Math worksheet featuring word problems on H.C.F. and L.C.M. for practice.

Let's solve each problem step by step.

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Problem 1:

Find the least length of a rope which can be cut into whole number of pieces of lengths 45 cm, 75 cm, and 81 cm.

#### Solution:
To find the least length of the rope, we need to determine the Least Common Multiple (LCM) of the given lengths: 45 cm, 75 cm, and 81 cm.

1. Prime Factorization:
- \( 45 = 3^2 \times 5 \)
- \( 75 = 3 \times 5^2 \)
- \( 81 = 3^4 \)

2. LCM Calculation:
- The LCM is found by taking the highest power of each prime factor present in the factorizations.
- For \( 3 \): The highest power is \( 3^4 \).
- For \( 5 \): The highest power is \( 5^2 \).

Therefore, the LCM is:
\[
\text{LCM} = 3^4 \times 5^2 = 81 \times 25 = 2025
\]

#### Final Answer:
\[
\boxed{2025}
\]

---

Problem 2:

Find the greatest number of 4-digits which is exactly divisible by 40, 48, and 60.

#### Solution:
To find the greatest 4-digit number divisible by 40, 48, and 60, we first need to determine the Least Common Multiple (LCM) of these numbers.

1. Prime Factorization:
- \( 40 = 2^3 \times 5 \)
- \( 48 = 2^4 \times 3 \)
- \( 60 = 2^2 \times 3 \times 5 \)

2. LCM Calculation:
- For \( 2 \): The highest power is \( 2^4 \).
- For \( 3 \): The highest power is \( 3^1 \).
- For \( 5 \): The highest power is \( 5^1 \).

Therefore, the LCM is:
\[
\text{LCM} = 2^4 \times 3 \times 5 = 16 \times 3 \times 5 = 240
\]

3. Finding the Greatest 4-Digit Number Divisible by 240:
- The greatest 4-digit number is 9999.
- Divide 9999 by 240 to find the largest multiple of 240 that is less than or equal to 9999:
\[
9999 \div 240 \approx 41.6625
\]
- Take the integer part: \( 41 \).
- Multiply 41 by 240 to get the largest multiple of 240:
\[
41 \times 240 = 9840
\]

#### Final Answer:
\[
\boxed{9840}
\]

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Problem 3:

What is the least number of saplings that can be arranged in rows of 12, 15, or 40 in each row?

#### Solution:
To find the least number of saplings, we need to determine the Least Common Multiple (LCM) of the given row sizes: 12, 15, and 40.

1. Prime Factorization:
- \( 12 = 2^2 \times 3 \)
- \( 15 = 3 \times 5 \)
- \( 40 = 2^3 \times 5 \)

2. LCM Calculation:
- For \( 2 \): The highest power is \( 2^3 \).
- For \( 3 \): The highest power is \( 3^1 \).
- For \( 5 \): The highest power is \( 5^1 \).

Therefore, the LCM is:
\[
\text{LCM} = 2^3 \times 3 \times 5 = 8 \times 3 \times 5 = 120
\]

#### Final Answer:
\[
\boxed{120}
\]

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Problem 4:

210 oranges, 252 apples, and 294 pears are equally packed in cartons so that no fruit is left. What is the biggest possible number of cartons needed?

#### Solution:
To find the biggest possible number of cartons, we need to determine the Greatest Common Divisor (GCD) of the quantities of fruits: 210, 252, and 294.

1. Prime Factorization:
- \( 210 = 2 \times 3 \times 5 \times 7 \)
- \( 252 = 2^2 \times 3^2 \times 7 \)
- \( 294 = 2 \times 3 \times 7^2 \)

2. GCD Calculation:
- The common factors are \( 2 \), \( 3 \), and \( 7 \).
- The lowest powers of these common factors are \( 2^1 \), \( 3^1 \), and \( 7^1 \).

Therefore, the GCD is:
\[
\text{GCD} = 2 \times 3 \times 7 = 42
\]

3. Number of Cartons:
- Each carton will contain 42 fruits.
- Total number of cartons needed is the sum of the quantities divided by the GCD:
\[
\text{Total cartons} = \frac{210}{42} + \frac{252}{42} + \frac{294}{42} = 5 + 6 + 7 = 18
\]

#### Final Answer:
\[
\boxed{18}
\]

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Problem 5:

Find the greatest number of 5-digits which on being divided by 9, 12, 24, and 45 leaves 3, 6, 18, and 39 as remainders respectively.

#### Solution:
We are given:
- When divided by 9, the remainder is 3.
- When divided by 12, the remainder is 6.
- When divided by 24, the remainder is 18.
- When divided by 45, the remainder is 39.

Notice that in each case, the remainder is 6 less than the divisor:
- \( 9 - 3 = 6 \)
- \( 12 - 6 = 6 \)
- \( 24 - 18 = 6 \)
- \( 45 - 39 = 6 \)

This suggests that the number \( N \) can be expressed as:
\[
N = k \times \text{LCM}(9, 12, 24, 45) - 6
\]
where \( k \) is an integer.

1. Prime Factorization:
- \( 9 = 3^2 \)
- \( 12 = 2^2 \times 3 \)
- \( 24 = 2^3 \times 3 \)
- \( 45 = 3^2 \times 5 \)

2. LCM Calculation:
- For \( 2 \): The highest power is \( 2^3 \).
- For \( 3 \): The highest power is \( 3^2 \).
- For \( 5 \): The highest power is \( 5^1 \).

Therefore, the LCM is:
\[
\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360
\]

3. Form of \( N \):
\[
N = 360k - 6
\]

4. Finding the Greatest 5-Digit Number:
- The greatest 5-digit number is 99999.
- We need \( N \leq 99999 \):
\[
360k - 6 \leq 99999
\]
\[
360k \leq 100005
\]
\[
k \leq \frac{100005}{360} \approx 277.79
\]
- Take the integer part: \( k = 277 \).

5. Calculate \( N \):
\[
N = 360 \times 277 - 6 = 99720 - 6 = 99714
\]

#### Final Answer:
\[
\boxed{99714}
\]

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Summary of Answers:

1. \(\boxed{2025}\)
2. \(\boxed{9840}\)
3. \(\boxed{120}\)
4. \(\boxed{18}\)
5. \(\boxed{99714}\)