with different constant speeds, how long (in hours) did Benjamin take to travel from Town B to Town A?


6 h : 4 h


A -- A (6 a.m.) --------------- M (12 noon) -------------- B (4 p.m.)


B – A (?) ---------------------- M (12 noon) -------------- B (6 a.m.)


6/4 x 6 : 6 h


= 9 h


9 + 6 = 15 h

NMOS 2010 Q4












      [   ]
[   ][   ][   ][   ]
      [   ]
      [   ]


3 + 4 + 5 + 6 + 7 + 8 + 9 = 42 -- even number

So thee number in the box at the intersection will have to be an even number -- 4, 6 or 8.

42 + 4 = 46

46 / 2 = 23

4 + 6 + 5 + 8 = 23     4 + 3 + 7 + 9 =  23

Digit 4 is fixed, so only 1 way.

5, 6 and 8 --> 3 x 2 x 1 = 6 ways

3, 7 and 9 --> 3 x 2 x 1 = 6 ways

5, 6, 8 and 3, 7, 9 can be switched, i.e. vertical becomes horizontal and horizontal becomes vertical

So total number of ways --> 1 x 6 x 6 x 2 = 72


42 + 6 = 48

48/2 = 24

6 + 7 + 3 + 8 = 24,   6 + 4 + 5 + 9 = 24


Number of ways --> 72 ways
   
42 + 8 = 50

50 / 2 = 25

8 + 7 + 4 + 6 = 25,   8 + 3 + 5 + 9 = 25

Number of ways --> 72 ways



Total number of ways --> 72 x 3 = 216.

NMOS 2010 Q5












An(1 h) + Bo(1 h) --> 1/16.8


An (1 h) + Ch(1 h) --> 1/21


Bo(1 h) + Ch(1 h) --> 1/28



Bo(1 h) - Ch(1 h) --> 1/16.8 - 1/21 = 1/84

2 Bo(1 h) --> 1/84 + 1/28 = 1/21
Bo(1 h) --> 1/2 x 1/21 = 1/42

1/42 wall --> 1 h
1 wall --> 42 x 1 = 42 h ---- B

An(1 h) --> 1/16.8 - 1/42 = 1/28
1 wall --> 28 x 1 = 28 h ----- A

Ch(1 h) --> 1/28 - 1/42 = 1/84
1 wall --> 84 x 1 = 84 h ---- C

A + B + C = 28 + 42 + 84
                 = 154 h

NMOS 2010 Q6














To walk on the same side, Peter would need to walk 150 m more than Queenie first.


50 – 47 = 3 m/min


3 m --> 1 min


150 m --> 150/3 x 1 = 50 min






50 m/min x 50 min = 2500 m


2500 / 300 = 8 r 100 m



To be on the same side, Peter would need to travel another 200 m more.  [since he was already 100 m on one side while Queenie was 300 m ahead (i.e. 100 m on another side of the triangle]


200 m / 50 m/min = 4 min


50 + 4 = 54 min

NMOS 2010 Q7



1/2 x 2 x 6 = 6

1/2 x 3 x 1 = 1.5

1/2 x 2 x 4 = 4

1/2 x 7 x 1 = 3.5

6 + 1.5 + 4 + 3.5 = 15


5 x 2 = 10

2 x 1 = 2

4 x 3 = 12

4 x 4 = 16

10 + 2 + 12 + 16 = 40


40 - 15 = 25 cm 2.

NMOS 2010 Q8












Shift triangle ADE to CDF (2 sides same and right angle)

16 = 4 x 4    DEBF is a square

½ x 4 x 2 = 4


16 – 4 = 12 cm²

NMOS 2010 Q9

Q9. a, b, c and d are four distinct positive integers such that the product of any two of them is a multiple of 12. Find the least possible value of a+b+c+d.


2 x 6, 2 x 12, 2 x 18 --> 2 + 6 + 12 + 18 = 38

a + b + c + d = 38

NMOS 2010 Q10

Q10. A shop owner sold only red dresses and blue dresses. The number of red dresses sold on Tuesday was 10% more than that sold on Monday. The number of blue dresses sold on Tuesday was 10% less than that sold on Monday. The total number of dresses sold on Tuesday was 5% more than that sold on Monday. What percentage of the dresses sold on Monday was red?


M --> 10 R + 10 B = 100%


T --> 11 R + 9 B = 105%



90 R + 90 B --> 900%


110 R + 90 B --> 1050%



20 R --> 1050% - 900% = 150%


10 R --> 10/20 x 150% = 75%

NMOS 2010 Q11

Q11. When each of the numbers 2010, 2377 and 3111 is divided by a number D, they give the same remainder R. Find the largest possible value of D+R.


2377 – 2010 = 367 -- D


2010 / 367 = 5 R 175


2377 / 367 = 6 R 175


3111 / 367 = 8 R 175



D + R = 367 + 175 = 542

NMOS 2010 Q12

Q12. We define the ending zeroes of an integer as the number of zeroes at the end of the integer. For example, 2010 has one ending zero whereas 10010000 has four ending zeroes. How many ending zeroes are there when the value of 16x17x18x…..x209x210 is simplified?


25, 75, 175  = ..00 --> 2 x 3 = 6


125 = ..000  --> 3

35, 45, 55, 65, 85, 95, 105, 115, 135, 145, 155, 165, 185, 195, 205 --> 15


50, 150 = ..00 --> 2 x 2 = 4

100, 200 --> 2 x 2 = 4


20, 30, 40, 60, 70, 80, 90, 110, 120, 130, 140, 160, 170, 180, 190, 210 --> 16



6 + 3 +  15 + 4 + 4 + 16 = 48 zeros

NMOS 2010 Q13












a + c = 2 b

b + (b + c) + (a + b) = 180



5 b = 180


b = 180/5 = 36


a + c = 2 x 36 = 72



72 + 36 = 108 degree

NMOS 2010 Q14

Q14. The ratio of male students to female students in a primary school is 5 : 4. The ratio of students who wear spectacles to students who do not is 11:1. Given that 64% of the students who do not wear spectacles are female, find the percentage of female students who do not wear spectacles.


M : F = 5 : 4 = 20 : 16


S : N = 11 : 1 = 33 : 3



64/100 x 3 = 1.92 u



1.92/16 x 100% = 12% -- Female who do not wear spectacles.

NMOS 2010 Q15

Q15. a, b, c and d are letters that represent four different positive integers such that axa+bxb=cxc+dxd. Find the least possible value of axa+bxb.


a x a + b x b = c x c + d x d


a x a – c x c = d x d – b x b


(a + c) x (a - c) = (d + b) x (d - b)


(8 + 7) x (8 - 7) = (4 + 1) x (4 - 1)



8 x 8 + 1 x 1 = 7 x 7 + 4 x 4 = 65

NMOS 2010 Q16

Q16. There are five members in John’s family: Father, Mother, John, elder sister Helen, and younger sister Lucy. Just before Lucy was born , the average age of the family was 19 years old, and the age of John’s father is the same as the sum of the age of the rest of the family. Two years ago, the average age of the family was 35.2 years old, and the age of John’s mother was eight years more than the sum of the ages of John and Lucy. The current average age of the family, excluding Helen, is 39.5 years old. Find the present age of John.


F, M, H, J, L

L = 0, F + M + J + H = 19 x 4 = 76


F = M + J + H = 76 / 2 = 38


2 years ago, F + M + J + H + L = 35.2 x 5 = 176


M = J + L + 8



Now, F + M + H + J + L = 176 + 5 x 2 = 186


M = J + L + 8 – 2 = J + L + 6



186 – 76 = 110


110 / 5 = 22 years – L



F = 38 + 22 = 60


F + M + J + L = 39.5 x 4 = 158




M + J = 158 – 60 – 22 = 76


M = J + 22 + 6 = J + 28



J = (76 – 28)/2 = 24 years old

NMOS 2010 Q17












A1 = A2

B1 = B2

A2 + B2 = 2 C

Total = 5 C

So 2 shaded parts are 

2/5 x 100 = 40 cm²

NMOS 2010 Q18

Q18 Albert had a basket of fruits which contained 35 cherries and 50 strawberries, while Calvin had another basket of fruits which contained 45 cherries and 10 strawberries. Albert gave Calvin some cherries and strawberries so that eventually, 40% of Albert’s fruit basket contained cherries and 30% of Calvin’s fruit basket contained strawberries. What was the total number of cherries and strawberries that Albert gave to Calvin?


A --> 35 C + 50 S


C --> 45 C + 10 S


3 x (35 – X) --> 2 x (50 – Y)


105 – 3 X --> 100 – 2 Y


3 X – 2 Y --> 105 – 100 = 5



7 x (10 + Y) --> 3 x (45 + X)


70 + 7 Y --> 135 + 3 X


7 Y – 3 X --> 135 – 70 = 65


5 Y --> 5 + 65 = 70


Y --> 70/5 = 14


3 X --> 5 + 2 x 14 = 33


X --> 33/3 = 11



X + Y --> 11 + 14 = 25 cherries and strawberries.

NMOS 2010 Q19

Q19. A, B and C are letters that represent different odd digits such that ABC is a 3-digit number. If ABC is a multiple of A, B and C , how many different 3-digit numbers ABC are there?


ABC = A x = B x = C x


A, B, C – odd digits

Odd digits – 1, 3, 5, 7, 9


5 x 4 x 3 = 60 different numbers



Numbers that meet the above conditions are


135 = 1 x 135, 3 x 45, 5 x 27


175 = 1 x 175, 7 x 25, 5 x 35


315 = 3 x 105, 1 x 315, 5 x 63


735 = 7 x 105, 3 x 245, 5 x 147



There are 4 different 3-digit numbers.

NMOS 2010 Q20

Q20 Twenty students took a test worth 100 marks. All their results were positive integers and none of them had the same score. Their teacher ranked them from the highest score (1st) to the lowest score (20th). The top 10 students had an average of 92 marks and the next 10 students have an average of 80 marks. The top 5 students had an average that was 12 marks higher than average of the next 15 students. Find the lowest possible score of the student in the 20th rank.


92 --> 91, 90, 89, 88, 87 (lowest)


           93, 94, 95, 96, 97

 80 --> 86, 85, 84, 83, 82, 81, 80, 79, 78, ? (lowest)

86 + 85 + 84 + 83 + 82 + 81 + 80 + 79 + 78 = 738

80 x 10 = 800

800 - 738 = 62 (lowest)


93 + 94 + 95 + 96 + 97 = 475


475 / 5 = 95


91 + 90 + 89 + 88 + 87 = 445


800 + 445 = 1245


1245 / 15 = 83




95 – 83 = 12 (as given in question)

So lowest possible score is 62.