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Mathematics Class 9 NCERT SOLUTIONS (Detailed Step-Wise as per CBSE Guidelines) Number Systems π© Question 1. Is zero rational number? Can you write it in the form πͺ , where p and q are integers and q β 0? 0 0 0 p Ans. Yes, zero is a rational number as 0 = 1 or 2 or β1. This is in the form q , q β 0. Question 2. Find six rational numbers between 3 and 4. Ans. For six rational numbers between 3 and 4, 3 = and 4 22 are 7 , 23 7 , 24 7 , 25 7 , 26 7 27 , 7 21 7 and 4 = 28 7 . Six rational numbers between 3 or another set is 3.1, 3.2 , 3.3 , 3.4 , 3.5 , 3.6 , There can be other set of rational numbers also. Question 3. Find five rational numbers between Ans. We have 3 5 18 30 = and 4 5 = 24 30 π π and π π . . Five rational numbers between 3 5 and 4 5 are 19 30 , 20 30 , 21 30 , 22 30 23 , 30 .There can be other set of rational numbers. also . 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. Ans. (i) True , as the set of whole numbers contains all the natural numbers. (ii) Flase, as negative integer , e. g. ., -2 is not a whole number. 2 (iii) Flase , as 3 is a rational number but not a whole number. www.practiceguru.in Question 5 . State whether the following statements are true or false. Justify your answers. (i) Every irrational numbers is a real numbers. (ii) Every point on the number line is of the form π , where m is a natural numbers. (iii) Every real numbers is an irrational number. Ans . (i) true , as real number consist of rational and irrational numbers. (ii) False, as 3 2 on the number line cannot be a squard root of a natural number. Also, a negative, as π represents a positive value. (iii)Flase, as 2 is a real number but not an irrational mumber. Question 6. Are the square roots of all positive integers irrational ? if not , give an example of the square root of a number that is a tational number. Ans . No . for example, 9 =3, 16 = 4 , etc ., 3 , 4 are rational numbers. Question 7. Show how π can be represented on the number line. Ans . for 5 , we have 5 = 5 × 1. On a number line , taken o at position 0 and OA = 5 and OB = 1. With AB as diameter draw a semicircle. Draw OP ! AB meeting semicircle at P . With O as centre and OP radius an are is drawn meeting the number line at Q. Then , OQ = OP = 5 and Q represents 5 on the number line. www.practiceguru.in Question 8 . Classroom activity (Constructing the βsquare root spiralβ); Take a large sheet of paper and construct the βsquare root spiralβ in the following fashion . Start with a point O and draw a line segment OP 1 of unit length. Draw a line segment P1 P2 perpendicular to OP 1 of unit length (se figure). Now draw a line segment p 2 P3 perpendicular op 2 of unit length . Than , draw a line segment P3 P4 perpendicular to op 3 of unit length . Continuing in this maner , you can get the line segment Pn - 1P n by drawing a line segment of unit length perpendicular to OP n β 1. In this manner , you will have created the points P2β P 3 β¦β¦., P n , β¦β¦.., and joined them to create a beautiful spiral depicting 2 , 3 , 4 , β¦β¦ . Ans. Question .9 Write the following in decimal form and say what kind of decimal xansion each has: ππ πππ π ππ (i) (ii) Ans. (i) (ii) (iii) 36 100 (ii) (vi) π ππ πππ πππ π (iii) 4 π (iv) π ππ . = 0.36 , terminating decimal expansion. 1 11 1 48 = 0.090909 β¦. = = 33 8 , non β terminating prating decimal expansion. 0.09 = 4.125 , terminating decimal expansion. www.practiceguru.in 3 = 0.230769230769 β¦.. = 0.230769 , non β terminating repeating decimal expansion. 13 2 =0.181818 β¦.. = 0.18 , non β terminating repeating decimal expansion. 11 329 = 0.8225 , terminating decimal expansion. 400 (iv) (v) (vi) π Question . 10 You know that π = 0.142857. Can you predict what the decimal expansions of π π , π π , π π π π , π π are without actually doing the long division ? If so , how? [ Hint : Study the remainders while funding the value of π π carefully . ] Ans. Yes , we can predict the required decimal expansions . we are given , π π = 0.142857 On dividing 1 by 7, we find that the remainders repeat after six divisions, therefore, the quotient has a repeating 1 π π π π block of six digits in the decimal expansion of 7. So, to obtain decimal expansions of π , π , π , π π and π : we multiply 142857 by 2, 3, 4, 5 and 6 respectively, to get the integral part and in the decimal part, we take block of six repeating digits in each case. Hence, we get π π = 2 x 7 = 0.285714 7 7 1 π π = 3 x 7 = 0.428571 7 7 π π = 4 x 7 = 0. 571428 7 7 π π =5x 1 1 π and π 1 7 = 0.714285 1 = 6 x 7 = 0.857142. Question . 11 Express the form p/q , where p and q are integers and q β 0 : (i) 0.6 Ans. (i) Let x = 0.6 or x = 0.666... 10 x = 6.666β¦ => 9x = 6 (ii) 0.47 (iii) 0.001. β¦.(ii) [On multiplying (i) by 10] [Subtracting (i) form (ii) ] www.practiceguru.in , 2 β΄ x=3. (ii) Let x = 0.47 or x = 0.4777 β¦. Multiplying (i) by 10 , we get 10 x = 4.777 β¦.. Subtracting equation (ii) form equation (iii) ,we get 90 x = 43 β΄ x= (iii) 43 . 90 Let or x = 0.001 x = 0.001001β¦ β¦(i) β 1000 x = 1.001001β¦ β¦(ii) [On multiplying (i) by 1000] β β΄ 999x 1 x = [ On subtracting (i) from (ii)] 1 . 999 π π Question .12 Express 0.99999... in the form . Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense. Ans. Let x = 0.99999β¦. β β 10 x 9.99999β¦. 9 x =9 β΄ β¦ (ii) [On multiplying (i) by 10] [On Subtracting (i) form (ii)] x = 1. Yes, we are surprised by our answer. Question .13 What can the maximum number of digits be in the repeating block lf digits in the decimal expansion of π ? ππ Perform the division to check your answer. Ans . www.practiceguru.in 1 Thus, 17 = 0.0588235294117647 Hence, the required number of digits in the repeating block is 16. Question.14 . Look at several examples of rational numbers in the form π π (q β 0), where p and q are integrs with no common factors other than 1 and having terminating dcimal representations (expansions ). Can you guess whatproperty q must satisfy ? www.practiceguru.in 3 4 7 9 Ans . examples are 4 ,5 , 8 , 10 , etc. q have only powers of 2 or powers of 5 or both . Question .15 Write three numbers whose decimal expansions are non β terminating non β recurring. Ans . (i) 2.01011011101111011111β¦β¦. (ii) 0.03003000300003β¦β¦. (ii) 4.12112111211112β¦.. Question.16 Find three different irrational numbers between the rational numbers Ans Given : 5 7 = 0.714285 and 9 11 π π and π ππ . = 0.81 We can have irrational numbers as 0.72072007200070β¦.; 0.801001800018β¦β¦.; 0.74301010010001β¦β¦; Question 17. Classify the following numbers as rational or irrational: ππ (ii) πππ (iii) 0.3796 (IV) 7.478478β¦.. (iii) π. πππππππππππππππ β¦ β¦ Ans . (i) 23 . As it is square root of a prime number , so , irrational number . (I) (ii) (iii) (iv) (v) 225 = 15 , rational number . 0.3796 , terminating decimal , so rational number . 7.478478β¦..= 7.478 , non β terminating repeating (recurring ) , so rational number. 1.101001000100001β¦.. non β terminating non β repeating,so irrational number. Question 18. Visualize 3.765 on the number line, using successive magnification. Ans . (i) We notice the number 3.7 lies between 3 and 4. So, first we locate numbers 3 and 4 on number line and divide the portion into ten equal parts and locate 3.7 and 3.8. [Refer step 2J (ii) Further 3.76 lies between 3.7 and 3.8. So, we magnify 3.7 and 3.8 and divide the portion into ten equal parts and locate 3.76 and 3.77. [Refer step 3J (iii) Further 3.765 lies between 3.76 and 3.77. So, we magnify 3.76 and 3.77 and divide the portion into ten equal parts and locate 3.765. [Refer step 4) www.practiceguru.in Point O in step 4 reprints the number 3.765 on the number one. Question 19. Visualize 4.26 on the number line, up to 4 decimal places. Ans . 4.26 = 4.2626262626 ......... (H) Visualize 4 and 5 as 4.26 lies between 4 and 5 and divide portion in ten equal parts and locate 4.2. [Refer step 2J (ii) Visualize 4.2 and 4.3 as 4.26 lies between 4.2 and 4.3 and divide portion in ten equal parts and locate 4.26. [Refer step 3] (iii) Visualize 4.26 and 4.27 as 4.262 lies between 4.26 and 4.27 and divide portion in ten equal parts and locate 4.262. [Refer step 41 (iv) Visualize 4.262 and 4.263 as 4.2626 lies between 4.262 and 4.263 and divide portion in ten equal parts and locate 4.2626. [Refer step 5] www.practiceguru.in Point 0 in step 5 represents the number 4.26 on the number line. Question .20 Classify the following numbers as rational or irrational: (i) 2 - π (ii) (3 + ππ)- ππ (iii) π π π π (iv) π π (v) 2π . (i) 2 - π is an irrational number, as difference of a rational and an irrational number is irrational. (ii) (3 + 23)- 23 = 3 + 23 - 23 = 3, is a rational number. Ans . (iii) (iv) 1 2 = 2 7 7 7 2 2 2 = 7, is a rational number. an irrational number, as divisors of an irrational number by a non-zero rational number is irrational. (V) 2π, irrational number, as π is an irrational number and multiplication of a rai >nal and an irrational number is irrational. QUESTION .21 Simplify each of the following expressions: (I) (3 + π) (2 + π ) (ii) (3 + π ) (3 - π ) (II) ( π + π )2 (iv) ( π - π) ( π + π) ANS. (I) (3 + 3)(2 + 2 ) =6 +3 2 + 2 3 + 6 . www.practiceguru.in (II) (3 + 3 ) (3 - 3 ) = (3)2 - ( 3)2 = 9 β 3 = 6. (III) ( 5 + 2 )2 = ( 5)2 + 2. 5 . 2 +( 2 )2 =5 +2 10 + 2 =7 +2 10 . 5 - 2) ( 5 + 2) = ( 5 )2 β( 2 )2 = 5 β 2 =3. (IV) Question. 22 Recall, π is defined as the ratio of the circumference (sayc) of a circle to its diameter π (say d). That is, π = π .This seems to contradict the fact that k is irrational. How will you resolve this contradiction? Ans .On measuring c with any device, we get only approximate measurement. Therefore, n is an irrational. Question . 23 Represent π. π on the number line. Ans . 9.3 = 9.3 × 1 Let position 0 be represented by O on the number line. Let OA = 9.3 and OB = 1. With AB as diameter draw a semicircle. Draw OP perpendicular to AB, meeting the semicircle at P. Then OP = 9.3 . With O as centre and OP as radius draw an arc to meet the number line at Q on the positive side. Then, OQ = 9.3 and the point Q thus obtained represents 9.3 . Question .24 Rationalise the denominators of the following: (i) π π Ans . (i) (ii) (iii) (iv) (ii) 1 7 1 1 7 = 7β 6 1 π (iii) π+ π (iv) π . πβπ 7 . 7 = = 7+ 6 7β6 = 7 + 6 . ( 7 β 6 )( 7+ 6) = 5+ 2 1 7β2 = π πβ π 7 × 7 7+ 6 5β 2 = 5β 2 5β2 5β 2 3 = . ( 5+ 2 )( 5β 2) = 7 +2 7 β2 ( 7 +2) 7 +2 7β4 = Question 25. Find: (i) 641/2 (ii) 321/5 (iii) 125 1/3. Ans. www.practiceguru.in = 7 +2 . 3 Question 26. Find: (i) 93/2 Ans. (ii) 322/5 (iii) 163/4 (iv) 125-1/3 Question 27. Ans. www.practiceguru.in