Menata Hati, Mengasah Minda

P6 Math: Mr. Tan tarts

2026-05-01T23:30:00.003+08:00

Mr. Tan packed 508 tarts into large and small boxes to sell. He filled each large box with 20 tarts and each small box with 12 tarts. All the boxes were full and there was no left over. What was the least number of boxes used by Mr. Tan ?

Solution:

Since the problem asks for the least number of boxes, then Mr. Tan should use as many as large boxes to pack.

Now assume all tarts are packed with large boxes.

508 / 20 = 25R8, there is reminder of 8 tarts --> 8 tarts will not fill a small box !

Try to reduce number of large boxes, one by one

Number of L boxes	Number of tarts inside L box	Extra tarts	Number of S boxes	Fit
25	500	8	1	No
24	480	28	2	No
23	460	48	4	Yes

Hence, we found the solution with 23 large boxes and 4 small boxes. The least amount of boxes is 23 + 4 = 27.

If we try to solve by assuming all tarts are packed in small boxes, the iteration process is quite tedious and may find few solutions (where tarts can fit in the boxes) but not the smallest number of boxes. For example, the whole tarts can fit in 39 small boxes and 2 large boxes, a total of 41 boxes !

P6 Math: Counters

2026-04-27T00:51:00.000+08:00

Gerald, Michael, and Pritam had 288 counters altogether. After Gerald gave Michael some counters, Michael had twice his original number of counters. Then Michael gave Pritam some counters and Pritam had twice his original number of counters. All the boys had an equal number of counters in the end. How many counters did Gerald have at first?

Solution:

There are one key informations here: each boys has same number of counters at the end.

288 / 3 = 96 counters in the end for each boy.

Let's start with Pritam, he had twice his original number of counters.

2p = 96 --> p = 48, at first Pritam had 48 counters.

48 additional counters Pritam had, was given by Michael.

2m - 48 = 96 --> 2m = 144 --> m = 72, at first Michael had 72 counters

72 additional counters Michael had, was given by Gerald.

g - 72 = 96 --> g = 96 + 72 = 168, at first Gerald has 168 counters.

TO DO: create the model to visualize it.

P6 Math: Money Donation

2026-04-26T23:52:00.005+08:00

Tanya, Mariah, and Devi donated some money to charity. Tanya donated 2/3 as much as the total donation made by Mariah and Devi. Mariah donated 1/2 as much as than as the total donation made by Tanya and Devi. Devi donated $138 less than Tanya. How much did Mariah donate?

Solution:

Solve by Ratio method.

In this situation, do take note that the total donation is unchanged.

T : M + D = 2u:3u, remove u and insert the total there to become
T : M + D : T + M + D = 2 : 3 : 5 (ratio 1)

M : T + D = 1u : 2u, remove u and insert the total there to become
M : T + D : T + M + D = 1 : 2 : 3 (ratio 2)

Since total is unchanged, then we can rewrite the two ratios such as:

T : M + D : T + M + D = 6 : 9 : 15
M : T + D : T + M + D = 5 : 10 : 15

Rearrange, to find D:

D = (T + D) - T = 10u - 6u = 4u

T - D = 6u - 4u = 2u

2u = 138, u = 69

Mariah donated 5u = 5 x 69 = $345

P6 Math: Vivian Sweets

2026-04-26T22:45:00.006+08:00

Vivian has some sweets. If she shares the sweets among 4 friends, she will have 3 sweets left. If she shares the sweets among 5 friends, she will have 4 sweets left. If she shares the sweets among 9 friends, she will have 8 sweets left. What is the smallest possible number of sweets she has?

Solution:

Remember Vivian shares with friends, so the number of children in the calculation is always +1.

E.g Vivian shares with 4 friends --> number of children = 4 + 1 = 5.

Method 1: Guess and Check (as taught in Primary 3)

x5: 5, 10, 15, 20, 25, 30, 35

+3: 8, 13, 18, 23, 28, 33, 38

x6: 6, 12, 18, 24, 30, 36, 42

+4: 10, 16, 22, 28, 34, 40, 46

x10: 10, 20, 30, 40, 50, 60, 70

+8: 18, 28, 38, 48, 58, 68, 78

Check the common multiplication there: 28, 58, ...

Method 2:

Two steps:

1. Find the LCM of 5, 6, 10 , which is 30.

2. Find multiple of 5, 6, or 10, which is less than 30. For example for 5:

5 --> 5 x 5 = 25 --> 25 + 3 = 28

Kakak Envisioning Board

2026-03-29T18:34:00.004+08:00

Insyaa Allah ,.. semoga tercapai ya kak.

P6 Math: Stickers

2026-03-24T07:12:00.002+08:00

In P6 Bukhara there are 15 boys and 10 girls student. Mdm Alwiyah gave same number of stickers for each student. Each boy then gave 2/3 of his share of stickers to the girls. As a result, each girl received 10 more stickers than each boy.

(a) How many stickers did the girls receive in the end?

(b) How many stickers did Mdm Alwiyah distribute altogether?

Solution:

Each boy gave 2/3 of his share to girls.
2/3u x 15 = 10u
Each boy now have u - 2/3u = 1/3u

Girls got 10u from boys.
Now, girls have 10u + 10u = 20u.
Each girl now have 20u / 10 = 2u.

Different share between girl and boy now = 2u - 1/3u = 5/3u, which is 10 more stickers.
5/3u = 10 --> u = 6

(a) At the end, girls received 20u = 20 x 6 =120

(b) Total distributed stickers remain unchanged = 15u + 10u = 25u = 150.

P5 Math: Flour and Sugar (A g a i n)

2025-10-27T07:34:00.002+08:00

Sakinah bought an equal amount of flour and sugar. She paid a total amount of $83.20. How many kilograms of flour did she buy?

Solution:

For Primary school student, the answer should come in a whole units as shown in the picture (e.g bags, sacks, ...) and not partial. It means mass of flour would be: 500, 1000, 1500, ... 10000 g, and mass of sugar would be: 1.2, 2.4, 3.6, ... 12 kg.

Find the common multiple of the two items.
Flour: 500, 1000, 1500, ... 5500, 6000
Sugar: 1.2, 2.4, 3.6, ... 4.8, 6.0

The first common multiple is 6000 g or 6 kg. It is the first set of common multiple, when amount (the mass) of flour and sugar is equal.

Price of 6000 g of flour = 12 x $1.8 = $21.6.
Price of 6 kg of sugar = 5 x $4 = $20.
Total both = $41.6.

However, Sakinah paid $83.20, it means she bought more than one set.

$83.20 / $41.60 = 2 --> it means two sets:

Flour: 2 x 6000 g = 12000 g
Sugar: 2 x 6 kg = 12 kg

So, she bought 12000 g or 12 kg of flour as the answer.

P5 Math: Arrange the chairs in a school hall

2025-10-26T23:26:00.001+08:00

The school attendant wanted to arrange the chairs in the school hall. If he had arranged the chairs in rows of 10, there would be 4 chairs left behind. If he had arranged the chairs in rows of 12, there would be 15 rows less than before.

How many chairs were there in the hall?

Solution:

We can use substitution method.

The model is assuming:

rows with 10 seats = r
rows with 12 seats = s

where r = s + 15

Total number of sets is constant.

10r + 4 = 12s --> 10(s + 15) + 4 = 12s --> 2s = 154 or s = 77.

Number of chair = 12 x 77 = 924.

P5 Math: Asmidah spent

2025-10-26T23:13:00.002+08:00

Asmidah spent 7/9 of her money on 6 files, 10 pens, and 15 rulers. She could buy 4 pens and 6 rulers with the remaining money but she spent all of it on files instead. How many files could Asmidah buy with the remaining money ?

Solution:

In a model, total is 9 units.

7 units for 6 files + 10 pens + 15 rulers --> 7u = 6f + 10p + 15r
2 units for 4 pens + 6 rulers --> 1u = 2p + 3r --> 5u = 10p + 15r

We can deduce from two facts above that 2u equals to 6f.

It means 2 units (which is the remaining money) can be used to buy 6 files.

P5 Math: Mdm Rohani lollipops and chocolates

2025-10-26T23:03:00.004+08:00

Mdm Rohani bought thrice as many lollipops as chocolates. On Children's Day she gave each of her pupils 4 chocolates and 3 lollipops. In the end she had 90 lollipops and 3 chocolates left.

(a) Find the number of pupils Mdm Rohani had

(b) Find the number of lollipops she bought

Solution:

Only algebraic solution available now (as discussed in FB page Maths Model Method)

If number of chocolates = u, then number of lollipops = 3u.

Let number of pupils = p.

Total choc = 4p + 3 = u --> 12p + 9 = 3u

Total lollipops = 3p + 90 = 3u

Elimination method on both:

12p + 9 = 3p + 90 --> 9p = 81, p = 9 i.e there are 9 pupils.

Number of lollipops = 3p + 90 = 27 + 90 = 117.

P5 Math: Tammy bought a blouse and a skirt

2025-10-26T22:54:00.007+08:00

Tammy bought a blouse and a skirt at 20% discount each. She was given a further 30% membership discount of the discounted price. She then paid %54.60 for the two items. If the skirt cost $52.50 before any discount, find the original price for the blouse.

Solution:

We have to work backwards.

$54.60 is after 30% discount i.e $54.60 is 70% --> before 30% discount = (100/70) x 54.60 = $78.

$78 is after 20% discount i.e $78 is 80% --> before 20% discount = (100/80) x 78 = $97.50.

$97.50 is price for both blouse and skirt before any discount.

So, price for the blouse before any discount = 97.50 - 52.50 = $45.

Belajar OSN SMP

2025-05-03T15:17:00.005+08:00

Latihan

P5 Math: Blue and Red beads

2025-05-03T11:53:00.002+08:00

A box contained blue beads and red beads. At first, there were 5 times as many as blue beads. After 28 blue beads and 28 red beads were removed, the difference in the number of blue beads and red beads left in the box was 260.

(a) How many blue beads were there in the box at first ?

(b) What was the total number of beads left in the box ?

Solution:

Please take note that 28 beads are fewer than one unit.

One unit = u + 28

From the model: 4 u + (4 * 28) = 260 --> u = 37

One unit = 37 + 28 = 65

By observing model above:

(a) At first there were 5 units = 5 * 65 = 325

(b) Total number of beads left in the box = 2u + 260 = 334

P5 Math: April and May

2025-04-23T23:18:00.004+08:00

In April, Amy, Betty, Clara, and Dione had a total savins of $324.

In May, they each saved an equal amount when Amy saved twice as much, Betty saved $4 less, Clara saved the same, and Dione saved $9 more than in April. How much did Betty and Dione save altogether in April ?

Solution:

Quite tricky sentences.

Two equations

(1) In April, A + B + C + D = 324

(2) In May, u + u + u + u

while

A = 1/2u
B = u + 4
C = u
D = u - 9

Solve (1) by substituting letters to values, 3.5u = 329, u = 94.

Now (B + D) = 2u - 5= 188 - 5 = 183

There is second method using model, that is not involving negative operation

From first picture, the facts are:

(i) B + D = 4u + 4 - 9

(ii) 7u = 324 - 4 + 9 = 320 + 9 = 329

u = 329 / 7 = 47

Solve (i)

B + D = (4 * 47) + 4 - 9 = 188 + 4 - 9 = 183

P5 Math: Age Problems

2025-02-23T11:47:00.008+08:00

Alice is 2 years old and her elder sister Bernice is 10 years old.
In how many years' time will Bernice be thrice as old as Alice?
Peter's age is 1/3 of Wendy's age now.
In 11 years, his age will be 7/10 of Wendy's age. How old is Peter now?
Mrs Adiba is 42 years now. Her nephew is 34 years younger than her.
In how many years' time will Mrs Adiba be thrice as old as her nephew?
Today, the age of Osman to the age of Faizal is 4/3.
15 years ago, Osman's age was thrice of Faizl's age.
How old is Faizal now ?
Mr Hamid is 3 times as old as his son now.
12 years ago, Mr. Hamid was 9 times as old as his son.
How old is Mr Hamid now?
Habib's age is 1/5 of his father's age.
In 10 years time, Habib's age will be 2/5 of his father's age.
How old is Habib now?
Elvira is 5 years younger than her sister.
In 4 years' time, his sister will be twice as old as her.
How old is Elvira's sister now?
Mr Fez is 5 times as old as Dimash now.
In how many years' time will Mr Fez be twice as old as Dimash if Dimash is 4 years old now?
Bukhara is 6 years older than his cousin. In 2 years' time, he will be twice his cousin's age. What is the ratio of Bukhara's age to his cousin's age now?
Twice Shalih's age now is the same as thrice of his age two years ago.
How old is Shalih now?
Adrian is 3 times as old as Bob now.
4 years ago, Adrian was 28 years older than Bob.
How old will Bob be in 4 years' time?
Mrs Intan is 50 years old.
Her daughter is 22 years old.
In how many years' time will Mrs Intan be twice of her daughter's age?
Uni is now 4 years older than Imah.
5 years ago Uni was twice as old as Imah.
How old is Uni now?
Umar is 56 years old.
His nephew, Yasir, is 4 years old.
In how many years time will Umar be five times as old as his nephew?

Solution:

Answer:

(14) 9 (video)

(13) 13

(12) 6

(11)

(10) 6

P5 Math: Jenny and Lily paper clips

2025-02-10T23:01:00.000+08:00

Jenny had 420 more paper clips than Lily. After Jenny gave 67 paper clips to Lily, Jenny still had 3 times as many paper clips as Lily. How many paper clips did each of them have at first ?

Solution:

Draw model

From the After Model picture, we can have

2u + (4 x 67) = 420

2u = 420 - 268 = 152

u = 76

Hence at first:

Jenny had (u + 420) = $496

Lily had u = $76

P5 Math: Library books

2025-02-10T22:30:00.000+08:00

4/9 of the books in a class library were non-fiction books, The rest were fiction books.

The student borrowed 1/2 of the non-fiction books and 2/5 of the fiction books.

There were 60 books left in the class library.

How many books were there altogether at first ?

Solution:

Working with model

Borrowed non-fiction books = 2u

Borrowed fiction books = 2u

Left non-fiction books = 4u - 2u = 2u

Left fiction books = 5u - 2u = 3u

Left books in total = 2u + 3u = 5u

5u = 60 or u = 12

At first, there were 9u = 108 books

P5 Math: Money at first

2025-02-09T22:45:00.005+08:00

Kakak had some money at first.

After spending $120 on a pair of jeans, she spent 4/7 of the remaining amount on a shirt.

She had 1/9 of his money left.

How much did Kakak have at first ?

Solution:

Working:

From the model, the clue is money left after shopping.

3u = 1/9 of total money. It means total money = 27u

It means, jeans portion is 27u - 4u - 3u = 20u
20u = 120, or u = 6
We can find total money at first = 27u = 27 x 6 = $142

Nss Primary Journey

2025-02-09T00:40:00.001+08:00

Alhamdulillaah rabbal 'alamiin,

Uni, putri sulung kami telah menamatkan tahap dasar menuntut ilmu akademis dan ukhrawi di madrasah semenjak awal 2019 hingga pengumuman hasil PSLE di akhir 2024. Semoga ilmu, adab, ketrampilan, dan pengalaman yang diperoleh menjadi bekal terbaik dalam menempuh kejayaan hidup di dunia dan di akhirat kelak. Man jadda wa jadda, Journey of thousand miles, starts from a little step.

Semoga Allah Swt mengaruniakan barakah atas kerja keras dan perjuangan uni, dan selalu mencurahkan rahmat Nya kepada anak-anak Papa dan Mama dimanapun dan kapanpun. Semoga Allah membalas pahala atas kebaikan, dukungan, dan doa dari orang-orang tercinta (keluarga, para guru dan asatizah, dan siapapun yang telah membersamai perjalanan uni).

S1 Math: Find p, q, ...

2025-01-25T20:43:00.001+08:00

Question 1

p and q are both prime numbers.
Find the values of p and q such that is an integer.

Question 2

One third of the product of 68 and 108 is the same as 2^px 3^q x 17^r

Find p, q, and r.

Question 3
Find smallest positive integer k for be a perfect cube

Solution:

Answer

(1) p = 11, q = 3

(2) p = 4, q = 2, r = 1

(3) k = (2 x 11)² = 484

S1 Math: Smallest positive number

2025-01-25T19:02:00.008+08:00

Written as a product of its prime factors, 600 = 2² x 3 x 5²

Find the smallest positive integer k such that is a perfect square

Solution:

In order to do that then the divisor must be 2 x 3 = 6, so the square is 2² x 5²= 100

S1 Math: Smallest positive number

2025-01-25T19:02:00.007+08:00

Written as a product of its prime factors, 540 = 2² x 3³ x 5

(i) Find the smallest positive integer k such that 540k is a perfect cube.

(ii) Give an example of a positive integer value of h such that is a perfect cube

S1 Math: Smallest positive number

2025-01-25T18:48:00.005+08:00

Find the smallest possible whole number k if 3575k is a multiple of 260.

S1 Math: Products of prime factors

2025-01-25T18:45:00.001+08:00

The numbers of 700 and 1050, written as the products of their prime factors, are

700 = 2² x 5² x 7,

1050 = 2 x 3 x 5² x 7

Find

(i) The smallest positive integer h for which 700h is a multiple of 1050

(ii) The smallest positive integer k for which is a factor of 700

Nahu I'rab

2024-11-04T20:03:00.007+08:00

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