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

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

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers MCQs
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Test Yourself

Very-Short Answer Questions
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, where 0 ≤ r < b.
Here, we call ‘a’ as dividend, b as divisor, q is quotient and r as remainder.
Dividend = (Divisor x Quotient) + Remainder

Question 2.
Solution:
Every composite number can be uniquely expressed as a product of two primes, except for the order in which these prime factors occurs.

Question 3.
Solution:
360 = 2 x 2 x 2 x 3 x 3 x 5 = 2³ x 3² x 5

Question 4.
Solution:
We know that HCF of two primes is
HCF (a, b) = 1

Question 5.
Solution:
a and b are two prime numbers then their
LCM = Product of these two numbers
LCM(a, b) = a x b = ab.

Question 6.
Solution:
We know that product of two numbers is equal to their HCF x LCM
LCM = \(\frac { Product of two numbers }{ HCF }\)
= \(\frac { 1050 }{ 25 }\) = 42
LCM of two numbers = 42

Question 7.
Solution:
A composite number is a number which is not a prime. In other words, a composite number has more than two factors.

Question 8.
Solution:
a and b are two primes, then their
HCF will be 1
HCF of a and b = 1

Question 9.
Solution:
\(\frac { a }{ b }\) is a rational number and it has terminating decimal
b will in the form 2^m x 5ⁿ where m and n are some non-negative integers.

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
2ⁿ x 5ⁿ = (2 x 5)ⁿ = (10)ⁿ
Which always ends in a zero
There is no value of n for which (2ⁿ x 5ⁿ) ends in 5

Question 13.
Solution:
We know that HCF is always a factor is its LCM
But 25 is not a factor of 520
It is not possible to have two numbers having HCF = 25 and LCM = 520

Question 14.
Solution:
Let two irrational number be (5 + √3) and (5 – √3).
Now their sum = (5 + √3) + (5 – √3) = 5 + √3 + 5 – √3 = 10
Which is a rational number.

Question 15.
Solution:
Let the two irrational number be (3 + √2) and (3 – √2)
Now, their product = (3 + √2) (3 – √2)
= (3)² – (√2)² {(a + b) (a – b) = a² – b²}
= 9 – 2 = 7
Which is a rational number.

Question 16.
Solution:
a and b are relative primes
their HCF = 1

Question 17.
Solution:
LCM of two numbers = 1200
and HCF = 500
But we know that HCF of two numbers divides their LCM.
But 500 does not divide 1200 exactly
Hence, 500 is not their HCF whose LCM is 1200.

Short-Answer Questions
Question 18.
Solution:
Let x = 0.4 = 0.444
Then 10x = 4.444….
Subtracting, we get
9x = 4 => x = \(\frac { 4 }{ 9 }\)
\(\bar { 0.4 }\) = \(\frac { 1 }{ 2 }\) which is in the simplest form.

Question 19.
Solution:
\(\bar { 0.23 }\)
Let x = \(\bar { 0.23 }\) = 0.232323…….
and 100x = 23.232323……
Subtracting, we get
99x = 23 => x = \(\frac { 23 }{ 99 }\)
\(\bar { 0.23 }\) = \(\frac { 23 }{ 99 }\) which is in the simplest form.

Question 20.
Solution:
0.15015001500015
It is non-terminating non-repeating decimal.
It is an irrational number.

Question 21.
Solution:
\(\frac { \surd 2 }{ 3 }\) = \(\frac { 1 }{ 3 }\) √2
Let \(\frac { 1 }{ 3 }\) √2 is a rational number
Product of two rational numbers is a rational
\(\frac { 1 }{ 3 }\) is rational and √2 is rational contradicts
But it contradicts the fact
\(\frac { \surd 2 }{ 3 }\) or \(\frac { 1 }{ 3 }\) √2 is irrational.

Question 22.
Solution:
√3 and 2.
√3 = 1.732 and 2.000
A rational number between 1.732 and 2.000 can be 1.8 or 1.9
Hence, 1.8 or 1.9 is a required rational.

Question 23.
Solution:
\(\bar { 3.1416 }\)
It is non-terminating repeating decimal.
It is a rational number.

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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

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

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F

Rationalise the denominator of each of the followings :

Question 1.
Solution:
Here,RF of √7 is √7

Question 2.
Solution:
Here RF √3 is √3

Question 3.
Solution:
Here RF of \(\frac { 1 }{ \left( { 2+ }\sqrt { 3 } \right) }\) is \(\frac { 1 }{ \left( { 2- }\sqrt { 3 } \right) }\)

Question 4.
Solution:
Here RF is √5 + 2

Question 5.
Solution:
Here RF is 5 – 3√2

Question 6.
Solution:
Here RF is √6 + √5

Question 7.
Solution:
Here RF = √7 – √3

Question 8.
Solution:
Here RF = √3 – 1

Question 9.
Solution:
Here RF = (3-2√2)

Find the values of a and b in each of the following :

Question 10.
Solution:
\(\frac { \sqrt { 3 } +1 }{ \sqrt { 3 } -1 } \), RF = √3+1
(Multiplying and dividing by √3+1)

Question 11.
Solution:
\(\frac { 3+\sqrt { 2 } }{ 3-\sqrt { 2 } } \), RF is 3+√2
(Multiplying and dividing by 3+√2)

Question 12.
Solution:
In \(\frac { 5-\sqrt { 6 } }{ 5+\sqrt { 6 } } \), RF is (5-√6)
(Multiplying and dividing by 5-√6)

Question 13.
Solution:
In \(\frac { 5+2\sqrt { 3 } }{ 7+4\sqrt { 3 } } \), RF is 7-4√3
(Multiplying and dividing by 7-4√3)

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
x = (4-√15)

Question 17.
Solution:
x = 2+√3

Question 18.
Solution:

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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

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

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
(i) Draw a line segment AB = 3.2 units (cm) and extend it to C such that BC = 1 unit.

(ii) Find the mid-point O of AC.
(iii) With centre O and OA as radius draw a semicircle on AC
(iv) Draw BD ⊥ AC meeting the semicircle at D.
(v) Join BD which is √3.2 units.
(vi) With centre B and radius BD, draw an arc meeting AC when produced at E.
Then BE = BD = √3.2 units. Ans.

Question 6.
Solution:
(i) Draw a line segment AB = 7.28 units and produce is to C such that BC = 1 unit (cm)

(ii) Find the mid-point O of AC.
(iii) With centre O and radius OA, draw a semicircle on AC.
(iv) Draw a perpendicular BD at AC meeting the semicircle at D
Then BD = √7.28 units.
(v) With centre B and radius BD, draw an arc which meet AC produced at E.
Then BE = BD = √7.28 units.

Question 7.
Solution:
(A) For Addition
(i) Closure property: The sum of two real numbers is always a real number.
(ii) Associative Law : (a + b) + c = a + (b + c), for all values of a, b and c.
(iii) Commutative Law : a + b = b + a for all real values of a and b.
(iv) Existance of Additive Identity : 0 is the real number such that: 0 + a = a + 0 = afor every real value of a.
(v) Existance of addtive inverse : For each real value of a, there exists a real value (-a) such that a + (-a) = (-a) + a = 0, Then (a) and (-a) are called the additive inverse of each other.
(v) Existence of Multiplicative Inverse. For each non zero real number a, there exists a real number \(\frac { 1 }{ a }\) such that a . \(\frac { 1 }{ a }\) = \(\frac { 1 }{ a }\) . a = 1
a and \(\frac { 1 }{ a }\) are called multiplicative inverse or reciprocal of each other.
(B) Multiplication
(i) Closure property: The product of two real numbers is always a real number.
(ii) Associative law : ab(c) = a(bc) for all real values of a, b and c
(iii) Commutative law : ab=ba for all real numbers a and b
(iv) Existance of Multiplicative Identity: clearly is a real number such that 1.a = a.1 = a for every value of a.

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.