GSEB std 10 science solution for Gujarati check Subject Chapters Wise::
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
9q2 = 3m, 9q2 + 6q + 1 = 3(3q2+ 2q) + 1 = 3m + 1,
9q2 + 12q + 4 = 3(3q2+ 4q + 1) + 1 = 3m + 1.
જવાબ : if 4n ends with 0, then it must have 5 as a factor. But, (4)n = (22)n = 22n i.e., the only prime factor
of 4n is 2. Also, we know from the fundamental theorem of arithmetic that the prime factorization of each number is unique.
∴ 4n can never end with 0.
જવાબ : 98710500 = 47500 and 500 = 22 × 53, so it has terminating decimal expansion.
જવાબ : As 1.7351 is a terminating decimal number, so q must be of the form 2m 5n, 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 33 × 5 and 32 × 52 = 32 × 5 = 45
જવાબ : Product of two numbers = Product of their LCM and HCF
⇒ 1800 = 12 × LCM
⇒ LCM = 180012 = 150.
જવાબ : HCF of 33 × 5 and 32 × 52 = 32 × 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 = 32, 12 = 22 × 3, 15 = 3 × 5
LCM = 22 × 32 × 5 = 4 × 9 × 5 = 180 minutes or 3 hours
They will next toll together after 3 hours.
જવાબ : LCM (3 × 52, 32 × 72) = 32 × 52 × 72 = 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 × 72
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 9n is given by 9n= (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 9n ends with the digit zero
જવાબ : Since prime factorization of 9n is given by 9n= (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 9n 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.
Then a = 3q + r for some integer q ≥ 0 And r = 0, 1, 2 because 0 ≤ r < 3 Therefore, a = 3q or 3q + 1 or 3q + 2 Or,
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
(i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600
જવાબ : Note: If the denominator has only factors of 2 and 5 or in the form of 2m × 5n 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
(ii) 0.120120012000120000. . . Solution: (i) 43.123456789Since 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 6n = (2 × 3)n Therefore, the prime factorization of 6n doesn’t contain prime number 5. Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.જવાબ : 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.(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
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 = 22 × 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 = 22 × 13 × 3 (iii) 3825
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 = 32 × 52 × 17 (iv) 5005
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; x2 = (3q)2 = 9q2 = 3.3q2 Let 3q2 = m Therefore, x2 = 3m ………………….(1) x2 = (3q + 1)2 = (3q)2 + 12 + 2 × 3q × 1 = 9q2 + 1 + 6q = 3(3q2 + 2q) + 1 Substitute, 3q2+2q = m, to get, x2 = 3m + 1 ……………………………. (2) x2 = (3q + 2)2 = (3q)2 + 22 + 2 × 3q × 2 = 9q2 + 4 + 12q = 3(3q2 + 4q + 1) + 1 Again, substitute, 3q2 + 4q + 1 = m, to get, x2 = 3m + 1…………………………… (3) Hence, from eq. 1, 2 and 3, we conclude that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.જવાબ : 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 : 4n = (22)n = 22n
The only prime in the factorization of 4n is 2.
There is no other prime in the factorization of 4n = 22n
(By uniqueness of the Fundamental Theorem of Arithmetic).
5 does not occur in the prime factorization of 4n for any n.
Therefore, 4n 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 = 32 × 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 = 25 × 3
360 = 23 × 32 × 5
LCM = 25 × 32 × 5 = 32 × 9 × 5 = 1440
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