Samacheer Kalvi Class 6 Maths Solutions Term 2 Chapter 1 Numbers Exercise 1.2

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.2

Question 1. Fill in the blanks
(i) The HCF of 45 and 75 is .......
(ii) The HCF of two successive even numbers is .......
(iii) If the LCM of 3 and 9 is 9, then their HCF is .......
(iv) The LCM of 26, 39 and 52 is ........
(v) The least number that should be added to 57 so that the sum is exactly divisible by 2, 3, 4 and 5 is .........
Answer:
(i) 15
(ii) 2
(iii) 3
(iv) 156
(v) 3
In simple words: This question tests your knowledge of HCF and LCM. For HCF, you find the largest common factor. For LCM, you find the smallest common multiple. The last part asks you to find a number that makes 57 divisible by many numbers.

🎯 Exam Tip: Remember that the HCF of two successive even numbers is always 2, as 2 is the only common prime factor. For the least number to add, find the LCM of the divisors and then determine the remainder when 57 is divided by that LCM.

Question 2. Say True or False
(i) The numbers 57 and 69 are co-primes.
(ii) The HCF of 17 and 18 is 1.
(iii) The LCM of two successive numbers is the product of the numbers.
(iv) The LCM of two co-primes is the sum of the numbers.
(v) The HCF of two numbers is always a factor of their LCM.
Answer:
(i) False
(ii) True
(iii) True
(iv) False
(v) True
In simple words: Co-prime numbers only have 1 as a common factor. The HCF of numbers that come right after each other is always 1. The LCM of two numbers that are next to each other is found by multiplying them. For co-primes, their LCM is their product, not their sum. The HCF of any two numbers will always divide their LCM perfectly.

🎯 Exam Tip: Co-prime numbers are pairs that have only 1 as their Highest Common Factor (HCF). This is an important property to remember for true/false questions and calculations.

Question 3. Find the HCF of each set of numbers using the prime factorisation method.
(i) 18, 24
(ii) 51, 85
(iii) 61, 76
(iv) 84, 120
(v) 27, 45, 81
(vi) 45, 55, 95
Answer:
(i) 18, 24
Factors of 18: \( 2 \times 3 \times 3 \)
Factors of 24: \( 2 \times 2 \times 2 \times 3 \)
The common prime factors are 2 and 3.
HCF = \( 2 \times 3 = 6 \)

(ii) 51, 85
Factors of 51: \( 3 \times 17 \)
Factors of 85: \( 5 \times 17 \)
The common prime factor is 17.
HCF = 17

(iii) 61, 76
Factors of 61: \( 1 \times 61 \) (61 is a prime number)
Factors of 76: \( 2 \times 2 \times 19 \)
The only common factor is 1.
HCF = 1

(iv) 84, 120
Factors of 84: \( 2 \times 2 \times 3 \times 7 \)
Factors of 120: \( 2 \times 2 \times 2 \times 3 \times 5 \)
The common prime factors are two 2s and one 3.
HCF = \( 2 \times 2 \times 3 = 12 \)

(v) 27, 45, 81
Factors of 27: \( 3 \times 3 \times 3 \)
Factors of 45: \( 3 \times 3 \times 5 \)
Factors of 81: \( 3 \times 3 \times 3 \times 3 \)
The common prime factors for all three numbers are two 3s.
HCF = \( 3 \times 3 = 9 \)

(vi) 45, 55, 95
Factors of 45: \( 3 \times 3 \times 5 \)
Factors of 55: \( 5 \times 11 \)
Factors of 95: \( 5 \times 19 \)
The common prime factor for all three numbers is 5.
HCF = 5
In simple words: To find the HCF using prime factorization, first break each number down into its prime factors. Then, look for all the prime factors that are shared by every number in the group. Multiply these common prime factors together to get the HCF. If there are no common prime factors, the HCF is 1.

🎯 Exam Tip: Always make sure you break down numbers into their *prime* factors. For the HCF of more than two numbers, a prime factor must be present in *all* the numbers to be included in the HCF calculation.

Question 4. Find the LCM of each set of numbers using the prime factorisation method.
(i) 6, 9
(ii) 8, 12
(iii) 10, 15
(iv) 14, 42
(v) 30, 40, 60
(vi) 15, 25, 75
Answer:
(i) 6, 9
Factors of 6: \( 2 \times 3 \)
Factors of 9: \( 3 \times 3 \)
To find the LCM, take all prime factors that appear, and for each factor, use its highest power.
LCM of 6 and 9 = \( 2 \times 3 \times 3 = 18 \)

(ii) 8, 12
Factors of 8: \( 2 \times 2 \times 2 \)
Factors of 12: \( 2 \times 2 \times 3 \)
LCM of 8 and 12 = \( 2 \times 2 \times 2 \times 3 = 24 \)

(iii) 10, 15
Factors of 10: \( 2 \times 5 \)
Factors of 15: \( 3 \times 5 \)
LCM of 10 and 15 = \( 2 \times 3 \times 5 = 30 \)

(iv) 14, 42
Factors of 14: \( 2 \times 7 \)
Factors of 42: \( 2 \times 3 \times 7 \)
LCM of 14 and 42 = \( 2 \times 3 \times 7 = 42 \)

(v) 30, 40, 60
Factors of 30: \( 2 \times 3 \times 5 \)
Factors of 40: \( 2 \times 2 \times 2 \times 5 \)
Factors of 60: \( 2 \times 2 \times 3 \times 5 \)
LCM of 30, 40, 60 = \( 2 \times 2 \times 2 \times 3 \times 5 = 120 \)

(vi) 15, 25, 75
Factors of 15: \( 3 \times 5 \)
Factors of 25: \( 5 \times 5 \)
Factors of 75: \( 3 \times 5 \times 5 \)
LCM of 15, 25, 75 = \( 3 \times 5 \times 5 = 75 \)
In simple words: To find the LCM using prime factorization, first break down each number into its prime factors. Then, write down all unique prime factors that appear in any of the numbers. For each unique prime factor, choose the highest power it appears with. Multiply these highest powers together to get the LCM.

🎯 Exam Tip: A common mistake in LCM is to only pick common factors. Remember, LCM includes all prime factors that appear in *any* of the numbers, raised to their highest individual power.

Question 5. Find the HCF and the LCM of the numbers 154, 198, 286
Answer:
First, find the prime factors for each number:
Factors of 154: \( 2 \times 7 \times 11 \)
Factors of 198: \( 2 \times 3 \times 3 \times 11 \)
Factors of 286: \( 2 \times 11 \times 13 \)

Now, calculate the HCF:
The common prime factors present in all three numbers are 2 and 11.
HCF = \( 2 \times 11 = 22 \)

Next, calculate the LCM:
We take all unique prime factors (2, 3, 7, 11, 13) and their highest powers.
Highest power of 2 is \( 2^1 \)
Highest power of 3 is \( 3^2 \) (from 198)
Highest power of 7 is \( 7^1 \)
Highest power of 11 is \( 11^1 \)
Highest power of 13 is \( 13^1 \)
LCM = \( 2 \times 3 \times 3 \times 7 \times 11 \times 13 \)
LCM = \( 2 \times 9 \times 7 \times 11 \times 13 \)
LCM = \( 18 \times 7 \times 11 \times 13 \)
LCM = \( 126 \times 11 \times 13 \)
LCM = \( 1386 \times 13 \)
LCM = 18018
In simple words: To find the HCF, look for the prime numbers that all three numbers share. Multiply these common prime numbers together. To find the LCM, list all the unique prime numbers from all the numbers, then multiply them using their highest power that appears in any of the factorizations. This gives you the smallest number that all three can divide into.

🎯 Exam Tip: When finding HCF and LCM for multiple numbers, it's a good practice to list the prime factorization for each number clearly, then systematically identify common factors for HCF and all unique factors (with highest powers) for LCM.

Question 6. What is the greatest possible volume of a vessel that can be used to measure exactly the volume of milk in cans (in full capacity) of 80 litres, 100 litres and 120 litres?
Answer: This problem asks for the greatest possible volume that can measure all given quantities exactly. This means we need to find the Highest Common Factor (HCF) of 80, 100, and 120.

First, find the prime factors of each volume:
Factors of 80: \( 2 \times 2 \times 2 \times 2 \times 5 \)
Factors of 100: \( 2 \times 2 \times 5 \times 5 \)
Factors of 120: \( 2 \times 2 \times 2 \times 3 \times 5 \)

Next, identify the common prime factors:
All three numbers share two 2s and one 5.
HCF = \( 2 \times 2 \times 5 = 20 \)

So, the greatest possible volume of the vessel is 20 litres.
In simple words: When you need to find the largest size that can measure several different amounts perfectly, you should calculate the HCF of those amounts. We found the prime factors of 80, 100, and 120, then multiplied their common prime factors to get 20. This means a 20-litre vessel can measure exactly 80, 100, and 120 litres.

🎯 Exam Tip: Words like "greatest possible," "largest," or "maximum" in a word problem often indicate that you need to find the HCF. Always ensure your answer includes the correct units, such as "litres" in this case.

Question 7. The traffic lights at three different road junctions change after every 40 seconds, 60 seconds, and 72 seconds respectively. If they changed simultaneously together at 8 a.m at the junctions, at what time will they simultaneously change together again?
Answer: This is a problem where we need to find when events will happen together again. This requires calculating the Least Common Multiple (LCM) of the given time intervals.

First, find the prime factors of 40, 60, and 72:
Factors of 40: \( 2 \times 2 \times 2 \times 5 \)
Factors of 60: \( 2 \times 2 \times 3 \times 5 \)
Factors of 72: \( 2 \times 2 \times 2 \times 3 \times 3 \)

Now, calculate the LCM:
LCM = \( 2 \times 2 \times 2 \times 3 \times 3 \times 5 \)
LCM = \( 8 \times 9 \times 5 \)
LCM = \( 72 \times 5 \)
LCM = 360 seconds

Convert 360 seconds into minutes:
360 seconds \( = \frac{360}{60} \) minutes \( = 6 \) minutes

The lights changed together at 8:00 a.m. They will change together again after 6 minutes.
So, they will change together again at 8:06 a.m.
In simple words: To find when things will happen together again, we need to find the smallest number that all the given times can divide into. This is called the LCM. We found the LCM of 40, 60, and 72 seconds, which was 360 seconds. Then we changed 360 seconds into 6 minutes. Since they started at 8 a.m., they will flash together again 6 minutes later, at 8:06 a.m.

🎯 Exam Tip: Problems that ask for "when will they next occur together" or "least number of" are usually LCM problems. Always convert the units if necessary (e.g., seconds to minutes or hours) to provide the answer in the most sensible format.

Question 8. The LCM of two numbers is 210 and their HCF is 14. How many such pairs are possible.
Answer: We know that for any two positive integers, the product of the two numbers is equal to the product of their HCF and LCM.
Let the two numbers be \( x \) and \( y \).
So, \( x \times y = \text{LCM} \times \text{HCF} \)
\( x \times y = 210 \times 14 \)
\( x \times y = 2940 \)

Also, both numbers \( x \) and \( y \) must be multiples of their HCF, which is 14. So we can write \( x = 14a \) and \( y = 14b \), where \( a \) and \( b \) are co-prime integers.
\( (14a) \times (14b) = 2940 \)
\( 196ab = 2940 \)
\( ab = \frac{2940}{196} \)
\( ab = 15 \)

Now we need to find pairs of co-prime integers \( (a, b) \) whose product is 15.
The pairs of factors of 15 are:
1. \( (1, 15) \): Here, 1 and 15 are co-prime.
If \( a=1, b=15 \), then \( x = 14 \times 1 = 14 \) and \( y = 14 \times 15 = 210 \).
Check: HCF(14, 210) = 14, LCM(14, 210) = 210. (This pair works)
2. \( (3, 5) \): Here, 3 and 5 are co-prime.
If \( a=3, b=5 \), then \( x = 14 \times 3 = 42 \) and \( y = 14 \times 5 = 70 \).
Check: HCF(42, 70) = 14, LCM(42, 70) = 210. (This pair works)

Other pairs of factors like (15, 1) and (5, 3) would give the same pairs of numbers, just in a different order.
So, there are two such pairs of numbers possible: (14, 210) and (42, 70).
In simple words: The key rule here is that when you multiply two numbers, you get the same answer as when you multiply their HCF and LCM. We found that the product of the two numbers is 2940. Also, both numbers must be multiples of their HCF (14). So, we looked for pairs of numbers whose product is 2940 and whose HCF is 14. We found two such pairs.

🎯 Exam Tip: Remember the fundamental relation: Product of two numbers = HCF \( \times \) LCM. When finding pairs, express the numbers as multiples of HCF and ensure the other factors are co-prime.

Question 9. The LCM of two numbers is 6 times their HCF. If the HCF is 12 and one of the numbers is 36, then find the other number.
Answer: Given:
HCF = 12
LCM = 6 \( \times \) HCF = 6 \( \times \) 12 = 72
One number = 36
Let the other number be \( x \).

Using the formula: Product of two numbers = HCF \( \times \) LCM
\( 36 \times x = 12 \times 72 \)
To find \( x \), divide both sides by 36:
\( x = \frac{12 \times 72}{36} \)
\( x = \frac{12 \times (2 \times 36)}{36} \)
\( x = 12 \times 2 \)
\( x = 24 \)

Therefore, the other number is 24.
In simple words: We know that the LCM is 6 times the HCF. Since HCF is 12, the LCM is 72. We also know that if you multiply two numbers, you get the same result as multiplying their HCF and LCM. One number is 36, and we want to find the other. So, we set up the equation: 36 multiplied by the unknown number equals 12 multiplied by 72. Solving this, we find the other number is 24.

🎯 Exam Tip: This question tests your ability to apply the relationship between HCF, LCM, and the product of two numbers. Clearly list the known values first to avoid confusion in calculation.

Question 10. Which of the following pairs is co-prime?
(a) 51, 63
(b) 52, 91
(c) 71, 81
(d) 81, 99
Answer: (c) 71, 81
In simple words: Co-prime numbers are pairs that only share 1 as a common factor. Let's check each option: (a) Both 51 and 63 are divisible by 3. (b) Both 52 and 91 are divisible by 13 (52 = 4x13, 91 = 7x13). (c) 71 is a prime number, and 81 = 3x3x3x3. They don't share any common factors other than 1. (d) Both 81 and 99 are divisible by 9. So, only 71 and 81 are co-prime.

🎯 Exam Tip: To quickly check if numbers are co-prime, try dividing them by small prime numbers (2, 3, 5, 7, 11, 13...). If they share any prime factor, they are not co-prime. If one number is prime, only check if the other number is a multiple of that prime number.

Question 11. The greatest four-digit number which is exactly divisible by 8, 9, and 12 is
(a) 9999
(b) 9996
(c) 9696
(d) 9936
Answer: (d) 9936
In simple words: To find a number exactly divisible by 8, 9, and 12, we first need to find the LCM of these three numbers. The LCM of 8, 9, and 12 is 72. Now, we need to find the largest four-digit number that is a multiple of 72. The largest four-digit number is 9999. If you divide 9999 by 72, you get a remainder. Subtract this remainder from 9999 to find the greatest four-digit number exactly divisible by 72. \( 9999 \div 72 = 138 \) with a remainder of 63. So, \( 9999 - 63 = 9936 \).

🎯 Exam Tip: To find a number exactly divisible by several numbers, first find their LCM. Then, if asked for the largest/smallest N-digit number, divide the largest/smallest N-digit number by the LCM and adjust based on the remainder.

Question 12. The HCF of two numbers is 2 and their LCM is 154. If the difference between the numbers is 8, then the sum is
(a) 26
(b) 36
(c) 46
(d) 56
Answer: (b) 36
In simple words: Let the two numbers be \( x \) and \( y \). We know their HCF is 2 and LCM is 154. So, their product \( x \times y \) will be \( 2 \times 154 = 308 \). We also know that their difference \( x - y \) is 8. We need to find their sum \( x + y \). We can use the identity \( (x+y)^2 = (x-y)^2 + 4xy \). Plugging in the values, \( (x+y)^2 = 8^2 + 4 \times 308 = 64 + 1232 = 1296 \). Taking the square root, \( x+y = \sqrt{1296} = 36 \). Thus, the sum of the numbers is 36.

🎯 Exam Tip: For problems involving HCF, LCM, sum, and difference of two numbers, the identity \( (x+y)^2 = (x-y)^2 + 4xy \) is very useful. Also, remember that both numbers must be multiples of their HCF.

Question 13. Which of the following cannot be the HCF of two numbers whose LCM is 120?
(a) 60
(b) 40
(c) 80
(d) 30
Answer: (c) 80
In simple words: A basic rule in number theory is that the HCF of any two numbers must always be a factor of their LCM. This means the LCM must be perfectly divisible by the HCF. Here, the LCM is 120. Let's check which of the options does not divide 120 evenly. 120 is divisible by 60 (120/60 = 2). 120 is divisible by 40 (120/40 = 3). 120 is NOT divisible by 80 (120/80 = 1.5, not a whole number). 120 is divisible by 30 (120/30 = 4). Since 80 does not divide 120, 80 cannot be the HCF of two numbers whose LCM is 120.

🎯 Exam Tip: Always remember the fundamental relationship: HCF always divides LCM exactly. If a given number does not divide the LCM, it cannot be the HCF.