Samacheer Kalvi Class 6 Maths Solutions Term 2 Chapter 1 Numbers Exercise 1.3

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.3

Miscellaneous Practice Problems

 

Question 1. Every even number greater than 2 can be expressed as the sum of two prime numbers. Verify this statement for every even number upto 16.
Answer: This statement is known as Goldbach's Conjecture. We need to check it for all even numbers from 4 to 16.
Even numbers greater than 2 up to 16 are: 4, 6, 8, 10, 12, 14, and 16.
\( 4 = 2 + 2 \) (Here, 2 is a prime number.)
\( 6 = 3 + 3 \) (Here, 3 is a prime number.)
\( 8 = 3 + 5 \) (Here, 3 and 5 are prime numbers.)
\( 10 = 3 + 7 \) or \( 5 + 5 \) (Here, 3, 5, and 7 are prime numbers.)
\( 12 = 5 + 7 \) (Here, 5 and 7 are prime numbers.)
\( 14 = 7 + 7 \) or \( 3 + 11 \) (Here, 3, 7, and 11 are prime numbers.)
\( 16 = 5 + 11 \) or \( 3 + 13 \) (Here, 3, 5, 11, and 13 are prime numbers.)
The statement is verified for all even numbers up to 16.
In simple words: For any even number bigger than 2, you can always find two prime numbers that add up to it. We showed this by checking numbers like 4, 6, 8, all the way to 16.

🎯 Exam Tip: When verifying a statement for a range of numbers, ensure you check every number in that specified range to provide a complete answer.

 

Question 2. Is 173 a prime? Why?
Answer: Yes, 173 is a prime number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. We can check for divisibility by small prime numbers (like 2, 3, 5, 7, 11, 13) up to the square root of 173 (which is about 13.15). Since 173 is not divisible by any of these primes, it is a prime number.
In simple words: Yes, 173 is a prime number because you can only divide it evenly by 1 and 173 itself. No other whole number can divide it without leaving a remainder.

🎯 Exam Tip: To quickly check if a number is prime, test its divisibility by prime numbers only up to its square root. If none divide it, the number is prime.

 

Question 3. For which of the numbers, from n = 2 to 8. Is \( 2n - 1 \) a prime?
Answer: We need to substitute values of \( n \) from 2 to 8 into the expression \( 2n - 1 \) and check if the result is a prime number.
When \( n = 2 \), we calculate \( 2n - 1 \):
\( 2 \times 2 - 1 \)
\( = 4 - 1 \)
\( = 3 \) (This is a prime number.)
When \( n = 3 \), we calculate \( 2n - 1 \):
\( 2 \times 3 - 1 \)
\( = 6 - 1 \)
\( = 5 \) (This is a prime number.)
When \( n = 4 \), we calculate \( 2n - 1 \):
\( 2 \times 4 - 1 \)
\( = 8 - 1 \)
\( = 7 \) (This is a prime number.)
When \( n = 5 \), we calculate \( 2n - 1 \):
\( 2 \times 5 - 1 \)
\( = 10 - 1 \)
\( = 9 \) (This is not a prime number because 9 is divisible by 3.)
When \( n = 6 \), we calculate \( 2n - 1 \):
\( 2 \times 6 - 1 \)
\( = 12 - 1 \)
\( = 11 \) (This is a prime number.)
When \( n = 7 \), we calculate \( 2n - 1 \):
\( 2 \times 7 - 1 \)
\( = 14 - 1 \)
\( = 13 \) (This is a prime number.)
When \( n = 8 \), we calculate \( 2n - 1 \):
\( 2 \times 8 - 1 \)
\( = 16 - 1 \)
\( = 15 \) (This is not a prime number because 15 is divisible by 3 and 5.)
So, for \( n = 2, 3, 4, 6, 7 \), the expression \( 2n - 1 \) results in a prime number.
In simple words: We put each number from 2 to 8 into the rule \( 2n - 1 \). If the answer was only divisible by 1 and itself, it was a prime number. This happened for n values of 2, 3, 4, 6, and 7.

🎯 Exam Tip: Carefully substitute each value of \( n \) and perform the calculation accurately. Remember the definition of a prime number to correctly identify the results.

 

Question 4. Explain your answer with the reason for the following statements
(a) A number is divisible by 9 if it is divisible by 3.
(b) A number is divisible by 6 if it is divisible by 12.
Answer: We need to verify each statement and provide a reason or example.
(a) This statement is False. If a number is divisible by 9, it must be divisible by 3, but the reverse is not always true. For example, the number 42 is divisible by 3 (since \( 4 + 2 = 6 \), which is divisible by 3), but 42 is not divisible by 9 (since \( 4 + 2 = 6 \), which is not divisible by 9).
(b) This statement is True. If a number is divisible by 12, it means it is divisible by both 3 and 4 (since \( 12 = 3 \times 4 \)). If a number is divisible by 12, it will automatically be divisible by its factors, which include 6. For example, 36 is divisible by 12 (since \( 36 \div 12 = 3 \)). Also, 36 is divisible by 6 (since \( 36 \div 6 = 6 \)).
In simple words: For part (a), just because a number can be divided by 3 doesn't mean it can also be divided by 9. Like 42 is divided by 3, but not 9. For part (b), if a number can be divided by 12, it means it can also be divided by 6 because 6 is a factor of 12. For example, 36 can be divided by 12, and it can also be divided by 6.

🎯 Exam Tip: Remember that divisibility rules are not always reversible. If a number is divisible by \(X\), it is also divisible by all factors of \(X\). However, if a number is divisible by a factor of \(X\), it is not necessarily divisible by \(X\).

 

Question 5. Find A as required
(i) The greatest 2 digit number 9A is divisible by 2.
(ii) The least number 567A is divisible by 3.
(iii) The greatest 3 digit number 9A6 is divisible by 6.
(iv) The number A08 is divisible by 4 and 9.
(v) The number 225A85 is divisible by 11.
Answer: We will find the value of A for each condition using divisibility rules.
(i) For a number to be divisible by 2, its last digit must be an even number (0, 2, 4, 6, 8). To make 9A the greatest 2-digit number divisible by 2, A must be the largest possible even digit, which is 8.
So, \( 9A = 98 \). Thus, \( A = 8 \).
(ii) For a number to be divisible by 3, the sum of its digits must be divisible by 3. For 567A, the sum of digits is \( 5 + 6 + 7 + A = 18 + A \).
To make \( 18 + A \) divisible by 3, and A the least possible digit, A can be 0 (since \( 18 + 0 = 18 \), which is divisible by 3).
So, \( 567A = 5670 \). Thus, \( A = 0 \).
(iii) For a number to be divisible by 6, it must be divisible by both 2 and 3.
The number 9A6 ends in 6, so it is already divisible by 2.
For it to be divisible by 3, the sum of its digits \( 9 + A + 6 = 15 + A \) must be divisible by 3.
To make 9A6 the greatest 3-digit number, A must be the largest possible digit that makes \( 15 + A \) divisible by 3. If \( A = 9 \), then \( 15 + 9 = 24 \), which is divisible by 3.
So, \( 9A6 = 996 \). Thus, \( A = 9 \).
(iv) For A08 to be divisible by 4, the number formed by its last two digits (08) must be divisible by 4. 08 is divisible by 4. So, this condition holds for any A.
For A08 to be divisible by 9, the sum of its digits \( A + 0 + 8 = A + 8 \) must be divisible by 9.
If \( A = 1 \), then \( 1 + 8 = 9 \), which is divisible by 9.
So, \( A08 = 108 \). Thus, \( A = 1 \).
(v) For a number to be divisible by 11, the difference between the sum of the digits at odd places and the sum of the digits at even places must be either 0 or a multiple of 11.
For 225A85:
Sum of digits at odd places (from right): \( 5 + A + 2 = A + 7 \)
Sum of digits at even places (from right): \( 8 + 5 + 2 = 15 \)
Difference: \( (A + 7) - 15 = A - 8 \)
For \( A - 8 \) to be 0 or a multiple of 11, and A being a single digit, \( A - 8 = 0 \) means \( A = 8 \).
So, \( 225A85 = 225885 \). Thus, \( A = 8 \).
In simple words: We used different divisibility rules to find the missing digit 'A' in each number. For example, to be divisible by 2, the last digit must be even. To be divisible by 3, the sum of digits must be a multiple of 3. For 11, we check the alternating sum of digits.

🎯 Exam Tip: Memorize the divisibility rules for 2, 3, 4, 5, 6, 9, 10, and 11, as they are essential for solving problems involving missing digits and checking divisibility.

 

Question 6. Numbers divisible by 4 and 6 are divisible by 24. Verify this statement and support your answer with an example.
Answer: This statement is False. A number divisible by both 4 and 6 is not necessarily divisible by 24. This is because 4 and 6 share a common factor (2).
For example, consider the number 12.
12 is divisible by 4 (since \( 12 \div 4 = 3 \)).
12 is also divisible by 6 (since \( 12 \div 6 = 2 \)).
However, 12 is not divisible by 24.
For a number to be divisible by 24, it must be divisible by both 3 and 8, because 3 and 8 are coprime factors of 24. Since 12 is divisible by 3 but not by 8, it is not divisible by 24. The rule only works if the two divisors are coprime (share no common factors other than 1).
In simple words: The idea that if a number can be divided by both 4 and 6, it can also be divided by 24 is not true. For example, 12 can be divided by both 4 and 6, but it cannot be divided by 24. This happens because 4 and 6 share a common factor (2).

🎯 Exam Tip: When checking divisibility by a composite number, ensure that the factors you use for checking are coprime. For example, to check divisibility by 24, check divisibility by 3 and 8 (since 3 and 8 are coprime and their product is 24), not by 4 and 6.

 

Question 7. The sum of any two successive odd numbers is always divisible by 4. Justify this statement with an example.
Answer: This statement is True. Let's consider two successive odd numbers.
If the first odd number is represented as \( 2n + 1 \), then the next successive odd number will be \( (2n + 1) + 2 = 2n + 3 \).
The sum of these two successive odd numbers is:
\( (2n + 1) + (2n + 3) = 4n + 4 \)
We can factor out 4 from this sum: \( 4n + 4 = 4(n + 1) \).
Since the sum can be written as \( 4 \times (\text{some integer}) \), it means the sum is always divisible by 4.
For example, let's take the successive odd numbers 3 and 5.
Their sum is \( 3 + 5 = 8 \).
8 is divisible by 4 (since \( 8 \div 4 = 2 \)).
Another example: 11 and 13.
Their sum is \( 11 + 13 = 24 \).
24 is divisible by 4 (since \( 24 \div 4 = 6 \)).
In simple words: If you take any two odd numbers that come right after each other and add them up, the answer will always be a number you can divide by 4. For example, 3 and 5 are successive odd numbers, and their sum is 8, which can be divided by 4.

🎯 Exam Tip: To prove a number property involving consecutive numbers, use algebraic expressions like \( 2n+1 \) for odd numbers or \( 2n \) for even numbers, then simplify the expression to show divisibility.

 

Question 8. Find the length of the longest rope that can be used to measure exactly the ropes of length 1 m 20cm, 3m 60 cm and 4 m.
Answer: To find the length of the longest rope that can measure all the given lengths exactly, we need to find the Highest Common Factor (HCF) of these lengths. First, convert all lengths to the same unit, which is centimeters (cm).
1 meter (m) = 100 centimeters (cm).
Length 1: \( 1 \text{ m } 20 \text{ cm } = 100 \text{ cm } + 20 \text{ cm } = 120 \text{ cm} \)
Length 2: \( 3 \text{ m } 60 \text{ cm } = 300 \text{ cm } + 60 \text{ cm } = 360 \text{ cm} \)
Length 3: \( 4 \text{ m } = 400 \text{ cm} \)
Now, we find the HCF of 120, 360, and 400 using prime factorization.

Prime factorization of 120: \( 2 \times 2 \times 2 \times 3 \times 5 \)
Prime factorization of 360: \( 2 \times 2 \times 2 \times 3 \times 3 \times 5 \)
Prime factorization of 400: \( 2 \times 2 \times 2 \times 2 \times 5 \times 5 \)
To find the HCF, we take the common prime factors raised to the lowest power they appear in any of the factorizations.
Common factors are \( 2^3 \) and \( 5^1 \).
\( \text{HCF} = 2 \times 2 \times 2 \times 5 = 8 \times 5 = 40 \)
So, the length of the longest rope is 40 cm.
In simple words: To find the longest rope that can measure all given ropes exactly, we need to find the biggest number that divides all their lengths. First, change all lengths to centimeters. Then, find the Highest Common Factor (HCF) of these lengths. The HCF of 120 cm, 360 cm, and 400 cm is 40 cm.

🎯 Exam Tip: Problems asking for the "largest," "greatest," or "maximum" size or quantity that can divide or measure a set of items usually require finding the HCF. Always convert to a common unit before calculating.

Challenge Problems

 

Question 9. The sum of three prime numbers is 80. The difference between the two of them is 4. Find the numbers.
Answer: Let the three prime numbers be \( p_1, p_2, p_3 \).
We are given that their sum is 80: \( p_1 + p_2 + p_3 = 80 \).
Since the sum is an even number (80), and we are adding three prime numbers, one of the prime numbers must be an even number. The only even prime number is 2.
So, let one of the prime numbers, say \( p_1 \), be 2.
Then, \( 2 + p_2 + p_3 = 80 \)
\( p_2 + p_3 = 80 - 2 \)
\( p_2 + p_3 = 78 \)
We are also given that the difference between the other two primes is 4: \( p_3 - p_2 = 4 \) (assuming \( p_3 > p_2 \)).
Now we have a system of two equations with two variables:
1. \( p_2 + p_3 = 78 \)
2. \( p_3 - p_2 = 4 \)
Add the two equations:
\( (p_2 + p_3) + (p_3 - p_2) = 78 + 4 \)
\( 2p_3 = 82 \)
\( p_3 = \frac{82}{2} \)
\( p_3 = 41 \)
Now substitute \( p_3 = 41 \) into the first equation:
\( p_2 + 41 = 78 \)
\( p_2 = 78 - 41 \)
\( p_2 = 37 \)
We check if 37 and 41 are prime numbers. Both 37 and 41 are prime numbers.
So, the three prime numbers are 2, 37, and 41.
In simple words: We know the sum of three prime numbers is 80 (an even number). This means one of them must be the number 2, which is the only even prime number. So, the other two primes add up to 78. We also know these two primes have a difference of 4. By solving these, we found the two other prime numbers are 37 and 41. So, the three numbers are 2, 37, and 41.

🎯 Exam Tip: When dealing with sums of prime numbers, especially when the sum is even, always consider if 2 must be one of the prime numbers. This often simplifies the problem significantly.

 

Question 10. Find the sum of all the prime numbers between 10 and 20 and check whether that sum is divisible by all the single-digit numbers.
Answer: First, let's list all the prime numbers between 10 and 20.
Prime numbers are numbers greater than 1 that have only two factors: 1 and themselves.
The prime numbers between 10 and 20 are: 11, 13, 17, and 19.
Next, we find the sum of these prime numbers:
\( \text{Sum} = 11 + 13 + 17 + 19 = 60 \)
Now, we need to check if this sum (60) is divisible by all the single-digit numbers. The single-digit numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9.
- Divisible by 1: Yes, 60 is divisible by 1 (any number is divisible by 1).
- Divisible by 2: Yes, 60 is an even number, so it is divisible by 2.
- Divisible by 3: Yes, the sum of digits of 60 is \( 6 + 0 = 6 \), which is divisible by 3.
- Divisible by 4: Yes, the number formed by the last two digits (60) is divisible by 4 (since \( 60 \div 4 = 15 \)).
- Divisible by 5: Yes, 60 ends in 0, so it is divisible by 5.
- Divisible by 6: Yes, 60 is divisible by both 2 and 3, so it is divisible by 6.
- Divisible by 7: No, \( 60 \div 7 \) leaves a remainder of 4.
- Divisible by 8: No, \( 60 \div 8 \) leaves a remainder of 4.
- Divisible by 9: No, the sum of digits of 60 is 6, which is not divisible by 9.
So, the sum (60) is divisible by 1, 2, 3, 4, 5, and 6, but not by 7, 8, and 9.
In simple words: First, we listed all the prime numbers between 10 and 20, which are 11, 13, 17, and 19. We added them up to get 60. Then, we checked if 60 could be divided evenly by all single-digit numbers (1 to 9). We found that 60 can be divided by 1, 2, 3, 4, 5, and 6, but not by 7, 8, or 9.

🎯 Exam Tip: When listing prime numbers, be careful to exclude composite numbers (like 15) and numbers outside the specified range. For divisibility, apply each rule methodically.

 

Question 11. Find the smallest number which is exactly divisible by all the numbers from 1 to 9.
Answer: To find the smallest number that is exactly divisible by all numbers from 1 to 9, we need to find the Least Common Multiple (LCM) of these numbers.
The numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9.
We can find the LCM by listing the prime factors for each number:
\( 1 = 1 \)
\( 2 = 2^1 \)
\( 3 = 3^1 \)
\( 4 = 2^2 \)
\( 5 = 5^1 \)
\( 6 = 2 \times 3 \)
\( 7 = 7^1 \)
\( 8 = 2^3 \)
\( 9 = 3^2 \)
To find the LCM, we take the highest power of each prime factor that appears in any of the numbers:
Highest power of 2 is \( 2^3 \) (from 8)
Highest power of 3 is \( 3^2 \) (from 9)
Highest power of 5 is \( 5^1 \) (from 5)
Highest power of 7 is \( 7^1 \) (from 7)
Now, multiply these highest powers together:
\( \text{LCM} = 2^3 \times 3^2 \times 5^1 \times 7^1 \)
\( \text{LCM} = 8 \times 9 \times 5 \times 7 \)
\( \text{LCM} = 72 \times 35 \)
\( \text{LCM} = 2520 \)
Thus, the smallest number exactly divisible by all numbers from 1 to 9 is 2520.
In simple words: To find the smallest number that can be divided evenly by all numbers from 1 to 9, we need to find their Least Common Multiple (LCM). We list the prime factors for each number and then multiply the highest power of each unique prime factor. The smallest number is 2520.

🎯 Exam Tip: When asked for the "smallest number exactly divisible by" a set of numbers, it's a clear indication that you need to calculate the Least Common Multiple (LCM) of those numbers.

 

Question 12. The product of any three consecutive numbers is always divisible by 6. Justify this statement with an example.
Answer: This statement is True. The product of any three consecutive numbers is always divisible by 6.
We can justify this by considering the properties of consecutive integers:
1. In any three consecutive integers, at least one of them must be an even number. This means the product will always be divisible by 2.
2. In any three consecutive integers, exactly one of them must be a multiple of 3. This means the product will always be divisible by 3.
Since the product is always divisible by both 2 and 3, and 2 and 3 are coprime (they have no common factors other than 1), the product must also be divisible by their product, which is \( 2 \times 3 = 6 \).
Example: Let's take the three consecutive numbers 5, 6, and 7.
Their product is \( 5 \times 6 \times 7 = 30 \times 7 = 210 \).
Now, let's check if 210 is divisible by 6:
\( 210 \div 6 = 35 \)
Since 210 is exactly divisible by 6, the statement is justified.
In simple words: If you multiply any three numbers that come one after another, the answer will always be a number you can divide by 6. This is because among any three consecutive numbers, one will always be even (divisible by 2) and one will always be a multiple of 3. Since it's divisible by both 2 and 3, it must be divisible by 6. For example, \( 5 \times 6 \times 7 = 210 \), and 210 can be divided by 6.

🎯 Exam Tip: To prove a divisibility rule for a product of consecutive integers, explain how the product guarantees divisibility by the prime factors of the divisor. For divisibility by 6, ensure divisibility by 2 and 3.

 

Question 13. Malarvizhi, Karthiga, and Anjali are friends and natives of the same village. They work in different places. Malarvizhi comes to her home once in 5 days. Similarly, Karthiga and Anjali come to their homes once in 6 days and 10 days respectively, Assuming that they met each other on the 1st of October, when will all the three meet again?
Answer: To find out when all three friends will meet again, we need to find the Least Common Multiple (LCM) of the number of days each friend takes to come home. The periods are 5 days, 6 days, and 10 days.
We will find the LCM of 5, 6, and 10.

To find the LCM, multiply the divisors and the remaining numbers at the bottom:
\( \text{LCM} = 5 \times 2 \times 1 \times 3 \times 1 = 30 \)
So, they will all meet again after 30 days.
If they met on the 1st of October, they will meet again 30 days later.
Since October has 31 days, 30 days after October 1st is October 31st.
Thus, all three will meet again on October 31st.
In simple words: To find when the three friends will meet again, we need to find the smallest number of days that is a multiple of all their visiting schedules (5, 6, and 10 days). This is called the Least Common Multiple (LCM). The LCM of 5, 6, and 10 is 30. So, they will meet again 30 days after October 1st, which is October 31st.

🎯 Exam Tip: Problems involving cycles, repetition, or meeting points (like bells ringing together or people meeting again) are typically solved by finding the Least Common Multiple (LCM).

 

Question 14. In an apartment consisting of 108 floors, two lifts A & B starting from the ground floor, stop at every 3rd and 5th floors respectively. On which floors, will both of them stop together?
Answer: To find the floors where both lifts stop together, we need to find the common multiples of 3 and 5. Since we are looking for the floors they stop at together, we need to find the Least Common Multiple (LCM) first.
Lift A stops at multiples of 3: 3, 6, 9, 12, 15, 18, ...
Lift B stops at multiples of 5: 5, 10, 15, 20, 25, ...
The LCM of 3 and 5 is \( 3 \times 5 = 15 \) (since 3 and 5 are prime numbers, their LCM is their product).
This means both lifts will stop together at every 15th floor.
We need to list these common stopping floors up to 108 floors:
Multiples of 15:
\( 15 \times 1 = 15 \)
\( 15 \times 2 = 30 \)
\( 15 \times 3 = 45 \)
\( 15 \times 4 = 60 \)
\( 15 \times 5 = 75 \)
\( 15 \times 6 = 90 \)
\( 15 \times 7 = 105 \)
\( 15 \times 8 = 120 \) (This is greater than 108, so we stop here.)
So, the lifts will both stop together on floors 15, 30, 45, 60, 75, 90, and 105.
In simple words: Lift A stops every 3 floors and Lift B stops every 5 floors. To find out when they stop together, we need to find the common floors, which means finding the Least Common Multiple (LCM) of 3 and 5. The LCM is 15. So, they will both stop at floors that are multiples of 15, like 15, 30, 45, and so on, up to 108 floors.

🎯 Exam Tip: When two or more events occur at regular intervals, and you need to find when they happen simultaneously, use the concept of LCM. Remember to list all common multiples within the given range.

 

Question 15. The product of 2 two-digit numbers is 300 and their HCF is 5. What are the numbers?
Answer: Let the two two-digit numbers be A and B.
We are given:
1. Product of the numbers \( A \times B = 300 \)
2. Highest Common Factor (HCF) of A and B is 5.
We know a property that states: For any two numbers, the product of the numbers is equal to the product of their HCF and LCM.
So, \( A \times B = \text{HCF} (A, B) \times \text{LCM} (A, B) \)
\( 300 = 5 \times \text{LCM} (A, B) \)
\( \text{LCM} (A, B) = \frac{300}{5} = 60 \)
Now we need to find two two-digit numbers whose HCF is 5 and LCM is 60.
Since the HCF is 5, both numbers must be multiples of 5.
Let \( A = 5x \) and \( B = 5y \), where \( x \) and \( y \) are coprime integers (meaning their HCF is 1).
We know that \( A \times B = 300 \):
\( (5x) \times (5y) = 300 \)
\( 25xy = 300 \)
\( xy = \frac{300}{25} \)
\( xy = 12 \)
Now we need to find pairs of coprime factors \( (x, y) \) for 12.
Possible pairs for \( xy = 12 \) are:
- \( (1, 12) \): Here, HCF(1, 12) = 1.
- \( (3, 4) \): Here, HCF(3, 4) = 1.
(Note: (2, 6) is not a coprime pair as HCF(2, 6) = 2).
Let's use these pairs to find A and B:
Case 1: If \( x = 1, y = 12 \)
\( A = 5x = 5 \times 1 = 5 \)
\( B = 5y = 5 \times 12 = 60 \)
Here, 5 is a single-digit number, but the question asks for two two-digit numbers. So, this pair is not valid.
Case 2: If \( x = 3, y = 4 \)
\( A = 5x = 5 \times 3 = 15 \)
\( B = 5y = 5 \times 4 = 20 \)
Both 15 and 20 are two-digit numbers.
Let's verify:
Product: \( 15 \times 20 = 300 \) (Correct)
HCF(15, 20): Factors of 15 are 1, 3, 5, 15. Factors of 20 are 1, 2, 4, 5, 10, 20. The common factors are 1, 5. The HCF is 5. (Correct)
Thus, the two numbers are 15 and 20.
In simple words: We know that for any two numbers, if you multiply them, you get the same result as multiplying their HCF and LCM. We used this to find the LCM of the two numbers, which was 60. Since their HCF is 5, both numbers must be multiples of 5. We then looked for two multiples of 5 whose product is 300 and whose HCF is 5. These numbers turned out to be 15 and 20.

🎯 Exam Tip: Remember the fundamental relationship: `Product of two numbers = HCF × LCM`. This formula is crucial for solving problems where the HCF, LCM, or product is given, and you need to find the numbers themselves.

 

Question 16. Find whether the number 564872 is divisible by 88. (use of the test of divisibility rule for 8 and 11 will help)
Answer: To check if a number is divisible by 88, we can use the divisibility rules for its coprime factors. Since \( 88 = 8 \times 11 \), and 8 and 11 are coprime (meaning they share no common factors other than 1), if a number is divisible by both 8 and 11, it is divisible by 88.
Let's test the number 564872 for divisibility by 8 and 11.
**Divisibility by 8:**
A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
For 564872, the last three digits form the number 872.
\( 872 \div 8 = 109 \)
Since 872 is divisible by 8, the number 564872 is divisible by 8.
**Divisibility by 11:**
A number is divisible by 11 if the difference between the sum of the digits at odd places and the sum of the digits at even places (from right to left) is either 0 or a multiple of 11.
For 564872:
Sum of digits at odd places (2nd, 4th, 6th from right): \( 2 + 8 + 6 = 16 \)
Sum of digits at even places (1st, 3rd, 5th from right): \( 7 + 4 + 5 = 16 \)
Difference = \( 16 - 16 = 0 \)
Since the difference is 0, the number 564872 is divisible by 11.
Since 564872 is divisible by both 8 and 11, it is divisible by 88.
In simple words: To check if 564872 can be divided by 88, we need to see if it can be divided by both 8 and 11. For 8, we check the last three digits (872), which are divisible by 8. For 11, we check the sum of digits at odd places and even places, and their difference is 0, which means it's divisible by 11. Since it passes both tests, it is divisible by 88.

🎯 Exam Tip: When checking divisibility by a composite number, break it down into its coprime factors (e.g., \( 88 = 8 \times 11 \)). If the number is divisible by all coprime factors, it is divisible by their product.

 

Question 17. Wilson, Mathan, and Guna can complete one round of a circular track in 10, 15, and 20 minutes respectively. If they start together from at the starting point of 7 am, at what time will they meet together again at the same starting point?
Answer: To find out when Wilson, Mathan, and Guna will meet together again at the starting point, we need to find the Least Common Multiple (LCM) of the time each person takes to complete one round. The times are 10 minutes, 15 minutes, and 20 minutes.
We will find the LCM of 10, 15, and 20.

To find the LCM, multiply the divisors and the remaining numbers at the bottom:
\( \text{LCM} = 5 \times 2 \times 1 \times 3 \times 2 = 60 \)
So, they will all meet together again after 60 minutes.
They started together at 7:00 am.
Adding 60 minutes to 7:00 am:
\( 7:00 \text{ am} + 60 \text{ minutes} = 7:00 \text{ am} + 1 \text{ hour} = 8:00 \text{ am} \)
Therefore, they will meet together again at the same starting point at 8:00 am.
In simple words: To find when the three people running on a track will meet again, we need to find the smallest number of minutes that is a multiple of their individual lap times (10, 15, and 20 minutes). This is the Least Common Multiple (LCM), which is 60 minutes. Since they started at 7:00 am, they will meet again 60 minutes later, which is 8:00 am.

🎯 Exam Tip: For problems involving objects or people moving in cycles and asking when they'll return to a common starting point, the solution typically involves finding the Least Common Multiple (LCM) of their respective cycle times.