Synchronization problems are the most intuitive LCM application:

When will two cyclic events happen at the same time?
How long until a pattern repeats?

We will explore classic scenarios like:

Scenario	The 'Cycle'
Runners	Time for one full lap
Bells	Interval between rings
Traffic Lights	Seconds between green signals

But this section ends with a twist.

A problem that is deliberately placed among LCM examples but actually requires HCF.

This final challenge tests whether you are:

Genuinely thinking about the problem structure
Just applying the current section's operation blindly

1. Filter — Circular Track with Interpretation

🏃 The Classic Synchronization Problem

Here's a scenario you might encounter in real life:

Two runners start at the same time from the same point on a circular track.

Runner A completes one lap in 16 minutes
Runner B completes one lap in 20 minutes

They keep running at their constant speeds, lap after lap.

✍️ Question

Your Challenge 🎯

Answer all three parts:

(a) After how many minutes will they meet at the starting point for the first time?
(b) How many laps will each runner have completed by then?
(c) Why is LCM the correct operation here (and not HCF)?

Let me build the model step by step.

Runner A completes one lap every $16$ minutes. So Runner A is at the starting point at: $t = 0, 16, 32, 48, 64, 80, 96, 112, . . .$

✍️ FIB

Fill in the blank

If Runner A is at the starting point at

64

minutes, what is the next time they reach the start?

Runner B completes one lap every $20$ minutes. So Runner B is at the starting point at: $t = 0, 20, 40, 60, 80, 100, 120, . . .$

✍️ MCQ

Choose one

What is the first common time after

t=0

for both runners?

We need the first POSITIVE time that appears in BOTH lists. Looking at the lists: the first common time is 80.

A timeline makes it easy to see when each runner returns to the start — and where those moments first coincide.

Mathematically: the meeting time must be a multiple of 16 AND a multiple of 20. The smallest such positive number is LCM(16, 20).

✍️ MCQ

Choose one

What is the prime factorization of

16

?

$16 = 2^{4}$ $20 = 2^{2} \times 5$

$LCM = 2^{4} \times 5 = 16 \times 5 = 80 minutes$

✍️ FIB

Fill in the blank

How many laps has Runner A completed in

80

minutes? (One lap takes

16

minutes)

Runner A: $80 \div 16 = 5 laps$ Runner B: $80 \div 20 = 4 laps$

Why not HCF?

Maybe you're wondering: "Why did we use LCM and not HCF?"

Let's test that idea. $HCF (16, 20) = 4$ . If the answer was the HCF, it would mean the runners meet at $t = 4$ minutes.

✍️ Yes/No

Yes or No?

If Runner A takes

16

minutes for one lap, would they be back at the start after

4

minutes?

At $t = 4$ :

Runner A: has completed $4 / 16 = 1 / 4$ of a lap — not at the starting point.
Runner B: has completed $4 / 20 = 1 / 5$ of a lap — also not at the starting point.

Multiple vs. Factor

HCF gives a factor of the cycle times, not a multiple.

The Rule: The meeting time must be a MULTIPLE (larger than or equal to both cycle times), not a factor (smaller than both).

✍️ MCQ

Choose one

When we need a result that is larger than or equal to the given numbers, we use:

Let's verify why HCF doesn't work here — at t = 4 minutes, neither runner has returned to START. The diagram shows exactly where each runner actually is.

2. Three Bells Problem with Clock Time

📋 Given Info

Three Bells Problem 🔔🔔🔔

Here is an extension of our synchronization problems — now with three events and the added challenge of converting to actual clock time.

Bell 1: Rings every 6 minutes
Bell 2: Rings every 8 minutes
Bell 3: Rings every 12 minutes

Starting condition: All three bells ring together at 12:00 noon.

✍️ Question

Two-part question:

(a) At what time will all three bells next ring together?
(b) How many times does each bell ring between 12:00 and 12:24 PM (inclusive of both endpoints)?

Show your working for both parts.

Part (a): When do all three bells ring together?

Earlier we saw how runners meet at the starting point using multiples. This bell problem works the exact same way! We are looking for the next time all three cyclic events coincide.

✍️ MCQ

Choose one

To find when the bells ring together, we need to find a:

Let's list out when each bell rings after noon:

Bell 1 rings at: 0, 6, 12, 18, 24, 30, 36, ... minutes after noon.
Bell 2 rings at: 0, 8, 16, 24, 32, ... minutes after noon.

✍️ MCQ

Choose one

What is the sequence of ringing times for Bell 3 (every 12 minutes)?

Bell 3 rings at: 0, 12, 24, 36, ... minutes after noon.

The first positive time that appears in all three lists is 24 minutes.

A timeline makes it easy to see exactly when each bell rings and where their schedules first overlap after noon.

✍️ FIB

Fill in the blank

Looking at the three timelines, at which minute after

t=0

do all three bells first ring together?

Formal Method: Prime Factorization

Mathematically, we find LCM(6, 8, 12) by breaking them into their prime building blocks first:

$6 = 2 \times 3$
$8 = 2^{3}$
$12 = 2^{2} \times 3$

✍️ MCQ

Choose one

Which power of

2

is the highest among the factors of

6, 8,

and

12

?

Now, we combine the highest powers:

$LCM = 2^{\max (1, 3, 2)} \times 3^{\max (1, 0, 1)} = 2^{3} \times 3 = 8 \times 3 = 24 minutes.$

✍️ FIB

Fill in the blank

What is the clock time

24

minutes after

12:00

noon?

Clock time:

12:00 noon + 24 minutes = 12:24 PM.

Part (b): Counting individual rings from 12:00 to 12:24.

Bell 1 (every 6 minutes):

Times: 12:00, 12:06, 12:12, 12:18, 12:24
Minutes: 0, 6, 12, 18, 24.
Count: 5 times.

The Formula: $\frac{24}{6} + 1 = 4 + 1 = 5$

The '+1' is because we include the starting ring at 12:00.

✍️ MCQ

Choose one

Why is the

+1

added to the calculation

\frac{24}{6} + 1

?

Bell 2 (every 8 minutes):

Times: 12:00, 12:08, 12:16, 12:24
Minutes: 0, 8, 16, 24.
Count: 4 times.
Formula: $\frac{24}{8} + 1 = 3 + 1 = 4$

Bell 3 (every 12 minutes):

Times: 12:00, 12:12, 12:24
Minutes: 0, 12, 24.
Count: 3 times.
Formula: $\frac{24}{12} + 1 = 2 + 1 = 3$

✍️ FIB

Fill in the blank

If a bell rings every

5

minutes, how many times would it ring from

12:00

to

12:20

(including both)?

Note the '+1'

When both endpoints (12:00 and 12:24) are included, the count is:

$Count = \frac{LCM}{interval} + 1$

✍️ MCQ

Choose one

What does the '

+1

' in the formula represent?

If you forget the +1, you will undercount by 1.

✍️ Yes/No

Yes or No?

If a problem asks for the number of rings after the initial ring at 12:00, do you still need to add

+1

?

3. Trap — Distribution Problem Disguised as LCM

⚠️ Careful — This One's Tricky!

This is the most important problem in this section. It appears among LCM examples, but looks can be deceiving.

This tests whether you're actually thinking about the problem structure — or just applying whatever operation the current section is about.

📋 Given Info

The Problem

A class teacher wants to distribute 30 books and 28 notebooks to students such that:

Each student gets the same number of books
Each student gets the same number of notebooks
No books or notebooks should be left over

What is the maximum number of students?

✍️ Question

Think carefully: 🤔

(a) Does this problem require HCF or LCM?
(b) Find the answer.
(c) Why would LCM give a nonsensical answer here?

The Distribution Trap

This is the textbook's deliberate trap. Let me show you why it is HCF, not LCM.

Suppose there are $k$ students. Each gets the same number of books and notebooks:

Books per student = $\frac{30}{k}$

For this to be a whole number, $k$ must divide $30$ .

✍️ Yes/No

Yes or No?

If we have

30

books, could we distribute them equally among

7

students?

It is the same logic for the notebooks:

Notebooks per student = $\frac{28}{k}$

For this to be a whole number, $k$ must divide $28$ .

So, $k$ divides both $30$ and $28$ .

✍️ MCQ

Choose one

Which tool finds the largest common divisor?

We want the MAXIMUM number of students $k$ .

Therefore, we need to find: $HCF (30, 28)$

Let's compute the prime factorisation for both numbers:

$30 = 2 \times 3 \times 5$ $28 = 2^{2} \times 7$

✍️ MCQ

Choose one

Which prime factor is common to both

30

and

28

?

Only common prime: 2 (min exponent = 1).

HCF = 2

✍️ FIB

Fill in the blank

What is the maximum number of students?

Maximum 2 students.

Each gets $30 / 2 = 15$ books
Each gets $28 / 2 = 14$ notebooks

Now let's update our diagram to show the complete solution — with 2 students each receiving 15 books and 14 notebooks, making the HCF result visually concrete.

Why LCM is wrong

Suppose we tried to use the LCM for this problem.

$LCM (30, 28) = 420$

✍️ Yes/No

Yes or No?

If we have only

30

books, is it possible to give

420

students at least one whole book each?

If there were 420 students, each would get: $\frac{30}{420} \approx 0.07 books$

You cannot give a fraction of a book to a student!

420 is LARGER than both 30 and 28, so it is impossible to divide 30 books among 420 students with whole-number results.

The Key Test

How do we decide between HCF and LCM? Ask yourself this Key Test:

Does the answer (e.g., number of students) divide the given numbers? If yes, the answer is a divisor $\to$ HCF.
Or do the given numbers divide the answer? If yes, the answer is a multiple $\to$ LCM.

✍️ MCQ

Choose one

In the 'Key Test', if the answer must be a divisor of the given values, which should we calculate?

Applying the Test

Let's see how the same numbers can lead to different tools depending on the question:

Students dividing 30 and 28: The students must divide the total items into groups. The answer is a divisor $\to$ HCF.
Finding a total to buy: If the question was 'what is the least number of items to buy so each student gets exactly 30 books and 28 notebooks', the total items would be a multiple $\to$ LCM.

Important Strategy: Never choose HCF or LCM based on which section you are reading. Always base it on what the problem actually asks.

✍️ Yes/No

Yes or No?

Is it a reliable strategy to choose between HCF and LCM based on the heading of the textbook chapter?

4. Summary Table — All CL-07 Problem Types

📋 Given Info

Now that you have seen every problem type in this cluster, let's test whether you can classify a new problem correctly by mapping it to the complete framework.

The Challenge Problem

A farmer has a field of length 360 m and width 240 m. He wants to divide it into the largest possible square plots.

Find:

What should be the side length of each square plot?
How many plots will he get?

✍️ Question

Your Task 🎯

Classify this problem (which pattern from CL-07 does it match?), solve it, and explain your reasoning.

This is a tiling problem — a variant of the 'maximum common measure' HCF pattern.

The field is 360 m long and 240 m wide. We want to divide it into square plots (all the same size), with nothing left over.

✍️ MCQ

Choose one

Should the side length of the small squares be a divisor or a multiple of the field's dimensions?

If each square has side length $s$ :

Along the length: $360 / s$ squares (must be a whole number $\to$ $s$ divides 360)
Along the width: $240 / s$ squares (must be a whole number $\to$ $s$ divides 240)

✍️ MCQ

Choose one

If

s

divides both

360

and

240

,

s

is a:

So $s$ must divide both 360 and 240. We want the LARGEST $s$ (largest possible square plots): HCF(360, 240).

Now that we've found the answer, let me update our diagram to show exactly how the field divides into square plots — this makes the HCF solution concrete and visual.

To find the HCF of $360$ and $240$ , we first look at their prime factorisations:

$360 = 2^{3} \times 3^{2} \times 5$ $240 = 2^{4} \times 3 \times 5$

✍️ MCQ

Choose one

Which power of

2

should be used for the HCF?

Common primes with minimum exponents:

$2 \to \min (3, 4) = 3$
$3 \to \min (2, 1) = 1$
$5 \to \min (1, 1) = 1$

$HCF = 2^{3} \times 3^{1} \times 5^{1} = 8 \times 3 \times 5 = 120 m$

✍️ FIB

Fill in the blank

Number of plots along the length

= 360 \div 120 =

?

Now we calculate the total number of plots:

Along length: $360 / 120 = 3$ plots
Along width: $240 / 120 = 2$ plots

$Total plots = 3 \times 2 = 6 plots$

Seeing the Pattern

This is exactly the same pattern as the tank problem (maximum container capacity), the marching problem (maximum columns), and the burfi problem (maximum stack size).

✍️ Yes/No

Yes or No?

Can the answer be larger than the given quantities in these scenarios?

The Common Thread

The answer DIVIDES the given quantities, and we want the MAXIMUM such answer.

✍️ MCQ

Choose one

Which statement is true for these 'Maximum Common Measure' problems?

Summary of Word Problem Patterns

With this, you have seen all the HCF/LCM word problem patterns in Chapter 1. Let's look at the HCF patterns first:

Remainder-adjusted HCF: (subtract remainders, then find the HCF).
Real-world HCF: (equal grouping, maximum measure, tiling, maximum columns).

✍️ MCQ

Choose one

Which tool finds the maximum number of equal groups or rows?

Now, let's review the LCM patterns:

LCM + remainder: (surplus pattern).
LCM - deficit: (constant-deficit pattern).
Greatest/least n-digit number: (LCM + boundary adjustment).
Real-world LCM: (circular tracks, bells, synchronization).

✍️ MCQ

Choose one

Which concept applies to synchronization problems like ringing bells?

Real-World LCM: Tracks, Bells, and Distribution Traps

1. Filter — Circular Track with Interpretation

🏃 The Classic Synchronization Problem

Your Challenge 🎯

Why not HCF?

Multiple vs. Factor

2. Three Bells Problem with Clock Time

Three Bells Problem 🔔🔔🔔

Part (a): When do all three bells ring together?

Formal Method: Prime Factorization

Part (b): Counting individual rings from 12:00 to 12:24.

Note the '+1'

3. Trap — Distribution Problem Disguised as LCM

⚠️ Careful — This One's Tricky!

The Problem

Think carefully: 🤔

The Distribution Trap

Why LCM is wrong

The Key Test

Applying the Test

4. Summary Table — All CL-07 Problem Types

The Challenge Problem

Your Task 🎯

Seeing the Pattern

The Common Thread

Summary of Word Problem Patterns