PSLE Mathematic 2009 Questions and Suggested Answers – Chocolates and Sweets by Jim and Ken (includes list of mathematic questions)
Primary School Leaving Examination (PSLE)
The Primary School Leaving Examination (PSLE) is a national examination taken by all students in Singapore near the end of their sixth year in primary … (reference from Wikipedia)
“Jim bought some chocolates and gave half of it to Ken.
Ken bought some sweets and gave half of it to Jim.
Jim ate 12 sweets and Ken ate 18 chocolates.
The ratio of Jim’s sweets to chocolates became 1:7
The ratio of Ken’s sweets to chocolates became 1:4.
How many sweets did Ken buy?”
Solution by using algebra!
Let x be the number of sweets bought
Let y be the number of chocolates bought
From Jim:
Since he has x/2 -12 sweets, y/2 chocolates and there are 7 times as many chocolates as there are sweets,
7 (x/2 – 12) : y/2
14 (x/2 – 12) : y
7x – 168 : y
7x – 168 = y
From Ken:
Since he has x/2 sweets, y/2 -18 chocolates and there are 4 times as many chocolates as there are sweets,
y/2 – 18 : 4(x/2)
y/2 – 18 : 2x
y – 36 : 4x
y = 4x + 36
Solving simultaneously and eliminating y,
Make y the subject of formula for both equations, i.e.
7x – 168 = y
and
4x + 36 = y
Therefore,
7x – 168 = 4x + 36
3x = 204
x = 68
Solution by guessing without any algebra!
This method is thinking like a young school kid who doesn’t know how to do it in algebra.
Very simple and easy. However, this need to guessing when both Jim and Ken becomes equal.
For Jim:
(1+12) : (7 x 1) = 13 : 7
(2+12) : (7 x 2) = 14 :14
(3+12) : (7 x 3) = 15 : 21
:
(21+12) : (7 x 21) = 33 : 147
(22+12) : (7 x 22) = 34 : 154
For Ken
(1) : (4 x 1 + 18) = 1 : 22
(2) : (4 x 2 + 18) = 2 : 26
(3) : (4 x 3 + 18) = 3 : 34
:
(33) : (4 x 33 + 18) = 33 : 150
(34) : (4 x 34 + 18) = 34 : 154
Once both Jim and Ken reaches the same ratio, which is 34:154, then sum them together!
Summary:
34 + 34 = 68 (total sweets)
154 + 154 = 308 (total chocolates)
Solution by doing it mentally!
Jim and Ken had the same number of chocolates after Jim gave half to Ken. And they had the same number of sweets after Ken gave half to Jim.
Jim had 12 fewer sweets than Ken after he ate 12. So Ken’s 1 unit of sweets = Jim’s 1 unit + 12. And so Ken’s 1 : 4 = Jim’s 1 unit + 12 sweets : Jim’s 4 units + 48.
After Ken ate 18 chocolates, Jim had 18 more chocolates than Ken. Jim has 7 units of chocolates, Ken has 4 units + 48. So 3 units = 48 + 18. And so 1 unit = 16 + 6 = 22.
Ken had 22 + 12 = 34 sweets after giving half to Jim. So he bought 34 x 2 = 68 sweets.
Solution by using modeling!
Reference from mycarforum.com (lol what does this do with cars?)
Below is the Leslielai’s elaboration of the diagram above. (on how it works to solve the problem)
1. We start out by plotting the diagram for Jim, with a unit to indicate the amt of sweets he has left + the dotted area indicating that he has consumed 12 sweets.
2. Next, since we know that the amount of chocolates he has left is 7 times the sweets he has left. Therefore, 7 units
3. Following that, we can now plot Ken’s diagram. Bear in mind that he did not consume any sweet at all and that what he has should be what Jim had initially (inclusive of the dotted area representing the 12 he consumed). This time round we use a regular line to include 12, not dotted line as it still exists. In doing so, we can obviously tell that each unit of Ken’s sweet or chocolates for that matter is equal to one unit of Jim’s, PLUS 12.
4. Follow through by drawing 4 times of the above.
Recall that since the amt of chocolates Jim has (which remains unchanged) should be equal to the amount Ken had at the first. We’ll extend the chocolate diagram for Ken to the same length as Jim’s. This represents the 18 Ken has consumed.
Since in step 3, the units for Ken is drawn in such a way that every unit of his sweets/chocolate is EQUAL to each unit of Jim’s PLUS 12. Therefore, the difference in terms of units between Jim and Ken should equate to all the numerical differences denoted by 12, 12, 12, 12 and 18.
The key here is in mapping Ken’s model. Each unit in Ken’s term is one unit of Jim’s plus 12.
############################################
OTHER PSLE MATHEMATIC QUESTIONS
The above figure is a square. The angle question answer above is 150 degrees. Because the bottom triangle is equilateral triangle. (draw a circular curve through all 3 lines and they become the radii of the incomplete circle otherwise known as the arc)
360 – 60 = 300 degree
300 degree / 4 = 75 degree
75 degree x 2 = 150 degree
Wei Ren’s Answer
For those who haven’t figured out the angle question:
There are 2 isosceles triangles that are REALLY out of shape.
If you look at the kite, the 2 angles at the side will add up to the answer.
By taking 360 (total of the angles in the kite) – 60 (angle at bottom of kite),
you get 300 degrees, which is the total of the 2 sides plus the answer.
Since the 2 sides add up to the answer, you can say that 300 degrees = 2 * answer. 300/2 = 150.TO CLARIFY: Lines BD, AD, CD and BC are EQUAL in length.
BD and ACD are both isosceles triangles. You can see that angles ABD and BAD are equal, and angles ACD and CAD are equal. This means that angles ABD + ACD = angle BAC (answer).
The angles at A, B, C and D all add up to 360. By taking 360 – angle BDC (60), you get angles ABD + ACD + BAC, which equals to BAC + BAC (since ABD + ACD = BAC).
Since BAC + BAC = 300, angle BAC = 150.Lim Jeck’s Solution
Given: ABCD is a square, EG=EF=EH. Find angle GFH.
Since ABCD is a square, AB=BC=CD=DA
As EF=BC=DA and AB=GH, hence EG=EH=AB=GH.
Therefore, triangle EGH is equilateral since EG=GH=EH.
angle GEH=60°
angle GEF=60/2=30°
Triangle EGF is isosceles. Hence, angle FGE= angle GFE = (180-30)/2 = 75°
angle GFE = angle HFE = 75°
Therefore angle GFH= 75X2=150°
Renting Computers
6 friends decided to rent computers from 2pm to 4.30pm. While 4 of them were playing, the other two would watch. If the cycle continues, and each of them played for equal number of minutes, how many minutes will each person get to play? Answer: 100 minutes
Renting Computer Solution
VanillaCake’s Answer
From 2 pm to 4.30 pm -> 2 h 30 min = 150 min
4 x 150 = 600 (Total for 4 computers since there were 4 players)
600 / 6 = 100 min
Amount of playtime that each person get was 100 minChua’s Answer
Renting Computer
From 2pm to 4:30pm
total time needed=4.30-2
=2hr 30min
=2*60 + 30
=150minSince 4 people can play at the same time hence there are 4 comuters
so 150*4=600 hours
time used per user=600/6
=100min
=1hr 40minConclusion: it is not healthy to use computer for such a long hour.
Bicycle Race
Mei and Lin were in a bicycle race. Mei was travelling at a constant speed of 20km/hr and they both did not change their speed. When Lin completed half the race, Mei was 3.5km ahead. Mei completed the race at 10.45am. What time did Lin complete the race? Answer: 11.06am
Bicycle Race Solution
Guan Hui’s Answer
Suggested Answer:
3.5 x 2 = 7km(how much ahead is mei if she keep cycling when Lin finish the race)speed= 20km/h
7km= 7/20 X 1
=0.35 hr
=21 mins (time needed to be 7km ahead for mei)
Time when Lin complete the race = 10.45am+ 21mins
= 11.06amPuppylover’s Answer
3.5 / 20 x 60 = 10.5 min
Mei took 10.5 min to travel 3.5 km (distance beyond half way mark).
So when Mei reached the half way mark,
Lin would need another 10.5 min to reach the half way mark,
i.e Lin was slower than Mei by 10.5 minutes for half the distance.
Hence for the race, Lin would be slower by 2 x 10.5 = 21 min.
21 min past 10.45 a.m. would be 11.06 a.m. (or use timeline)
Lin reach the finishing line at 11.06 a.m.
Chocolate and Almond Cookies
Sally baked some cookies to sell. 3/4 of them were chocolate cookies and the remaining were almond cookies. After she sold 5/6 of chocolate cookies and 210 almond cookies, she had 1/5 of the cookies left. How many cookies did she sell? Answer: 960
Chocolate and Almond Cookies Solution
Mr Jackie Lim’s Answer
960 cookies were sold.
convert chocolate 3/4 to 30/40 and almond 1/4 to 10/40.
7 units of almond is 210. 1 unit is 30.
Number of units sold is 32u, hence it’s 960.
Chocolates and Sweets by Jim and Ken
Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim. Jim ate 12 sweets and Ken ate 18 chocolates. The ratio of Jim’s sweets to chocolates became 1 : 7 and the ratio of Ken’s sweets to chocolates became 1:4. How many sweets did Ken buy?. Answer: 68 sweets (Check the solution above)
Curry Puff
Curry puff shop sells puffs at $0.80 each. Offer:Buy 3 get 4th half price. If customer has $50, how many puffs can he buy if he buys as many as he can. Answer: 71 puff
Curry Puff Solution
eugenemon’s Answer
$2.80 = 4 pieces
50 / $2.80 = 17
17 x 4 = 68 pieces
Left $2.40 can buy 3 pieces.
68 + 3 = 71 pieces
Balloon’s String
String of 2 big balloons is 90cm
String of 5 small balloons is is 1.2m
If both strings are of the same length, there would be 105 more small balloons than the big balloons.
How many balloons are there altogether? Answer: 345
Balloon’s String Solution
eugenemon’s Answer
2.6667 big = 1.2m
5 small = 1.2m
5 / 2.66667 = 1.8749766
big : small = 1 : 1.8749766
105 / 0.8749766 = 120
Total balloons = 120 x 2.8749766 = 345
Rice for Sale
Shop A has 156kg of rice, while Shop B has 72kg of rice. Each shop sold an equal amount of rice and after that the ratio of the amount of rice left in their shop was 4:1. Find out how much rice has been sold. (3 marks) Answer: 44kg
Paras’s Answer
Since, they sold an equal amount of rice, the original difference remains the same.
The difference is 156-72=84kg. Now the ratio difference is 4-1=3units.
Now, that 3units is = 84 kg. Sho B now has left 1unit so it has 84/3 = 28 kg.
To find out the rice sold , 72kg(what it had before)-28kg(what it has now)=44 kg.kofplayer’s Answer
Just take 72 – [(156-72)/3]Pros’s Answer
some of the answers do not sounded logical…
A had 156kg, B has 72 kg.
Same amount of rice sold, hence the deduction from above will be same.
After sales, ratio is 4:1, hence for the 3 parts it is (156-72)kg = 84kg, hence each part is 28kg.
They left a total of 5 parts = 28 x 5 = 140 kg.
As a result, amount of rice sold = (156 + 72) – 140 = 88kg.
Chairs in the Hall
6/14 of the chairs in the hall are in rows of 13. Half of the chairs in the hall are in rows of 7. There are 112 more chairs in rows of 7. The rest of the chairs are stacked up. Find out the total number of chairs in the hall (5 marks) Answer: 1568
Ananto’s Answer
Most Of The Information Given Is To Fool You… All It Says Is6/14 chairs are in group (You Dont Need To Worry What Are They In A Group Of)
1/2 chairs are in another Group (1/2 = 7/14)
6/14 + 7/14 = 13/14There are 112 more chairs left.
Therefore the remaining 112 chairs must be 1/14 of all the Chairs…
So All You Do Is Multiply That By 14
So That The Fraction Becomes 14/14 Or In Other Words 100% of The ChairsIn Simple: 112 x 14 = 1568
Ayumilove 
