In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 1 Real Numbers EX 1A Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 1 Real Numbers EX 1A Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers EX 1A Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 1 |

Chapter Name | Real Numbers |

Exercise | 1A |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A**

**Questions 1:**

For any two given positive integers a and b there exist unique whole numbers q and r such that

Here, we call ‘a’ as dividend, b as divisor, q as quotient and r as remainder.

Dividend = (divisor quotient) + remainder

**Questions 2:**

By Euclid’s Division algorithm we have:

Dividend = (divisor * quotient) + remainder

= (61 * 27) + 32 = 1647 + 32 = 1679

**Questions 3:**

By Euclid’s Division Algorithm, we have:

Dividend = (divisor quotient) + remainder

**Questions 4:**

(i) On dividing 2520 by 405, we get

Quotient = 6, remainder = 90

2520 = (405 x 6) + 90

Dividing 405 by 90, we get Quotient = 4, Remainder = 45

405 = 90 x 4 + 45

Dividing 90 by 45 We get Quotient = 2, remainder = 0

90 = 45 x 2

H.C.F. of 405 and 2520 is 45

(ii) Dividing 1188 by 504, we get Quotient = 2, remainder = 180

1188 = 504 x 2+ 180

Dividing 504 by 180 Quotient = 2, remainder = 144

504 = 180 x 2 + 144

Dividing 180 by 144, we get Quotient = 1, remainder = 36

Dividing 144 by 36

Quotient = 4, remainder = 0

H.C.F. of 1188 and 504 is 36

(iii) Dividing 1575 by 960, we get

Quotient = 1, remainder = 615

1575 = 960 x 1 + 615

Dividing 960 by 615, we get Quotient = 1, remainder = 345

960 = 615 x 1 + 345

Dividing 615 by 345 Quotient = 1, remainder = 270

615 = 345 x 1 + 270

Dividing 345 by 270, we get Quotient = 1, remainder = 75

345 = 270 x 1 + 75

Dividing 270 by 75, we get Quotient = 3, remainder =45

270 = 75 x 3 + 45

Dividing 75 by 45, we get Quotient = 1, remainder = 30

75 = 45 x 1 + 30

Dividing 45 by 30, we get Remainder = 15, quotient = 1

45 = 30 x 1 + 15

Dividing 30 by 15, we get Quotient = 2, remainder = 0

H.C.F. of 1575 and 960 is 15**Questions 5:**

(i) By prime factorization, we get

(ii) By prime factorization. We get

(iii) By prime factorization, we get**Questions 6:**

(i) By prime factorization, we get

(ii) By prime factorization, we get

(iii) By prime factorization, we get**Questions 7:****Questions 8:**

H.C.F. of two numbers = 11, their L.C.M = 7700

One number = 275, let the other number be b

Now, 275 x b = 11 x 7700**Questions 9:**

By going upward

5 x 11= 55

55 x 3 = 165

165 x 2 = 330

330 x 2 = 660**Questions 10:**

Subtracting 6 from each number:

378 – 6 = 372, 510 – 6 = 504

Let us now find the HCF of 372 and 504 through prime factorization:

372 = 2 x 2 x 3 x 31

504 = 2 x 2 x 2 x 3 x 3 x 7

The required number is 12.**Questions 11:**

Subtracting 5 and 7 from 320 and 457 respectively:

320 – 5 = 315,

457 – 7 = 450

Let us now find the HCF of 315 and 405 through prime factorization:

The required number is 45.**Questions 12:****Questions 13:**

The prime factorization of 42, 49 and 63 are:

42 = 2 x 3 x 7, 49 = 7 x 7, 63 = 3 x 3 x 7

Therefore, H.C.F. of 42, 49, 63 is 7

Hence, greatest possible length of each plank = 7 m**Questions 14:**

7 m = 700cm, 3m 85cm = 385 cm

12 m 95 cm = 1295 cm

Let us find the prime factorization of 700, 385 and 1295:

Greatest possible length = 35cm.**Questions 15:**

Let us find the prime factorization of 1001 and 910:

1001 = 11 x 7 x 13

910 = 2 x 5 x 7 x 13

H.C.F. of 1001 and 910 is 7 x 13 = 91

Maximum number of students = 91**Questions 16:**

Let us find the HCF of 336, 240 and 96 through prime factorization:

Each stack of book will contain 48 books

Number of stacks of the same height**Questions 17:****Questions 18:**

Let us find the LCM of 64, 80 and 96 through prime factorization:

L.C.M of 64, 80 and 96 =

Therefore, the least length of the cloth that can be measured an exact number of times by the rods of 64cm, 80cm and 96cm = 9.6m**Questions 19:****Questions 20:**

Interval of beeping together = LCM (60 seconds, 62 seconds)

The prime factorization of 60 and 62:

60 = 30 x 2, 62 = 31 x 2

L.C.M of 60 and 62 is 30 x 31 x 2 = 1860 sec = 31min

electronic device will beep after every 31minutes

After 10 a.m., it will beep at 10 hrs 31 minutes**Questions 21:**

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class10**

**All Subject NCERT Solutions For Class 10**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.