GSEB Class 10 Maths Solutions Chapter 1 વાસ્તવિક સંખ્યાઓ Exercise 1.2

Question 1. નીચેની દરેક સંખ્યાને અવિભાજ્ય અવયવની ગુણાકાર સ્વરૂપે દર્શાવો:
(1) 140
Answer: અવયવ વૃક્ષની રીતે,

આમ, \( 140 = 2 \times 2 \times 5 \times 7 \)
\( = 2^2 \times 5 \times 7 \).
In simple words: To find the prime factors of 140, we break it down into smaller numbers until we only have prime numbers left. We start by dividing 140 by 2, which gives 70. Then, we divide 70 by 2 again, getting 35. Next, we divide 35 by 5, which results in 7. Since 7 is a prime number, we stop there. So, 140 can be written as \( 2 \times 2 \times 5 \times 7 \) or \( 2^2 \times 5 \times 7 \).

Exam Tip: Remember to stop when all factors are prime numbers. A prime number is a whole number greater than 1 that has only two divisors: 1 and itself.

Question 1. (2) 156
Answer: અવયવ વૃક્ષની રીતે,

આમ, \( 156 = 2 \times 2 \times 3 \times 13 \)
\( = 2^2 \times 3 \times 13 \).
In simple words: To find the prime factors of 156, we break it down. We divide 156 by 2 to get 78. Then, we divide 78 by 2 to get 39. Next, we divide 39 by 3, which gives us 13. Since 13 is a prime number, we stop. So, 156 can be written as \( 2 \times 2 \times 3 \times 13 \) or \( 2^2 \times 3 \times 13 \).

Exam Tip: Always verify your prime factorization by multiplying all the prime factors together to ensure you get the original number.

Question 1. (3) 3825
Answer: અવયવ વૃક્ષની રીતે,

આમ, \( 3825 = 3 \times 3 \times 5 \times 5 \times 17 \)
\( = 3^2 \times 5^2 \times 17 \).
In simple words: To find the prime factors of 3825, we start by dividing it by 3, which gives 1275. We divide 1275 by 3 again to get 425. Next, we divide 425 by 5, resulting in 85. We divide 85 by 5 once more, which gives 17. Since 17 is a prime number, we stop. So, 3825 can be written as \( 3 \times 3 \times 5 \times 5 \times 17 \) or \( 3^2 \times 5^2 \times 17 \).

Exam Tip: For numbers ending in 5 or 0, always start by checking divisibility by 5. For numbers where the sum of digits is divisible by 3, try dividing by 3.

Question 1. (4) 5005
Answer: અવયવ વૃક્ષની રીતે,

આમ, \( 5005 = 5 \times 7 \times 11 \times 13 \).
In simple words: To find the prime factors of 5005, we start by dividing it by 5, which results in 1001. Then, we divide 1001 by 7, giving us 143. Next, we divide 143 by 11, which gives 13. Since 13 is a prime number, we stop. So, 5005 can be written as \( 5 \times 7 \times 11 \times 13 \).

Exam Tip: When a number is not divisible by 2, 3, or 5, check for divisibility by prime numbers like 7, 11, 13, and so on.

Question 1. (5) 7429
Answer: અવયવ વૃક્ષની રીતે,

આમ, \( 7429 = 17 \times 19 \times 23 \).
In simple words: To find the prime factors of 7429, we start by dividing it by 17, which results in 437. Then, we divide 437 by 19, which gives us 23. Since 23 is a prime number, we stop. So, 7429 can be written as \( 17 \times 19 \times 23 \).

Exam Tip: For larger prime numbers, trial and error with common primes like 7, 11, 13, 17, 19, 23, etc., is essential to find the correct factors.

Question 2. નીચે આપેલ પૂર્ણાકોની જોડીના ગુ.સા.અ. અને લ.સા.અ. શોધો અને ગુ.સા.અ. × લ.સા.અ. = બંને પૂર્ણાકોનો ગુણાકાર થાય છે તેમ ચકાસોઃ
(1) 26 અને 91
Answer: અવયવ વૃક્ષની રીતે,

\( \therefore 26 = 2 \times 13 \) અને \( 91 = 7 \times 13 \)
હવે, લ.સા.અ. (26, 91) \( = 2 \times 7 \times 13 = 182 \)
અને ગુ.સા.અ. (26, 91) \( = 13 \)
હવે, લ.સા.અ. \( \times \) ગુ.સા.અ. \( = 182 \times 13 = 2366 \)
અને \( 26 \times 91 = 2366 \).
આમ, ગુ.સા.અ. \( \times \) લ.સા.અ. \( = \) બંને પૂર્ણાકોનો ગુણાકાર.
In simple words: First, we break down 26 into prime factors: \( 2 \times 13 \). Then, we break down 91 into prime factors: \( 7 \times 13 \). The Highest Common Factor (HCF) is the common prime factor, which is 13. The Least Common Multiple (LCM) is found by multiplying all unique prime factors, taking the highest power of each: \( 2 \times 7 \times 13 = 182 \). To check, we multiply HCF and LCM: \( 13 \times 182 = 2366 \). We also multiply the two original numbers: \( 26 \times 91 = 2366 \). Since both results are the same, the verification is complete.

Exam Tip: Remember the fundamental theorem: For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b. This is crucial for verifying your answers.

Question 2. (2) 510 અને 92
Answer: અવયવ વૃક્ષની રીતે,

\( \therefore 510 = 2 \times 3 \times 5 \times 17 \)
અને \( 92 = 2 \times 2 \times 23 = 2^2 \times 23 \).
હવે,
લ.સા.અ. (510, 92) \( = 2^2 \times 3 \times 5 \times 17 \times 23 = 23,460 \)
અને ગુ.સા.અ. (510, 92) \( = 2 \)
હવે, લ.સા.અ. \( \times \) ગુ.સા.અ. \( = 23,460 \times 2 = 46,920 \)
અને \( 510 \times 92 = 46,920 \).
આમ, ગુ.સા.અ. \( \times \) લ.સા.અ. \( = \) બંને પૂર્ણાકોનો ગુણાકાર.
In simple words: First, we find the prime factors for 510: \( 2 \times 3 \times 5 \times 17 \). Then, for 92: \( 2^2 \times 23 \). The HCF is the common prime factor with the lowest power, which is 2. The LCM is found by taking all unique prime factors with their highest powers: \( 2^2 \times 3 \times 5 \times 17 \times 23 = 23,460 \). To verify, we multiply HCF and LCM: \( 2 \times 23,460 = 46,920 \). Multiplying the original numbers gives \( 510 \times 92 = 46,920 \). Since these match, the relationship is proven.

Exam Tip: When finding LCM, ensure you include every prime factor present in either number, raised to its highest power. For HCF, use only common prime factors raised to their lowest power.

Question 2. (3) 336 અને 54
Answer: અવયવ વૃક્ષની રીતે,

\( \therefore 336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 = 2^4 \times 3 \times 7 \)
અને \( 54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3 \).
હવે, ગુ.સા.અ. (336, 54) \( = 2 \times 3 = 6 \)
લ.સા.અ. (336, 54) \( = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024 \)
હવે, લ.સા.અ. \( \times \) ગુ.સા.અ. \( = 3024 \times 6 = 18,144 \)
અને \( 336 \times 54 = 18,144 \).
આમ, ગુ.સા.અ. \( \times \) લ.સા.અ. \( = \) બંને પૂર્ણાકોનો ગુણાકાર.
In simple words: First, we find the prime factors for 336: \( 2^4 \times 3 \times 7 \). For 54, the prime factors are \( 2 \times 3^3 \). The HCF is found by taking the common prime factors with their lowest powers: \( 2 \times 3 = 6 \). The LCM is found by taking all unique prime factors with their highest powers: \( 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024 \). To verify the relation, we multiply HCF and LCM: \( 6 \times 3024 = 18,144 \). Multiplying the original numbers gives \( 336 \times 54 = 18,144 \). Both products are equal, confirming the relationship.

Exam Tip: Systematically listing prime factors with their powers helps avoid errors when calculating HCF and LCM, especially for larger numbers.

Question 3. નીચે આપેલ પૂર્ણાકોના અવિભાજ્ય અવયવની રીતે ગુ.સા.અ. અને લ.સા.અ. શોધોઃ
(1) 12, 15 અને 21
Answer: અવયવ વૃક્ષની રીતે,

\( \therefore 12 = 2 \times 2 \times 3 = 2^2 \times 3 \)
\( 15 = 3 \times 5 \)
અને \( 21 = 3 \times 7 \).
હવે,
લ.સા.અ. (12, 15, 21) \( = 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420 \)
અને ગુ.સા.અ. (12, 15, 21) \( = 3 \).
In simple words: First, we find the prime factors: for 12 it is \( 2^2 \times 3 \), for 15 it is \( 3 \times 5 \), and for 21 it is \( 3 \times 7 \). The HCF (Highest Common Factor) is the prime factor common to all three numbers, which is 3. The LCM (Least Common Multiple) includes all unique prime factors raised to their highest power: \( 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420 \).

Exam Tip: When finding HCF and LCM for three numbers, remember that the HCF must be a common factor to *all* numbers, while the LCM must be a multiple of *all* numbers.

Question 3. (2) 17, 23 અને 29
Answer: આપેલ દરેક સંખ્યા અવિભાજ્ય સંખ્યા હોવાથી,
\( 17 = 17 \times 1 \)
\( 23 = 23 \times 1 \)
અને \( 29 = 29 \times 1 \).
હવે,
લ.સા.અ. (17, 23, 29) \( = 17 \times 23 \times 29 = 11,339 \)
અને ગુ.સા.અ. (17, 23, 29) \( = 1 \).
In simple words: Since 17, 23, and 29 are all prime numbers, their only factors are 1 and themselves. Therefore, their HCF (Highest Common Factor) is 1. The LCM (Least Common Multiple) of prime numbers is simply their product: \( 17 \times 23 \times 29 = 11,339 \).

Exam Tip: The HCF of any set of prime numbers is always 1, and their LCM is always their product.

Question 3. (3) 8, 9 અને 25
Answer: અવયવ વૃક્ષની રીતે,

\( \therefore 8 = 2 \times 2 \times 2 = 2^3 \)
\( 9 = 3 \times 3 = 3^2 \)
અને \( 25 = 5 \times 5 = 5^2 \).
હવે,
લ.સા.અ. (8, 9, 25) \( = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800 \)
અને ગુ.સા.અ. (8, 9, 25) \( = 1 \).
In simple words: First, we find the prime factors: for 8 it is \( 2^3 \), for 9 it is \( 3^2 \), and for 25 it is \( 5^2 \). Since these numbers share no common prime factors (they are pairwise coprime), their HCF is 1. The LCM is found by multiplying all the distinct prime factors raised to their highest powers: \( 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800 \).

Exam Tip: If numbers have no common prime factors other than 1, they are called coprime, and their HCF is 1, while their LCM is simply their product.

Question 4. જો ગુ.સા.અ. (306, 657) = 9 આપેલ હોય, તો લ.સા.અ. (306, 657) શોધો.
Answer: આપણે જાણીએ છીએ કે, લ.સા.અ. (a, b) \( = \frac { a \times b }{ \text{ગુ.સા.અ.} (a, b) } \)
અહીં \( a = 306 \) અને \( b = 657 \).
લ.સા.અ. (306, 657) \( = \frac { 306 \times 657 }{ \text{ગુ.સા.અ.} (306, 657) } \)
\( = \frac { 306 \times 657 }{ 9 } \)
\( = 34 \times 657 \)
\( = 22,338 \).
આમ, લ.સા.અ. (306, 657) \( = 22,338 \).
In simple words: We know a special rule for two numbers: if you multiply their HCF and LCM, you get the same answer as multiplying the numbers themselves. So, to find the LCM, we divide the product of the two numbers (306 and 657) by their given HCF (9). When we calculate \( (306 \times 657) / 9 \), we get 22,338.

Exam Tip: This formula, HCF(a, b) × LCM(a, b) = a × b, is a fundamental property for two positive integers and is often used to find one value if the other two are known.

Question 5. કોઈક પ્રાકૃતિક સંખ્યા n માટે \( 6^n \) નો અંતિમ અંક શૂન્ય થાય કે નહીં તે ચકાસો.
Answer: જો કોઈ સંખ્યાનો અંતિમ અંક 0 હોય, તો તે સંખ્યા 5 તેમજ 2 બંને વડે વિભાજ્ય હોય. એટલે કે, અંતિમ અંક 0 હોય તેવી સંખ્યાના અવિભાજ્ય અવયવીકરણમાં 5 તેમજ 2 બંનેનો સમાવેશ થાય.
હવે, \( 6^n = (2 \times 3)^n = 2^n \times 3^n \), જ્યાં n એ કોઈ પ્રાકૃતિક સંખ્યા છે. આમ, \( 6^n \) ને 2 અને 3 એમ ફક્ત બે જ અવિભાજ્ય અવયવો છે. આમ, \( 6^n \) ના અવિભાજ્ય અવયવીકરણમાં 5 નો સમાવેશ થતો ન હોવાથી કોઈક પ્રાકૃતિક સંખ્યા n માટે \( 6^n \) નો અંતિમ અંક 0 ન જ થાય.
In simple words: For a number to end with the digit 0, it needs to have both 2 and 5 as its prime factors. When we look at \( 6^n \), its prime factors are only 2 and 3 because 6 is \( 2 \times 3 \). Since 5 is not a prime factor of \( 6^n \), it's not possible for \( 6^n \) to end with the digit 0 for any natural number n.

Exam Tip: For a number \( x^n \) to end with the digit 0, its base 'x' must have both 2 and 5 as prime factors. If either is missing, it cannot end in 0.

Question 6. સમજાવો કે, \( 7 \times 11 \times 13 + 13 \) અને \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \) એ શા માટે વિભાજ્ય સંખ્યાઓ છે?
Answer: પ્રથમ સંખ્યા:
\( 7 \times 11 \times 13 + 13 \)
\( = 13 (7 \times 11 + 1) \)
\( = 13 (77 + 1) \)
\( = 13 (78) \)
\( = 13 \times 2 \times 3 \times 13 \) (કારણ કે \( 78 = 2 \times 3 \times 13 \)).
આમ, \( 7 \times 11 \times 13 + 13 \) ને ભિન્ન અવિભાજ્ય સંખ્યાઓના ગુણાકાર તરીકે દર્શાવી શકાય છે. આથી \( 7 \times 11 \times 13 + 13 \) એ વિભાજ્ય સંખ્યા છે.
બીજી સંખ્યા:
\( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \)
\( = 5 (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) \)
\( = 5 (1008 + 1) \)
\( = 5 \times 1009 \).
આમ, \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \) ને ભિન્ન અવિભાજ્ય સંખ્યાઓના ગુણાકાર તરીકે દર્શાવી શકાય છે. આથી \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \) એ વિભાજ્ય સંખ્યા છે.
In simple words: A composite number is a positive integer that has at least one divisor other than 1 and itself. For the first expression, \( 7 \times 11 \times 13 + 13 \), we can take 13 as a common factor, which simplifies to \( 13 \times (7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78 \). Since 78 can be further factored into \( 2 \times 3 \times 13 \), the entire number is \( 13 \times 2 \times 3 \times 13 \). Because it has factors other than 1 and itself (namely 2, 3, and 13), it's a composite number. For the second expression, \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \), we can take 5 as a common factor, simplifying it to \( 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times (1008 + 1) = 5 \times 1009 \). Since it has factors 5 and 1009 (both greater than 1), it is also a composite number.

Exam Tip: To show a number is composite, factor it to express it as a product of two numbers greater than 1. Finding a common factor is a quick way to achieve this.

Question 7. એક રમતના મેદાનમાં વર્તુળાકાર માર્ગ છે. સોનિયાને તેનું એક પરિભ્રમણ પૂર્ણ કરતાં 18 મિનિટ લાગે છે, જ્યારે રવિને તેનું એક પરિભ્રમણ પૂર્ણ કરતાં 12 મિનિટ લાગે છે. ધારો કે બંને એક જ સમયે, એક જ બિંદુએથી, એક જ દિશામાં પરિભ્રમણ કરવાનું પ્રારંભ કરે છે, તો કેટલી મિનિટ બાદ બંને ફરી પ્રારંભબિંદુ પર ભેગા થાય?
Answer: અહીં, સોનિયાને અને રવિને એક પરિભ્રમણ પૂર્ણ કરતાં લાગતા સમય (મિનિટમાં) નો લ.સા.અ. એ આપેલ પ્રશ્નનો જવાબ થાય.
હવે, \( 12 = 2 \times 2 \times 3 = 2^2 \times 3 \)
અને \( 18 = 2 \times 3 \times 3 = 2 \times 3^2 \).
માટે, લ.સા.અ. (12, 18) \( = 2^2 \times 3^2 = 4 \times 9 = 36 \).
આથી જો સોનિયા અને રવિ એક જ સમયે, એક જ બિંદુએથી, એક જ દિશામાં પરિભ્રમણ કરવાનું પ્રારંભ કરે, તો 36 મિનિટ, બાદ બંને ફરી પ્રારંભબિંદુ પર ભેગા થાય.
In simple words: To find when Soniya and Ravi will meet again at the starting point, we need to find the Least Common Multiple (LCM) of the time each person takes to complete one round. Soniya takes 18 minutes, and Ravi takes 12 minutes. The prime factors of 12 are \( 2^2 \times 3 \), and for 18 are \( 2 \times 3^2 \). The LCM is \( 2^2 \times 3^2 = 4 \times 9 = 36 \). So, they will meet again after 36 minutes.

Exam Tip: Problems involving finding when events will recur simultaneously (like bells ringing together, or people meeting on a track) typically require calculating the Least Common Multiple (LCM).