### RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers

Here you can get solutions of RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers. These Solutions are part of RS Aggarwal Solutions Class 10. we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers download pdf.

## RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers

### Exercise 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:**

### Exercise 1B

**Questions 1:****Questions 2:****Questions 3:****Questions 4:**

(i) 53.123456789 is a rational number since it is a terminating decimal.

(ii) is a rational number because it is a non – terminating repeating decimal.

(iii) 0.12012001200012…… is not a rational number as it is a non-terminating, non – repeating decimal.

### Exercise 1C

**Questions 1:****Questions 2:****Questions 3:****Questions 4:****Questions 5:****Questions 6:**

(i) The sum of two rationals is always rational – True

(ii) The product of two rationals is always rational – True

(iii) The sum of two irrationals is an irrational – False

(iv) The product of two irrationals is an irrational – False

(v) The sum of a rational and an irrational is irrational – True

(vi) The product of a rational and an irrational is irrational – True

**Complete RS Aggarwal Solutions Class 10**

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