# GSEB Solutions for ધોરણ ૧૦ English

#### GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

Find the [HCF X LCM] for the numbers 100 and 190.

જવાબ : HCF x LCM = one number x another number
= 100 x 190 = 19000

An army contingent of 1000 members is to march behind an army band of 56 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

જવાબ : 1000 =2x2x2x5x5x5
56 = 2x2x2x7
HCF of 1000 and 56 = 8
Maximum number of columns = 8.

Find the LCM and HCF of 120 and 144 by using Fundamental Theorem of Arithmetic.

જવાબ : 120 = 23 x 3 x 5
144 = 24 x 32
∴ HCF = 23 x 3 = 24
LCM = 24 x 5 x 32 = 720

Three bells tolls at intervals of 12 minutes, 15 minutes and 18 minutes respectively. If they start tolling together, after what time will they next toll together?

જવાબ : LCM of 12, 15, 18 = 2²x 3² x 5
=4x9x5 = 180
So, next time the bells will ring together after 180 minutes.

Two tankers contain 850 liters and 680 liters of petrol respectively. Find the maximum capacity of a container which can measure the petrol of either tanker in exact number of times.

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

Find the [HCF and LCM] for the numbers 105 and 120. [All India]

જવાબ : 105 = 5 x 7 x 3
120 = 2x2x2x3x5
HCF = 3 X 5 = 15
LCM = 5x7x3x2x2x2 = 840

State whether 0.120120012000. . . is a rational or an irrational number.

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

All the rational and irrational numbers together form _______ numbers.

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

State whether true or false:

(i) The sum or difference of a rational number and an irrational number is irrational.

(ii) Addition, subtraction, multiplication or division of two irrational numbers always results in an irrational number again.

(iii) The product of a non-zero rational number with an irrational number is always irrational.

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

Write the decimal representation of the rational number 8/27.

જવાબ : Decimal representation of number 8/27 = 0.296

Write three rational numbers lying between 0 and 0.1.

જવાબ : 0.01, 0.02 and 0.03

If x is a rational number whose decimal expansion terminates then it can be expressed in the form p/q, where p and q are _________ (prime/co prime).

જવાબ : Coprime

State whether zero is a rational or an irrational number.

જવાબ : Zero is a rational number.

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

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

Define rational numbers.

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

“The product of three consecutive positive integers is divisible by 6”. Is this statement true or false? Justify your answer.

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.

What is the least number that is divisible by all the numbers from 1 to 10?

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

Explain why 3 × 5 × 7 + 7 is a composite number.

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

The product of two consecutive integers is divisible by 2. Is this statement true or false? Give reason.

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

Write a rational number between √3 and √5.

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

Can two numbers have 18 as their HCF and 380 as their LCM? Give reason.

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

Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer.

જવાબ : No, because any positive integer can be written as 3q, 3q + 1, 3q + 2, therefore, square will be
9q= 3m, 9q+ 6q + 1 = 3(3q2+ 2q) + 1 = 3m + 1,
9q2 + 12q + 4 = 3(3q2+ 4q + 1) + 1 = 3m + 1.

Can the number 4n, n be a natural number, end with the digit 0? Give reason.

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.

Without actually performing the long division, find 98710500 will have terminating or non-terminating repeating decimal expansion. Give reason for your answer.

જવાબ : 98710500 = 47500 and 500 = 22 × 53, so it has terminating decimal expansion.

A rational number in its decimal expansion is 1.7351. What can you say about the prime factors of q when this number is expressed in the form pq? Give reason.

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

What are the possible values of remainder r, when a positive integer a is divided by 3?

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.

A number N when divided by 14 gives the remainder 5. What is the remainder when the same number is divided by 7?

જવાબ : 5, because 14 is multiple of 7.
Therefore, remainder in both the cases is same.

What is the HCF of 33 × 5 and 32 × 52?

જવાબ : HCF of 33 × 5 and 32 × 52 = 32 × 5 = 45

If HCF of a and b is 12 and product of these numbers is 1800. Then what is LCM of these numbers?

જવાબ : Product of two numbers = Product of their LCM and HCF
1800 = 12 × LCM
LCM = 180012 = 150.

What is the HCF of 33 × 5 and 32 × 52?

જવાબ : HCF of 33 × 5 and 32 × 52 = 32 × 5 = 45

If HCF of a and b is 12 and product of these numbers is 1800. Then what is LCM of these numbers?

જવાબ : Product of two numbers = Product of their LCM and HCF
1800 = 12 × LCM
LCM = 180012 = 150.

What is the HCF of the smallest composite number and the smallest prime number?

જવાબ : Smallest composite number = 4
Smallest prime number = 2
So, HCF (4, 2) = 2

Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together? (2013)

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.

Find LCM of numbers whose prime factorization are expressible as 3 × 52 and 32 × 72. (2014)

જવાબ : LCM (3 × 52, 32 × 72) = 32 × 52 × 72 = 9 × 25 × 49 = 11025

Find HCF and LCM of 13 and 17 by prime factorization method. (2013)

જવાબ : 13 = 1 × 13; 17 = 1 × 17
HCF = 1 and LCM = 13 × 17 = 221

HCF and LCM of two numbers is 9 and 459 respectively. If one of the numbers is 27, find the other number. (2012)

જવાબ : We know, 1st number × 2nd number = HCF × LCM
27 × 2nd number = 9 × 459
2nd number = 9×45927 = 153

If the HCF of 408 and 1032 is expressible in the form 1032 × 2 + 408 × p, then find the value of p.

જવાબ : HCF of 408 and 1032 is 24.
1032 × 2 + 408 × (p) = 24
408p = 24 – 2064
p = -5

Express 98 as a product of its primes.

જવાબ : 2 × 72

Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11, and 15 respectively.

જવાબ : Algorithm
398 – 7 = 391, 436 – 11 = 425, 542 – 15 = 527
HCF of 391, 425, 527 = 17

The decimal expansion of the rational number 432453 will terminate after how many places of decimals? (2013)

જવાબ : The decimal representation of 61250 will terminate after how many places of decimal?

જવાબ : This representation will terminate after 4 decimal places

Examine whether 1730 is a terminating decimal or not.

જવાબ : Since the denominator has 3 as its factor.
1730 is a non4ermznatlng decimal.

Find the sum of જવાબ : If HCF of 144 and 180 is expressed in the form 13m – 3, find the value of m.

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

Find HCF of 378,180 and 420 by prime factorization method. Is HCF x LCM of three numbers equal to the product of the three numbers?

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.

By using, Euclid’s algorithm, find the largest number which divides 650 and 1170.

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.

Prove that the product of any three consecutive positive integers is divisible by 6. Solution: Let three consecutive numbers are n, n + 1, n + 2

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.

Determine the values o, t, p and q so that the prime factorization of2520 is expressible as 23 X y X q x 7.

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.

Show that 9n can not end with digit 0 for any natural number n.

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

By using Euclid’s algorithm, find the largest number which divides 650 and 1170.

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.

Find HCF of 65 and 117 and find a pair of integral values of m and n such that HCF = 65m + 117n.

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.

Find the HCF of 255 and 867 by Euclid’s division algorithm

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.

8. Find the HCF (865, 255) using Euclid’s division lemma.
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.

Find the HCF (865, 255) using Euclid’s division lemma.

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.

Given that HCF (306, 657) = 9, find LCM (306, 657).

જવાબ : Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12,15 and 21 (ii) 17,23 and 29 (iii) 8,9 and 25

જવાબ : Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

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 k1k2, 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.

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(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.

(i) 13/3125

Factorizing the denominator, we get,

3125 = 5 × 5 × 5 × 5 × 5 = 55

Or

= 20 × 55

Since the denominator is of the form 2m × 5n then, 13/3125 has a terminating decimal expansion.

(ii) 17/8

Factorizing the denominator, we get,

8 = 2× 2 × 2 = 23

Or

= = 23 × 50

Since the denominator is of the form 2m × 5n then, 17/8 has a terminating decimal expansion.

(iii) 64/455

Factorizing the denominator, we get,

455 = 5 × 7 × 13

Since the denominator is not in the form of 2m × 5n, therefore 64/455 has a non-terminating repeating decimal expansion.

(iv) 15/1600

Factorizing the denominator, we get,

1600 = 2× 52

Since the denominator is in the form of 2m × 5n, 15/1600 has a terminating decimal expansion.

The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p/q what can you say about the prime factors of q?

જવાબ : (i) 43.123456789

(ii) 0.120120012000120000. . .

Solution:

(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.

Check whether 6n can end with the digit 0 for any natural number n.

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 6= (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.

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

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.

Prove that 3 + 2√5 is irrational.

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.

Express each number as a product of its prime factors:
(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 = 2× 5 × 7

(ii) 156
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 × 5× 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

Given that HCF (306, 657) = 9, find LCM (306, 657).

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) = 22338

Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

જવાબ : 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+ 2 × 3q × 2

= 9q2 + 4 + 12q

= 3(3q2 + 4q + 1) + 1

Again, substitute, 3q+ 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.

Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons. (2017 OD)

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.

Check whether 4n can end with the digit 0 for any natural number n. (2015)

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.

Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number? (2015)

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.

Show that 3√7 is an irrational number. (2016)

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.

Find the prime factorization of the denominator of rational number expressed as 6.12¯ in simplest form. (2014)

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

Find the largest number which divides 70 and 125 leaving the remainder 5 and 8 respectively. (2015)

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

Prove that 2 + 3√5 is an irrational number. (2014)

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.

The decimal expansion of the rational number 432453 will terminate after how many places of decimals? (2013)

જવાબ : Find the LCM of 96 and 360 by using the fundamental theorem of arithmetic. (2012)

જવાબ : 96 = 25 × 3
360 = 23 × 32 × 5
LCM = 25 × 32 × 5 = 32 × 9 × 5 = 1440 ### Take a Test

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