{"id":31494,"date":"2026-05-11T16:45:23","date_gmt":"2026-05-11T11:15:23","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31494"},"modified":"2026-05-11T16:46:23","modified_gmt":"2026-05-11T11:16:23","slug":"introduction-to-linear-polynomials-ncert-solutions-class-9-ganita-manjari","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/introduction-to-linear-polynomials-ncert-solutions-class-9-ganita-manjari\/","title":{"rendered":"Introduction to Linear Polynomials\u00a0&#8211; NCERT Solutions Class 9 Ganita Manjari"},"content":{"rendered":"\n<p>Introduction to Linear Polynomials &#8211; NCERT Solutions Class 9 Maths Ganita Manjari includes all the questions with solutions given in NCERT Class 9 <a href=\"https:\/\/ncert.nic.in\/textbook.php?iemh1=0-8\">Ganita Manjari<\/a> textbook.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">NCERT Solutions Class 9<\/h2>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-english-kaveri\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >English Kaveri<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-hindi-ganga\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Hindi Ganga<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-sanskrit-sharada\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Sanskrit Sharada<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-maths-ganita-manjari\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Maths Ganita Manjari<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-science-exploration\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Science Exploration<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-social-understanding-society\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Social Understanding Society<\/a>\n\n\n\n<h2 class=\"wp-block-heading\">Introduction to Linear Polynomials \u2013 NCERT Solutions<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.1:<\/p>\n\n\n\n<p>Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Perimeter of a square {tex}=4 \\times{\/tex} side<br>Now calculate:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Side = {tex}1 {~cm} \\rightarrow{\/tex} Perimeter = {tex}4 \\times 1=4 {~cm}{\/tex}<\/li>\n\n\n\n<li>Side {tex}=1.5 {~cm} \\rightarrow{\/tex} Perimeter {tex}=4 \\times 1.5=6 {~cm}{\/tex}<\/li>\n\n\n\n<li>Side {tex}=2 {~cm} \\rightarrow{\/tex} Perimeter {tex}=4 \\times 2=8 {~cm}{\/tex}<\/li>\n\n\n\n<li>Side {tex}=2.5 {~cm} \\rightarrow{\/tex} Perimeter {tex}=4 \\times 2.5=10 {~cm}{\/tex}<\/li>\n\n\n\n<li>Side {tex}=3 {~cm} \\rightarrow{\/tex} Perimeter {tex}=4 \\times 3=12 {~cm}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>So, the perimeters are:<\/p>\n\n\n\n<p>{tex} 4 {~cm}, 6 {~cm}, 8 {~cm}, 10 {~cm}, 12 {~cm} {\/tex}<br>Observation:<br>Each time the side increases by 0.5 cm, the perimeter increases by 2 cm.<br>If the side increases by 0.5 cm, the perimeter increases by a constant value of 2 cm (linear pattern).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.2:<\/p>\n\n\n\n<p>Using the expression 2n &#8211; 1, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given expression for number of tiles:<\/p>\n\n\n\n<p>{tex} 2 n-1 {\/tex}<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Number of tiles:<br>For 15th stage: {tex} 2(15)-1=30-1=29 {\/tex} For 26th stage: {tex} 2(26)-1=52-1=51 {\/tex}<\/li>\n\n\n\n<li>Find stage number:<br>For 21 tiles: {tex} 2 n-1=21 {\/tex}<br>{tex} 2 n=22 {\/tex}<br>{tex} n=11 {\/tex} For 47 tiles: {tex} 2 n-1=47 {\/tex}<br>{tex} 2 n=48 {\/tex}<br>{tex} n=24 {\/tex}<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.3:<\/p>\n\n\n\n<p>Differentiate between the graphs of the equations y = 3x + 1, and y = -3x + 1.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given equations:{tex} y=3 x+1 \\text { and } y=-3 x+1 {\/tex}<br>Difference:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Slope:<br>In {tex}y=3 x+1{\/tex}, slope {tex}=3{\/tex} (positive) \u2192 line rises from left to right (linear growth)<br>In {tex}y=-3 x+1{\/tex}, slope {tex}=-3{\/tex} (negative) \u2192 line falls from left to right (linear decay)<\/li>\n\n\n\n<li>Direction:<br>{tex}y=3 x+1{\/tex}: increasing straight line<br>{tex}y=-3 x+1{\/tex}: decreasing straight line<\/li>\n\n\n\n<li>{tex}y{\/tex}-intercept:<br>Both have same y-intercept {tex}=1{\/tex} Both lines cut the y-axis at point {tex}(0,1){\/tex}<\/li>\n<\/ol>\n\n\n\n<p>Both lines pass through the same point (0, 1), but one slopes upward (positive slope) and the other slopes downward (negative slope).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.4:<\/p>\n\n\n\n<p>Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes.<br>From the graphs, we can conclude:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>When {tex}a{\/tex} is fixed, the slope remains the same, so all lines have the same inclination.<\/li>\n\n\n\n<li>When {tex}b{\/tex} varies, the y -intercept changes, so the lines shift up or down.<\/li>\n\n\n\n<li>Therefore, all such lines are parallel to each other.<\/li>\n<\/ul>\n\n\n\n<p>Conclusion:<br>For the equation {tex}y=a x+b{\/tex}, if {tex}a{\/tex} is fixed and {tex}b{\/tex} varies, the lines are parallel and only their position changes (they cut the {tex}y{\/tex}-axis at different points).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.5:<\/p>\n\n\n\n<p>Find the degrees of the following polynomials:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>2x<sup>2<\/sup>\u00a0&#8211; 5x + 3<\/li>\n\n\n\n<li>y<sup>3<\/sup>\u00a0+ 2y &#8211; 1<\/li>\n\n\n\n<li>-9<\/li>\n\n\n\n<li>4z &#8211; 3<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Degree of a polynomial is the highest power of the variable.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>2x<sup>2<\/sup>\u00a0&#8211; 5x + 3<br>Highest power of x = 2<br>\u2234 Degree = 2<\/li>\n\n\n\n<li>y<sup>3<\/sup>\u00a0+ 2y &#8211; 1<br>Highest power of y = 3<br>\u2234 Degree = 3<\/li>\n\n\n\n<li>-9<br>This is a constant polynomial (no variable)<br>\u2234 Degree = 0<\/li>\n\n\n\n<li>4z &#8211; 3<br>Highest power of z = 1<br>\u2234 Degree = 1<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.6:<\/p>\n\n\n\n<p>Write polynomials of degrees 1, 2 and 3.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>A polynomial of degree 1 (linear polynomial):<br>Example: 2x + 5<\/p>\n\n\n\n<p>A polynomial of degree 2 (quadratic polynomial):<br>Example: x<sup>2<\/sup>&nbsp;+ 3x + 1<\/p>\n\n\n\n<p>A polynomial of degree 3 (cubic polynomial):<br>Example: x<sup>3<\/sup>&nbsp;&#8211; 2x<sup>2<\/sup>&nbsp;+ x + 4<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.7:<\/p>\n\n\n\n<p>What are the coefficients of x<sup>2<\/sup>&nbsp;and x<sup>3<\/sup>&nbsp;in the polynomial x<sup>4<\/sup>&nbsp;&#8211; 3x<sup>3<\/sup>&nbsp;+ 6x<sup>2<\/sup>&nbsp;&#8211; 2x + 7?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given polynomial: x<sup>4<\/sup>&nbsp;&#8211; 3x<sup>3<\/sup>&nbsp;+ 6x<sup>2<\/sup>&nbsp;&#8211; 2x + 7<br>Coefficient of x<sup>3<\/sup>&nbsp;= -3<br>Coefficient of x<sup>2<\/sup>&nbsp;= 6<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.8:<\/p>\n\n\n\n<p>What is the coefficient of z in the polynomial 4z<sup>3<\/sup>&nbsp;+ 5z<sup>2<\/sup>&nbsp;&#8211; 11?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The given polynomial is 4z<sup>3<\/sup>&nbsp;+ 5z<sup>2<\/sup>&nbsp;&#8211; 11.<br>There is no term containing z<sup>1<\/sup>&nbsp;(i.e., z).<br>Hence, the coefficient of z is 0.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.9:<\/p>\n\n\n\n<p>What is the constant term of the polynomial 9x<sup>3<\/sup>&nbsp;+ 5x<sup>2<\/sup>&nbsp;\u2212 8x \u2212 10?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The constant term is the term without any variable.<br>In the polynomial 9x<sup>3<\/sup>&nbsp;+ 5x<sup>2<\/sup>&nbsp;&#8211; 8x &#8211; 10, the constant term is -10.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.10:<\/p>\n\n\n\n<p>Find the value of the linear polynomial 5x &#8211; 3 if:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>x = 0<\/li>\n\n\n\n<li>x = -1<\/li>\n\n\n\n<li>x = 2<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given polynomial: 5x &#8211; 3<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>x = 0<br>= 5(0) &#8211; 3 = -3<\/li>\n\n\n\n<li>x = -1<br>= 5(-1) &#8211; 3 = -5 &#8211; 3 = -8<\/li>\n\n\n\n<li>x = 2<br>= 5(2) &#8211; 3 = 10 &#8211; 3 = 7<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.11:<\/p>\n\n\n\n<p>Find the value of the quadratic polynomial 7s<sup>2<\/sup>&nbsp;&#8211; 4s + 6 if:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>s = 0<\/li>\n\n\n\n<li>s = -3<\/li>\n\n\n\n<li>s = 4<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>For s = 0<br>7s<sup>2<\/sup>\u00a0&#8211; 4s + 6<br>= 7(0)<sup>2<\/sup>\u00a0&#8211; 4(0) + 6 = 6<\/li>\n\n\n\n<li>For s = -3<br>7s<sup>2<\/sup>\u00a0&#8211; 4s + 6<br>= 7(-3)<sup>2<\/sup>\u00a0&#8211; 4(-3) + 6<br>= 7(9) + 12 + 6<br>= 63 + 12 + 6 = 81<\/li>\n\n\n\n<li>For s = 4<br>7s<sup>2<\/sup>\u00a0&#8211; 4s + 6<br>= 7(4)<sup>2<\/sup>\u00a0&#8211; 4(4) + 6<br>= 7(16) &#8211; 16 + 6<br>= 112 &#8211; 16 + 6 = 102<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.12:<\/p>\n\n\n\n<p>The present age of Salil\u2019s mother is three times Salil\u2019s present age. After 5 years, their ages will add up to 70 years. Find their present ages.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let Salil\u2019s present age = x years<br>Mother\u2019s present age = 3x years<\/p>\n\n\n\n<p>After 5 years:<br>Salil\u2019s age = x + 5<br>Mother\u2019s age = 3x + 5<\/p>\n\n\n\n<p>According to question: (x + 5) + (3x + 5) = 70<br>\u21d2 4x + 10 = 70<br>\u21d2 4x = 60<br>\u21d2 x = 15<br>Therefore, Salil\u2019s age = 15 years<br>Mother\u2019s age = 45 years<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.13:<\/p>\n\n\n\n<p>The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the integers be 2x and 5x<br>Difference: 5x &#8211; 2x = 63<br>\u21d2 3x = 63<br>\u21d2 x = 21<\/p>\n\n\n\n<p>Integers: 2x = 42<br>\u21d2 5x = 105<br>Required integers are 42 and 105.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.14:<\/p>\n\n\n\n<p>Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total \u20b988, how many coins does she have of each type?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the number of five-rupee coins = x<br>Number of two-rupee coins = 3x<\/p>\n\n\n\n<p>Total value: 5x + 2(3x) = 88<br>\u21d2 5x + 6x = 88<br>\u21d2 11x = 88<br>\u21d2 x = 8<br>Hence:<br>Five-rupee coins = 8<br>Two-rupee coins = 24<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.15:<\/p>\n\n\n\n<p>A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let shorter piece = x feet<br>Longer piece = 4x feet<\/p>\n\n\n\n<p>Total: x + 4x = 300<br>\u21d2 5x = 300<br>\u21d2 x = 60<\/p>\n\n\n\n<p>Therefore:<br>Shorter piece = 60 feet<br>Longer piece = 240 feet.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.16:<\/p>\n\n\n\n<p>If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let width = x cm<br>Length = 2x + 3 cm<\/p>\n\n\n\n<p>Perimeter = 2(length + width)<br>\u21d2 2[(2x + 3) + x] = 24<br>\u21d2 2(3x + 3) = 24<br>\u21d2 6x + 6 = 24<br>\u21d2 6x = 18<br>\u21d2 x = 3<\/p>\n\n\n\n<p>Therefore:<br>Width = 3 cm<br>Length = 2(3) + 3 = 9 cm<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.17:<\/p>\n\n\n\n<p>A student has \u20b9500 in her savings bank account. She gets \u20b9150 every month as pocket money. How much money will she have at the end of every month from the second month onwards?&nbsp;Find a linear expression to represent the amount she will have in the n<sup>th<\/sup>&nbsp;month.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Initial amount in bank = \u20b9500<br>Monthly pocket money = \u20b9150<\/p>\n\n\n\n<p>At the end of:<br>2nd month = 500 + 2(150) = \u20b9800<br>3rd month = 500 + 3(150) = \u20b9950<br>4th month = 500 + 4(150) = \u20b91100<br>and so on\u2026<\/p>\n\n\n\n<p>Let the amount in the n\u1d57\u02b0 month be A<sub>n<\/sub><br>Then, A<sub>n<\/sub> = 500 + 150n<br>Thus, the required linear expression is:<br>A<sub>n<\/sub> = 150n + 500<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.18:<\/p>\n\n\n\n<p>A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3, \u2026 hours? Find a linear expression to represent the number of members at the end of the n<sup>th<\/sup> hour.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Initial members = 120<br>Members leaving each hour = 9<br>After:<br>1 hour = 120 &#8211; 9 = 111<br>2 hours = 120 &#8211; 18 = 102<br>3 hours = 120 &#8211; 27 = 93<\/p>\n\n\n\n<p>Let number of members after n hours = M<sub>n<\/sub><br>M<sub>n<\/sub> = 120 &#8211; 9n<br>Thus, required linear expression is M<sub>n<\/sub>&nbsp;= 120 &#8211; 9n.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.19:<\/p>\n\n\n\n<p>Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area of the rectangle.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Length = 13 cm<br>Area = Length \u00d7 Breadth<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Breadth = 12 cm<br>Area = 13 \u00d7 12 = 156 cm\u00b2<\/li>\n\n\n\n<li>Breadth = 10 cm<br>Area = 13 \u00d7 10 = 130 cm\u00b2<\/li>\n\n\n\n<li>Breadth = 8 cm<br>Area = 13 \u00d7 8 = 104 cm\u00b2<\/li>\n<\/ol>\n\n\n\n<p>Let breadth = x cm<br>Area = 13x<br>Thus, linear pattern is A = 13x.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.20:<\/p>\n\n\n\n<p>Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern representing the volume of the rectangular box.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Length = 7 cm, Breadth = 11 cm<br>Volume = Length \u00d7 Breadth \u00d7 Height<br>= 7 \u00d7 11 \u00d7 h = 77h<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Height = 5 cm<br>Volume = 77 \u00d7 5 = 385 cm<sup>3<\/sup><\/li>\n\n\n\n<li>Height = 9 cm<br>Volume = 77 \u00d7 9 = 693 cm<sup>3<\/sup><\/li>\n\n\n\n<li>Height = 13 cm<br>Volume = 77 \u00d7 13 = 1001 cm<sup>3<\/sup> Let height = h cm<br>Volume = 77h<\/li>\n<\/ol>\n\n\n\n<p>Thus, linear pattern is V = 77h.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.21:<\/p>\n\n\n\n<p>Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Total pages = 500<br>Pages read per day = 20<br>Pages read in 15 days = 20 \u00d7 15 = 300<br>Pages left = 500 &#8211; 300 = 200<br>Let pages left after n days = P<sub>n<\/sub><br>So, P<sub>n<\/sub> = 500 &#8211; 20n<br>Thus, linear pattern is P<sub>n<\/sub>&nbsp;= 500 &#8211; 20n.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.22:<\/p>\n\n\n\n<p>Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Find the height after 7 months.<\/li>\n\n\n\n<li>Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.<\/li>\n\n\n\n<li>Find an expression that relates h and t, and explain why it represents linear growth.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<br>Initial height = 1.75 feet<br>Growth per month = 0.5 feet<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Height growth in each month = 0.5 feet<br>So, Height after 7 months:<br>h = 1.75 + (0.5 \u00d7 7)<br>= 1.75 + 3.5<br>= 5.25 feet<\/li>\n\n\n\n<li>Table of values: <strong>t (months)<\/strong> <strong>h (feet)<\/strong> 0 1.75 1 2.25 2 2.75 3 3.25 4 3.75 5 4.25 6 4.75 7 5.25 8 5.75 9 6.25 10 6.75<\/li>\n\n\n\n<li>Expression:<br>Let height after t months = h<br>h = 1.75 + 0.5t<br>This represents linear growth because height increases by a constant amount (0.5 feet) every month.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.23:<\/p>\n\n\n\n<p>A mobile phone is bought for \u20b910,000. Its value decreases by \u20b9800 every year.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Find the value of the phone after 3 years.<\/li>\n\n\n\n<li>Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.<\/li>\n\n\n\n<li>Find an expression that relates v and t, and explain why it represents linear decay.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<br>Initial value = \u20b910,000<br>Decrease per year = \u20b9800<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Decrease in value after 1 year = \u20b9800<br>So, Value after 3 years:<br>v = 10000 \u2212 (800 \u00d7 3)<br>= 10000 \u2212 2400<br>= \u20b97600<\/li>\n\n\n\n<li>Table of values: <strong>t (years)<\/strong> <strong>v (P)<\/strong> 0 10000 1 9200 2 8400 3 7600 4 6800 5 6000 6 5200 7 4400 8 3600<\/li>\n\n\n\n<li>Expression:<br>Let value after t years = v<br>v = 10000 \u2212 800t<br>This represents linear decay because the value decreases by a constant amount (\u20b9800) every year.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.24:<\/p>\n\n\n\n<p>The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Find the population of the village after 6 years.<\/li>\n\n\n\n<li>Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.<\/li>\n\n\n\n<li>Find an expression that relates P and t, and explain why it represents linear growth.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Population after 6 years:<br>P = 750 + (50 \u00d7 6)<br>= 750 + 300<br>= 1050<\/li>\n\n\n\n<li>Table of values: <strong>t (years)<\/strong> <strong>P (population)<\/strong> 0 750 1 800 2 850 3 900 4 950 5 1000 6 1050 7 1100 8 1150 9 1200 10 1250<\/li>\n\n\n\n<li>Expression:<br>Let population after t years = P<br>P = 750 + 50t<br>This represents linear growth because the population increases by a constant number (50 people) every year.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.25:<\/p>\n\n\n\n<p>A telecom company charges \u20b9600 for a certain recharge scheme. This prepaid balance is reduced by \u20b915 each day after recharge.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.<\/li>\n\n\n\n<li>After how many days will the balance run out?<\/li>\n\n\n\n<li>Make a table of values for x varying from 1 to 10 days and show how the balance b(x) reduces with time.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Expression:<br>Let remaining balance after x days = b(x)<br>b(x) = 600 &#8211; 15x<br>This represents linear decay because the balance decreases by a constant amount (\u20b915) every day.<\/li>\n\n\n\n<li>When balance runs out:<br>b(x) = 0<br>600 &#8211; 15x = 0<br>15x = 600<br>x = 40<br>So, the balance will run out after 40 days.<\/li>\n\n\n\n<li>Table of values: <strong>x (days)<\/strong> <strong>{tex}b(x)(\u20b9){\/tex}<\/strong> 1 585 2 570 3 555 4 540 5 525 6 510 7 495 8 480 9 465 10 450<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.26:<\/p>\n\n\n\n<p>A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observed that when she accessed 10 modules, her bill was \u20b9400. When she accessed 14 modules, her bill was \u20b9500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<br>When x = 10, y = 400<br>When x = 14, y = 500<br>Using y = ax + b<br>For x = 10: 400 = 10a + b \u2026(1)<br>For x = 14: 500 = 14a + b \u2026(2)<br>Subtracting (1) from (2), we get:<br>500 &#8211; 400 = 14a &#8211; 10a<br>\u21d2 100 = 4a<br>\u21d2 a = 25<br>Substituting a = 25 in (1), we get:<br>400 = 10(25) + b<br>\u21d2 400 = 250 + b<br>\u21d2 b = 150<br>Therefore, a = 25 and b = 150.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.27:<\/p>\n\n\n\n<p>A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was \u20b9800. When she used it for 15 hours, her bill was \u20b91100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<br>When x = 10, y = 800<br>When x = 15, y = 1100<br>Using y = ax + b<br>For x = 10: 800 = 10a + b \u2026(1)<br>For x = 15: 1100 = 15a + b \u2026(2)<br>Subtracting (1) from (2), we have:<br>1100 &#8211; 800 = 15a &#8211; 10a<br>\u21d2 300 = 5a<br>\u21d2 a = 60<br>Substitute a = 60 in (1), we get:<br>800 = 10(60) + b<br>\u21d2 800 = 600 + b<br>\u21d2 b = 200<br>Therefore, a = 60 and b = 200.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.28:<\/p>\n\n\n\n<p>Consider the relationship between temperature measured in degrees Celsius (\u00b0C) and degrees Fahrenheit (\u00b0F), which is given by \u00b0C = a\u00b0F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given relation: \u00b0C = a\u00b0F + b<br>Using the point (\u00b0F, \u00b0C) = (32, 0), we have:<br>0 = 32a + b \u2026(1)<br>Using the point (\u00b0F, \u00b0C) = (212, 100), we have:<br>100 = 212a + b \u2026(2)<br>Subtracting (1) from (2), we get:<br>100 &#8211; 0 = 212a &#8211; 32a<br>\u21d2 100 = 180a<br>\u21d2 a = 100\/180<br>\u21d2 a = 5\/9<br>Substitute a = 5\/9 in (1), we get:<br>0 = 32(5\/9) + b<br>\u21d2 0 = 160\/9 + b<br>\u21d2 b = -160\/9<br>Therefore, a = 5\/9 and b = -160\/9.<br>Hence, the linear relationship is \u00b0C = (5\/9)\u00b0F &#8211; 160\/9.<br>It can also be written as \u00b0C = (5\/9)(\u00b0F &#8211; 32).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.29:<\/p>\n\n\n\n<p>Draw the graph&nbsp;of the&nbsp;set&nbsp;of lines, y = 4x, y = 2x, y = x. Reflect on the role of \u2018a\u2019 and \u2018b\u2019.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>All lines are of the form {tex}y=a x(b=0){\/tex}.<br>Observation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>All lines pass through the origin {tex}(0,0){\/tex}.<\/li>\n\n\n\n<li>The value of &#8216;a&#8217; (slope) determines steepness.<\/li>\n\n\n\n<li>Larger &#8216;{tex}a{\/tex}&#8217; \u2192 steeper line.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777022793-47cz4r.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.30:<\/p>\n\n\n\n<p>Draw the graph&nbsp;of the&nbsp;set&nbsp;of lines, y = &#8211; 6x, y = &#8211; 3x, y = -\u2009x. Reflect on the role of \u2018a\u2019 and \u2018b\u2019.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>All lines are of the form {tex}y=a x(b=0){\/tex}.<br>Observation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>All lines pass through the origin.<\/li>\n\n\n\n<li>Negative &#8216;a&#8217; means lines slope downward.<\/li>\n\n\n\n<li>Larger magnitude of &#8216;a&#8217; \u2192 steeper downward slope.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777022945-jhgwz2.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.31:<\/p>\n\n\n\n<p>Draw the graph&nbsp;of the&nbsp;set&nbsp;of lines, y = 5x, y = -5x. Reflect on the role of \u2018a\u2019 and \u2018b\u2019.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Observation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Both Lines pass through the origin.<\/li>\n\n\n\n<li>{tex}y=5 x{\/tex} slopes upward, {tex}y=-5 x{\/tex} slopes downward.<\/li>\n\n\n\n<li>Same magnitude of &#8216;a&#8217; \u2192 same steepness but opposite direction.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777023300-bugwrp.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.32:<\/p>\n\n\n\n<p>Draw the graph&nbsp;of the&nbsp;set&nbsp;of lines, y = 3x &#8211; 1, y = 3x, y = 3x + 1. Reflect on the role of \u2018a\u2019 and \u2018b\u2019.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>All lines have same slope ({tex}a=3{\/tex}).<br>Observation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Lines are parallel (same slope).<\/li>\n\n\n\n<li>Different values of &#8216;b&#8217; shift the line up or down.<\/li>\n\n\n\n<li>{tex}{b}=-1 \\rightarrow{\/tex} Line below origin<\/li>\n\n\n\n<li>{tex}{b}=0 \\rightarrow{\/tex} passes through origin<\/li>\n\n\n\n<li>{tex}{b}=+1 \\rightarrow{\/tex} Line above origin.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777023808-xe7z2j.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.33:<\/p>\n\n\n\n<p>Draw the graph&nbsp;of the&nbsp;set&nbsp;of lines, y = -2x &#8211; 3, y = -2x, y = 2x + 3. Reflect on the role of \u2018a\u2019 and \u2018b\u2019.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Observation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}y=-2 x-3{\/tex} and {tex}y=-2 x{\/tex} have same slope {tex}(-2) \\rightarrow{\/tex} parallel lines.<\/li>\n\n\n\n<li>{tex}y=2 x+3{\/tex} has positive slope \u2192 different direction.<\/li>\n\n\n\n<li>&#8216;b&#8217; changes vertical position of Line.<\/li>\n<\/ul>\n\n\n\n<p>Conclusion:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>&#8216;a&#8217; (coefficient of x) controls slope (steepness and direction).<\/li>\n\n\n\n<li>&#8216;b&#8217; (constant term) controls vertical shift (y-intercept).<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777024049-pryymt.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.34:<\/p>\n\n\n\n<p>Write a polynomial of degree 3 in the variable x, in which the coefficient of the x<sup>2<\/sup> term is -7.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>A polynomial of degree 3 has the general form: ax<sup>3<\/sup> + bx<sup>2<\/sup>&nbsp;+ cx + d<br>Given that coefficient of x<sup>2<\/sup>&nbsp;is -7, so b = -7<br>One such polynomial is x<sup>3<\/sup>&nbsp;&#8211; 7x<sup>2<\/sup>&nbsp;+ 2x + 1.<br>(Any polynomial of degree 3 with -7 as the coefficient of x<sup>2<\/sup>&nbsp;is correct)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.35:<\/p>\n\n\n\n<p>Find the values of the following polynomials at the indicated values of the variables.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>5x<sup>2<\/sup>\u00a0&#8211; 3x + 7 if x = 1<\/li>\n\n\n\n<li>4t<sup>3<\/sup> &#8211; t\u00b2 + 6 if t = a<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>5x<sup>2<\/sup>\u00a0&#8211; 3x + 7 if x = 1<br>Substitute x = 1:<br>= 5(1)<sup>2<\/sup>\u00a0&#8211; 3(1) + 7<br>= 5 &#8211; 3 + 7<br>= 9<\/li>\n\n\n\n<li>4t<sup>3<\/sup> &#8211; t<sup>2<\/sup>\u00a0+ 6 if t = a<br>Substitute t = a:<br>= 4a<sup>3<\/sup>\u00a0&#8211; a<sup>2<\/sup>\u00a0+ 6<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.36:<\/p>\n\n\n\n<p>If we multiply a number by {tex}\\frac{5}{2}{\/tex} and add {tex}\\frac{2}{3}{\/tex} to the product, we get {tex}\\frac{-7}{12}{\/tex}. Find the number.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the number be x.<br>According to the question:<\/p>\n\n\n\n<p>{tex} (5 \/ 2) x+2 \/ 3=-7 \/ 12 {\/tex}<br>Subtract {tex}2 \/ 3{\/tex} from both sides, we get:<\/p>\n\n\n\n<p>{tex} (5 \/ 2) x=-7 \/ 12-2 \/ 3 {\/tex}<br>{tex} \\Rightarrow(5 \/ 2) x=-7 \/ 12-8 \/ 12 {\/tex}<br>{tex} \\Rightarrow(5 \/ 2) x=-15 \/ 12 {\/tex}<br>{tex} \\Rightarrow(5 \/ 2) x=-5 \/ 4 {\/tex}<\/p>\n\n\n\n<p>Now multiply both sides by {tex}2 \/ 5{\/tex}, we get:<\/p>\n\n\n\n<p>{tex} x=(-5 \/ 4) \\times(2 \/ 5) {\/tex}<br>{tex} \\Rightarrow x=-1 \/ 2 {\/tex}<br>Therefore, the number is {tex}-1 \/ 2{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.37:<\/p>\n\n\n\n<p>A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the smaller number be x.<br>Then the larger number = 5x<br>After adding 21 to both numbers: New numbers are x + 21 and 5x + 21<br>According to the question: 5x + 21 = 2(x + 21)<br>\u21d2 5x + 21 = 2x + 42<br>\u21d2 5x \u2013 2x = 42 \u2013 21<br>\u21d2 3x = 21<br>\u21d2 x = 7<br>So, the smaller number = 7 and the larger number = 5 \u00d7 7 = 35<br>Therefore, the two numbers are 7 and 35.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.38:<\/p>\n\n\n\n<p>If you have \u20b9800 and you save \u20b9250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Initial amount = \u20b9800<br>Saving every month = \u20b9250<br>Linear pattern: Amount after n months = 800 + 250n<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Amount after 6 months:<br>= 800 + 250(6)<br>= 800 + 1500<br>= \u20b92300<\/li>\n\n\n\n<li>Amount after 2 years: 2 years = 24 months<br>Amount after 24 months:<br>= 800 + 250(24)<br>= 800 + 6000<br>= \u20b96800<\/li>\n<\/ol>\n\n\n\n<p>Therefore:<br>Amount after 6 months = \u20b92300<br>Amount after 2 years = \u20b96800<br>Linear pattern: A = 800 + 250n, where n is the number of months.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.39:<\/p>\n\n\n\n<p>The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the tens digit be x and the units digit be y.<br>Then the original number = 10x + y<br>The interchanged number = 10y + x<br>According to the question: (10x + y) + (10y + x) = 143<br>\u21d2 11x + 11y = 143<br>\u21d2 11(x + y) = 143<br>\u21d2 x + y = 13 \u2026(1)<br>The digits differ by 3.<br>So, x &#8211; y = 3 \u2026(2)<br>Now adding (1) and (2), we get:<br>2x = 16<br>\u21d2 x = 8<br>Substitute x = 8 in (1), we have:<br>8 + y = 13<br>\u21d2 y = 5<br>Therefore, the original number = 85 and the interchanged number = 58<br>Hence, the two numbers are 85 and 58.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.40:<\/p>\n\n\n\n<p>Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>y = -3x + 4<\/li>\n\n\n\n<li>2y = 4x + 7<\/li>\n\n\n\n<li>5y = 6x &#8211; 10<\/li>\n\n\n\n<li>3y = 6x &#8211; 11<\/li>\n<\/ol>\n\n\n\n<p>Are any of the lines parallel?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>y = -3x + 4<br>Comparing with y = ax + b:<br>Slope a = -3<br>y-intercept b = 4<br>So, the point, where the line cuts the y-axis, is (0, 4).<\/li>\n\n\n\n<li>2y = 4x + 7<br>Dividing both sides by 2: y = 2x + 7\/2<br>Slope a = 2<br>y-intercept b = 7\/2<br>So, the point, where the line cuts the y-axis, is (0, 7\/2).<\/li>\n\n\n\n<li>5y = 6x &#8211; 10<br>Dividing both sides by 5: y = (6\/5)x &#8211; 2<br>Slope a = 6\/5<br>y-intercept b = -2<br>So, the point, where the line cuts the y-axis is (0, -2).<\/li>\n\n\n\n<li>3y = 6x &#8211; 11<br>Dividing both sides by 3: y = 2x &#8211; 11\/3<br>Slope a = 2<br>y-intercept b = -11\/3<br>So, the point, where the line cuts the y-axis is (0, -11\/3).<\/li>\n<\/ol>\n\n\n\n<p>Parallel lines:<br>Two lines are parallel if they have the same slope.<br>Equation (ii): slope = 2<br>Equation (iv): slope = 2<br>Therefore, lines (ii) and (iv) are parallel.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777026640-hn4y87.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.41:<\/p>\n\n\n\n<p>If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y \u00b0F, the relation between the two systems of measurement of temperature is given by the linear equation y = (9\/5)(x &#8211; 273) + 32.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.<\/li>\n\n\n\n<li>If the temperature is 158 \u00b0F, then find the temperature in Kelvin.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>When x = 313 K, We have to find y:<br>Using equation y = (9\/5)(x \u2013 273) + 32, we have<br>y = (9\/5)(313 \u2013 273) + 32<br>\u21d2 y = (9\/5)(40) + 32<br>\u21d2 y = 72 + 32<br>\u21d2 y = 104<br>Therefore, the temperature is 104 \u00b0F.<\/li>\n\n\n\n<li>When y = 158 \u00b0F, we have to find x:<br>Using equation y = (9\/5)(x \u2013 273) + 32, we have<br>158 = (9\/5)(x \u2013 273) + 32<br>Subtracting 32 from both sides, we get:<br>158 \u2013 32 = (9\/5)(x \u2013 273)<br>\u21d2 126 = (9\/5)(x \u2013 273)<br>Multiply both sides by 5\/9, we get:<br>126 \u00d7 (5\/9) = x \u2013 273<br>\u21d2 70 = x \u2013 273<br>\u21d2 x = 343<br>Therefore, the temperature is 343 K.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.42:<\/p>\n\n\n\n<p>The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force.&nbsp;Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>We know that:<br>Work done = Force \u00d7 Distance<br>According to question, work done = w<br>Distance travelled = d<br>Constant force = 3 units<br>Therefore, w = 3d<br>This is the required linear equation in two variables.<br>If d = 2 units, force w = 3 \u00d7 2 = 6 units.<br>Taking w on y-axis and d on x-axis, we can plot the graph.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777027358-4yvfja.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p>The point (2, 6) lie on a straight line, so, it is verified by graph.<br>So, the Work done when distance travelled is 2 units: w = 3 \u00d7 2<br>Hence, w = 6 units<br>Verification from graph:<br>The point corresponding to d = 2 is (2, 6), so the work done is 6 units.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.43:<\/p>\n\n\n\n<p>The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Find the polynomial p(x).<\/li>\n\n\n\n<li>Find the coordinates where the graph of p(x) cuts the axes.<\/li>\n\n\n\n<li>Draw the graph of p(x) and verify your answers.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Let p(x) = ax + b<br>Since the graph passes through (1, 5),<br>a(1) + b = 5<br>\u21d2 a + b = 5 \u2026(1)<br>Since the graph passes through (3, 11),<br>a(3) + b = 11<br>\u21d2 3a + b = 11 \u2026(2)<br>Subtracting (1) from (2), we get:<br>3a + b &#8211; (a + b) = 11 \u2013 5<br>\u21d2 2a = 6<br>\u21d2 a = 3<br>Substituting a = 3 in (1), we have:<br>3 + b = 5<br>\u21d2 b = 2<br>Therefore, p(x) = 3x + 2.<\/li>\n\n\n\n<li>Coordinates where the graph cuts the axes:<br>To find the y-axis intercept: Put x = 0<br>y = p(0) = 3(0) + 2 = 2<br>So, the graph cuts the y-axis at (0, 2). To find the x-axis intercept: Put y = 0<br>3x + 2 = 0<br>\u21d2 3x = -2<br>\u21d2 x = -2\/3<br><br>So, the graph cuts the x-axis at (-2\/3, 0).<br> <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777027715-j285sd.jpg\"><\/li>\n\n\n\n<li>From the graph, it verifies that the line passes through the given points and cuts:<br>y-axis at (0, 2)<br>x-axis at (-2\/3, 0).<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.44:<\/p>\n\n\n\n<p>Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>p(0) = 5<\/li>\n\n\n\n<li>The polynomial p(x) \u2212 q(x) cuts the x-axis at (3, 0).<\/li>\n\n\n\n<li>The sum p(x) + q(x) is equal to 6x + 4 for all real x.<\/li>\n<\/ol>\n\n\n\n<p>Find the polynomials p(x) and q(x).<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let p(x) = ax + b and q(x) = cx + d<br>Using the condition (i), we have p(0) = 5<br>\u21d2 a(0) + b = 5<br>\u21d2 b = 5<br>So, p(x) = ax + 5<br>Now, using condition (iii), we have<br>p(x) + q(x) = 6x + 4<br>\u21d2 (ax + 5) + (cx + d) = 6x + 4<br>\u21d2 (a + c)x + (5 + d) = 6x + 4<br>Comparing coefficients, we get a + c = 6 \u2026(1)<br>5 + d = 4<br>\u21d2 d = -1<br>So, q(x) = cx &#8211; 1<br>Using condition (ii), p(x) &#8211; q(x) cuts x-axis at (3, 0)<br>\u21d2 p(3) &#8211; q(3) = 0<br>Here:<br>p(3) = 3a + 5<br>q(3) = 3c &#8211; 1<\/p>\n\n\n\n<p>So, (3a + 5) &#8211; (3c &#8211; 1) = 0<br>\u21d2 3a + 5 &#8211; 3c + 1 = 0<br>\u21d2 3a &#8211; 3c + 6 = 0<br>\u21d2 a &#8211; c = -2 \u2026(2)<\/p>\n\n\n\n<p>Solving equations<br>From (1): a + c = 6<br>From (2): a &#8211; c = -2<br>Adding both: 2a = 4 \u21d2 a = 2<br>Then: 2 + c = 6 \u21d2 c = 4<br>Hence:<br>p(x) = 2x + 5<br>q(x) = 4x &#8211; 1.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.45:<\/p>\n\n\n\n<p>Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777028877-w64fgf.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw the next two stages of the pattern. How many matchsticks will be required at these stages?<\/li>\n\n\n\n<li>Complete the following table. Stage Number 1 2 3 4 5 {tex}\\ldots{\/tex} {tex}n{\/tex} Number of matchsticks \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/li>\n\n\n\n<li>Find a rule to determine the number of matchsticks required for the n<sup>th<\/sup> stage.<\/li>\n\n\n\n<li>How many matchsticks will be required for the 15th stage of the pattern?<\/li>\n\n\n\n<li>Can 200 matchsticks form a stage in this pattern? Justify your answer.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>A single hexagon needs 6 matchsticks.<br>Since each new hexagon shares one side with the previous hexagon, only 5 new matchsticks are added at every new stage.<br>So the pattern is:<br>Stage 1 = 6<br>Stage 2 = 6 + 5 = 11<br>Stage 3 = 11 + 5 = 16<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Next two stages:<br>Stage 4: Number of matchsticks = 16 + 5 = 21<br>Stage 5: Number of matchsticks = 21 + 5 = 26<br>Therefore:<br>Stage 4 requires 21 matchsticks<br>Stage 5 requires 26 matchsticks<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777029158-z6pcwj.jpg\"><\/li>\n\n\n\n<li>Complete table: Stage Number 1 2 3 4 5 {tex}\\cdots{\/tex} {tex}n{\/tex} Number of matchsticks 6 11 16 21 26 \u00a0 {tex}5 n+1{\/tex}<\/li>\n\n\n\n<li>Rule for the n<sup>th<\/sup> stage:<br>The number of matchsticks forms an arithmetic pattern: 6, 11, 16, 21, 26, \u2026<br>First term = 6<br>Difference = 5<br>So, Number of matchsticks in the n\u1d57\u02b0 stage<br>= 6 + (n \u2013 1) \u00d7 5<br>= 6 + 5n \u2013 5<br>= 5n + 1<br>Hence, the rule is M<sub>n<\/sub>\u00a0= 5n + 1<\/li>\n\n\n\n<li>Matchsticks required for the 15th stage:<br>M<sub>15<\/sub> = 5(15) + 1<br>= 75 + 1<br>= 76<br>Therefore, 76 matchsticks will be required for the 15th stage.<\/li>\n\n\n\n<li>Can 200 matchsticks form a stage in this pattern?<br>For some stage n,<br>5n + 1 = 200<br>\u21d2 5n = 199<br>\u21d2 n = 199\/5<br>\u21d2 n = 39.8<br>Since n is not a whole number, 200 matchsticks cannot form any stage in this pattern.<br>Therefore, 200 matchsticks cannot form a stage in this pattern.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.46:<\/p>\n\n\n\n<p>Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The graph of p(x) passes through the points (2, 3) and (6, 11).<\/li>\n\n\n\n<li>The graph of q(x) passes through the point (4, -1).<\/li>\n\n\n\n<li>The graph of q(x) is parallel to the graph of p(x).<\/li>\n<\/ol>\n\n\n\n<p>Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<br>p(x) = ax + b<br>q(x) = cx + d<br>First, to find p(x).<br>Since p(x) passes through (2, 3) and (6, 11), its slope is m = (11 &#8211; 3)\/(6 &#8211; 2)<br>= 8\/4 = 2<br>So, p(x) = 2x + b<br>Using the point (2, 3), we have:<br>3 = 2(2) + b<br>\u21d2 3 = 4 + b<br>\u21d2 b = -1<br>Therefore, p(x) = 2x &#8211; 1.<\/p>\n\n\n\n<p>Now to find q(x).<br>Since q(x) is parallel to p(x), it has the same slope.<br>So, slope of q(x) = 2<br>Hence, q(x) = 2x + d<br>Since q(x) passes through (4, -1), so -1 = 2(4) + d<br>\u21d2 -1 = 8 + d<br>\u21d2 d = -9<br>Therefore, q(x) = 2x &#8211; 9<\/p>\n\n\n\n<p>Now we have to find where these lines meet the x-axis.<br>A line meets the x-axis where y = 0.<br>For p(x) = 2x &#8211; 1: 0 = 2x &#8211; 1<br>\u21d2 2x = 1<br>\u21d2 x = 1\/2<br>So, p(x) meets the x-axis at (1\/2, 0).<\/p>\n\n\n\n<p>For q(x) = 2x &#8211; 9: 0 = 2x &#8211; 9<br>\u21d2 2x = 9<br>\u21d2 x = 9\/2<br>So, q(x) meets the x-axis at (9\/2, 0).<\/p>\n\n\n\n<p>Hence, p(x) = 2x &#8211; 1 and q(x) = 2x &#8211; 9.<br>The x-axis intercepts are:<br>For p(x): (1\/2, 0)<br>For q(x): (9\/2, 0)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.47:<\/p>\n\n\n\n<p>What do all linear functions of the form f(x) = ax + a, a &gt; 0, have in common?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given: f(x) = ax + a, where a &gt; 0<br>We can write it as: f(x) = a(x + 1)<br>Common properties of all such linear functions are:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Slope:<br>The slope is a, and since a > 0, all the lines have positive slope.<br>So, all these lines rise from left to right.<\/li>\n\n\n\n<li>y-intercept:<br>Putting x = 0,<br>f(0) = a<br>So, the y-intercept is (0, a).<br>Since a > 0, all the lines cut the y-axis above the origin.<\/li>\n\n\n\n<li>x-intercept:<br>To find where the line cuts the x-axis, put f(x) = 0<br>ax + a = 0<br>a(x + 1) = 0<br>Since a > 0, a is not zero.<br>So, x + 1 = 0<br>or x = -1<br>Thus, every line cuts the x-axis at the same point (-1, 0).<\/li>\n<\/ol>\n\n\n\n<p>Therefore, all linear functions of the form f(x) = ax + a, a &gt; 0, have the following in common:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>all have positive slope<\/li>\n\n\n\n<li>all cut the y-axis above the origin<\/li>\n\n\n\n<li>all pass through the fixed point (-1, 0).<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Introduction to Linear Polynomials &#8211; NCERT Solutions Class 9 Maths Ganita Manjari includes all the questions with solutions given in NCERT Class 9 Ganita Manjari textbook. 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