SEBA Class 10 Maths Chapter 1 Real Numbers

SEBA Class 10 Maths Chapter 1 Real Numbers Solutions English Medium As Per New Syllabus. to each chapter is provided in the list so that you can easily browse through different chapters SEBA Class 10 Maths Chapter 1 Real Numbers Notes and select need one. SEBA Class 10 Maths Chapter 1 Real Numbers Question Answers Download PDF. SEBA Class 10 General Mathematics Solutions.

SEBA Class 10 Maths Chapter 1 Real Numbers

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per State Council of Educational Research and Training (SCERT) Book guidelines. SEBA Class 10 Maths Chapter 1 Real Numbers are part of All Subject Solutions. Here we have given SEBA Solutions For Class 10 Maths for All Chapters, You can practice these here.

Real Numbers

Chapter: 1

Exercise 1.1

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

(i) 140

Ans: 

∴ 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7 

(ii) 156

Ans: 

(iii) 3825 

Ans: 

∴ 3825 = 3 × 3 × 5 × 17

(iv) 5005 

Ans: 

∴ 5005 = 5 × 7 × 11 × 13

(v) 7429

Ans:  

∴ 7429 = 17 × 19 × 23 

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

Ans: we have 26 and 91 

Thus LCM (26, 91) = 2 × 7 × 13 = 182 

HCF (26, 91) = 13 

Now LCM (26, 91) × HCF (26, 91 ) = 182 × 13 = 2366

And product of the numbers = 26 x 91 = 2366.

Hence the LCM × HCF = product of two numbers.

(ii) 510 and 92

Ans: We have 510 and 92

510 = 2 × 3⁵

92 = 2² × 23 

∴ LCM (510, 92 ) = 2² × 3⁵ × 23 = 23460

HCF (510, 92) = 2 

Now LCM (510, 92) × HCF (510, 92) = 23460 × 2 = 46920

Product of two numbers = 510 × 92 = 46920 

Hence LCM x HCF = Product of two numbers. 

(iii) 336 and 54

Ans: We have 336 = 2⁴ × 3 × 7  54 = 2 × 3³  

HCF (336, 92) = 2 × 3 = 6

Now LCM (336, 54) × HCF (336, 54) = 3024 × 6 = 18,144 and product of two numbers = 336 × 54 

= 18,144

Hence LCM × HCF = product of two numbers.

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

Ans: (i) 12 = 2² × 3   15 = 3 × 5 and 21 = 3 × 7

∴ The HCF of these integers is 3

2², 3¹, 5¹ and 7¹ are the greatest powers involved in the prime factors of 12, 15 and 21.

So LCM (12, 15, 21) = 2² × 3¹ × 5¹ × 7¹ = 420

(ii) 17, 23 and 29 

Ans: 17, 23 and 29

There is no common factor as 17,23,29  they are primes. Hence HCF is 1.

LCM is the product of all prime factors 17, 23, 29 = 17 × 23 × 29 = 11339

∴ HCF (17, 23, 29) = 1

(iii) 8, 9 and 25

Ans: 8, 9 and 25

8 = 2 × 2 × 2 = 2³   9 = 3²   25 = 5²

∴ HCF (8, 9, 25) = 1  [∵ There is no common factor of 8, 9, 25]

LCM = 2³ × 3² × 5² = 1800

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

Ans: We have

HCF (306, 657) = 9

We know that, LCM × HCF = Product of two numbers.

306 × 657

LCM × 9 = 306 × 657

306 × 657/9 = 22338

LCM = 22338

5. Check whether 6ⁿ can end with the digit 0 for many natural numbers n.

Ans: If the number 6ⁿ ends with the digit zero, then it must have 5 as a factor.

But, we know that the only prime factors of 6ⁿ are 2 and 3.

Also, we know from the fundamental theorem of arithmetic, that the prime factorisation of each number is unique.

So, there is no value of n natural numbers for which 6ⁿ ends with the digit zero.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Ans: We have 7 × 11 × 13 + 13 = 1001 + 13 = 1014

1014 = 2 × 3 × 13 × 13

So, it is the product of prime factors, 2 × 3 × 13 × 13

Hence, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × 1009.

= 5045 Thus 5045 can be expressed as products of primes.

It is the of prime factor 5 × 1009

Hence it is a composite number.

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting  point?

Ans: Required number of minutes is the LCM of 18 and 12

We have, 18 =2 × 3²  12 = 2² × 3

∴ LCM of 18 and 12 is 2² × 3² = 36

Hence, Ravis and Sonia will meet again at the starting point after 36 minutes.

8. The soldiers in the regiment can be stood in some rows consisting of 15, 20 and 25 numbers of soldiers . find the least number soldiers in the regiment.

Ans: Least numbers of soldiers in the regiment = LCM (15, 20, 25)

Now, 15 = 3 × 5

20 = 2² × 5

25 = 5²

∴ LCM (15, 20, 25) = 3 × 2² × 5²

= 3 × 4 × 25  

= 300

∴ 300 Soldier in the regiment.

(ii) A bell rings at every 18 seconds, another bell rings at every 60 seconds. If these two bells ring simultaneously at an instant, then find after how many seconds will the bells ring simultaneously again.

Ans: Required second = LCM (18, 60)

Now, 18 = 2 × 3²

60 = 2² × 3 × 5

∴ LCM (18, 60) = 2² × 3² × 5

= 4 × 9 × 5

= 180

∴ After 180 seconds will the bells rings simultaneously again. 

(iii) A radio station plays ‘Assam Sangeet’ once every two days. Another radio station plays the same song once every three days. How many times in 30 days will both the radio station play the same song on the same day.

Ans: One radio’s station plays ‘Assam Sangeet’ once every 2 days another radio station plays the same songs once every 3 days

LCM of (2, 3) = 6

∴ In 30 days both the radio stations play the same song on the same day = 30/6 = 5 times.

(iv) 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?

Ans: Number of members in the army contingent = 616

Number of members in the army band = 32

Maximum numbers of columns will be the HCF of 616 and 32

Now,

616 = 2³ × 7 × 11

32 = 2⁵

HCF (616,32) = 2³ = 8

Therefore, maximum number of columns in which the army can march is 8.

(v) Himadri has a collection of 625 Indian postal stamps and 325 International postal stamps. She wants to display them in identical groups of Indian and International stamps with no stamp left out. What is greatest number of groups Himadri can display the stamps?

Ans: Number of Indian postal stamps = 625

Number of International postal stamps = 325

The greatest number of groups of postal stamps will be the HCF of 625 and 325.

Now,

625 = 5⁴

325 = 5² × 13

HCF (625,325) = 5² = 25

∴ Greatest number of groups in which Himadri can display the stamps is 25.

(vi) Two ropes are of length 64 cm and 80 cm. Both are to be cut into pieces of equal length. What should be the maximum length of the pieces be?

Ans: Length of first rope = 64 cm

Length of second rope = 80 cm

Now,

64 = 2⁶

80 = 2⁴ × 5

HCF (64,80) = 2⁴ = 16

Maximum length of each piece of the rope is 16 cm.

9. Find the greatest number of 4 digits which is exactly divisible by 18, 24 and 36.

Ans: Prime factorisation:

18 = 2 × 3²

24 = 2³ × 3

36 = 2² × 3²

LCM = 2³ × 3² = 72

Now, we divide the greatest 4 digit number i.e. 9999 by LCM, i.e. 72

The required number = 72 × 138

= 9936

∴ The greatest 4-digit number exactly divisible by 18, 24 and 36 is 9936.