NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2

Ex 1.2 Class 9 Maths Question 1.
State whether the following statements are true or false. Justify your answers.

1. Every irrational number is a real number.
2. Every point on the number line is of the form , where m is a natural number.
3. Every real number is an irrational number.

Solution:
(1) True (∵ Real numbers = Rational numbers + Irrational numbers.)
(2) False (∵ no negative number can be the square root of any natural number.)
(3) False (∵ rational numbers are also present in the set of real numbers.)

Ex 1.2 Class 9 Maths Question 2.
Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Solution:
No, the square roots of all positive integers are not irrational.
e.g., 
Here, ‘4’ is a rational number.

Ex 1.2 Class 9 Maths Question 3.
Show how  can be represented on the number line.
Solution:
We know that, = 


Draw of right angled triangle OQP, such that
OQ = 2 units
PQ= 1 unit
and   ∠OQP = 90°
Now, by using Pythagoras theorem, we have
OP2 =   OQ2 +PQ2=    22 +12= op=   = 
Now, take O as center OP = 45 as radius, draw an arc, which intersects the line at point R.
Hence, the point R represents .

Ex 1.2 Class 9 Maths Question 4.
A classroom activity (constructing the ‘square root spiral’).
Solution:
Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1P2 perpendicular to OP, of unit length (see figure).

Now, draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to 0P3. Continuing in this manner, you can get the line segment Pn-1Pn by drawing a line segment of unit length perpendicular to OPn-1. In this manner, you will have created the points P2, P3, ….,Pn…, and joined them to create a beautiful spiral depicting   ,………..

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NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.1

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.1

Ex 1.1 Class 9 Maths Question 1.
Is zero a rational number? Can you write it in the form , where p and q are integers and q ≠ 0?
Solution:
Yes, zero is a rational number.
Zero can be written in any of the following forms :

Thus, 0 can be written as , where p = 0 and q is any non-zero integer.
Hence, 0 is a rational number.

Ex 1.1 Class 9 Maths Question 2.
Find six rational numbers between 3 and 4.
Solution:
The rational numbers between 3 and 4.


Ex 1.1 Class 9 Maths Question 3.
Find five rational numbers between  and .
Solution:
Since we want 5 rational numbers between  and , so we write

Ex 1.1 Class 9 Maths Question 4.
State whether the following statements are true or false. Give reasons for your answers.

1. Every natural number is a whole number.
2. Every integer is a whole number.
3. Every rational number is a whole number.

Solution:
(1) True: Every natural number lies in the collection of whole numbers.
(2) False: -3 is not a whole number.
(3) False:   is not a whole number.

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4

Ex 1.4 Class 10 Maths Question 1.Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or non-terminating repeating decimal expansion:

Solution:

Ex 1.4 Class 10 Maths Question 2.Write down the decimal expansions of those rational numbers in the question 1, which have terminating decimal expansions.
Solution:

Ex 1.4 Class 10 Maths Question 3.The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational and of the form , what can you say about the prime factors of q ?
(i) 43. 123456789
(ii) 0.120120012000120000…
(iii) 43. 
Solution:(i) 43.123456789

Here, the denominator is of the form 2m5n.
Hence, the number is a rational number, specifically a terminating decimal.
(ii) Since the given decimal number is non­terminating non-repeating, it is not rational
(iii) Since the given decimal number is non­terminating repeating, it is rational, but the denominator is not of the form 2m5n

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

Ex 1.3 Class 10 Maths Question 1.Prove that √5 is irrational.
Solution:Let √5 =  be a rational number, where p and q are co-primes and q ≠ 0.
Then, √5q = p => 5q2=p2⇒  p2 – Sq2     … (i)
Since 5 divides p2, so it will divide p also.
Let p = 5r
Then p2 – 25r 2     [Squaring both sides]
⇒ 5q2 = 25r2     [From(i)]
⇒ q2 = 5r2Since 5 divides q2, so it will divide q also. Thus, 5 is a common factor of both p and q.
This contradicts our assumption that √5 is rational.
Hence, √5 is irrational. Hence, proved.

Ex 1.3 Class 10 Maths Question 2.
Show that 3 + √5 is irrational.
Solution:Let 3 + 2√5 =  be a rational number, where p and q are co-prime and q ≠ 0.
Then, 2√5 =  – 3 = 
⇒ √5 =  
since   is a rational number,
therefore, √5 is a rational number. But, it is a contradiction.
Hence, 3 + √5 is irrational. Hence, proved.

Ex 1.3 Class 10 Maths Question 3.Prove that the following are irrational.

Solution:
(i) Let   =  be a rational number,
where p and q are co-prime and q ≠ 0.
Then, √2 = 
Since  is rational, therefore, √2 is rational.
But, it is a contradiction that √2 is rational, rather it is irrational.
Hence,  is irrational.
Hence, proved.

(ii) Let 7√5 =  be a rational number, where p, q are co-primes and q ≠ 0.
Then, √5 = 
Since  is rational therefore, √5 is rational.
But, it is a contradiction that √5 is rational rather it is irrational.
Hence, 7√5 s is irrational.
Hence proved.

(iii) Let 6 + √2 =  be a rational number, where p, q are co-primes and q ≠ 0.
Then, √2 =  – 6 = 
Since  is rational therefore, √2 is rational.
But, it is a contradiction that √2 is rational, rather it is irrational.
Hence, 6 + √2 is irrational.

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

 Question 1.Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:(i) By prime factorization, we get:

(ii) By prime factorization, we get:

(iii) By prime factorization, we get:

(iv) By prime factorization, we get:

(v) By prime factorization, we get:

 Question 2.Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = Product of the two numbers:
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:(i) By prime factorisation, we get:
26 = 2 x 13
91 = 7 x 13
∴ HCF of 26 and 91 = 13
and LCM of 26 and 91 = 2 x 7 x 13
= 182
Now, HCF x LCM = 182 x 13 = 2366 … (i)
Product of numbers = 26 x 91 = 2366  … (ii)
From (i) and (ii), we get:
HCF x LCM = Product of number
Hence, verified.

(ii) By prime factorisation, we get:
510 = 2 x 3 x 5 x 17
92 = 2 x 2 x 23
∴ HCF of 510 and 92 = 2
and LCM of 510 and 92
= 22 x 3 x 5 x 17 x 23 = 23460
Now, HCF x LCM = 2 x 23460 = 46920 … (i)
Product of numbers
= 510 x 92 = 46920 … (ii)
From (i) and (ii), we get:
LCM x HCF = Product of numbers
Hence, verified.

iii) By prime factorisation, we get:
336 = 2 x 2 x 2 x 2 x 3 x 7
54 = 2 x 3 x 3 x 3
∴ HCF of 336 and 54 = 2 x 3 = 6
and LCM of 336 and 54 = 24 x 33 x 7
= 3024
Now, LCM x HCF = 3024 x 6 = 18144… (i)
Product of numbers
= 336 x 54 = 18144 … (ii)
From (i) and (ii), we get:
LCM x HCF = Product of number
Hence, verified.

Question 3.Find the LCM and HCF of the following integers by applying the prime factorisation method:
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:(i) By prime factorisation, we get:
12 = 2 x 2 x 3
15 = 3 x 5
21 = 3 x 7
∴ HCF of 12, 15 and 21 = 3
and LCM = 2 x 2 x 3 x 5 x 7 = 420.

(ii) By prime factorisation, we get:
17 = 17 x 1
23 = 23 x 1
29 = 29 x 1
∴ HCF of 17, 23 and 29 = 1
and LCM = 17 x 23 x 29 = 11339

(iii) By prime factorisation, we get:
8=2 x 2 x 2 x 1
9 = 3 x 3 x 1
25 = 5 x 5 x 1
∴  HCF of 8, 9 and 25 = 1
and LCM of 8, 9 and 25 = 23 x 32 x 52= 1800.

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

Hence, LCM (306, 657) = 22338.

 Question 5.Check whether 6n can end with the digit 0 for any natural number n.
Solution:6n will end with 0 if 5 is one of the primes of 6.
Prime factors of 6 = 2 and 3.
Since 5 is not a prime factor of 6,
therefore, 6n cannot end with the digit 0.

 Question 6.Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.
Solution:7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x 78
The given number has more than two factors.
Hence, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x 1+5
= 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1)
= 5 x 1009 x 1
The given number has more than two factors.
Hence, it is a composite number.

 Question 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?
Solution:
The minimum time when Sonia and Ravi meet again at starting point will be the LCM of 18 min and 12 min. LCM of 12 and 18 is 36.
Hence, they will meet each other at starting point after 36 min.

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

Ex1.1 Class 10 Maths 

Question 1.
Use Euclid’s Division Algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) By Euclid’s Division Algorithm, we have
225 = 135 x 1 + 90 135
= 90 x 1 + 45 90
= 45 x 2 + 0
∴ HCF (135, 225) = 45.

(ii) By Euclid’s Division Algorithm, we have
38220 = 196 x 195 + 0
196 = 196 x 1 + 0
∴  HCF (38220, 196) = 196.

(iii) By Euclid’s Division Algorithm, we have
867 = 255 x 3 + 102
255 = 102 x 2 + 51
102 = 51 x 2 + 0
∴ HCF (867, 255) = 51.

Question 2.Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let a be a positive odd integer. Also, let q be the quotient and r the remainder after dividing a by 6.
Then, a = 6q + r, where 0 ≤ r < 6.
Putting r = 0, 1, 2, 3, 4, and 5, we get:
a = 6q, a = 6q + 1, a = 6q + 2, a = 6q + 3, a = 6q + 4, a = 6q + 5
But a = 6q, a = 6q + 2 and a = 6q + 4 are even.
Hence, when a is odd, it is of the form 6q + 1, 6q + 3, and 6q + 5 for some integer q.
Hence proved.

 Question 3.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?
Solution:
Let n be the number of columns such that the value of n be maximum and it must divide both the numbers 616 and 32.
Then, n = HCF (616, 32)
By Euclid’s Division Algorithm, we have:
616 = 32 x 19 + 8 32 = 8 x 4 + 0
∴ HCF (616, 32) = 8
i. e., n = 8
Hence, the maximum number of columns is 8.

 Question 4.Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solution:Let a be a positive integer, q be the quotient and r be the remainder.
Dividing a by 3 using the Euclid’s Division Lemma,
we have:
a = 3q + r, where 0 ≤ r < 3
Putting r = 0, 1 and 2, we get:
a = 3q
⇒ a2 = 9q2
= 3 x 3q2
= 3m (Assuming m = q2)
Then, a = 3q + 1
⇒  a2 = (3q + l)2 = 9q2 + 6q + 1
= 3(3q 2 + 2q) + 1
= 3m + 1 (Assuming m = 3q2 + 2q)
Next, a = 3q + 2
⇒ a2 = (3q + 2)2 =9q2 + 12q + 4
= 9q2 + 12q + 3 + 1
= 3(3q2 + 4q + 1) + 1
= 3m + 1.   (Assuming m = 3q2 + 4q+l)
Therefore, the square of any positive integer (say, a2) is always of the form 3m or 3m + 1.
Hence, proved.

 Question 5.Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.
Solution:Let a be a positive integer, q be the quotient and r be the remainder.
Dividing a by 3 using the Euclid’s Division Algorithm, we have,
a = 3q + r, where 0 ≤ r < 3
Putting r = 0, 1 and 2, we get:
a = 3q, a = 3q + 1 and a = 3q + 2
If a = 3q, then a3 = 27q3 = 9(3q3) = 9m. (Assuming m = 3q3.)
If a = 3q + 1, then
a3 = (3q + l)3 = 27q3 + 9q(3q + 1) + 1 = 9(3q3 + 3q2 + q) + 1 = 9m + 1,  (Assuming m = 3q3 + 3q2 + q)
If a = 3q + 2, then a3 = (3q + 2)3= 27q3 + 18q(3q + 2) + (2)3= 9(3q3 + 6q2 + 4q) + 8
= 9m + 8, (Assuming m – 3q3 + 6q2 + 4q)
Hence, a3 is of the form 9m, 9m + 1 or 9m + 8.

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