{"id":4364,"date":"2023-05-17T17:20:23","date_gmt":"2023-05-17T11:50:23","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=4364"},"modified":"2023-05-17T17:34:30","modified_gmt":"2023-05-17T12:04:30","slug":"ap-board-9th-class-maths-solutions-chapter-6-ex-6-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-9th-class-maths-solutions-chapter-6-ex-6-3\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 6 Linear Equation in Two Variables Ex 6.3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.3<\/h2>\n

Question 1.
\nDraw the graph of each of the following linear equations.
\ni) 2y = – x + 1
\nii) – x + y = 6
\niii) 3x + 5y = 15
\niv) \\(\\frac{x}{2}-\\frac{y}{3}=3\\)
\nSolution:
\ni) 2y = – x + 1
\n\u21d2 x + 2y = 1
\nii) – x + y = 6
\niii) 3x + 5y = 15
\niv) \\(\\frac{x}{2}-\\frac{y}{3}=3\\)
\n\u21d2 \\(\\frac{3 x-2 y}{6}=3\\)
\n\u21d2 3x – 2y = 18<\/p>\n

\"AP
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\nDraw the graph of each of the following linear equations and answer the following questions.
\ni) y – x
\nii) y = 2x
\niii) y = – 2x
\niv) y = 3x
\nv) y = – 3x
\nSolution:
\ni) y = x<\/p>\n\n\n\n\n\n
x<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
(x, y)<\/td>\n(1, 1)<\/td>\n(2, 2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

ii) y = 2x<\/p>\n\n\n\n\n
x<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n2<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

iii) y = – 2x<\/p>\n\n\n\n\n
x<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n-2<\/td>\n-4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

iv) y = 3x<\/p>\n\n\n\n\n
x<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n3<\/td>\n6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

v) y = – 3x<\/p>\n\n\n\n\n
x<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n-3<\/td>\n-6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP<\/p>\n

\"AP<\/p>\n

i) Are all these equations of the form y = mx; where m is a real number ?
\nSolution:
\nYes. All the equations are of the form y = mx where m e R.<\/p>\n

ii) Are all these graphs passing through the origin ?
\nSolution:
\nYes. All these lines pass through the origin.<\/p>\n

iii) What can you conclude about these graphs ?
\nSolution:
\nAll lines of the form y = mx, pass through the origin.<\/p>\n

Question 3.
\nDraw the graph of the equation 2x + 3y = 11. Find from the graph value of y when x = 1.
\nSolution:<\/p>\n\n\n\n\n
x<\/td>\n1<\/td>\n4<\/td>\n<\/tr>\n
y<\/td>\n3<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\nFrom the graph; when x = 1 then y = 3.<\/p>\n

\"AP<\/p>\n

Question 4.
\nDraw the graph of the equation y – x = 2. Find from the graph
\ni) the value of y when x = 4
\nii) the value of x when y = – 3
\nSolution:
\nThe given equation is y – x = 2 or – x + y = 2<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n-2<\/td>\n<\/tr>\n
y<\/td>\n2<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\ni) If x = 4 then y = 6 (\u2235 from the graph)
\nii) When y = – 3 then x = – 5 (\u2235 from the graph)<\/p>\n

Question 5.
\nDraw the graph of the equation 2x + 3y = 12. Find the solutions from the graph,
\n(i) Whose y-coordinate is 3 (OR) Whose y-coordinate is 2
\n(ii) Whose x-coordinate is – 3
\nSolution:
\nThe given equation is 2x + 3y = 12<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n6<\/td>\n<\/tr>\n
y<\/td>\n4<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\ni) From the graph when y = 3 then 2x + 3(3) = 12 \u21d2 2x + 9 = 12 \u21d2 2x = 3 \u21d2 x = \\(\\frac{3}{2}\\) ; solution is (\\(\\frac{3}{2}\\) , 3) (OR) When y = 2 then 2x + 3(2) = 12 \u21d2 2x + 6 = 12 \u21d2 2x = 6 \u21d2 x = 3 solution is (3, 2),
\nii) From the graph when x = – 3 then y = 6; solution is (- 3, 6)<\/p>\n

\"AP<\/p>\n

Question 6.
\nDraw the graph of each of the equations given below and also find the coordinates of the points where the graph cuts the coordinate axes.
\ni) 6x – 3y = 12
\nSolution:
\n6x – 3y = 12<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n-4<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

From the graph the line cuts the X-axis at (2, 0) and Y-axis at (0, – 4).<\/p>\n

ii) -x + 4y = 8
\nSolution:<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n-8<\/td>\n<\/tr>\n
y<\/td>\n2<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

From the graph the line cuts the X-axis at (- 8, 0) and Y -axis at (0, 2).<\/p>\n

iii) 3x + 2y + 6 = 0
\nSolution:
\n3x + 2y + 6 = 0<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n-2<\/td>\n<\/tr>\n
y<\/td>\n-3<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

From the graph the line cuts the X-axis at (- 2, 0) and Y -axis at (0, -3).
\n\"AP<\/p>\n

Question 7.
\nRajiya and Preethi two students of class IX together collected \u20b91000 for the Prime Minister Relief Fund for victims of natural calamities. Write a linear equation and
\ndraw a graph to depict the statement. Clfp)
\nSolution:
\nLet Rajiya’s contribution to P.M.R.F be = \u20b9 x
\nPreethi’s contribution to P.M.R.F be = \u20b9 y
\nThen by problem x + y = 1000<\/p>\n\n\n\n\n\n
x + y = 1000<\/td>\n<\/tr>\n
x<\/td>\n200<\/td>\n300<\/td>\n<\/tr>\n
y<\/td>\n800<\/td>\n700<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP<\/p>\n

\"AP<\/p>\n

Question 8.
\nGopaiah sowed w heat and paddy in two fields of total area 5000 sq. meters. Write a linear equation and draw a graph to represent the same.
\nSolution: Let the wheat be sowed in a land equal to x sq.m,
\nand the paddy be sowed in a land equal to y sq.m.
\n\u2234 By problem x + y = 5000<\/p>\n\n\n\n\n\n
x + y = 5000<\/td>\n<\/tr>\n
x<\/td>\n1000<\/td>\n2000<\/td>\n<\/tr>\n
y<\/td>\n4000<\/td>\n3000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP<\/p>\n

Question 9.
\nThe force applied on a body of mass 6 kg. is directly proportional to the acceleration produced in the body. Write an equation to express this observation and draw the graph of the equation.
\nSolution:
\nLet the lorce = f; mass = 6 kg; acceleration = a
\nBy problem f \u221d a or f = m . a \u21d2 f = 6a<\/p>\n\n\n\n\n\n
f = 6a<\/td>\n<\/tr>\n
a<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n
f<\/td>\n12<\/td>\n18<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP<\/p>\n

\"AP<\/p>\n

Question 10.
\nA stone is falling from a mountain. The velocity of the stone is given by v = 9.8t.
\nDraw its graph and find the velocity of the stone 4 seconds after start.
\nSolution:
\nGiven that, the velocity of the stone v = 9.8 t<\/p>\n\n\n\n\n\n
v = 9.8t<\/td>\n<\/tr>\n
v<\/td>\n49<\/td>\n98<\/td>\n<\/tr>\n
t<\/td>\n5<\/td>\n10<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\nThe velocity after 4 seconds = v = 9.8 \u00d7 4 = 39.2 m\/sec2<\/sup>.<\/p>\n

Question 11.
\nIn an election 60 % of voters cast their votes. Form an equation and draw the graph for this data. Find the following from the graph.
\ni) The total number of voters, if 1200 voters cast their votes,
\nii) The number of votes cast, if the total number of voters are 800.
\n[Hint: If the number of voters who cast their votes be ‘x’ and the total number of voters be ‘y’ then x = 60 % of y.]
\nSolution:
\nLet the total number of votes be = y
\nThen the number of voters who cast their votes = x
\nBy problem x = 60 % of y<\/p>\n\n\n\n\n
x<\/td>\n1200<\/td>\n480<\/td>\n<\/tr>\n
y<\/td>\n2000<\/td>\n800<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\ni) From the graph when x = 1200, then y = 2000
\nii) From the graph when y = 800 then x = 480<\/p>\n

\"AP<\/p>\n

Question 12.
\nWhen Rupa was born, her father was 25 years old. Form an equation and draw a graph for this data. From the graph find
\ni) The age of the father when Rupa is 25 years old.
\nii) Rupa’s age when her father is 40 years old.
\nSolution:
\nLet her father’s age be = x years.
\nand Rupa’s age be = y years
\nBy problem x – y = 25 years<\/p>\n\n\n\n\n
x<\/td>\n40<\/td>\n50<\/td>\n<\/tr>\n
y<\/td>\n15<\/td>\n25<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\ni) From the graph age of the father when Rupa is 25 years is 50 years.
\nii) Rupa’s age when her father is 40 years is 15 years.<\/p>\n

Question 13.
\nAn auto charges \u20b915 for first kilometre and \u20b9 8 each for subsequent kilometre. For a distance of x km. an amount of \u20b9y is paid. Write the linear equation representing this information and draw the graph. With the help of graph find the distance travelled if the fare paid is \u20b955. How much would have to be paid for 7 kilometres?
\nSolution:
\nCharge for the first kilometre = \u20b915
\nCharge for the subsequent kilometres = \u20b9 8 per km.
\nAmount paid = \u20b9 y when the distance travelled is x km
\n\u2234 By problem y = 15 + 8x
\n\u2234 8x – y + 15 = 0<\/p>\n\n\n\n\n
x<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n
y<\/td>\n31<\/td>\n23<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\ni) When y = 55 then x = 5
\nii) From the graph when x = 7 then y = 71<\/p>\n

\"AP<\/p>\n

Question 14.
\nA lending library has fixed charge for the first three days and an additional charges for each day thereafter. John paid \u20b9 27 for a book kept for seven days. If the fixed charges be \u20b9 x and subsequent per day charges be \u20b9 y; then write the linear equation representing the above information and draw the graph of the same. From the graph if the fixed charge is \u20b9 7, find the subsequent per day charge. And if the per day charge is \u20b9 4, find the fixed charge, (charge is \u20b97)
\nSolution:
\nJohn kept a book for 7 days. He paid \u20b9 27
\nFor first three days = \u20b9 x (fixed)
\nFor the last four days = \u20b9 4y (? y for a day)
\nBy problem x + 4y = 27<\/p>\n\n\n\n\n
x<\/td>\n3<\/td>\n11<\/td>\n7<\/td>\n<\/tr>\n
y<\/td>\n6<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\nWhen x = 7 then y = 5
\nWhen y = 4 then x = 11<\/p>\n

Question 15.
\nThe parking charges of a car in Hyderabad Railway station for first two hours is \u20b9 50 and \u20b910 for each subsequent hour. Write down an equation and draw the graph. Find the following charges from the graph.
\ni) For three hours ii) For six hours iii) How many hours did Rekha park her car if she paid \u20b9 80 as parking charges ?
\nSolution: Let the total money paid be = \u20b9 y
\nParking charges for the first two hours = \u20b9 50
\nParking charges for total x hours @ \u20b910 per hour = 50 + (x – 2) 10
\n= 50 + 10x – 20
\n= 10x + 30
\n\u2234 By problem y = 10x + 30<\/p>\n\n\n\n\n
x<\/td>\n3<\/td>\n5<\/td>\n6<\/td>\n<\/tr>\n
y<\/td>\n60<\/td>\n80<\/td>\n90<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\ni) For three hours = 50 + 10 \u00d7 1 = \u20b9 60
\n[ \u2235 From the graph we see the same]
\nii) For six hours = 50 + 10 \u00d7 4 = 50 + 40 = \u20b9 90
\niii) Rekha parked her car for 5 hours.<\/p>\n

\"AP<\/p>\n

Question 16.
\nSameera was driving a car with uniform speed of 60 kmph. Draw distance – time graph. From the graph find the distance travelled by Sameera in
\ni) \\(\\frac { 1 }{ 2 }\\) hours
\nii) 2 hours
\niii) 3\\(\\frac { 1 }{ 2 }\\) hours
\nSolution:
\nSpeed of the car = 60 kmph
\nLet the time taken be = x hours
\nThen the total distance travelled be = y hours
\nBy problem 60x = y or 60x – y = 0<\/p>\n\n\n\n\n
x<\/td>\n2<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n
y<\/td>\n120<\/td>\n240<\/td>\n300<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\nii) Distance travelled in 1\\(\\frac { 1 }{ 2 }\\) hours = 120 km
\nii) Distance travelled in 2 hours = 120 km
\niii) Distance travelled in 3\\(\\frac { 1 }{ 2 }\\) hours = 210 km<\/p>\n

Question 17.
\nThe ratio of molecular weight of Hydrogen and Oxygen in water is 1 : 8. Set up an equation between Hydrogen and Oxygen and draw its graph. From the graph find the quantity of Hydrogen if Oxygen is 12 grams. And quantity of Oxygen if
\nHydrogen is \\(\\frac { 3 }{ 2 }\\) grams.
\n[Hint : If the quantities of hydrogen and oxygen ‘x’ and ‘y’ respectively, then
\nx : y = 1 : 8 \u21d2 8x = y]
\nSolution:
\nLet the quantity of Hydrogen = x grams
\nAnd the quantity of Oxygen = y grams
\nBy problem 8x = y \u21d2 8x – y = 0<\/p>\n\n\n\n\n
x<\/td>\n1<\/td>\n2<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n
y<\/td>\n8<\/td>\n16<\/td>\n32<\/td>\n40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\nFrom the graph, the quantity of Hydrogen if Oxygen is 12 gm = \\(\\frac { 3 }{ 2 }\\) g.
\nFrom the graph, the quantity of Oxygen if Hydrogen is \\(\\frac { 3 }{ 2 }\\) g = 12 g.<\/p>\n

\"AP<\/p>\n

Question 18.
\nIn a mixture of 28 litres, the ratio of milk and water is 5 : 2. Set up the equation between the mixture and milk. Draw its graph. By observing the graph find the quantity of milk in the mixture.
\n[Hint: Ratio between mixture and milk = 5 + 2:5 = 7:5]
\nSolution:
\nLet the quantity of milk in the mixture be = x lit.
\nquantity of the mixture = y lit.
\nRatio of the milk and water = 5:2
\nSum of the terms of the ratio = 5 + 2 = 7
\nQuantity of milk x = \\(\\frac { 5 }{ 7 }\\) y lit.
\n7x = 5y \u21d2 7x – 5y = 0<\/p>\n\n\n\n\n
x<\/td>\n10<\/td>\n20<\/td>\n25<\/td>\n<\/tr>\n
y<\/td>\n14<\/td>\n28<\/td>\n35<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\nFrom the graph quantity of milk in the mixture = 20 lit.<\/p>\n

\"AP<\/p>\n

Question 19.
\nIn countries like U.S.A. and Canada temperature is measured in Fahrenheit whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius F = \\(\\frac { 9 }{ 5 }\\) C + 32
\ni) Draw the graph of the above linear equation having Celsius on X-axis and Fahrenheit on Y-axis.
\nii) If the temperature is 30\u00b0C, what is the temperature in Fahrenheit ?
\niii) If the temperature is 95\u00b0F, what is the temperature in Celsius ?
\niv) Is there a temperature that has numerically the same value in both Fahrenheit and Celsius ? If yes, find it.
\nSolution:
\ni) Given that F = \\(\\frac { 9 }{ 5 }\\) C + 32<\/p>\n\n\n\n\n
C<\/td>\n20<\/td>\n30<\/td>\n35<\/td>\n-40<\/td>\n<\/tr>\n
F<\/td>\n68<\/td>\n86<\/td>\n95<\/td>\n-40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP
\nIf C = 20, F = \\(\\frac { 9 }{ 5 }\\) \u00d7 20 + 32 = 68
\nIf c = 30, F = \\(\\frac { 9 }{ 5 }\\) \u00d7 30 + 32 = 86
\nIf C = 35, F = \\(\\frac { 9 }{ 5 }\\) \u00d7 35 + 32 = 95
\nIf C = -40, F = \\(\\frac { 9 }{ 5 }\\) \u00d7 (-40) + 32 = -40<\/p>\n

ii) From the graph 30\u00b0C = 86\u00b0F
\niii) 95\u00b0F = 35\u00b0C
\niv) When C = – 40 then F = – 40<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.3 Question 1. Draw the graph of each of the following linear equations. i) 2y = – x + … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4364"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=4364"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4364\/revisions"}],"predecessor-version":[{"id":40510,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4364\/revisions\/40510"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=4364"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=4364"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=4364"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}