{"id":41034,"date":"2019-06-07T08:46:55","date_gmt":"2019-06-07T08:46:55","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=41034"},"modified":"2019-06-07T08:46:55","modified_gmt":"2019-06-07T08:46:55","slug":"ncert-solutions-for-class-9-maths-chapter-1","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ncert-solutions-for-class-9-maths-chapter-1\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 1 Number Systems"},"content":{"rendered":"

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems are part of\u00a0NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 1 Number Systems.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems<\/h2>\n

NCERT Solutions for Class 9 Maths Chapter 1\u00a0Number Systems Ex 1.1<\/h3>\n

Question 1.<\/strong><\/span>
\nIs zero a rational number? Can you write it in the form \\(\\frac { p }{ q }\\),where p and q are integers and q \u22600?<\/strong>
\nSolution:<\/strong><\/span>
\nYes, write \\(\\frac { 0 }{ 1 }\\)(where 0 and 1 are integers and q = 1 which is not equal to zero).<\/p>\n

Question 2.<\/strong><\/span>
\nFind six rational numbers between 3 and 4.<\/strong>
\nSolution:<\/strong><\/span>
\nThere can be infinitely many rationals between 3 and 4, one way is
\n\"NCERT<\/p>\n

Question 3.<\/strong><\/span>
\nFind five rational numbers between \\(\\frac { 3 }{ 5 }\\) and \\(\\frac { 4 }{ 5 }\\).<\/strong>
\nSolution:<\/strong><\/span>
\n\"NCERT
\n\"NCERT<\/p>\n

Question 4.<\/strong><\/span>
\nState whether the following statements are true or false. Give reasons for your answers. ‘<\/strong>
\n(i) Every natural number is a whole number.<\/strong>
\n(ii) Every integer is a whole number.<\/strong>
\n(iii) Every rational number is a whole number.<\/strong>
\nSolution:<\/strong><\/span>
\n(i) True, because natural numbers are 1, 2, 3, 4 ……….\u221e and whole
\nnumbers are 0,1, 2, 3, 4, 5, ……..,\u221e
\nor
\nThe collection of whole numbers contain all the natural numbers.
\n(ii) False (\u2235 negative integers are not included in the list of whole numbers.)
\n(iii) False [\u2235 \\(\\frac { 1 }{ 3 }\\), \\(\\frac { 6 }{ 7 }\\), \\(\\frac { 10 }{ 19 }\\) )are not whole numbers.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 1\u00a0Number Systems Ex 1.2<\/h3>\n

Question 1.<\/strong><\/span>
\nState whether the following statements are true or false. Justify your answers.<\/strong>
\n(i) Every irrational number is a real number.<\/strong>
\n(ii) Every point on the number line is of the form \u221am , where m is a natural number.<\/strong>
\n(iii) Every real number is an irrational number.<\/strong>
\nSolution:<\/strong><\/span>
\n(i) True (\u2235 Real numbers = Rational numbers + Irrational numbers.)
\n(ii) False (\u2235 no negative number can be the square root of any natural number.)
\n(iii) False (\u2235 rational numbers are also present in the set of real numbers.)<\/p>\n

Question 2.<\/strong><\/span>
\nAre 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.<\/strong>
\nSolution:<\/strong><\/span>
\nNo, the square roots of all positive integers are not irrational.
\ne.g., \u221al6 = 4
\nHere, \u20184\u2019 is a rational number.<\/p>\n

Question 3.<\/strong><\/span>
\nShow how \u221a5 can be represented on the number line.<\/strong>
\nSolution:<\/strong><\/span>
\n\"NCERT
\nNow, take O as centre OP = \u221a5 as radius, draw an arc, which intersects the line at point R. .
\nHence, the point R represents \u221a5.<\/p>\n

Question 4.<\/strong><\/span>
\nClassroom activity (constructing the ‘square root spiral’).<\/strong>
\nSolution:<\/strong><\/span>
\nTake a large sheet of paper and construct the \u2018square root spiral\u2019 in the following fashion. Start with a point O and draw a line segment OP1<\/sub>, of unit lengths Draw a line segment P1<\/sub>, P2<\/sub>\u00a0perpendicular to OP1<\/sub>\u00a0of unit length (see figure).
\n\"NCERT
\nNow, draw a line segment P2<\/sub>P3<\/sub>\u00a0perpendicular to OP2<\/sub>. Then draw a line segment P3<\/sub>P4<\/sub>\u00a0perpendicular to OP3<\/sub>. Continuing in this manner, you can get the line segment Pn-1<\/sub>\u00a0Pn<\/sub>\u00a0by drawing a line segment of unit length perpendicular to
\nOPn-1<\/sub>. In this manner, you will have created the points P2<\/sub>, P3<\/sub>,…… Pn<\/sub>,….. and
\njoined them to create a beautiful spiral depicting \u221a2,\u221a3,\u221a4,……<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 1\u00a0Number Systems Ex 1.3<\/h3>\n

Question 1.<\/strong><\/span>
\nWrite the following in decimal form and say what kind of decimal expansion each has<\/strong>
\n\"NCERT
\nSolution:<\/strong><\/span>
\n\"NCERT
\n\"NCERT
\n\"NCERT<\/p>\n

Question 2.<\/strong><\/span>
\nYou know that \\(\\frac { 1 }{ 7 }\\) = \\(\\bar { 0.142857 }\\). Can you predict what the decimal expansions of \\(\\frac { 2 }{ 7 }\\) , \\(\\frac { 13 }{ 7 }\\) , \\(\\frac { 4 }{ 7 }\\) , \\(\\frac { 5 }{ 7 }\\) , \\(\\frac { 6 }{ 7 }\\) are , without actually doing the long division? If so, how?<\/strong>
\nSolution:<\/strong><\/span>
\n\"NCERT<\/p>\n

Question 3.<\/strong><\/span>
\nExpress the following in the form \\(\\frac { p }{ q }\\)where p and q are integers and q \u2260 0.<\/strong>
\n(i) 0.\\(\\bar { 6 }\\)<\/strong>
\n(ii) 0.4\\(\\bar { 7 }\\)<\/strong>
\n(iii) 0.\\(\\overline { 001 }\\)<\/strong>
\nSolution:<\/strong><\/span>
\n(i)<\/strong>Let x= 0.\\(\\bar { 6 }\\) = 0.666… ….(i)
\nMultiplying Eq. (i) by 10, we get
\n10x = 6.666.. ….(ii)
\nOn subtracting Eq. (ii) from Eq. (i), we get
\n(10x- x)=(6.666…) – (0.666…)
\n9x = 6
\nx= 6\/9
\n\u21d2 x=2\/3
\n
\n(ii)<\/strong>\u00a0Let x = 0.4\\(\\bar { 7 }\\) = 0.4777… …(iii)
\nMultiplying Eq. (iii) by 10. we get
\n10x = 4.777… . …(iv)
\nMultiptying Eq. (iv) by 10, we get
\n100x = 47.777 ….. (v)
\nOn subtracting Eq. (v) from Eq. (iv), we get
\n(100 x – 10x)=(47.777….)-(4.777…)
\n90x =43
\n\u21d2 x = \\(\\frac { 43 }{ 90 }\\)
\n
\n(iii)<\/strong>\u00a0Let x = 0.\\(\\overline { 001 }\\)= 0.001001001… …(vI)
\nMultiplying Eq. (vi) by (1000), we get
\n1000x = 1.001001001… .. .(vii)
\nOn subtracting Eq. (vii) by Eq. (vi), we get
\n(1000x\u2014x)=(1.001001001….) – (0.001001001……)
\n999x = 1
\n\u21d2 x = \\(\\frac { 1 }{ 999 }\\)<\/p>\n

Question 4.<\/strong><\/span>
\nExpress 0.99999… in the form \\(\\frac { p }{ q }\\)Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.<\/strong>
\nSolution:<\/strong><\/span>
\nLet x = 0.99999… ………..(i)
\nMultiplying Eq. (i) by 10, we get
\n10x = 9.99999… …(ii)
\nOn subtracting Eq. (ii) by Eq. (i), we get
\n(10 x – x) = (9.99999..) – (0.99999…)
\n9x = 9
\n\u21d2 x = \\(\\frac { 9 }{ 9 }\\)
\nx = 1<\/p>\n

Question 5.<\/strong><\/span>
\nWhat can the maximum number of digits be in the repeating block of digits in the decimal expansion of \\(\\frac { 1 }{ 17 }\\)? Perform the division to check your answer.<\/strong>
\nSolution:<\/strong><\/span>
\nThe maximum number of digits in the repeating block of digits in the decimal expansion of \\(\\frac { 1 }{ 17 }\\) is 17-1 = 16 we have,
\n\"NCERT
\nThus,\\(\\frac { 1 }{ 17 }\\) = 0.\\(\\overline { 0588235294117647….., }\\)a block of 16-digits is repeated.<\/p>\n

Question 6.<\/strong><\/span>
\nLook at several examples of rational numbers in the form \\(\\frac { p }{ q }\\) (q\u2260 0). Where, p and q are integers with no common factors other that 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?<\/strong>
\nSolution:<\/strong><\/span>
\nConsider many rational numbers in the form \\(\\frac { p }{ q }\\) (q\u2260 0). where p and q are integers with no common factors other that 1 and having terminating decimal representations.
\nLet the various such rational numbers be \\(\\frac { 1 }{ 2 }\\), \\(\\frac { 1 }{ 4 }\\), \\(\\frac { 5 }{ 8 }\\), \\(\\frac { 36 }{ 25 }\\), \\(\\frac { 7 }{ 125 }\\), \\(\\frac { 19 }{ 20 }\\), \\(\\frac { 29 }{ 16 }\\) etc.
\nIn all cases, we think of the natural number which when multiplied by their respective denominators gives 10 or a power of 10.
\n\"NCERT
\nFrom the above, we find that the decimal expansion of above numbers are terminating. Along with we see that the denominator of above numbers are in the form 2m x 5n, where m and n are natural numbers. So, the decimal representation of rational numbers can be represented as a terminating decimal.<\/p>\n

Question 7.<\/strong><\/span>
\nWrite three numbers whose decimal expansions are non-terminating non-recurring.<\/strong>
\nSolution:<\/strong><\/span>
\n0.74074007400074000074…
\n0.6650665006650006650000…
\n0.70700700070000…<\/p>\n

Question 8.<\/strong><\/span>
\nFind three different irrational numbers between the rational numbers \\(\\frac { 5 }{ 7 }\\) and \\(\\frac { 9 }{ 11 }\\) .<\/strong>
\nSolution:<\/strong><\/span>
\nTo find irrational numbers, firstly we shall divide 5 by 7 and 9 by 11,
\nso,
\n\"NCERT<\/p>\n

Question 9.<\/strong><\/span>
\nClassify the following numbers as rational or irrational<\/strong>
\n\"NCERT
\nSolution:<\/strong><\/span>
\n(i) \\( \\sqrt{23} \\) (irrational \u2235 it is not a perfect square.)
\n(ii) \\( \\sqrt{225} \\) = 15 (rational) (whole number.)
\n(iii) 0.3796 = rational (terminating.)
\n(iv) 7.478478… =7.\\(\\bar { 478 }\\) = rational (non-terminating repeating.)
\n(v) 1.101001000100001… = irrational (non-terminating non-repeating.)<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 1\u00a0Number Systems Ex 1.4<\/h3>\n

Question 1.<\/strong><\/span>
\nVisualise 3.765 on the number line, using successive magnification.<\/strong>
\nSolution:<\/strong><\/span>
\nWe know that, 3.765 lies between 3 and 4. So, let us divide the part of the number line between 3 and 4 into 10 equal parts and look at the portion between 3.7 and 3.8 through a magnifying glass. Now 3.765 lies between 3.7 and 3.8 [Fig. (i)]. Now, we imagine to divide this again into ten equal parts. The first mark will represent 3.71, the next 3.72 and soon. To see this clearly,we magnify this as shown in [Fig. (ii)].<\/p>\n

Again 3.765 lies between 3.76 and 3.77 [Fig. (ii)]. So, let us focus on this portion of the number line [Fig. (iii)] and imagine to divide it again into ten equal parts [Fig. (iii)]. Here, we can visualise that 3.761 is the first mark and 3.765 is the 5th mark .in these subdivisions. We call this process of visualisation of representation of numbers on the number line through a magnifying glass as the process of successive magnification.<\/p>\n

So, we get seen that it is possible by sufficient successive magnifications of visualise the position (or representation) of a real number with a terminating decimal expansion on the number line.
\n\"NCERT<\/p>\n

Question 2.<\/strong><\/span>
\nVisualise 4.\\(\\bar { 26 }\\) on the number line, upto 4 decimal places.<\/strong>
\nSolution:<\/strong><\/span>
\nWe adopt process by successive magnification and successively decreasethe lengths of the portion of the number line in which 4.\\(\\bar { 26 }\\) is located. Since 4.\\(\\bar { 26 }\\) is located between 4 and 5 and is divided into 10 equal parts [Fig. (i)]. In further, we locate 4.26between 4.2 and 4.3 [Fig. (ii)].<\/p>\n

To get more accurate visualisation of the representation, we divide this portion into 10 equal parts and use a magnifying glass to visualise that 4.\\(\\bar { 26 }\\) lies between 4.26 and 4.27. To visualise 4.\\(\\bar { 26 }\\) more clearly we divide again between 4.26 and 4.27 into 10 equal parts and visualise the repsentation of 4.\\(\\bar { 26 }\\) between 4.262 and 4.263 [Fig. (iii)].<\/p>\n

Now, for a much better visualisation between 4.262 and 4.263 is agin divided into 10 equal parts [Fig. (iv)]. Notice that 4.\\(\\bar { 26 }\\) is located closer to 4.263 then to 4.262 at 4.2627.
\n\"NCERT<\/p>\n

Remark:<\/strong>\u00a0We can adopt the process endlessly in this manner and simultaneously imagining the decrease in the length of the number line in which 4.\\(\\bar { 26 }\\) is located.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 1\u00a0Number Systems Ex 1.5<\/h3>\n

Question 1.<\/strong><\/span>
\nClassify the following numbers as rational or irrational.<\/strong>
\n\"NCERT
\nSolution:<\/strong><\/span>
\n(i)<\/strong>\u00a0Irrational \u2235 2 is a rational number and \u221a5 is an irrational number.
\n\u2234 2.\u221a5 is an irrational number.
\n(\u2235The difference of a rational number and an irrational number is irrational)
\n(ii)<\/strong>\u00a03 + \\( \\sqrt{23} \\) – \\( \\sqrt{23} \\) = 3 (rational)
\n(iii)<\/strong>\u00a0\\(\\frac { 2\\sqrt { 7 } }{ 7\\sqrt { 7 } }\\) (rational)
\n(iv)<\/strong>\u00a0\\(\\frac { 1 }{ \\sqrt { 2 } }\\)(irrational) \u2235 1 \u2260 0 is a rational number and \\( \\sqrt{2} \\)\u2260 0 is an irrational number.
\n\u2234 \\(\\frac { 1 }{ \\sqrt { 2 } }\\) is an irrational number. 42
\n(\u2235 The quotient of a non-zero rational number with an irrational number is irrational).
\n(v)<\/strong>\u00a02\u03c0 (irrational) \u2235 2 is a rational number and \u03c0 is an irrational number.
\n\u2234 2x is an irrational number. ( \u2235The product of a non-zero rational number with an irrational number is an irrational)<\/p>\n

Question 2.<\/strong><\/span>
\nSimplify each of the following expressions<\/strong>
\n\"NCERT
\nSolution:<\/strong><\/span>
\n\"NCERT<\/p>\n

Question 3.<\/strong><\/span>
\nRecall, \u03c0 is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is \u03c0 = \\(\\frac { c }{ d }\\). This seems to contradict the fact that n is irrational. How will you resolve this contradiction?<\/strong>
\nSolution:<\/strong><\/span>
\nActually \\(\\frac { c }{ d }\\) = \\(\\frac { 22 }{ 7 }\\),which is an approximate value of \u03c0.<\/p>\n

Question 4.<\/strong><\/span>
\nRepresent \\( \\sqrt{9.3} \\) on the number line.<\/strong>
\nSolution:<\/strong><\/span>
\nFirstly we draw AB = 9.3 units. Now, from S, mark a distance of 1 unit. Let this point be C. Let O be the mid-point of AC. Now, draw aemi-circle with centre O and radius OA. Let us draw a line perpendicular to AC passing through point B and intersecting the semi-circle at point D.
\n\"NCERT
\n\u2234 The distance BD = \\( \\sqrt{9.3} \\)
\nDraw an arc with centre B and radius BD, which intersects the number line at point E, then the point E represents \\( \\sqrt{9.3} \\) .<\/p>\n

Question 5.<\/strong><\/span>
\nRationalise the denominator of the following<\/strong>
\n\"NCERT
\nSolution:<\/strong><\/span>
\n\"NCERT<\/a><\/p>\n

NCERT Solutions for Class 9 Maths Chapter 1\u00a0Number Systems Ex 1.6<\/h3>\n

Question 1.<\/strong><\/span>
\nFind:<\/strong>
\n\"NCERT
\nSolution:<\/strong><\/span><\/p>\n

\"NCERT<\/p>\n

Question 2.<\/strong><\/span>
\nFind:<\/strong>
\n\"NCERT
\nSolution:<\/strong><\/span>
\n\"NCERT<\/p>\n

Question 3.<\/strong><\/span>
\nSimplify:<\/strong>
\n\"NCERT
\nSolution:<\/strong><\/span><\/p>\n

\"NCERT<\/p>\n

We hope the NCERT Solutions for Class 9 Maths Chapter 1 Number Systems, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 1 Number Systems, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems are part of\u00a0NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 1 Number Systems. NCERT Solutions for Class 9 Maths Chapter 1 Number Systems NCERT Solutions for Class 9 Maths Chapter 1\u00a0Number Systems Ex 1.1 Question 1. […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[6805],"tags":[55431],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 1 Number Systems<\/title>\n<meta name=\"description\" content=\"Get FREE NCERT Solutions for Class 9 Maths Chapter 1 Number Systems. 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