જવાબ : HCF x LCM = one number x another number

= 100 x 190 = 19000

જવાબ : 1000 =2x2x2x5x5x5

56 = 2x2x2x7

HCF of 1000 and 56 = 8

Maximum number of columns = 8.

જવાબ : 120 = 23 x 3 x 5

144 = 24 x 32

∴ HCF = 23 x 3 = 24

LCM = 24 x 5 x 32 = 720

જવાબ : LCM of 12, 15, 18 = 2²x 3² x 5

=4x9x5 = 180

So, next time the bells will ring together after 180 minutes.

જવાબ : Maximum capacity of a container, which can measure the petrol in exact number of times.

જવાબ : 105 = 5 x 7 x 3

120 = 2x2x2x3x5

HCF = 3 X 5 = 15

LCM = 5x7x3x2x2x2 = 840

જવાબ : 0.120120012000 . . . . is an irrational number as it has non-terminating and non-repeating decimal expansion.

જવાબ : All the rational and irrational numbers together form real numbers.

જવાબ : (i) True (ii) False; it may be rational or irrational. (iii) True

જવાબ : Decimal representation of number 8/27 = 0.296

જવાબ : 0.01, 0.02 and 0.03

જવાબ : Coprime

જવાબ : Zero is a rational number.

જવાબ : Every composite number can be factorized as a product of __primes__, and this factorization is unique.

જવાબ : A number ‘r’ is called a rational number, if it can be written in the form p/q, where p and q are integers and q ≠ 0

જવાબ : True, because n(n + 1) (n + 2) will always be divisible by 6, as at least one of the factors will be divisible by 2 and at least one of the factors will be divisible by 3.

જવાબ : Required number = LCM of 1, 2, 3 … 10 = 2520

જવાબ : 3 × 5 × 7 + 7 = 7(3 × 5 + 1) = 7 × 16, which has more than two factors.

જવાબ : True, because n(n + 1) will always be even, as one out of the n or n+ 1 must be even.

જવાબ : A rational number between √3 and √5 is √3.24 = 1.8 = 1810 = 95

જવાબ : No, because here HCF (18) does not divide LCM (380).

જવાબ : No, because any positive integer can be written as 3q, 3q + 1, 3q + 2, therefore, square will be

9q^{2 }= 3m, 9q^{2 }+ 6q + 1 = 3(3q^{2}+ 2q) + 1 = 3m + 1,

9q^{2} + 12q + 4 = 3(3q^{2}+ 4q + 1) + 1 = 3m + 1.

જવાબ : if 4^{n} ends with 0, then it must have 5 as a factor. But, (4)^{n} = (2^{2})^{n} = 2^{2n} i.e., the only prime factor

of 4^{n} is 2. Also, we know from the fundamental theorem of arithmetic that the prime factorization of each number is unique.

∴ 4^{n} can never end with 0.

જવાબ : 98710500 = 47500 and 500 = 2^{2} × 5^{3}, so it has terminating decimal expansion.

જવાબ : As 1.7351 is a terminating decimal number, so q must be of the form 2^{m} 5^{n}, where in, n are natural numbers.

જવાબ : According to Euclid’s division lemma

a = 3q + r, where O r < 3 and r is an integer.

Therefore, the values of r can be 0, 1 or 2.

જવાબ : 5, because 14 is multiple of 7.

Therefore, remainder in both the cases is same.

જવાબ : HCF of 3^{3} × 5 and 3^{2} × 5^{2} = 3^{2} × 5 = 45

જવાબ : Product of two numbers = Product of their LCM and HCF

⇒ 1800 = 12 × LCM

⇒ LCM = 180012 = 150.

જવાબ : HCF of 3^{3} × 5 and 3^{2} × 5^{2} = 3^{2} × 5 = 45

જવાબ : Product of two numbers = Product of their LCM and HCF

⇒ 1800 = 12 × LCM

⇒ LCM = 180012 = 150.

જવાબ : Smallest composite number = 4

Smallest prime number = 2

So, HCF (4, 2) = 2

જવાબ : 9 = 3^{2}, 12 = 2^{2} × 3, 15 = 3 × 5

LCM = 2^{2} × 3^{2} × 5 = 4 × 9 × 5 = 180 minutes or 3 hours

They will next toll together after 3 hours.

જવાબ : LCM (3 × 5^{2}, 3^{2} × 7^{2}) = 3^{2} × 5^{2} × 7^{2} = 9 × 25 × 49 = 11025

જવાબ : 13 = 1 × 13; 17 = 1 × 17

HCF = 1 and LCM = 13 × 17 = 221

જવાબ : We know,
1st number × 2nd number = HCF × LCM

⇒ 27 × 2nd number = 9 × 459

⇒ 2nd number = 9×45927 = 153

જવાબ : HCF of 408 and 1032 is 24.

1032 × 2 + 408 × (p) = 24

408p = 24 – 2064

p = -5

જવાબ : 2 × 7^{2}

જવાબ : Algorithm

398 – 7 = 391, 436 – 11 = 425, 542 – 15 = 527

HCF of 391, 425, 527 = 17

જવાબ :

જવાબ : This representation will terminate after 4 decimal places

જવાબ :

Since the denominator has 3 as its factor.

∴ 1730 is a non4ermznatlng decimal.

જવાબ :

જવાબ : On applying Euclid’s division algorithm,

180 = 144 x 1 + 36

144 = 36 x 4 + 0

At the last stage, the divisor is 36.

∴ HCF of 144 and 180 is 36.

∵ 36 = 13 x 3 – 3

So, m = 3

જવાબ : 378 = 2 x 33 x 7

180 = 22 x 32 x 5

420 = 22 x 3 x 5 x 7

∴ HCF (378, 180, 420) = 2 x 3 = 6.

No. HCF (p, q, r) x LCM (p, q, r) ≠ p x q x r. where p, q, r are positive integers.

જવાબ : Given numbers are 650 and 1170.

On applying Euclid’s division algorithm,

we get 1170 = 650 x 1 + 520

650 = 520 x 1 + 130

520 = 130 x 4 + 0

∵ At the last stage, the divisor is 130.

∴ The HCF of 650 and 1170 is 130.

જવાબ : 1st Case: If n is even

This means n + 2 is also even.

Hence n and n + 2 are divisible by 2

Also, product of n and (n + 2) is divisible by 2.

.’. n (n + 2) is divisible by 2.

This conclude n (n + 2) (n + 1) is divisible by 2 … (i)

As, n, n + 1, n + 2 are three consecutive numbers. n (n + 1) (n + 2) is a multiple of 3.

This shows n (n + 1) (n + 2) is divisible by 3. … (ii)

By equating (i) and (ii) we can say

n (n + 1) (n + 2) is divisible by 2 and 3 both.

Hence, n (n + 1) (n + 2) is divisible by 6.

2nd Case: When n is odd.

This show (n + 1) is even

Hence (n + 1) is divisible by 2. … (iii)

This concludes n (n + 1) (n + 2) is an even number and divisible by 2.

Also product of three consecutive numbers is a multiple of 3.

n (n + 1) (n + 2) is divisible by 3. … (iv)

Equating (iii) and (iv) we can say

n (n + 1) (n + 2) is divisible by both 2 and 3 Hence, n(n + 1)(n + 2) is divisible by 6.

જવાબ : Prime factorization of 2520 is given by

2520 = 23 x 32 x 5 x 7

Given that 2520 = 23 x 3p x q x 7

On comparing both factorization we get p = 2 and q = 5.

જવાબ : Since prime factorization of 9^{n} is given by 9^{n}= (3 x 3)^{ n} = 3271.

Prime factorization of 9″ contains only prime number 3.

9 may end with the digit 0 for some natural number V if 5 must be in its prime factorization, which is not present.

So, there is no natural number N for which 9^{n} ends with the digit zero

જવાબ : Since prime factorization of 9^{n} is given by 9^{n}= (3 x 3)^{n} = 3271.

Prime factorization of 9″ contains only prime number 3.

9 may end with the digit 0 for some natural number V if 5 must be in its prime factorization, which is not present.

So, there is no natural number N for which 9^{n} ends with the digit zero.

જવાબ : Given numbers are 65 and 117.

On applying Euclid’s division algorithm, we get

117 = 65 x 1 + 52

65 = 52 x 1 + 13

52 = 13 x 4 + 0

∵At the last stage, the divisor is 13.

∴ The HCF of 65 and 117 is 13.

The required pair of integral values of m and n is

(2,-1) which satisfies the given relation HCF = 65m + 117n.

જવાબ : Given numbers are 255 and 867.

On applying Euclid’s division algorithm, we have

867 = 255 x 3 + 102

255 = 102 x 2 + 51

102 = 51 x 2 + 0

∵ At the last stage, the divisor is 51

∴ The HCF of 255 and 867 is 51.

Solution:

Given numbers are 255 and 865.

On applying Euclid’s division algorithm, we have

865 = 255 x 3 + 100

255 = 100 x 2 + 55

100 = 55 x 1 + 45

55 = 45 x 1 + 10

45 = 10 x 4 + 5

10 = 5 x 2 + 0

∵At the last stage, the divisor is 5

∴ The HCF of 255 and 865 is 5.

જવાબ : Given numbers are 255 and 865.

On applying Euclid’s division algorithm, we have

865 = 255 x 3 + 100

255 = 100 x 2 + 55

100 = 55 x 1 + 45

55 = 45 x 1 + 10

45 = 10 x 4 + 5

10 = 5 x 2 + 0

∵At the last stage, the divisor is 5

∴ The HCF of 255 and 865 is 5.

જવાબ :

જવાબ :

જવાબ : Let *a* be any positive integer and *b* = 3.

જવાબ : **Note: **If the denominator has only factors of 2 and 5 or in the form of 2^{m} × 5^{n} then it has a terminating decimal expansion.

If the denominator has factors other than 2 and 5 then it has a non-terminating repeating decimal expansion.

જવાબ : **(i) 43.123456789**

Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only. (ii) 0.120120012000120000. . .

Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number.

જવાબ : If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with a unit place as 0 or 5 is divisible by 5.

Prime factorization of 6જવાબ : HCF (616, 32) will give the maximum number of columns in which they can march.

We can use Euclid’s algorithm to find the HCF. 616 = 32 × 19 + 8 32 = 8 × 4 + 0 The HCF (616, 32) is 8. Therefore, they can march in 8 columns each.જવાબ : Let 3 + 2√5 be a rational number.

Then the co-primes x and y of the given rational number where (y ≠ 0) is such that: 3 + 2√5 = x/y Rearranging, we get, 2√5 = (x/y) – 3 √5 = 1/2[(x/y) – 3] Since x and y are integers, thus, 1/2[(x/y) – 3] is a rational number. Therefore, √5 is also a rational number. But this confronts the fact that √5 is irrational. Thus, our assumption that 3 + 2√5 is a rational number is wrong. Hence, 3 + 2√5 is irrational.જવાબ : (i) 140

Using the division of a number by prime numbers method, we can get the product of prime factors of 140.

Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2^{2 }× 5 × 7

Using the division of a number by prime numbers method, we can get the product of prime factors of 156. Hence, 156 = 2 × 2 × 13 × 3 = 2

Using the division of a number by prime numbers method, we can get the product of prime factors of 3825. Hence, 3825 = 3 × 3 × 5 × 5 × 17 = 3

Using the division of a number by prime numbers method, we can get the product of prime factors of 5005. Hence, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13 (v) 7429

Using the division of a number by prime numbers method, we can get the product of prime factors of 7429. Hence, 7429 = 17 × 19 × 23 = 17 × 19 × 23

જવાબ : As we know that,

HCF × LCM = Product of the two given numbers Therefore, 9 × LCM = 306 × 657 LCM = (306 × 657)/9 = 22338 Hence, LCM(306,657) = 22338જવાબ : Let x be any positive integer and y = 3.

By Euclid’s division algorithm; x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3) Therefore, x = 3q, 3q + 1 and 3q + 2 As per the given question, if we take the square on both the sides, we get; xજવાબ : No, LCM = Product of the highest power of each factor involved in the numbers.

HCF = Product of the smallest power of each common factor.

We can conclude that LCM is always a multiple of HCF, i.e., LCM = k × HCF

We are given that,

LCM = 175 and HCF = 15

175 = k × 15

⇒ 11.67 = k

But in this case, LCM ≠ k × HCF

Therefore, two numbers cannot have LCM as 175 and HCF as 15.

જવાબ : 4^{n} = (2^{2})^{n} = 2^{2n}

The only prime in the factorization of 4^{n} is 2.

There is no other prime in the factorization of 4^{n} = 2^{2n}

(By uniqueness of the Fundamental Theorem of Arithmetic).

5 does not occur in the prime factorization of 4^{n} for any n.

Therefore, 4^{n} does not end with the digit zero for any natural number n.

જવાબ : 17 × 5 × 11 × 3 × 2 + 2 × 11 …(i)

= 2 × 11 × (17 × 5 × 3 + 1)

= 2 × 11 × (255 + 1)

= 2 × 11 × 256

Number (i) is divisible by 2, 11 and 256, it has more than 2 prime factors.

Therefore (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number.

જવાબ : Let us assume, to the contrary, that 3√7 is rational.

That is, we can find coprime a and b (b ≠ 0) such that 3√7 = ab

Rearranging, we get √7 = a3b

Since 3, a and b are integers, a3b is rational, and so √7 is rational.

But this contradicts the fact that √7 is irrational.

So, we conclude that 3√7 is irrational.

જવાબ : Let x = 6.12¯ …(i)

100x = 612.12¯ …(ii)

…[Multiplying both sides by 100]

Subtracting (i) from (ii),

99x = 606

x = 60699 = 20233

Denominator = 33

Prime factorisation = 3 × 11

જવાબ : It is given that on dividing 70 by the required number, there is a remainder 5.

This means that 70 – 5 = 65 is exactly divisible by the required number.

Similarly, 125 – 8 = 117 is also exactly divisible by the required number.

65 = 5 × 13

117 = 3^{2} × 13

HCF = 13

Required number = 13

જવાબ : Let us assume, to the contrary, that 2 + 3√5 is rational.

So that we can find integers a and b (b ≠ 0).

Such that 2 + 3√5 = ab, where a and b are coprime.

Rearranging the above equation, we get

Since a and b are integers, we get a3b−23 is rational and so √5 is rational.

But this contradicts the fact that √5 is irrational.

So, we conclude that 2 + 3√5 is irrational.

જવાબ :

જવાબ : 96 = 2^{5} × 3

360 = 2^{3} × 3^{2} × 5

LCM = 2^{5} × 3^{2} × 5 = 32 × 9 × 5 = 1440

gseb maths textbook std 10

- Math Book for GSEB ધોરણ ૧૦
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- Biology Book for GSEB ધોરણ ૧૦
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The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.