{"id":27535,"date":"2020-07-04T10:46:08","date_gmt":"2020-07-04T05:16:08","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27535"},"modified":"2021-07-02T15:07:07","modified_gmt":"2021-07-02T09:37:07","slug":"kseeb-solutions-for-class-8-maths-chapter-8-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-8-maths-chapter-8-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 8 Linear Equations in One Variable Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions<\/h2>\n

Question 1.
\nChoose the correct answer
\na. The value of \u2018x\u2019 in the equation 5x – 35 = 0 is
\nA. 2
\nB. 7
\nC. 8
\nD. 11
\nSolution:
\n(B) 7<\/p>\n

b. If 14 is taken away from one fifth of a number, the result is 20. The equation expressing this statement is
\nA. \\(\\left(\\frac{x}{5}\\right)-14=20\\)
\nB. \\(x-\\left(\\frac{14}{5}\\right)=\\frac{20}{5}\\)
\nC. \\(x-14=\\left(\\frac{20}{5}\\right)\\)
\nD. \\(x+\\left(\\frac{14}{5}\\right)=20\\)
\nSolution:
\nA. \\(\\left(\\frac{x}{5}\\right)-14=20\\)<\/p>\n

c. If five times a number is increased by 8 is 53, the number is
\nA. 12
\nB. 9
\nC. 11
\nD. 2
\nSolution:
\nB. 9<\/p>\n

d. The value of x in the equation 5 (x – 2) = 3 (x – 3) is
\nA. 2
\nB. \\(\\frac{1}{2}\\)
\nC. \\(\\frac{3}{4}\\)
\nD. 0
\nSolution:
\nB. \\(\\frac{1}{2}\\)<\/p>\n

e. If the sum of two numbers is 84 and their difference is 30, the numbers are
\nA. – 57 and 27
\nB. 57 and 27
\nC. 57 and – 27
\nD. – 57 and – 27
\nSolution:
\n(B) 57 and 27<\/p>\n

f. If the area of a rectangle whose length is twice its breadth is 800 m2<\/sup>, then the length and breadth of the rect-angle are
\nA. 60 m and 20 m
\nB. 40 m and 20m
\nC. 80 m and 10 m
\nD. 100 m and 8m
\nSolution:
\n(B) 40 m and 20 m<\/p>\n

g. If the sum of three consecutive odd numbers is 249, the numbers are
\nA. 81, 83, 85
\nB. 79, 81, 83
\nC. 103, 105, 10
\nD. 95, 97, 99
\nSolution:
\n(A) 81, 83, 85<\/p>\n

h. If \\(\\frac{(x+0.7 x)}{x}\\) = 0.85 the value of x is
\nA. 2
\nB. 1
\nC. – 1
\nD. 0
\nSolution:
\nB. 1<\/p>\n

i. If 2x – (3x – 4) = 3x – 5 then x equals
\nA. \\(\\frac{4}{9}\\)
\nB. \\(\\frac{9}{4}\\)
\nC. \\(\\frac{3}{2}\\)
\nD. \\(\\frac{2}{3}\\)
\nSolution:
\nB. \\(\\frac{9}{4}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nSolve
\n(i) \\(\\frac{(3 x+24)}{(2 x+7)}=2\\)
\nSolution
\n\\(\\frac{(3 x+24)}{(2 x+7)}=2\\)
\n3x + 24 = 2( 2x + 7)
\n3x + 24 = 4x + 14
\n24 – 14 = 4x – 3x
\n10 = x
\nor x = 10<\/p>\n

(ii) \\(\\frac{(1-9 y)}{(11-3 y)}=\\frac{5}{8}\\)
\nSolution
\n\\(\\frac{(1-9 y)}{(11-3 y)}=\\frac{5}{8}\\)
\n8 (1 – 9y) = 5 ( 11 – 3y)
\n8 – = 55 – 15y
\n8 – 55 = 72 y – 15 y
\n– 47 = 57 y
\n\\(\\frac{-47}{57}\\) = y
\nor y = \\(\\frac{-47}{57}\\)<\/p>\n

Question 3.
\nThe sum of two numbers is 45 and their ratio is 7 : 8. Find the numbers.
\nSolution:
\nRatio of the number = 7 : 8
\nLet the numbers be 7x and 8x
\nsum of the numbers = 45
\n7x + 8x = 45
\n15x = 45
\nx = \\(\\frac{45}{15}\\)
\n7x = 7 x 3 = 21, 8x = 8 x 3 = 24
\n\u2234 The numbers are 21 and 24<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nShona\u2019s mother is four times as old as shona. After 5 years,her mother will be three times as old as shona (at that time). What are their present age?
\nSolution:
\nLet shona\u2019s present age be \u2018x\u2019 years.
\nHer mother\u2019s present age = 4x years
\n5 years later shona\u2019s age = x + 5 and
\nshona\u2019s mother\u2019s age = 4x + 5
\nGiven that 4x + 5 = 3 (x + 5)
\n4x + 5 = 3x + 15
\n4x – 3x = 15 – 5
\nx = 10
\nShonas present age = x = 10 years
\nHer mothers present age = 4x = 4 x 10 = 40 years.<\/p>\n

Question 5.
\nThe sum of 3 consecutive even numbers is 336. Find them.
\nSolution:
\n\u2018 Let the first even number be \u2018x\u2019
\nThe consecutive even numbers are x + 2 and x + 4
\nTheir sum = 336
\nx + x + 2 + x + 4 = 335
\n3x + 6 = 336
\n3x = 336 – 6
\n3x = 330
\nx = \\(\\frac{330}{3}\\) = 110
\nx = 110, x + 2 = 110 + 2
\n= 112, x + 4 = 110 + 4 = 114.
\n\u2234 The numbers are 11, 112, 114.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nTwo friend\u2019s A and B start a joint business with a capital Rs. 60,000. If As share is twice that of B how much have each invested?
\nSolution:
\nLet \u2018B\u2019 Invest Rs. x. Then \u2018A\u2019s investment is 2x.
\nTotal Investment = 60,000
\nx + 2x = 60,000
\n3x = 60,000
\nx = \\(\\frac{60000}{3}\\)
\nx = 20,000
\n\u2018B\u2019s investment = x = Rs. 20,000
\nAs investment = 2x = 2 x 20,000 = Rs. 40,000<\/p>\n

Question 7.
\nWhich is the number when 40 is subtracted gives one third of the original number.
\nSolution:
\nLet the number be \u2018x\u2019. One third the
\n\\(\\frac{1}{3}\\) x number is x = \\(\\frac{x}{3}\\)
\nGiven that x – 40 = \\(\\frac{x}{3}\\)
\n3 (x – 40) = x
\n3x – 120 = x
\n3x – x = 120
\n2x = 120
\nx = \\(\\frac{120}{2}\\)
\nx = 60
\n\u2234 The original number is 60<\/p>\n

Question 8.
\nFind the number whose 6th part exceeds its eighth part by 3.
\nSolution:
\nLet the number be \u2018x\u2019. Its 6th part is \\(\\frac{x}{6}\\) and 8th part is \\(\\frac{x}{8}\\)
\n\"KSEEB
\n\"KSEEB
\nx = 72, The number is 72<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nA house and a garden together cost Rs. 8,40,000. The price of the garden is \\(\\frac{5}{12}\\) times the total cost. Find the price of the house and the garden.
\nSolution:
\nThe price of the garden is
\n\\(\\frac{5}{12}\\) \u00d7 8,40,000 . 12
\n= 5 \u00d7 70,000
\n= Rs. 3,50,000
\nThe cost of the house is 8,40,000 – 3,50,000 = 4,90,000
\nThe cost of the house is Rs. 4,90,000<\/p>\n

Question 10.
\nTwo farmers A and B together own a stock of grocery. They agree to divide it by value. Farmer A takes 72 bags while B takes 92 bags and gives Rs. 8,000 to A. What is the cost o each bag?
\nSolution:
\nLet the cost of each bag be \u2018x’.
\nA takes 72 bags and Receives Rs. 8000
\n\u2234 The value of the stock = 72x + 8000
\nB takes 92 bags and gives Rs. 8000
\nThe value of the stock = 92x – 8000
\n\u2234 72x + 8000 = 92x – 8000
\n8000 + 8000 = 92x – 7x
\n16000 = 20x
\n\u2234 x = \\(\\frac{16000}{20}\\)
\nx = 800
\n\u2234 The cost of each bag is Rs. 800<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nA fathers age is four times that of his son. After 5 year s, it will be three times that of his son. How many more years will take if fathers age is to be twice that of his son?
\nSolution:
\nLet the sons age be x. Fathers age is 4x .
\nAfter 5 years sons age = x + 5 years and fathers age = 4x + 5
\nGiven (4x + 5) = 3 (x + 5)
\n4x + 5 = 3x + 15 4x – 3x = 15 – 5 x = 10
\nSons present age = 10 years
\nFathers present age = 4 \u00d7 x = 4 \u00d7 10 = 40 years.
\nLet in \u2018y\u2019 years fathers age will be twice as his sons age.
\nAfter ‘y’ years son will be (10 + y) years and father will be (40 + y) years.
\n40 + y = 2 (10 + y)
\n40 + y = 20 + 2y
\n40 – 20 = 2y – y
\n20 = y
\ny = 20 \u2234 20 – 5 = 15 years
\nIn 15 years fathers age wi11 be twice his sons age.<\/p>\n

Question 12.
\nFind a number which when multiplied by 7 is as much above 132 as it was origina11y below it.
\nSolution:
\nLet the number be x. When its multi, plied by 7 the product is 7x. This number is above 132 by (7x – 12). This is same as it was origina11y below it i.e., (132 – x)
\n\u2234 7x – 132 = 132 – x
\n7x + x = 132 + 132
\n8x = 264
\nx = \\(\\frac{264}{8}\\) ; x = 33
\nThe number is 33<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nA person buys 25 pens worth Rs. 250, each of equal cost. He wants to keep 5 pens for himself and se11 the remaining to recover his money what should be the price of each pen?
\nSolution:
\nNumber pens bought = 25
\nThe cost of 25 pens = Rs. 250
\n\u2234 The cost of each pen
\n\\(\\frac{250}{25}\\) = Rs. 10
\nHe keeps 5 pens with him and se11s the remaining (25 – 5) 20 pens to recover the same amount. Let the S.R of each pen be x.
\n20x = 250
\nx = \\(\\frac{250}{20}\\) = Rs. 12.5
\nThe price of each pen should be Rs. 12.50<\/p>\n

Question 14.
\nThe sum of the digits of a two digit number is 12. If the new number formed by reversing the digits is greater than the original number by 18. Find the original number. Check your solution.
\nSolution:
\nLet the digit in the units place be \u2018x\u2019 and digit in the ten\u2019s place be \u2018y
\nSum of the digits = 12
\nx + y = 12
\ny = 12 – x
\n\u2234 The original number is 10 y + x
\nThe number got by reversing the digits is 10 x + y
\nGiven that 10 x + y = 10 y + x + 18
\n10x + (12 – x ) = 10 (12 – x) + x 18
\n10x + 12 – x = 120 – 10 x + x + 18
\n9x + 12 = 138 – 9x
\n9x + 9x = 138 – 12
\n18x = 126
\nx = \\(\\frac{126}{18}\\) = 7
\ny = 12 – x
\n= 12 – y = 5
\n\u2234 The original number is (10 x 5) +7 = 50 + 7 = 57<\/p>\n

Verification :
\nOriginal number = 57
\nThe new number = 75
\n75 – 57 = 18<\/p>\n

\"KSEEB<\/p>\n

Question 15.
\nThe distance between two stations is 340 km. Two trains start simaltaneously from these stations on para11el tracks and cross each other. The speed of one of them is greater than that of the other by 5km \/ hr. If the distance between two trains after 2 hrs of their start is 30 km find the speed of each train.
\nSolution:
\nLet the speed of one train be x km \/hr. The speed of the other train is (x + 5) km \/ hr
\nDistance covered by the trains in 2 hrs are 2x and 2 (x + 5) respectively. After crossing each other the distance between the trains is 30 km. Therefore the total distance covered = 340 + 30 = 370 km
\n2x + 2 (x + 5) = 370
\n2x + 2x + 10 = 370
\n4x + 10 = 370
\n4x = 370 – 10
\n4x = 360
\nx = \\(\\frac{360}{4}\\) = 90
\nx = 90 km \/ hr
\nThe speeds of the trains are 90 km\/hr and 90 + 5 = 95 km\/ hr.<\/p>\n

Question 16.
\nA steamer goes down stream and coveres the distance between two ports in 4 hrs while it coveres the same distance upstream in 5 hrs. If the speed of the steamer up stream is 2km \/ hr find the speed of the steamer in sti11 water.
\nSolution
\nSpeed upstream = 2km \/ hr
\nTime taken up stream = 5 hr
\n\u2234 Distance between two ports
\n= 2 x 5 = 10 km<\/p>\n

\"KSEEB<\/p>\n

Time taken down stream = 4 hrs Distance between two ports = 10 km
\nspeed down stream<\/p>\n

\"KSEEB<\/p>\n

Let the speed of the steamer in sti11 water be \u2018x\u2019 km \/ hr and speed of the stream be y km \/ hr.<\/p>\n

\"KSEEB
\nx = 2.25 km \/ hr
\n\u2234 The spped of the steamer in sti11 water is 2.25 km \/ hr.<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nThe numerator of rational number is less than its denaminator by 3. If the numerator becomes 3 times and denominator is increased by 20 the new number becomes \\(\\frac{1}{8}\\). Find the original number.
\nSolution:
\nLet the denominator be \u2018x. The numera-tor is x – 3. The original fraction is \\(\\frac{x-3}{x}\\)
\nIf the numerator becomes 3 times it is 3 (x – 3) and denominator is increased by 20 it is x + 20 new number is – \\(\\frac{1}{8}\\)<\/p>\n

\"KSEEB
\n8(3x – 9) = x + 20
\n24x – 72 = x + 20
\n24x – x = 20 + 72
\n23x = 92
\nx = \\(\\frac{92}{23}\\)
\nx = 4
\n\u2234 original fraction = \\(\\frac{x-3}{x}=\\frac{4-3}{4}=\\frac{1}{4}\\)<\/p>\n

Question 18.
\nThe digit at the tens place of a two digit number is three times the digit at the units place. If the sum of this number and the number formed by reversing the digits is 88 find the number.
\nSolution:
\nLet the digit in the units place be x and tens place be y.
\nGiven y = 3x \u2234 The original number is 10 y + x
\nThe number formed by reversing the digits is 10 x + y.
\nSum of the numbers = 88
\n10 y + x + 10 x + y = 88
\n11y + 11x = 88
\nbut y = 3x
\n\u2234 11 (3x) + 11x = 88
\n33x + 11x = 88
\n44x = 88
\nx = \\(\\frac{88}{44}\\)
\nx = 2
\ny = 3x = 3 x 2 = 6
\nThe original number is 10 y + x = 10 x 6 + 2 = 62<\/p>\n

\"KSEEB<\/p>\n

Question 19.
\nThe altitude of a triangle is five thirds the length of its corresponding base. If the altitude is increased by 4cm and the base decreased by 2 cm the area of the triangle would remain the same. Find the base and altitude of the triangle.
\nSolution:
\nLet the base of the triangle be \u2018x\u2019 cm
\nThe altitude is \\(\\frac{5x}{3}\\) cm
\nArea of the traingle = \\(\\frac{1}{2}\\) x base x height<\/p>\n

\"KSEEB<\/p>\n

If the height is increased by 4cm it is \\(\\left(\\frac{5 x^{2}}{3}+4\\right)\\) and the base is decreased by 2 cm it is (x – 2) the area remains the same.
\n\"KSEEB
\nx(5x + 12)- 2(5x +12) = 6 x \\(\\frac{5 x^{2}}{6}\\)
\n5x2<\/sup> + 12x – 10x – 24 = 5x2<\/sup>
\n5x2<\/sup> + 2x – 24 – 5x2<\/sup> = 0
\n2x = 24
\nx = \\(\\frac{24}{2}\\) = 12
\nx = 12
\n\\(\\frac{5 x}{3}=\\frac{5 \\times 12}{3}=\\frac{60}{3}=20 \\mathrm{cm}\\)
\nThe base of the triangle is 12 cm and altitude is 20 cm<\/p>\n

Question 20.
\nOne of the angles of a triangle is equal to the sum of the other two angles. If the ratio of the other two angles of the triangle is 4 : 5 find the angles of the triangles.
\nSolution:
\nThe ratio of the two angles is 4 : 5 Let the angles be 4x and 5x. Sum of the two angles = 4x + 5x = 9x. The third angle is equal to sum of the other , two angles i.e., 9x.
\nSum of the angles of a traingle
\n= 180\u00b0
\n9x + 9x = 180
\n18x = 180
\n\u2234 x = \\(\\frac{180}{8}\\)
\nx = 10\u00b0
\n4x = 4 x 10 = 40\u00b0
\n5x = 5 x 10 = 50\u00b0
\n9x = 9 x 10 = 90\u00b0 .
\n\u2234 The angles of the triangle are 90\u00b0, 40\u00b0 and 50\u00b0.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 8 Linear Equations in One Variable Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions Question …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27535"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=27535"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27535\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27535"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27535"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27535"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}