RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2

RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2

Question 1.
Assuming that x, y, z are positive real numbers, simplify each of the following:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q1.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q1.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q1.3
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q1.4
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q1.5
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q1.6

Question 2.
Simplify:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q2.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q2.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q2.3
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q2.4
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q2.5
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q2.6

Question 3.
Prove that:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.3
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.4
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.5
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.6
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.7
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.8
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.9
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.10
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.11
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.12
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q3.13

Question 4.
Show that:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q4.1
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q4.2
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q4.3
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q4.4
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q4.5
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q4.6
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q4.7
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q4.8
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q4.9
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q4.10

Question 5.
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q5.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q5.2

Question 6.
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q6.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q6.2

Question 7.
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q7.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q7.2

Question 8.
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q8.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q8.2

Question 9.
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q9.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q9.2

Question 10.
Find the values of x in each  of the following:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q10.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q10.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q10.3
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q10.4
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q10.5
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q10.6
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q10.7
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q10.8

Question 11.
If x = 21/3 + 22/3, show that x3 – 6x = 6.
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q11.1
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q11.2

Question 12.
Determine (8x)x, if 9x+ 2 = 240 + 9x.
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q12.1

Question 13.
If 3x+1 = 9x-2, find the value of 21 +x.
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q13.1

Question 14.
If 34x = (81)-1 and 101/y = 0.0001, find the value of 2-x+4y
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q14.1

Question 15.
If 53x = 125 and 10y = 0.001 find x and y.
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q15.1

Question 16.
Solve the following equations:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q16.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q16.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q16.3
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q16.4
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q16.5
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q16.6

Question 17.
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q17.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q17.2

Question 18.
If a and b are different positive primes such that
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q18.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q18.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q18.3

Question 19.
If 2x x 3y x 5z = 2160, find x, y and z. Hence, compute the value of 3x x 2-y x 5-z.
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q19.1

Question 20.
If 1176 = 2a x 3b x 7c, find the values of a, b and c. Hence, compute the value of 2a x 3b x 7-c as a fraction.
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q20.1

Question 21.
Simplify:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q21.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q21.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q21.3

Question 22.
Show that:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q22.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q22.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q22.3

Question 23.
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q23.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q23.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.2 Q23.3

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RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B

RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B.

Other Exercises

Question 1.
Solution:
We know that a fraction \(\frac { p }{ q } \) is terminating if prime factors of q are 2 and 5 only.
Hence.
(i) \(\frac { 13 }{ 80 } \) and \(\frac { 16 }{ 125 } \) are the terminating decimals.

Question 2.
Solution:
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 1
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 2
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 3
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 4
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 5

Question 3.
Solution:
(i) Let,x = 0.\(\overline { 3 } \) = 0.3333…(i)
Then, 10x = 3.3333….
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 6
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 7
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 8
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 9
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 10
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 11
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1B 12

Question 4.
Solution:
(i) True, because set of natural numbers is a subset of whole number.
(ii) False, because the number 0 does not belong to the set of natural numbers.
(iii) True, because a set of integers is a subset of a rational numbers.
(iv) False, because the set of rational numbers is not a subset of whole numbers.
(v) True, because rational number can be expressed as terminating or repeating decimals.
(vi) True, because every rational number can be express as repeating decimals.
(vii) True, because 0 = \(\frac { 0 }{ 1 } \), which is a rational number Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C

RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1C.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 1
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 2
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 3
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 4
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 5

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 6
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 7
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 8
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 9

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 10
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 11
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 12
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 13
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1C 14

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RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1

RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1

Other Exercises

Question 1.
Add the following rational numbers:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 1
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 2
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 3

Question 2.
Add the following rational numbers:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 4
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 5
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 6
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 7
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 8
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 9
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 10

Question 3.
Simplify:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 11
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 12
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 13
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 14
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 15
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 16
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 17
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 18
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 19

Question 4.
Add and Express the sum as mixed fraction:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 20
Solution:
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 21
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 22
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 23
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 24
RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1 25

 

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RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1

RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1

Question 1.
Simplify the following:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q1.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q1.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q1.3

Question 2.
If a = 3 and b =-2, find the values of:
(i) aa+ bb
(ii) ab + ba
(iii) (a+b)ab
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q2.1

Question 3.
Prove that:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q3.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q3.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q3.3
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q3.4

Question 4.
Prove that
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q4.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q4.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q4.3

Question 5.
Prove that
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q5.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q5.2

Question 6.
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q6.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q6.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q6.3

Question 7.
Simplify the following:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q7.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q7.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q7.3

Question 8.
Solve the following equations for x:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q8.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q8.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q8.3
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q8.4
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q8.5

Question 9.
Solve the following equations for x:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q9.1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q9.2
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q9.3

Question 10.
If 49392 = a4b2V3, find the values.of a, b and c, where a, b and c are different positive primes.
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q10.1

Question 11.
If 1176 = 2a x 3b x Tc, find a, 6 and c.
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q11.1

Question 12.
Given 4725 = 3a5b7c, find:
(i) the integral values of a, b and c
(ii) the value of 2-a 3b 7c
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q12.1

Question 13.
If a = xyp-1, b = xy q-1 and c = xyr-1, prove that aq-r br-p cp-q = 1
Solution:
RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 Q13.1

Hope given RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers Ex 2.1 are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B

RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Question 1.
Solution:
(i) 36, 84
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 1
36 = 2 x 2 x 3 x 3 = 2² x 3²
84 = 2 x 2 x 3 x 7 = 2² x 3 x 7
HCF = 2² x 3 = 2 x 2 x 3 = 12
LCM = 2² x 3² x 7 = 2 x 2 x 3 x 3 x 7 = 252
Now HCF x LCM = 12 x 252 = 3024
and product of number = 36 x 84 = 3024
HCF x LCM = Product of given two numbers.
(ii) 23, 31
23 = 1 x 23
31 = 1 x 31
HCF= 1
and LCM = 23 x 31 = 713
Now HCF x LCM = 1 x 713 = 713
and product of numbers = 23 x 31 = 713
HCF x LCM = Product of given two numbers
(iii) 96, 404
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 2
96 = 2 x 2 x 2 x 2 x 2 x 3 = 25 x 3
404 = 2 x 2 x 101 = 2² x 101
HCF = 2² = 2 x 2 = 4
LCM = 25 x 3 x 101 = 32 x 3 x 101 = 9696
Now HCF x LCM = 4 x 9696 = 38784
and product of two numbers = 96 x 404 = 38784
HCF x LCM = Product of given two numbers
(iv) 144, 198
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 3
144 = 2 x 2 x 2 x 2 x 3 x 3 = 24 x 32
198 = 2 x 3 x 3 x 11 = 2 x 3² x 11
HCF = 2 x 32 = 2 x 3 x 3 = 18
LCM = 24 x 3² x 11 = 16 x 9 x 11 = 1584
and product of given two numbers = 144 x 198 = 28512
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 4
and HCF x LCM = 18 x 1584 = 28512
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 5
HCF x LCM = Product of given two numbers
(v) 396, 1080
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 6
396 = 2 x 2 x 3 x 3 x 11 = 2² x 3² x 11
1080 = 2 x 2 x 2 x 3 x 3 x 3 x 5 = 23 x 33 x 5
HCF = 2² x 3² = 2 x 2 x 3 x 3 = 36
LCM = 23 x 33 x 11 x 5 = 2 x 2 x 2 x 3 x 3 x 3 x 5 x 11 = 11880
Now HCF x LCM = 36 x 11880 = 427680
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 7
Product of two numbers = 396 x 1080 = 427680
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 8
HCF x LCM = Product of two given numbers.
(vi) 1152, 1664
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 9
1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 = 27 x 3²
1664 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 13 = 27 x 13
HCF = 27 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128
LCM = 27 x 3² x 13 = 128 x 9 x 13 = 128 x 117= 14976
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 10
Now HCF x LCM = 128 x 14976= 1916928
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 11
and product of given two numbers = 1152 x 1664 = 1916928
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 12
HCF x LCM = Product of given two numbers.

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 13
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 14
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 15
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 16

Question 3.
Solution:
HCF of two numbers = 23
LCM =1449
One number = 161
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 17
Second number = 207

Question 4.
Solution:
HCF of two numbers = 145
LCM = 2175
One number = 725
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 18
Second number = 435

Question 5.
Solution:
HCF of two numbers = 18
and product of two numbers = 12960
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 19
LCM of two numbers = 720

Question 6.
Solution:
HCF= 18
LCM = 760
HCF always divides the LCM completely
760 – 18 = 42 and remainder 4
Hence, it is not possible.

Question 7.
Solution:
(a) \(\frac { 69 }{ 92 }\)
HCF of 69 and 92 = 23
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 20
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 21
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 22
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 23

Question 8.
Solution:
Numbers are 428 and 606 and remainder in each case = 6
Now subtracting 6 from each number, we get 438 – 6 = 432
and 606 – 6 = 600
Required number = HCF of 432 and 600 = 24
The largest required number is 24
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 24

Question 9.
Solution:
The numbers are 320 and 457
and remainders are 5 and 7 respectively
320 – 5 = 315 and 457 – 7 = 450
Now the required greatest number of 315 and 450 is their HCF
Now HCF of 315 and 450 = 45
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 25

Question 10.
Solution:
The numbers are given = 35, 56, 91 and the remainder = 7 in each case,
Now the least number = LCM of 35, 56, 91 = 3640
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 26
LCM = 7 x 5 x 8 x 13 = 3640
Required least number = 3640 + 7 = 3647

Question 11.
Solution:
Given numbers are 28 and 32
Remainders are 8 and 12 respectively
28 – 8 = 20
32 – 12 = 20
Now, LCM of 28 and 32 = 224
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 27
LCM = 2 x 2 x 7 x 8 = 224
Least required number = 224 – 20 = 204

Question 12.
Solution:
The given numbers are 468 and 520
Now LCM of 468 and 520 = 4680
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 28
LCM = 2 x 2 x 13 x 9 x 10 = 4680
When number 17 is increase then required number = 4680 – 17 = 4663

Question 13.
Solution:
LCM of 15, 24, 36 = 360
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 29
Required number = 9999 – 279 = 9720

Question 14.
Solution:
Greatest number of 4 digits is 9999
LCM of 4, 7 and 13 = 364
On dividing 9999 by 364, remainder is 171
Greatest number of 4 digits divisible by 4, 7 and 13 = (9999 – 171) = 9828
Hence, required number = (9828 + 3) = 9831

Question 15.
Solution:
LCM of 5, 6, 4 and 3 = 60
On dividing 2497 by 60, the remainder is 37
Number to be added = (60 – 37) = 23

Question 16.
Solution:
We can represent any integer number in the form of: pq + r, where ‘p is divisor, ‘q is quotient’, r is remainder*.
So, we can write given numbers from given in formation As :
43 = pq1 + r …(i)
91 = pq2 + r …(ii)
And 183 = pq3 + r …(iii)
Here, we want to find greatest value of ‘p’ were r is same.
So, we subtract eq. (i) from eq. (ii), we get
Pq2 – Pq1 = 48
Also, subtract eq. (ii) from eq. (iii), we get
pq3 – pq2 = 92
Also, subtract eq. (i) from eq. (iii), we get
Pq3 – Pq1 = 140
Now, to find greatest value of ‘p’ we find HCF of 48, 92 and 140 as,
48 = 2 x 2 x 2 x 2 x 3
92 = 2 x 2 x 2 x 2 x 2 x 3
and 140 = 2 x 2 x 5 x 7
So, HCF (48, 92 and 140) = 2 x 2 = 4
Greatest number that will divide 43, 91 and 183 as to leave the same remainder in each case = 4.

Question 17.
Solution:
Remainder in all the cases is 6, i.e.,
20 – 14 = 6
25 – 19 = 6
35 – 29 = 6
40 – 34 = 6
The difference between divisor and the corresponding remainder is 6.
Required number = (LCM of 20, 25, 35, 46) – 6 = 1400 – 6 = 1394

Question 18.
Solution:
Number of participants in Hindi = 60
Number of participants in English = 84
Number of participants in Mathematics =108
Minimum number of participants in one room = HCF of 60, 84 and 108 = 12
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 30
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 31

Question 19.
Solution:
Number of books in English = 336
Number of books in Mathematics = 240
Number of books in Science = 96
Minimum number of books of each topic in a stack = HCF of 336, 240 and 96 = 48
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 32

Question 20.
Solution:
Length of first piece of timber = 42 m
Length of second piece of timber = 49 m
and length of third piece of timber = 63 m
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 33
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 34

Question 21.
Solution:
Lengths are given as 7 m, 3 m 85 cm and 12 m 95 cm = 700 cm, 385 cm and 1295 cm
Greatest possible length that can be used to measure exactly = HCF of 700, 385, 1295 = 35 cm
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 35

Question 22.
Solution:
Number of pens =1001
and number of pencils = 910
Maximum number of pens and pencils equally distributed to the students = HCF of 1001 and 910 = 91
Number of students = 91
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 36

Question 23.
Solution:
Length of the room = 15 m 17 cm = 1517 cm
and breadth = 9 m 2 cm = 902 cm
Maximum side of square tile used = HCF of 1517 and 902 = 41 cm
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 37

Question 24.
Solution:
Measures of three rods = 64 cm, 80 cm and 96 cm
Least length of cloth that can be measured an exact number of times
= LCM of 64, 80, 96
= 960 cm
= 9 m 60 cm
= 9.6 m
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 38
LCM = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3 = 960

Question 25.
Solution:
Beep made by first devices after every = 60 seconds
Second device after = 62 seconds
Period after next beep together = LCM of 60, 62
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 39
LCM = 2 x 30 x 31 = 1860 = 1860 seconds = 31 minutes
Time started beep together, first time together = 10 a.m.
Time beep together next time = 10 a.m. + 31 minutes = 10 : 31 a.m.

Question 26.
Solution:
The traffic lights of three roads change after
48 sec., 72 sec. and 108 sec. simultaneously
They will change together after a period of = LCM of 48 sec., 72 sec. and 108 sec.
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 40
= 7 minutes, 12 seconds
First time they light together at 8 a.m. i.e., after 8 hr.
Next time they will light together = 8 a.m. + 7 min. 12 sec. = 8 : 07 : 12 hrs.

Question 27.
Solution:
Tolling of 6 bells = 2, 4, 6, 8, 10, 12 minutes
They take time tolling together = LCM of 2, 4, 6, 8, 10, 12 = 120 minutes = 2 hours
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 41
LCM of 2 x 2 x 2 x 3 x 5 = 120 min. (2 hr)
They will toll together after every 2 hours Total time given = 30 hours
Number of times, there will toll together in 30 hours = \(\frac { 30 }{ 2 }\) = 15 times
Total numbers of times = 15 + 1 (of starting time) = 16 times

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RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS

RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS

Other Exercises

the correct alternative in each of the following:
Question 1.
Which one of the following is a correct statement?
(a) Decimal expansion of a rational number is terminating
(b) Decimal expansion of a rational number is non-terminating
(c) Decimal expansion of an irrational number is terminating
(d) Decimal expansion of an irrational number is non-terminating and non-repeating
Solution:
Decimal expansion of an irrational number is non-terminating and non-repeating . (d)

Question 2.
Which one of the following statements is true?
(a) The sum of two irrational numbers is always an irrational-number            
(b) The sum of two irrational numbers is always a rational number
(c) The sum of two irrational numbers may be a rational number or an irrational number
(d) The sum of two irrational numbers is always an integer
Solution:
The sum of two irrational numbers may be a rational number or an irrational number (c)

Question 3.
Which of the following is a correct statement?
(a) Sum of two irrational numbers is always irrational
(b) Sum of a rational and irrational number is always an irrational number
(c) Square of an irrational number is always a rational number
(d) Sum of two rational numbers can never be an integer
Solution:
Sum of a rational and irrational number is always an irrational number         (b)

Question 4.
Which of the following statements is true?
(a) Product of two irrational numbers is always irrational
(b) Product of a rational and an irrational number is always irrational
(c) Sum of two irrational numbers can never be irrational
(d) Sum of an integer and a rational number can never be an integer
Solution:
Product of a rational and an irrational number is always irrational    (b)

Question 5.
Which of the following is irrational?
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q5.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q5.2

Question 6.
Which of the following is irrational?
(a) 14
(b)  0.14\(\overline { 16 }\)
(c)   0.\(\overline { 1416 }\)                  
(d)  0.1014001400014
Solution:
0.1014001400014…….. is irrational as it is non-terminating nor repeating decimal, (d)

Question 7.
Which of the following is rational?
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q7.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q7.2

Question 8.
The number 0.318564318564318564… is:
(a) a natural number
(b) an integer
(c) a rational number
(d) an irrational number
Solution:
The number = 0.318564318564318564…………
= 0.\(\overline { 318564 }\)
∵ The decimal is non-terminating and recurring
∴ It is rational number.   (c)

Question 9.
If n is a natural number, then \(\sqrt { n } \)is
(a) always a natural number
(b) always a rational number
(c) always an irrational number
(d) sometimes a natural number and sometimes an irrational number
Solution:
If n is a natural number then \(\sqrt { n } \) may sometimes a natural number and sometime an irrational number e.g.
If n = 2 then \(\sqrt { n } \) =\(\sqrt { 2 } \) which is are irrational and if n = 4, then \(\sqrt { n } \)= \(\sqrt { 4 } \) =  2 which is a rational number.       (d)

Question 10.
Which of the following numbers can be represented as non-terminating, repeating decimals?
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q10.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q10.2

Question 11.
Every point on a number line represents
(a) a unique real number
(b) a natural number
(c) a rational number
(d) an irrational number
Solution:
Every point on a number line represents a unique real number.         (a)

Question 12.
Which of the following is irrational?
(a) 0.15                     
(b) 0.01516
(c) 0.\(\overline { 1516 }\)                
(d) 0.5015001500015..
Solution:
As it is non-terminating non-repeating decimals while others are terminating or non-terminating repeating decimals. (d)

Question 13.
The number 1.\(\overline { 27 }\) in the form \(\frac { p }{ q }\)  , where p and q are integers and q ≠ 0, is
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q13.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q13.2

Question 14.
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q14.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q14.2

Question 15.
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q15.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q15.2

Question 16.
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q16.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q16.2

Question 17.
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q17.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q17.2

Question 18.
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q18.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q18.2

Question 19.
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q19.1
Solution:
An irrational number between 2 and 2.5 is \(\sqrt { 5 } \) as it has approximate value 2.236… (b)

Question 20.
The number of consecutive zeros in 23 x 34 x 54 x 7, is
(a) 3                            
(b) 2
(c) 4
(d) 5
Solution:
In 23 x 34 x 54 x 7, number of consecutive zero will be 3 as 23 x 54 = 2 x 2 x 2 x 5x 5 x 5 x 5 = 5000      (a)

Question 21.
The smallest rational number by which \(\frac { 1 }{ 3 }\) should be multiplied so that its decimal expansion terminates after one place of decimal, is
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q21.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems MCQS Q21.2

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

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

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A.

Other Exercises

Question 1.
Solution:
For any two given positive integers a and b, there exist unique whole numbers q and r such that
a = bq + r, when 0 ≤ r < b.
Here, a is called dividend, b as divisor, q as quotient and r as remainder.
Dividend = (Divisor x Quotient) + Remainder.

Question 2.
Solution:
Using Euclid’s divison Lemma
Dividend = (Divisor x Quotient) + Remainder
= (61 x 27) + 32
= 1647 + 32
= 1679
Required number = 1679

Question 3.
Solution:
Let the required divisor = x
Then by Euclid’s division Lemma,
Dividend = (Divisor x Quotient) + remainder
1365 = x x 31 + 32
=> 1365 = 31x + 32
=> 31x= 1365 – 32 = 1333
x = \(\frac { 1331 }{ 31 }\) = 43
Divisor = 43

Question 4.
Solution:
(i) 405 and 2520
HCF of 405 and 2520 = 45
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A 1
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A 2

Question 5.
Solution:
Let n be an arbitrary positive integer.
On dividing n by 2, let m be the quotient and r be the remainder, then by Euclid’s division lemma
n = 2 x m + r = 2m + r, 0 ≤ r < 2
n = 2m or 2m + 1 for some integer m.
Case 1 : When n = 2m, then n is even
Case 2 : When n = 2m + 1, then n is odd.
Hence, every positive integer is either even or odd.

Question 6.
Solution:
Let n be a given positive odd integer.
On dividing n by 6, let m be the quotient and r be the remainder, then by Euclid’s division Lemma.
n = 6m + r, where 0 ≤ r < 6 => n = 6m + r, where r = 0, 1, 2, 3, 4, 5
=> n = 6m or (6m + 1) or (6m + 2) or (6m + 3) or (6m + 4) or (6m + 5)
But n = 6m, (6m + 2) and (6m + 4) are even.
Thus when n is odd, it will be in the form of (6m + 1) or (6m + 3) or (6m + 5) for some integer m.

Question 7.
Solution:
Let n be an arbitrary odd positive integer.
On dividing by 4, let m be the quotient and r be the remainder.
So by Euclid’s division lemma,
n = 4m + r, where 0 ≤ r < 4
n = 4m or (4m + 1) or (4m + 2) or (4m + 3)
But 4m and (4m + 2) are even integers.
Since n is odd, so n ≠ 4m or n ≠ (4m + 2)
n = (4m + 1) or (4m + 3) for some integer m.
Hence any positive odd integer is of the form (4m + 1) or (4m + 3) for some integer m.

Question 8.
Solution:
Let a = n3 – n
=> a = n (n2 – 1)
=> a = n (n – 1) (n + 1) [(a2 – b2) = (a – b) (a + b)]
=> a = (n – 1 ) n (n + 1)
We know that,
(i) If a number is completely divisible by 2 and 3, then it is also divisible by 6.
(ii) If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
(iii) If one of the factor of any number is an even number, then it is also divisible by 2.
a = (n – 1) n (n + 1) [From Eq. (i)]
Now, sum of the digits
= n – 1 + n + n + 1 = 3n
= Multiple of 3, where n is any positive integer.
and (n – 1) n (n +1) will always be even, as one out of (n – 1) or n or (n + 1) must be even.
Since, conditions (ii) and (iii) is completely satisfy the Eq. (i).
Hence, by condition (i) the number n3 – n is always divisible by 6, where n is any positive integer.
Hence proved.

Question 9.
Solution:
Let x = 2m + 1 and y = 2m + 3 are odd positive integers, for every positive integer m.
Then, x2 + y2 = (2m + 1)2 + (2m + 3)2
= 4m2 + 1 + 4 m + 4m2 + 9 + 12m [(a + b)2 = a2 + 2ab + b2]
= 8m2 + 16m + 10 = even
= 2(4m2 + 8m + 5) or 4(2m2 + 4m + 2) + 1
Hence, x2 + y2 is even for every positive integer m but not divisible by 4.

Question 10.
Solution:
We find HCF (1190, 1145) using the following steps:
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A 3
(i) Since 1445 > 1190, we divide 1445 by 1190 to get 1 as quotient and 255 as remainder.
By Euclid’s division lemma, we get
1445 = 1190 x 1 + 255 …(i)
(ii) Since the remainder 255 ≠ 0, we divide 1190 by 255 to get 4 as a quotient and 170 as a remainder.
By Euclid’s division lemma, we get
1190 = 255 x 4 + 170 …(ii)
(iii) Since the remainder 170 ≠ 0, we divide 255 by 170 to get 1 as quotient and 85 as remainder.
By Euclid’s division lemma, we get
255 = 170 x 1 +85 …(iii)
(iv) Since the remainder 85 ≠ 0, we divide 170 by 85 to get 2 as quotient and 0 as remainder.
By Euclid’s division lemma, we get
170 = 85 x 2 + 0 …(iv)
The remainder is now 0, so our procedure steps
HCF (1190, 1445) = 85
Now, from (iii), we get
255 = 170 x 1 + 85
=> 85 = 255 – 170 x 1
= (1445 – 1190) – (1190 – 255) x 4
= (1445 – 1190) – (1190 – 255) x 4
= (1445 – 1190) x 2 + (1445 – 1190) x 4
= 1445 – 1190 x 2 + 1445 x 4 – 1190 x 4
= 1445 x 5 – 1190 x 6
= 1190 x (-6) + 1445 x 5
Hence, m = -6, n = 5

Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.6

RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.6

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.6

Question 1.
Visualise 2.665 on the number line, using successive magnification.
Solution:
2.665
∵ It lies between 2 and 3
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.6 Q1.1

Question 2.
Visualise the representation of 5.3\(\overline { 7 }\) on the number line upto 5 decimal places, that is upto 5.37777.         [NCERT]
Solution:
5.37777.
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.6 Q2.1

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RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5

RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5

Question 1.
Complete the following sentences:
(i) Every point on the number line corresponds to a … number which many be either … or
(ii) The decimal form of an irrational number is neither … nor …
(iii) The decimal representation of a rational number is either … or …
(iv) Every real number is either … number or … number.
Solution:
(i) Every point on the number line corresponds to a real number which many be either rational or irrational.
(ii) The decimal form of an irrational number is neither terminating nor repeating.
(iii) The decimal representation of a rational number is either terminating or non­terminating, recurring.
(iv) Every real number is either rational number or an irrational number.

Question 2.
Find whether the following statements are true or false:
(i)  Every real number is either rational or irrational.
(ii) π is an irrational number.
(iii) Irrational numbers cannot be represented by points on the number line.
Solution:
(i) True. (Value of π = 3.14)
(ii) False : we can represent irrational number also.

Question 3.
Represent \(\sqrt { 6 } \), \(\sqrt { 7 } \), \(\sqrt { 8 } \) on the number line.
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q3.1
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q3.2
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q3.3
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q3.4
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q3.5
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q3.6
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q3.7
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q3.8

Question 4.
Represent \(\sqrt { 3.5 } \) , \(\sqrt { 9.4 } \)and \(\sqrt { 10.5 } \) on the real number line.
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q4.1
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q4.2
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q4.3
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q4.4
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q4.5
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.5 Q4.6

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RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4

RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4

Question 1.
Define an irrational number.
Solution:
A number which cannot be expressed in the form of \(\frac { p }{ q }\) where p and q are integers and q ≠ 0 is called an irrational number.

Question 2.
Explain, how irrational numbers differ from rational numbers?
Solution:
A rational number can be expressed in either terminating decimal or non-terminating recurring decimals but an irrational number is expressed in non-terminating non-recurring decimals.

Question 3.
Examine, whether the following numbers are rational or irrational:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q3.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q3.2
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q3.3
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q3.4
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q3.5

Question 4.
Identify the following as rational or irrational numbers. Give the decimal representation of rational numbers:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q4.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q4.2
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q4.3

Question 5.
In  the following equation, find which variables x, y, z etc. represent rational or irrational numbers:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q5.1
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q5.2
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q5.3
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q5.4

Question 6.
Given two rational numbers lying between 0.232332333233332… and 0.212112111211112.
Solution:
Two rational numbers lying between 0.232332333233332… and 0.212112111211112… will be 0.232 and 0.212

Question 7.
Give two rational numbers lying between 0.515115111511115… and 0.5353353335…
Solution:
Two rational numbers lying between 0.515115111511115… and 0.535335333533335… will be 0.515, 0.535

Question 8.
Find one irrational numbers between 0.2101 and 0.2222… = 0.\(\overline { 2 }\).
Solution:
One irrational number lying between 0.2101 and 0.2222… = 0.\(\overline { 2 }\) will be 2201.0010001…

Question 9.
Find a rational number and also an irrational number lying between the numbers, 0.3030030003… and 0.3010010001…
Solution:
Between two numbers 0.3030030003… and 0.3010010001…, a rational will be 0.301 and irrational number will be 0.3020020002…

Question 10.
Find three different irrational numbers between the rational numbers \(\frac { 5 }{ 7 }\) and \(\frac { 9 }{ 11 }\). [NCERT]
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q10.1
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q10.2
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q10.3

Question 11.
Give an example of each, of two irrational numbers whose:
(i) difference is a rational number.
(ii) difference is an irrational number.
(iii) sum is a rational number.
(iv) sum is an irrational number.
(v) product is a rational number.
(vi) product is an irrational number.
(vii) quotient is a rational number.
(viii) quotient is an irrational number.
Solution:
(i) Two numbers whose difference is also a rational number, e.g. \(\sqrt { 2 } \)
, \(\sqrt { 2 } \)  which are irrational numbers.
∴ Difference = \(\sqrt { 2 } \) – \(\sqrt { 2 } \) = 0 which is also a rational number.
(ii) Two numbers whose difference is an irrational number.
e.g. \(\sqrt { 3 } \)  and \(\sqrt { 2 } \)  which are irrational numbers.
Now difference = \(\sqrt { 3 } \)  –\(\sqrt { 2 } \)  which is also an irrational number.
(iii) Let two irrational numbers be \(\sqrt { 3 } \)  and –\(\sqrt { 2 } \)  which are irrational numbers.
Now sum = \(\sqrt { 3 } \)  + (-\(\sqrt { 3 } \)) = \(\sqrt { 3 } \)
– \(\sqrt { 3 } \)  = 0 Which is a rational number.
(iv) Let two numbers be \(\sqrt { 5 } \)  , \(\sqrt { 3 } \) which are irrational numbers.
Now sum = \(\sqrt { 5 } \) + \(\sqrt { 3 } \)  which is an irrational number.
(v) Let numbers be \(\sqrt { 3 } \)  +\(\sqrt { 2 } \)and  \(\sqrt { 3 } \)  –\(\sqrt { 2 } \)which are irrational numbers.
Now product = (\(\sqrt { 3} \)  +\(\sqrt { 2 } \) ) (\(\sqrt { 3 } \) –\(\sqrt { 2 } \))
= 3-2 = 1 which is a rational number.
(vi) Let numbers be \(\sqrt { 3 } \) and \(\sqrt { 5 } \) , which are irrational number.
Now product = \(\sqrt { 3 } \) x \(\sqrt { 5 } \)  = \(\sqrt { 3×5 } \)
= \(\sqrt { 15 } \)
which is an irrational number.
(vii) Let numbers be 6 \(\sqrt { 2 } \)  and 2 \(\sqrt { 2 } \) which are irrational numbers.
Quotient =\(\frac { 6\sqrt { 2 } }{ 2\sqrt { 2 } }\) = 3 which is a rational number.
(viii) Let numbers be \(\sqrt { 3 } \)and \(\sqrt { 5 } \) which are irrational numbers.
Now quotient =\(\frac { \sqrt { 3 } }{ \sqrt { 5 } }\) = \(\sqrt { \frac { 3 }{ 5 } }\) which is an  irrational number.

Question 12.
Find two irrational numbers between 0.5 and 0.55.
Solution:
Two irrational numbers between 0.5 and 0.55 will be 0.51010010001… and 52020020002…

Question 13.
Find two irrational numbers lying betwee 0.1 and 0.12.
Solution:
Two irrational numbers lying between 0.1 and 0.12 will be 0.1010010001… and 0.1020020002…

Question 14.
Prove that \(\sqrt { 3 } \)+\(\sqrt { 5 } \) is an irrational number.
Solution:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q14.1
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.4 Q14.2

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