Question

can you give some questions of class 10 maths chapter 1 to 5 plz​

Answer 1

Answer:

Question 2.

Write the decimal form of 129275775

Solution:

Non-terminating non-repeating.

Question 3.

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

Solution:

Algorithm

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

HCF of 391, 425, 527 = 17

Question 4.

Express 98 as a product of its primes.

Solution:

2 × 72

Question 5.

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

Solution:

HCF of 408 and 1032 is 24.

1032 × 2 + 408 × (p) = 24

408p = 24 – 2064

p = -5

Question 6.

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

Solution:

We know,

1st number × 2nd number = HCF × LCM

⇒ 27 × 2nd number = 9 × 459

⇒ 2nd number = 9×45927 = 153

Question 7.

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

Solution:

13 = 1 × 13; 17 = 1 × 17

HCF = 1 and LCM = 13 × 17 = 221

Answer 2

Step-by-step explanation:

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.

Solution:

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

Substituting 3q2+2q = m we get,

x2 = 3m + 1 ……………………………. (2)

x2 = (3q + 2)2

= (3q)2 + 22 + 2 × 3q × 2

= 9q2 + 4 + 12q

= 3(3q2 + 4q + 1) + 1

Again, substituting 3q2 + 4q + 1 = m, we get,

x2 = 3m + 1…………………………… (3)

Hence, from eq. 1, 2 and 3, we conclude that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

Q.2: Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solution:

(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

(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 × 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