RBSE Solutions Class 10 Maths Chapter 2 Real Numbers Exercise 2.2

Question 1. Express each number as a product of its prime factors:
(i) 468
(ii) 945
(iii) 140
(iv) 3825
(v) 20570
Answer:
(i) \( 468 = 2 \times 2 \times 3 \times 3 \times 13 = 2^{2} \times 3^{2} \times 13 \)
(ii) \( 945 = 3 \times 3 \times 3 \times 5 \times 7 = 3^{3} \times 5 \times 7 \)
(iii) \( 140 = 2 \times 2 \times 5 \times 7 = 2^{2} \times 5 \times 7 \)
(iv) \( 3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^{2} \times 5^{2} \times 17 \)
(v) \( 20570 = 2 \times 5 \times 11 \times 11 \times 17 = 2 \times 5 \times 11^{2} \times 17 \)
In simple words: To find prime factors, we divide the original number by the smallest prime numbers possible (like 2, 3, or 5) until we are left with 1. The list of all the prime numbers we used as divisors is the product of prime factors. 📝 Teacher's Note: Guide students to use the vertical division method or a factor tree to keep their work organized. Emphasize starting with the smallest prime divisor and moving upwards to ensure no factors are missed. 🎯 Exam Tip: Always double-check your final list of prime factors by multiplying them together to see if the result equals the original number. Presenting your answer in power notation (like \( 2^{2} \)) makes the final result look neat and accurate.

Question 2. Find the HCF and LCM of the following numbers and verify that HCF \( \times \) LCM = Product of the two numbers:
(i) 96 and 404
(ii) 336 and 54
(iii) 90 and 144
Answer:
(i) For 96 and 404:
Prime factorization of 96:
\[ \begin{array}{r|l} 2 & 96 \\ \cline{2-2} 2 & 48 \\ \cline{2-2} 2 & 24 \\ \cline{2-2} 2 & 12 \\ \cline{2-2} 2 & 6 \\ \cline{2-2} 3 & 3 \\ \cline{2-2} {} & 1 \end{array} \]
So, \( 96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3 \)
Prime factorization of 404:
\[ \begin{array}{r|l} 2 & 404 \\ \cline{2-2} 2 & 202 \\ \cline{2-2} 101 & 101 \\ \cline{2-2} {} & 1 \end{array} \]
So, \( 404 = 2 \times 2 \times 101 = 2^2 \times 101 \)
To find the HCF, we take the smallest power of the common prime factors:
H.C.F. = \( 2^2 = 4 \)
To find the LCM, we take the highest power of all prime factors:
L.C.M. = \( 2^5 \times 3^1 \times 101^1 = 32 \times 3 \times 101 = 9696 \)
Verification:
Product of the two integers = \( 96 \times 404 = 38784 \)
H.C.F. \( \times \) L.C.M. = \( 4 \times 9696 = 38784 \)
Hence, H.C.F. \( \times \) L.C.M. = Product of the two integers.

(ii) For 336 and 54:
Prime factorization of 336:
\( 336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 = 2^4 \times 3 \times 7 \)
Prime factorization of 54:
\( 54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3 \)
To find the HCF, we take the minimum power of common prime factors:
H.C.F. = \( 2^1 \times 3^1 = 2 \times 3 = 6 \)
To find the LCM, we take the highest power of all prime factors:
L.C.M. = \( 2^4 \times 3^3 \times 7^1 = 16 \times 27 \times 7 = 3024 \)
Verification:
Product of the two integers = \( 336 \times 54 = 18144 \)
H.C.F. \( \times \) L.C.M. = \( 6 \times 3024 = 18144 \)
Hence, H.C.F. \( \times \) L.C.M. = Product of the two integers.

(iii) For 90 and 144:
Prime factorization of 90:
\[ \begin{array}{r|l} 2 & 90 \\ \cline{2-2} 3 & 45 \\ \cline{2-2} 3 & 15 \\ \cline{2-2} 5 & 5 \\ \cline{2-2} {} & 1 \end{array} \]
So, \( 90 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5 \)
Prime factorization of 144:
\[ \begin{array}{r|l} 2 & 144 \\ \cline{2-2} 2 & 72 \\ \cline{2-2} 2 & 36 \\ \cline{2-2} 2 & 18 \\ \cline{2-2} 3 & 9 \\ \cline{2-2} 3 & 3 \\ \cline{2-2} {} & 1 \end{array} \]
So, \( 144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2 \)
To find the HCF, we take the minimum power of common prime factors:
H.C.F. = \( 2^1 \times 3^2 = 2 \times 9 = 18 \)
To find the LCM, we take the highest power of all prime factors:
L.C.M. = \( 2^4 \times 3^2 \times 5^1 = 16 \times 9 \times 5 = 720 \)
Verification:
Product of the two integers = \( 90 \times 144 = 12960 \)
H.C.F. \( \times \) L.C.M. = \( 18 \times 720 = 12960 \)
Hence, H.C.F. \( \times \) L.C.M. = Product of the two integers.
In simple words: First, find all the prime factors for each number. The HCF is made of the common prime factors raised to their smallest power. The LCM is made of all prime factors (common and uncommon) raised to their largest power. To check your work, multiply the HCF and LCM together. This answer should be the same as multiplying the two original numbers.

🎯 Exam Tip: Remember that for two positive integers 'a' and 'b', the relationship HCF(a, b) × LCM(a, b) = a × b always holds true. This is an important property for verification.

Question 3. Find the HCF and LCM of the following integers by the prime factorization method:
(i) 12, 15 and 21
(ii) 24, 15 and 36
(iii) 17, 23 and 29
(iv) 6, 72 and 120
(v) 40, 36 and 126
(vi) 8, 9 and 25
Answer:
(i) For 12, 15 and 21:
Prime factorization:
\( 12 = 2 \times 2 \times 3 = 2^2 \times 3 \)
\( 15 = 3 \times 5 \)
\( 21 = 3 \times 7 \)
To find HCF, we take the minimum power of common prime factors:
H.C.F. = \( 3^1 = 3 \)
To find LCM, we take the maximum power of all prime factors:
L.C.M. = \( 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420 \)

(ii) For 24, 15 and 36:
Prime factorization:
\( 24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3 \)
\( 15 = 3 \times 5 \)
\( 36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2 \)
To find HCF, we take the minimum power of common prime factors:
H.C.F. = \( 3^1 = 3 \)
To find LCM, we take the maximum power of all prime factors:
L.C.M. = \( 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360 \)

(iii) For 17, 23 and 29:
Prime factorization:
\( 17 = 1 \times 17 \)
\( 23 = 1 \times 23 \)
\( 29 = 1 \times 29 \)
Since 17, 23, and 29 are all prime numbers, their only common factor is 1.
H.C.F. = \( 1 \)
To find LCM, we take the maximum power of all prime factors:
L.C.M. = \( 1 \times 17 \times 23 \times 29 = 11339 \)

(iv) For 6, 72 and 120:
Prime factorization:
\( 6 = 2 \times 3 \)
\( 72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2 \)
\( 120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5 \)
To find HCF, we take the minimum power of common prime factors:
H.C.F. = \( 2^1 \times 3^1 = 6 \)
To find LCM, we take the maximum power of all prime factors:
L.C.M. = \( 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360 \)

(v) For 40, 36 and 126:
Prime factorization:
\( 40 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5 \)
\( 36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2 \)
\( 126 = 2 \times 3 \times 3 \times 7 = 2 \times 3^2 \times 7 \)
To find HCF, we take the minimum power of common prime factors:
H.C.F. = \( 2^1 = 2 \)
To find LCM, we take the maximum power of all prime factors:
L.C.M. = \( 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520 \)

(vi) For 8, 9 and 25:
Prime factorization:
\( 8 = 2 \times 2 \times 2 = 2^3 \)
\( 9 = 3 \times 3 = 3^2 \)
\( 25 = 5 \times 5 = 5^2 \)
These numbers have no common prime factors, so their common factor is 1.
H.C.F. = \( 1 \)
To find LCM, we take the maximum power of all prime factors:
L.C.M. = \( 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800 \)
In simple words: For HCF, find the prime factors common to all numbers and multiply them, taking the lowest power for each. For LCM, list all prime factors from all numbers and multiply them, taking the highest power for each. Remember, if numbers have no common prime factors, their HCF is 1.

🎯 Exam Tip: When finding HCF and LCM for more than two numbers, the product rule (HCF \( \times \) LCM = Product of numbers) does not always apply. Use the prime factorization method carefully for each case.

Question 4. There is a circular path around a sports field. Raman takes 18 minutes to drive one round of the field while Anupriya takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point.
Answer:
Time taken by Raman to complete one round = 18 minutes.
Time taken by Anupriya to complete one round = 12 minutes.
To find when they will meet again at the starting point, we need to find the Least Common Multiple (LCM) of their individual timings.
Prime factors of 18: \( 18 = 2 \times 3 \times 3 = 2 \times 3^2 \)
Prime factors of 12: \( 12 = 2 \times 2 \times 3 = 2^2 \times 3 \)
Now, we find the LCM of 18 and 12:
L.C.M. (18, 12) = \( 2^2 \times 3^2 = 4 \times 9 = 36 \)
So, Raman and Anupriya will meet again at the starting point after 36 minutes.
In simple words: Raman takes 18 minutes and Anupriya takes 12 minutes to go around a field. To find when they will next meet at the starting spot, we need to find the smallest number that both 18 and 12 can divide into perfectly. This number is 36 minutes.

🎯 Exam Tip: Questions involving objects moving in circles and meeting at the starting point always require finding the Least Common Multiple (LCM) of their travel times.

Question 5. In a school event, there are 60 participants from English, 84 from Science, and 108 from Maths. If all participants need to be seated in rooms with an equal number of participants from the same subject, find the minimum number of rooms required.
Answer:
Number of participants from English = 60
Number of participants from Science = 84
Number of participants from Maths = 108
To find the maximum equal number of participants that can sit in each room, we need to find the Highest Common Factor (HCF) of 60, 84, and 108.
Prime factorization of 60: \( 60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3 \times 5 \)
Prime factorization of 84: \( 84 = 2 \times 2 \times 3 \times 7 = 2^2 \times 3 \times 7 \)
Prime factorization of 108: \( 108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3 \)
H.C.F. (60, 84, 108) = \( 2^2 \times 3^1 = 4 \times 3 = 12 \)
This means that 12 participants can be seated in each room to ensure an equal number from each subject.
Now, we calculate the total number of participants:
Total participants = \( 60 + 84 + 108 = 252 \)
To find the minimum number of rooms required, divide the total participants by the number of participants per room:
Number of rooms required \( = \frac{\text{Total number of participants}}{\text{HCF}} = \frac{252}{12} = 21 \)
Therefore, a minimum of 21 rooms are required.
In simple words: We have 60, 84, and 108 students from different subjects. To put an equal number of students from the same subject in each room, we find the biggest number that divides all three counts, which is 12. This means 12 students can sit in each room. Since there are 252 students in total, we divide 252 by 12 to find that we need 21 rooms.

🎯 Exam Tip: When a question asks for the "maximum number of items" or "equal distribution" in the "minimum number of groups," it usually indicates an HCF problem. When it asks for "least common time" or "meeting again," it usually indicates an LCM problem.