# 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.

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# 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 min

Chua’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

= 11.06am

Puppylover’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

How do u score 4 a stars for psle? pls tell me

actually the exam was easy i gt 4a*s topped my school, with a score of 276 and now currently in raffles instituion ^^

actually the exam was easy i gt 4a*s topped my schoo, with a score of 276 and now currently in raffles instituion ^^

sorry not clear repeat

(156-x)/(72-x) = 4/1

cross multiply

1(156-x)/4(72-x) = (156-x)/(288-4x)

3x is 288 -156 = 132

1x= 44

44×2=88

Here you go your answer full marks will be given

the rice question is easy:

Take the rice sold as x

(156-x)

——- = 4/1

(72-x)

now cross multiply

1(156-x)

——— = (156-x)

4(72-x) ——–

(288-4x)

3x = 288-156=132

1x = 44

44×2= 88

Wala 88 this methord is simple and easy to comprehend full marks will be given.

I am current 12 and i tink that the rice for sale question is quite easy

A : 4 Unit + 1 Unknown = 156

B : 1 Unit + 1 Unknown = 72

So , 156 – 72 = 84

84 = 3 Units [ Since 4-1=3 and 1-1=0 ]

1 Unit Is Therefore = 84/3 = 28

To Get Unknown 72-1Unit[28] = 44

To Get 2 Unknowns = 44×2=88 😀

– SOLVED –

Erm.. i feel that maths are fucking !!

Heyys, yr algebra is far too complicated.. The teachers do not teach in primary sch! Hw do u expect the primary sch students to do?

I found the questions quite easy to solve! ☺

Dun act Smart.

dont brag.

errr…. do you have the answer for the question that have a graph and they say if a+b…b+c….a+c

It depend on whether the teacher teaches it or not, my teacher taught me that before, and i think a question like that came up last yr.

& anyway, the difficulty depends on the year.

Last year’s was easy, this yr’s is tough, and next yr would probably be easy.

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

hi where do you get 2.6667???? are u sure your ans is right? any kind soul pls help. thanks.

i find that the Jim and Ken sum not the most difficult. If you draw a proper model, you should be able to solve it. But I don’t understand the one about the balloons. so if an adult like me find it confusing, what more the poor kids!!!What exactly are you testing? English comprehension or Maths??? If the children have difficulty understanding the questions, how on earth can they do the questions, not to mention the time constraints and exam phobia. poor, poor kids…

the exam was soooooo difficult and i didn’t understand the questions at first… but now i do…thx!!!

i know… i agree wit you

difficult rite

hello..the exam was so difficult.. i feel like crying now!!!

OMG yar….im scared that i wont get good marks now….you guys this year p6 right?

Bicycle Race

total of distance ahead for the race = 3.5*2=7km

time needed for Mei to travel 7km = 7/20 hr

= 7/20 * 60 min

= 21min

Hence time that Lin completed the race = 10:45 + 0:21

= 11:06

Conclusion: Kids are not adviced cycle for too long or in constant speed 😛

Renting Computer

From 2pm to 4:30pm

total time needed=4.30-2

=2hr 30min

=2*60 + 30

=150min

Since 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 40min

Conclusion: it is not healthy to use computer for such a long hour. 😀

thanks for your solution, yours is much easier to understand 😀

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

If question is how much rice has been sold by each shop: 44kg. But if it’s how much rice has been sold: 44 X 2 = 88kg.

yeah , you ‘re right 😀

O.o this question is tought for primary school student.

Really can solve without algebra?

It take long time to solve it by using guessing.

what happen if the question get harder…

12SCCCC4818

S — sweat

C — chocolate

12S:CCCC48=1:4, 48=12×4

S:CCCC=1:4

S:CCCC4818=1:7

S=(48+18)/3=22

S+12=34

34×2=68

what kind of modelling is that o__O

i have not learned before lol.

where 48 came from? or 4818 came from?

which 1 is from Jim and which 1 is from Ken?

You could elaborate more on how this modeling works

http://app.reach.gov.sg/reach/YourSay/YourDiscussionCorner/tabid/117/ptid/414/page/3/totrecs/29/threadid/2690/forumtype/posts/Default.aspx

Sorry, I don’t know how to post drawing, using characters instead:

model Jim: (12+S)(CCCCCCC)

model ken: (12+S)(CCCC+48+18)

S — sweat

C — chocolate

(12+S) : (CCCC+48)=1:4, 48=12×4

S:CCCC=1:4

S: (CCCC+48+18)=1:7

S=(48+18)/3=22

S+12=34

34×2=68

PSLE doesn’t allow algebra. You wanna try again?

there is another method call guessing , but the guessing is accurate and logical one, i posted another one up there

i thought algebra can?

Let me get this right. Jim had 308 chocolates and gave half to Ken? Ken had 68 sweets and gave half to Jim? How did they have so much confectionery in the first place? Why was Jim so much more generous than Ken? Were the chocolates really that lousy? Or the sweets so fantastic?

These questions trouble me.

lol I think there were more chocolates in stock compare to sweets in the candy store 😀

I didnt understand this question much when I did it . Thx 🙂