NCERT Exemplar Class 9 Maths Solutions Chapter 1 Number Systems

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NCERT Exemplar Class 9 Maths Solutions Number Systems

Exercise 1.1: Multiple Choice Questions (MCQs)

Question 1:
Every rational number is
(a) a natural number    (b) an integer
(c) a real number          (d) a whole number
Solution:
(c) Since, real numbers are the combination of rational and irrational numbers.
Hence, every rational number is a real number.

Question 2:
Between two rational numbers
(a) there is no rational number
(b) there is exactly one rational number
(c) there are infinitely many rational numbers
(d) there are only rational numbers and no irrational numbers
Solution:
(c) Between two rational numbers, there are infinitely many rational numbers.

Question 3:
Decimal representation of a rational number cannot be
(a) terminating                             (b) non-terminating
(c) non-terminating repeating  (d) non-terminating non-repeating
Solution:
(d) Decimal representation of a rational number cannot be non-terminating non-repeating because the decimal expansion of rational number is either terminating or non-terminating recurring (repeating).

Question 4:
The product of any two irrational numbers is
(a) always an irrational number    (b) always a rational number
(c) always an integer                        (d) sometimes rational, sometimes irrational
Solution:
(d) We know that, the product of any two irrational numbers is sometimes rational and sometimes irrational.
e.g., √2 x √2 = 2 (rational) and √2 x √3 = √6 (irrational)

Question 5:
The decimal expansion of the number √2 is
(a) a finite decimal                          (b) 1.41421
(c) non-terminating recurring     (d) non-terminating non-recurring
Solution:

The decimal expansion of the number √2 is non-terminating non-recurring. Because √2 is an irrational number.
Also, we know that an irrational number is non-terminating non-recurring.

Question 6:
Which of the following is irrational?

Solution:
(c)

Question 7:
Which of the following is irrational?

Solution:
(d) An irrational number is non-terminating non-recurring which is 0.4014001400014….

Question 8:

Solution:

(c)

Question 9:
The value of 1.999… in the form of p/q, where p and q are integers and

Solution:
(c)

Question 10:
The value of 2√3 + √3 is
(a) 2√6     (b) 6     (c) 3√3     (d) 4√6
Solution:
2√3 + √3 = √3 (2 + 1) = 3√3

Question 11:
√10, √15 is equal to
(a) 6√5    (b) 5√6   (c) 7√5    (d) 10√5
Solution:
(b) √10,√15 = √2.5 √3.5 = √2 √5 √3 √5 = = 5  √6

Question 12:

Solution:

Question 13:

Solution:

Question 14:

Solution:

Question 15:

Solution:

Question 16:

Solution:

Question 17:

Solution:
(c)

Question 18:

Thinking Process
Take the LCM of indices of three irrational numbers. Then, convert all individually in the form whose index will be equal to LCM.
Solution:
(b)

Question 19:

Solution:

Question 20:
The value of (256)^0.16 x (256)^0.09 is
(a) 4      (b) 16           (c) 64             (d) 256.25
Solution:
(a)

Question 21:
Which of the following is equal to X?

Solution:

Exercise 1.2: Short Answer Type Questions

Question 1:
Let x and y be rational and irrational numbers, respectively. Is x+y necessarily an irrational number? Give an example in support of your answer.
Solution:
Yes, (x + y) is necessarily an irrational number.

Question 2:
Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example.
Solution:
No, (xy) is necessarily an irrational only when x ≠0.
Let x be a non-zero rational and y be an irrational. Then, we have to show that xy be an irrational. If possible, let xy be a rational number. Since, quotient of two non-zero rational number is a rational number.
So,(xy/x) is a rational number => y is a rational number.
But, this contradicts the fact that y is an irrational number. Thus, our supposition is wrong. Hence, xy is an irrational number. But, when x = 0, then xy = 0, a rational number.

Question 3:
State whether the following statements are true or false? Justify your answer.

Solution:
(i) False, here √2 is an irrational number and 3 is a rational number, we know that when we divide irrational number by non-zero rational number it will always give an irrational number.
(ii) False, because between two consecutive integers (likel and 2), there does not exist any other integer.
(iii) False, because between any two rational numbers there exist infinitely many rational numbers.
(iv) True, because there are infinitely many numbers which cannot be written in the form p/q, q ≠0. p,q both are integers and these numbers are called irrational numbers.

Question 4:
Classify the following numbers as rational or irrational with justification

Thinking Process
To classify, use the definition a rational number is in the form of p/q, where p and q are integers and q≠0 and otherwise it is an irrational number.
Solution:

Exercise 1.3: Short Answer Type Questions

Question 1:
Find which of the variables x, y, z and u represent rational numbers and which irrational numbers.
(i) x² =5
(ii) y² = 9
(iii) z² = 0.04
(iv) u² = 17/4
Solution:

Question 2:
Find three rational numbers between
(i) -1 and -2           (ii) 0.1 and 0.11
(iii) 5/7 and 6/7   (iv) 1/4 and 1/5
Thinking Process
Use the concept that three rational numbers between x and y are x+ d, x+2d and  x+3d, where d =(y-x)/(n+1)  , x<y and n=3.
Solution:

Question 3:
Insert a rational number and an irrational number between the following
(i) 2 and 3         (ii) 0 and 0.1                   (iii) 1/3 and 1/2
(iv) -2/5 and -1/2      (v) 0.15 and 0.16           (vi) √2 and √3
(vii) 2.357 and 3.121                   (viii) .0001 and .001
(ix) 3.623623 and 0.484848    (x) 3.375289 and 6.375738
Solution:
We know that, there are infinitely many rational and irrational values between any two numbers.
(i) A rational number between 2 and 3 is 2.1.
To find an irrational number between 2 and 3. Find a number which is non-terminating non-recurring lying between them.
Such number will be 2.040040004…………..
(ii) A rational number between 0 and 0.1 is 0.03.
An irrational number between 0 and 0.1 is 0.007000700007……….
(iii) A rational number between 1/3 and 1/2 is 5/12. An irrational number between 1/3 and 1/2 i.e., between 0-3 and 0.5 is 0.4141141114………….
(iv) A rational number between -2/5 and 1/2 is 0. An irrational number between -2/5 and 1/2  i.e., between – 0.4 and 0.5 is 0.151151115………..
(v) A rational number between 0.15 and 0.16 is 0.151. An irrational number between 0.15 and 0.16 is 0.1515515551…….
(vi) A rational number between √2 and √3 i.e.,, between 1.4142…… and 1.7320…… is 1.5.
An irrational number between √2 and √3 is 1.585585558……….
(vii) A rational number between 2.357 and 3.121 is 3. An irrational number between 2.357 and 3.121 is 3.101101110……..
(viii) A rational number between 0.0001 and 0.001 is 0.00011. An irrational number between 0.0001 and 0.001 is 0.0001131331333………..
(ix) A rational number between 3.623623 and 0.484848 is 1. An irrational number between 3.623623 and 0.484848 is 1.909009000……….
(x) A rational number between 6.375289 and 6.375738 is 6.3753. An irrational number between 6.375289 and 6.375738 is 6.375414114111………

Question 4:
Represent the following numbers on the number line 7, 7.2, -3/2 and -12/5 .
Solution:
Firstly, we draw a number line whose mid-point is 0. Marks a positive numbers on right
hand side of 0 and negative numbers on left hand side of 0.

(i) Number 7 is a positive number. So we mark a number 7 on the right hand side of 0, which is a 7 units distance from zero.
(ii) Number 7.2 is a positive number. So, we mark a number 7.2 on the right hand side of 0, which is a 7.2 units distance from zero.
(iii) Number -3/2 or -1.5 is a negative number. So, we mark a number 1.5 on the left hand side of 0, which is a 1.5 units distance from zero.
(iv) Number – 12/5 or -2.4 is a negative number. So, we mark a number 2.4 on the left hand side of 0, which is a 2.4 units distance from zero.

Question 5:
Locate √5, √10 and √17 on the number line.
Solution:

Question 6:
Represent geometrically the following numbers on the number line
(i) √ 4.5     (ii) √5.6
(iii) √ 8.1   (iv) √2.3
Thinking Process
(i) Firstly, we draw a line segment AB of length equal to the number inside the root and extend it to C such that BC=1
(ii) Draw a semi-circle with centre O (O is the mid-point of AC) and radius OA
(iii) Now, draw a perpendicular line from B to cut the semi-circle atD.

(iv) Further, draw an arc with centre B and radius BD meeting AC produced at £
Solution:
(i) Firstly, we draw a line segment AB = 4.5 units and extend it to C such that BC = 1 unit. Let O be the mid-point of AC.
Now, draw a semi-circle with centre O and radius OA Let us draw BD perpendicular to AC passing through point B and intersecting the semi-circle at point D.
Hence, the distance BD is √4.5 units.
Draw an arc with centre B and radius BD, meeting AC produced at E, then BE = BD = √4.5 units.

(ii) Firstly, we draw a line segment AB = 5.6 units and extend it to C such that BC = 1 unit. Let O be the mid-point of AC.
Now, draw a semi-circte with centre O and radius OA. Let us draw BD perpendicular to AC passing through point B and intersecting the semi-circle at point D.
Hence, the distance BD is √5.6 units.
Draw an arc with centre B and radius BD, meeting AC produced at E, then BE = BD = √5.6 units.

(iii) Firstly, we draw a line segment AB = 8.1 units and extend it toC such that SC = 1 unit. Let O be the mid-point of AC.
Now, draw a semi-circle with centre 0 and radius OA
Let us draw BD perpendicular to AC passing through point 6 intersecting the semi-circle at point D:
Hence, the distance BD is √8.1 units .
Draw an arc with centre Sand radius BD, meeting AC produced at E, then BE = BD = √8.1 units.

(iv) Firstly, we draw a line segment AS = 2.3 units and extend it to C such that SC = 1 unit. Let O be the mid-point of AC.
Now, draw a semi-circle with centre 0 and radius OA.
Let us draw BD perpendicular to AC passing through point 6 and intersecting the semi-circle at point D.
Hence, the distance BD is √2.3 units.
Draw an arc with centre 6 and radius BD, meeting AC produced at E, then BE = BD = √2.3 units.

Question 7:

Solution:

Question 8:
Show that 0.142857142857… = 1/7.
Solution:
Let x = 0.142857142857 …………………..(i)
On multiplying both sides of Eq. (i) by 1000000, we get
1000000 x = 142857.142857…………………………(ii)
On subtracting Eq. (i) from Eq. (ii), we get
1000000 x – x = (142857.142857…) – (0.142857..) => 999999 x = 142857
∴ x = 142857/999999 = 1/7  Hence proved.

Question 9:
Simplify the following

Solution:

Question 10:
Rationalise the denominator of the following

Solution:

Question 11:
Find the values of a and b in each of the following

Solution:

Question 12:

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Question 13:
Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 upto three places of decimal.

Solution:

Question 14:
Simplify

Solution:

Exercise 1.4: Long Answer Type Questions

Question 1:

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Question 2:

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Question 3:

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Question 4:

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Question 5:

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Question 6:

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Question 7:

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