This blog is managed by Song Hock Chye, author of Improve Your Thinking Skills in Maths (P1-P3 series), which is published and distributed by EPH.

Friday, October 03, 2008

Catholic High Sch 2006 PSLE Math Prelim Q48

The diagram below shows 3 figures formed by shaded and unshaded triangles.

Solution

Q(a)
Total number of triangles ----- 5 x 5 = 25
Number of shaded Triangles ----- 1, 3, 6, 10, 15

Q(b)
15 x 15 = 225

Q(c)
At Level 1 (top most) there is 1 shaded triangle.
At Level 2, there are 2 shaded triangles.

1 + 2 + 3 + …….. + 30
= (1 + 29) + (2 + 28) + (3 + 27) + …..+ (14 + 16) + 15 + 30
= 14 groups of 30 + 15 +30
= (14 x 30) + 45
= 420 + 45

Observer said...

Mr Song,

For (c), 1 + 2 + ....+ 30
you can just use the averaging method.

[(1 + 30)/2] x 30 = 465

(1 + 30) is the total of the 1st and last number. Then divide by 2 to get the average. Same apply to (2 and 29) and so on. So to get the sum of the series, you just use the ave x the total number in the series.

Excel Eduservice said...

Yes, that is correct. Averaging, then multiply by the total number is another way.

Just feel that the mechanics of 14 x 30, then adding 15 and 30 is easier for average students to understand than the mechanics of 31 x 30, then dividing it by 2, or 31 x 15.

It’s all a matter of preference.

keentolearn said...

Dear Mr Song

Ah Meng walks to school at an average speed of 4km/h everyday. One day, after walking a distance of 1km, he realised that his watch was slow. So, he ran at a constant speed and managed to reach the school just on time. Later, he calculated that if he had run at this speed right from the start of the journey, he would have reached school 5 minutes earlier.What was his running speed?

Excel Eduservice said...

Saw this Q only this morning. The Math PSLE paper is already over.

However, it can still benefit future P6 students. At this moment, I am busy and have not worked out the Q.

Where did you get this Q anyway?

keentolearn said...

Dear Mr Song

This Q is from another blog.

ME said...

Dear Keentolearn,
I will provide if slightly different question perhaps if you can solve this question you will be able to solve your own question. I hope this helps.
Ali walked home from school. For the 1st 2minutes he walked at a speed of 50m/min. When he realised he was going late by 5minutes he increased his speed by 10m/min. Thus, he was early by 2minutes. Find the distance between his home and school.

keentolearn said...

Dear ME

You are putting up another question for a very busy Mr Song.

Two Butterflies said...

Dear Mr Song

1) The average age of 5 adults is 25 years old. They are all of different ages. If the oldest person is 30 years old, find the smallest possible age of the youngest person.

2) 2/5 of the coins in a container are 10 cents coins. The rest are 50 cents coins. The coins add up to be less than \$70 but more than \$65.Write down two possible values of the coins in the container.

MathsGeek said...

Hi Keentolearn!,

Ah Meng walks to school at an average speed of 4km/h everyday. One day, after walking a distance of 1km, he realised that his watch was slow. So, he ran at a constant speed and managed to reach the school just on time. Later, he calculated that if he had run at this speed right from the start of the journey, he would have reached school 5 minutes earlier.What was his running speed?

Note: I will be using abbreviation such as speed of wallking is denoted as SW, difference in speed is denoted as Sd...

Solution

Given that:

SW = 4 km /h, Td = 1/12 h,

Not given:

SR = ? and Sd = ?

4 km -----> 1 h
1 km -----> 1/4 h

Td = TW - TR *
1/12 = 1/4 - TR
TR = 1/4 - 1/12
= 1/6 h to cover 1 km

1/6 h -----> 1 km
1 h -----> 6 km

Speed of running, SR = 6 km/h

*Note : The time taken to walk takes longer than the time taken to run.
eg, Td = TW - TR

**Note : The speed of running is faster than the speed of walking.
eg, Sd = SR - SW.

PS: Please do not mix them up. All these are logics.

Cheers :)

MathsGeek said...

Hi Two Butterflies,

1) The average age of 5 adults is 25 years old. They are all of different ages. If the oldest person is 30 years old, find the smallest possible age of the youngest person.

2) 2/5 of the coins in a container are 10 cents coins. The rest are 50 cents coins. The coins add up to be less than \$70 but more than \$65.Write down two possible values of the coins in the container.

The questions above were extracted from your thread and these questions that you have posted relates to secondary school topics, like inequalities.

Inequalities are taught in Secondary school only and algebra is involved.

This blog is dedicated to P6 only...As such, for the benefits of all P5 & 6 students, please do not confuse others who are making an effort to prepare themselves for the next year PSLE exam.

Thanks

Two Butterflies said...

Dear MathsGeek

These two questions are from P5 SA2 papers.

I am not sure why you relate them to inequalities in secondary Maths.

This is a very informative site and I wish to learn as much as possible from other readers.

Best Wishes

MathsGeek said...

Hi Two Butterflies!,

I did encounter this question in P5 and P6 papers as well as Secondary 1 and 2 papers.

In Secondary level, this can be linked to inequalities which was not taught and cover in Primary school syllabus.

Whichever method that you used are abled to derive the same answer.

Excel Eduservice said...

Q1) The average age of 5 adults is 25 years old. They are all of different ages. If the oldest person is 30 years old, find the smallest possible age of the youngest person.

For average of 5 to be 25 yr old, total age must be –- > 5 x 125 = 125.
Oldest is 30 therefore the total age for remaining 4 -- > 125 – 30 = 95.
For youngest to have smallest possible age, 3 of them must be 29, therefore -- > 95 – 29 – 29 – 29 = 8 (Ans)

Q2) 2/5 of the coins in a container are 10 cents coins. The rest are 50 cents coins. The coins add up to be less than \$70 but more than \$65.Write down two possible values of the coins in the container.

No. of coins
10 cents --- 2 units
50 cents --- 3 units

Value of coins
10 cents --- 2 units
50 cents --- 15 units (3 x 5 cos 50c is 5x 10c)
Total --- 17 units

4th multiple of 17 = 68 (value between 65 and 70)
1 unit --- \$68 divided by 17 = \$4

(10c) 2u -- \$8 -- 80 coins (Ans)
(50c) 15u -- \$60 -- 120 coins (Ans)

Anonymous said...

Dear Mr Song!,

Q1) The average age of 5 adults is 25 years old. They are all of different ages. If the oldest person is 30 years old, find the smallest possible age of the youngest person.

For average of 5 to be 25 yr old, total age must be –- > 5 x 125 = 125.
Oldest is 30 therefore the total age for remaining 4 -- > 125 – 30 = 95.
For youngest to have smallest possible age, 3 of them must be 29, therefore -- > 95 – 29 – 29 – 29 = 8 (Ans)

The question stated that the ages must be different, but base on your solution, you have indicated the remaining 3 to be of the same age, which is 29.

Please enlighten as i am confused.

Anonymous said...

Hi everyone!,

Can someone kind enough to solve the questions below.

1) Troy's monthly allowance is \$42 more than Earl's. Earl spends \$54 more than Troy every month. Earl's savings is 1/2 of Troy's savings. If Troy spends 3/7 of his allowance every month, what is his allowance for the entire year?

2) Siew Leng paid \$8.56 for some 26-cents, 30-cents and 50-cent stamps. She bought 4 more 30-cents stamps than 50-cent stamps. There were twice as many 26-cent stamps as 30-cent stamps. How many 26-cent stamps did she buy?

Thanks

Tilted Earth said...

Dear anonymous

Are your questions from P5 or P6?

Excel Eduservice said...

”The question stated that the ages must be different, but base on your solution, you have indicated the remaining 3 to be of the same age, which is 29.”

Then it should be

125 – 30 -- > 95 total for the other 4.

Then 95 –29 – 28 –27 = 11

Ans: 11 yrs old.

Anonymous said...

Thanks Mr Song!

But Mr Song, why you deduct by consecutive number of 27, 28 and 29? How you derive the 27, 28 and 29? Can we use different numbers?

The reason is because the question did not state that the different numbers must be in consecutive numbers...

Thanks

Anonymous said...

Dear Tilted earth!,

The below-mentioned questions are P6 problem sums. Each question is worth 5-marks. I am not keen of the answers but i really need to know how you derive the answers by providing your working step-by-step.

) Troy's monthly allowance is \$42 more than Earl's. Earl spends \$54 more than Troy every month. Earl's savings is 1/2 of Troy's savings. If Troy spends 3/7 of his allowance every month, what is his allowance for the entire year?

2) Siew Leng paid \$8.56 for some 26-cents, 30-cents and 50-cent stamps. She bought 4 more 30-cents stamps than 50-cent stamps. There were twice as many 26-cent stamps as 30-cent stamps. How many 26-cent stamps did she buy?

Thanks

Excel Eduservice said...

“But Mr Song, why you deduct by consecutive number of 27, 28 and 29? How you derive the 27, 28 and 29? Can we use different numbers?

The reason is because the question did not state that the different numbers must be in consecutive numbers...”

After ‘30’ is used (the eldest), the second eldest is 29 and so on. It must be remembered we want to have the youngest age to be as small as possible. Hence, if the others have larger numbers, by the law of averaging, the last number will of course be the smallest.

That is averaging, isn’t it?

If you or not convinced, try other numbers and see it for yourself. The youngest will be more than 11.

Anonymous said...

I really like this thread as it is informative where ideas and opinions are exchanged and discussed broadly.

But Sir, I came across the other thread with regard about the three-legged stools. Why 84 stools minus away 33 legs? Do you think it is correct?

Thanks

Tilted Earth said...

For every 9 stamps Alan has, Ben has 11 stickers. How many stickers do they have all together if Ben has 132 stamps.

Anonymous said...

Hi all,

Can someome please facilitate to solve this sum.

Peter started his journey from Town A to Town B. If he travelled at a speed of 80 km/h, he will arrive at Town B 1/3 hours late. If he travelled at a speed of 60 km/h, he will arrive at Town B 3/4 hours late.

(a) Find the distance from Town A to Town B

(b) How long will he take to cover the journey with a speed of 80 km/h?

Thanks

Anonymous said...

Hi all,

Can someome please facilitate to solve this sum.

Peter started his journey from Town A to Town B. If he travelled at a speed of 80 km/h, he will arrive at Town B 1/3 hours late. If he travelled at a speed of 60 km/h, he will arrive at Town B 3/4 hours late.

(a) Find the distance from Town A to Town B

(b) How long will he take to cover the journey with a speed of 90 km/h?

Thanks

MathsGeek said...

Hi Anonymous and all,

The question with accompanying solutions are stated below.

Extracted:

Peter started his journey from Town A to Town B. If he travelled at a speed of 80 km/h, he will arrive at Town B 1/3 hours late. If he travelled at a speed of 60 km/h, he will arrive at Town B 3/4 hours late.

(a) Find the distance from Town A to Town B

(b) How long will he take to cover the journey with a speed of 90 km/h?

Solution:

(a)

Difference in time = 3/4 - 1/3
= 5/12 hrs

Remaining distance yet to be covered with a speed of 60 km/h

= 5/12 hrs * 60 km/h
= 25 km

Difference in speed = 80 - 60
= 20 km/h more

20 km more -----> 1 h
25 km more -----> 25 / 20
= 5 / 4 hrs

It took 5/4 hrs to cover 25 km more. In other words, it took 5/4 hrs to cover the whole journey.

Total distance covered = 5/4 * 80
= 100 km

(b)

Time taken to cover the whole journey with a speed of 90 km/h

= Total distance / Speed
= 100 / 90
= 10 / 9

= 1 1/9 hrs

Hope that i have cleared all your doubts. If you still have difficulties understanding my working, perhaps you can ask Mr Song for assistance.

Thanks

Anonymous said...

1) Troy's monthly allowance is \$42 more than Earl's. Earl spends \$54 more than Troy every month. Earl's savings is 1/2 of Troy's savings. If Troy spends 3/7 of his allowance every month, what is his allowance for the entire year?

Troy's allowance is \$4032 per year.Is the answer correct?

Anonymous said...

2) Siew Leng paid \$8.56 for some 26-cents, 30-cents and 50-cent stamps. She bought 4 more 30-cents stamps than 50-cent stamps. There were twice as many 26-cent stamps as 30-cent stamps. How many 26-cent stamps did she buy?

Anonymous said...

Please facilitate to provide the working for this questions...

A 6-sided dice with equal chances has the numbers 1, 2, 3, 4. 5 and 6 written on its faces. The dice is tossed twice and the numbers obtained are recorded accordingly. What is the maximum number of ways for the sum of the two recorded numbers to be a number that has 2 factors only?

All i know is that the maximum for the sum of the dice is 12

and the lowest is 2.

How many factors does the number 3 has?

Anonymous said...

Whay do you mean by number that has 2 factors only?Any examples?

Anonymous said...

" What do you mean by the number that has 2 factors only? Any examples?"

MathsGeek said...

Extracted

A 6-sided dice with equal chances has the numbers 1, 2, 3, 4. 5 and 6 written on its faces. The dice is tossed twice and the numbers obtained are recorded accordingly. What is the maximum number of ways for the sum of the two recorded numbers to be a number that has 2 factors only?

Thinking process :

Numbers which have only 2 factors are numbers that can be divided by 1 and itself.

Lets take the number 2 for an instance. Although 2 is an even number but it is also a prime number because the number 2 can only have 1 and 2. eg 1 * 2 = 2

However, the number 9 is not a prime but is an odd number.

Because 9 has more than 2 factors, eg, 1 * 9, 3 * 3

Solution: Listing

Prime numbers are 2, 3, 5, 7, 11

For prime number 2:

1 way : 1 + 1

For prime number 3:

2 ways : 1 + 2
2 + 1

For prime number 5:

4 ways : 1 + 4
2 + 3
3 + 2
4 + 1

For prime number 7:

6 ways : 1 + 6
2 + 5
3 + 4
4 + 3
5 + 2
6 + 1

For prime number 11:

2 ways : 5 + 6
6 + 5

Total = 1 + 2 + 4 + 6 + 2
= 15 ways

MathsGeek said...

Extracted from the chatroom...

John's watch is faulty. His time on his watch is slow by 2 minutes for every 45 minutes. If the time on his watch is 3.50 pm now, what is the time on his watch after 6 hours?

Solution:

1 hour -----> 60 minutes
6 hours -----> 6 * 60
= 360 minutes

every 45 mins -----> 2 mins slow
360 mins -----> (360 * 2) / 45
= 16 mins slow

Actual taken = 3.50 pm + 6 hrs
= 9.50 pm

Since John's watch is 16 mins slow,
Time taken = 9.50 pm + 16 mins
= 10.06 pm.

MathsGeek said...

Extracted from chatroom:

Mrs Ang has some children. Each of her daughters has twice as many brothers as sisters. Each of her sons has the same number of brothers as sisters. How many daughters does Mrs Ang have?

Solution: Inferential

We assume that one of the daugher is Mary and one of the son is John.

For Mary:

Brother = 4
Sister = 3 - 1(exclude Mary)
= 2

Notice that the number of brothers Mary has is twice the number of sisters.

For John:

Brother = 4 - 1(exclude John)
= 3
Sister = 3

Notice that the number of brothers is the same as the number of sisters.

Therefore,

number of daughters = 3

number of children = 7

Thus, Mrs Ang has 3 daughters.

Anonymous said...

Q1)During a warehouse book sale, Sally spent 62.5% of her money on 24 books and 18 pens. She also spent 25% of her remaining money on 18 files. Each pen costs 8/9 as much as the price of one book. The file costs \$7.80 less than a book. Find the total cost of one book and one pen.

Q2)There are 600 children in Team A and 30% of them are boys. There are 400 children in Team B and 60% of them are boys. After some children are transferred from Team B to Team A , 40% of the children in Team A and 60% of the children in Team B are boys. How many children are transferred from Team B to Team A?

Anonymous said...