## NCERT Solutions Class 10 Maths Chapter 1 : Real Numbers## Exercise 1.1
**135 and 225****196 and 38220****867 and 255**
I. Since 225 > 135, we can apply Euclid's Division Algorithm a=bq + r; 0≤ r < b with a=225 and b=135. 225 = 135 × 1 + 90 where r=90 Since r≠0, we will re-apply the algorithm on b = 135, and r = 90 135 = 90×1 + 45 where r1=45 Since r1≠0, we will re-apply the algorithm on b1=90 and r1=45 90 = 45×2 + 0 where r2=0 We have r2=0 Therefore, the HCF of 135 and 225 is II. Since 38220 > 196, we can apply Euclid's Division Algorithm a=bq + r; 0≤ r < b with a=38220 and b=196. 38220 = 196×195 + 0 We have r=0 Therefore, the HCF of 196 and 38220 is III. Since 867 > 255, we can apply Euclid's Division Algorithm a=bq + r; 0≤ r < b with a=867 and b=255. 867 = 255 × 3 + 102 where r=102 Since r≠0, we will re-apply the algorithm on b = 255, and r = 102 255 = 102×2 + 51 where r1=51 Since r1≠0, we will re-apply the algorithm on b1=102 and r1=51 102 = 51×2 + 0 where r2=0 We have r2=0 Therefore, the HCF of 867 and 255 is
Let there be an odd positive integer a, such that a=6q + r; 0≤ r< 6 (By Euclid's Division Lemma) where q is the quotient and r is the remainder. Since 0≤ r< 6, we can put 0, 1, 2, 3, 4, or 5 in place of r. For each of the cases we get the following:
As given previously, a is an odd positive integer. Hence, r cannot be equal to 2 or 4 because a=6q, a=6q+2 and a=6q+4 present a as an even positive integer which is false. Hence, an odd positive integer can be only of the forms 6q+1, 6q+3 or 6q+5.
It is clear from the problem that we need to find the maximum number of columns, which is the HCF of 616 and 32. Since 616 > 32, we can apply Euclid's Division Algorithm a=bq + r; 0≤ r < b with a=867 and b=32. 616 = 32 × 19 + 8 where r=8 Since r≠0, we will re-apply the algorithm on b = 32, and r = 8 32 = 8×4 + 0 where r1=0 We have r1=0 Therefore, the HCF of 616 and 32 is 8. Hence, the maximum number of columns the army contingent of 616 members can march behind band of 32 members is 8.
Let there be a positive integer a. After applying Euclid's division lemma with b=3 we get a=3q + r; 0≤ r< 3 Therefore, a can be of the forms a=3q, a=3q+1 or a=3q+2 Squaring both sides in each equation we get a a a a a a a a a Hence, square of any positive integer is always of the form 3m or 3m+1.
Let a be positive integer. After applying Euclid's Division Lemma with b=3 we get a=3q+r; 0≤ r< 3 Therefore, a can be of the forms a=3q, a=3q+1 or a=3q+2 Cubing both sides of each equation we get a a a a a a a a a Hence, cube of any positive integer is always of the form 9m, 9m+1, or 9m+8. ## Exercise 1.2
**140****156****3825****5005****7429**
**26 and 91****510 and 92****336 and 54**
26 = 2 × 13 And 91 = 7 × 13 Therefore, HCF(26, 91)=13 and LCM(26, 91) = 2 × 7 × 13 = 182. Product of both numbers = 26 × 91 = 2366 LCM × HCF = 182 × 13= 2366. Hence, product of the two numbers has been verified to be equal to LCM × HCF.
510 = 2 × 3 × 5 × 17 And 92 = 2 × 2 × 23 Therefore, HCF(510, 92) = 2 and LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460. Product of both numbers = 510 × 92 = 46920 LCM × HCF = 23460 × 2 = 46920. Hence, product of the two numbers has been verified to be equal to LCM × HCF.
336 = 2 × 2 × 2 × 2 × 3 × 7 And 54 = 2 × 3 × 3 × 3 Therefore, HCF(336, 54) = 2 × 3 = 6 and LCM(336, 54) = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024. Product of both numbers = 336 × 54 = 18144 LCM × HCF = 3024 × 6 = 18144. Hence, product of the two numbers has been verified to be equal to LCM × HCF.
**12, 15, and 21****17, 23 and 29****8, 9 and 25**
12 = 2 × 2 × 15 = 21 = HCF (12, 15, 21) = 3 LCM (12, 15, 21) = 2 × 2 × 3 × 5 × 7 = 420
17 = 17 × 1 23 = 23 × 1 29 = 29 × 1 HCF (17, 23, 29) = 1 LCM (17, 23, 29) = 17 × 23 × 29 = 11339
8 = 2 × 2 × 2 9 = 3 × 3 × 3 25 = 5 × 5 HCF (8, 9, 25) = 1 LCM (8, 9, 25) = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800
Since, it is known that Product of two given numbers = LCM × HCF Therefore, 306 × 657 = LCM × 9 Solving the above equation, we get LCM = 22338 Hence, LCM (306, 657) = 22338
6 By using prime factorization, we have 6 = 2 × 3 Since, 5 is not a prime factor of 6, 6
Let 7 × 11 × 13 + 13 = a and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = b. a = 7 × 11 × 13 + 13 = 13 × (7 × 11 × 1 + 1) = 13 × (77 + 1) = 13 × 78 = 13 × 2 × 3 × 13 (Prime Factorization of 78) Since a can be expressed as a product of primes, it is a composite number. b = 7 × 6 × 5 × 4 × 3 × 2 + 5 = 5 × (7 × 6 × 1 × 4 × 3 × 2 + 1) =5 × (1008 + 1) = 5 × 1009 Since b can be expressed as a product of primes, it is a composite number. Hence, 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are both composite numbers.
Sonia takes 18 minutes while Ravi takes 12 minutes. In order to find the time elapsed (in minutes) when they meet again at the starting point, we need to calculate the LCM of 18 and 12 (time taken by each of them to complete a round). By Prime factorization, 18 = 2 × 3 × 3 12 = 2 × 2 × 3 LCM = 2 × 2 × 3 × 3 = 36 Hence, it will take them 36 minutes to meet again at the starting point. ## Exercise 1.3
Let us assume that √5 is a rational. Since, √5 is rational, there exists two coprime integers a and b; b ≠ 0 such that √5= a/b b√5= a Squaring both sides 5b From the equation, it can be deduced that a Therefore, we can write a = 5c where c is an integer. If we substitute this value of a in (I), we get 5b b From the equation, it can be deduced that b Hence, a and b are both divisible by 5. This contradicts the fact that a and b are coprime integers as they both have 5 as a common factor. The contradiction has arisen due to our wrong assumption that √5 is rational. Hence, √5 is irrational.
Let us assume that 3 + 2√5 is rational. Since, 3 + 2√5 is rational, there exists two coprime integers a and b; b ≠ 0 such that But √5 is irrational because of the following proof: ## (NOTE: It is not necessary for students to add the following proof when answering the questions unless the question awards more than 3 marks)Let us assume that √5 is a rational. Since, √5 is rational, there exists two coprime integers a and b; b ≠ 0 such that √5= a/b b√5= a Squaring both sides 5b From the equation, it can be deduced that a Therefore, we can write a = 5c where c is an integer. If we substitute this value of a in (I), we get 5b b From the equation, it can be deduced that b Hence, a and b are both divisible by 5. This contradicts the fact that a and b are coprime integers as they both have 5 as a common factor. The contradiction has arisen due to our wrong assumption that √5 is rational. Hence, √5 is irrational. From the above proof it is clear that √5 is irrational. This contradicts the assumption that 3 + 2√5 is rational. Hence, 3 + 2√5 is irrational.
**1/√2****7√5****6 + √2**
Since, √2 is rational, there exist two coprimes a and b; b ≠ 0 such that √2 = a/b. √2b = a Squaring both sides, we get 2b We can deduce that a Therefore, a = 2c for some integer c. Substituting value of a in (I) we get 2b b This implies that b Therefore, a and b have 2 as a common factor. This contradicts the fact that a and b are coprimes. The contradiction has arisen due to our wrong assumption that √2 is rational. Hence, √2 is irrational. This implies that 1/√2 is also irrational.
Since, √5 is rational, there exists two coprime integers a and b; b ≠ 0 such that √5= a/b b√5= a Squaring both sides 5b From the equation, it can be deduced that a Therefore, we can write a = 5c where c is an integer. If we substitute this value of a in (I), we get 5b b From the equation, it can be deduced that b Hence, a and b are both divisible by 5. This contradicts the fact that a and b are coprime integers as they both have 5 as a common factor. The contradiction has arisen due to our wrong assumption that √5 is rational. Hence, √5 is irrational. This implies that 7√5 is also irrational.
Since, 6 + √2 is rational, there exists two coprime integers a and b; b ≠ 0 such that 6 + √2 = a/b. Therefore, √2 is also rational. But √2 is irrational because of the following proof: ## (NOTE: It is not necessary for students to add the following proof when answering the questions unless the question awards more than 3 marks)Let us assume that √2 is a rational. Since, √2 is rational, there exists two coprime integers a and b; b ≠ 0 such that √2= a/b b√2= a Squaring both sides 2b From the equation, it can be deduced that a Therefore, we can write a = 2c where c is an integer. If we substitute this value of a in (I), we get 2b b From the equation, it can be deduced that b Hence, a and b are both divisible by 2. This contradicts the fact that a and b are coprime integers as they both have 2 as a common factor. The contradiction has arisen due to our wrong assumption that √2 is rational. Hence, √2 is irrational. From the above proof it is clear that √2 is irrational. This contradicts the assumption that 6 + √2 is rational. Hence, 6 + √2 is irrational. ## Exercise 1.4
**13/3125****17/8****64/455****15/1600****29/343****23/2**^{3}5^{2}**129/2**^{2}5^{7}7^{5}**6/15****35/50****77/210**
Since, the given number can be represented in the form of p/q where p and q are coprimes and q is of the form 2 Therefore, the given number has a terminating decimal expansion. Since, the given number can be represented in the form of p/q where p and q are coprimes and q is of the form 2 Therefore, the given number has a terminating decimal expansion. Since, the given number cannot be represented in the form of p/q where p and q are coprimes and q is of the form 2 Therefore, the given number has a non-terminating repeating decimal expansion. Since, the given number can be represented in the form of p/q where p and q are coprimes and q is of the form 2 Therefore, the given number has a terminating decimal expansion. Since, the given number cannot be represented in the form of p/q where p and q are coprimes and q is of the form 2 Therefore, the given number has a non-terminating repeating decimal expansion.
Therefore, the given number has a terminating decimal expansion. Since, the given number cannot be represented in the form of p/q where p and q are coprimes and q is of the form 2 Therefore, the given number has a non-terminating repeating decimal expansion.
Therefore, the given number has a terminating decimal expansion.
Therefore, the given number has a terminating decimal expansion.
Therefore, the given number has a non-terminating repeating decimal expansion.
Solution I, II, IV, VI, VIII, IX in question 1 have a terminating decimal expansion. These can be calculated as follows: Multiply and divide by 5 Multiply and divide by 5 Multiply and divide by 2
- 43.123456789
- 0.120120012000120000. . .
- 43.123456789
I. Since, the given decimal expansion is terminating, it is a rational. The denominator q can be expressed in the form of 2 II. Since, the given decimal expansion is non-terminating non-repeating, it is irrational and hence the denominator cannot be represented as 2 III. Since, the given decimal expansion is non-terminating repeating, it is rational but the denominator cannot be expressed in the form of 2 Next TopicClass 10 Maths Chapter 2 |