{"id":95336,"date":"2019-10-04T10:31:24","date_gmt":"2019-10-04T05:01:24","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=95336"},"modified":"2021-09-18T15:16:38","modified_gmt":"2021-09-18T09:46:38","slug":"ncert-solutions-for-class-8-maths-linear-equation-in-one-variable-ex-2-2","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-8-maths-linear-equation-in-one-variable-ex-2-2\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2"},"content":{"rendered":"
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2<\/strong><\/p>\n Ex 2.2 Class 8 Maths Question 1. Ex 2.2 Class 8 Maths\u00a0Question 2. Ex 2.2 Class 8 Maths\u00a0Question 3. <\/p>\n Ex 2.2 Class 8 Maths\u00a0Question 4. Ex 2.2 Class 8 Maths\u00a0Question 5. Ex 2.2 Class 8 Maths\u00a0Question 6. Ex 2.2 Class 8 Maths\u00a0Question 7. Ex 2.2 Class 8 Maths\u00a0Question 8. Ex 2.2 Class 8 Maths\u00a0Question 9. Ex 2.2 Class 8 Maths\u00a0Question 10. Ex 2.2 Class 8 Maths\u00a0Question 11. Ex 2.2 Class 8 Maths\u00a0Question 12. Ex 2.2 Class 8 Maths\u00a0Question 13. Ex 2.2 Class 8 Maths\u00a0Question 14. Ex 2.2 Class 8 Maths\u00a0Question 15. Ex 2.2 Class 8 Maths\u00a0Question 16. <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2 Class 8 Maths Linear Equations in One Variable Exercise 2.1 Class 8 Maths Linear Equations in One Variable Exercise 2.2 Class 8 Maths Linear Equations in One Variable Exercise 2.3 Class 8 Maths Linear Equations in One Variable Exercise 2.4 …<\/p>\n","protected":false},"author":27,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\n
\nIf you subtract \\(\\frac { 1 }{ 2 }\\) from a number and multiply the result by \\(\\frac { 1 }{ 2 }\\), you get \\(\\frac { 1 }{ 8 }\\). What is the number?
\nSolution:
\nLet the required number be x.
\n<\/p>\n
\nThe perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
\nSolution:
\nLet the breadth of the pool be x m.
\nCondition I: Length = (2x + 2) m.
\nCondition II: Perimeter = 154 m.
\nWe know that Perimeter of rectangle = 2 \u00d7 [length + breadth]
\n2 \u00d7 [2x + 2 + x] = 154
\n\u21d2 2 \u00d7 [3x + 2] = 154
\n\u21d2 6x + 4 = 154 (solving the bracket)
\n\u21d2 6x = 154 – 4 [Transposing 4 from (+) to (-)]
\n\u21d2 6x = 150
\n\u21d2 x = 150 \u00f7 6 [Transposing 6 from (\u00d7) to (\u00f7)]
\n\u21d2 x = 25
\nThus, the required breadth = 25 m
\nand the length = 2 \u00d7 25 + 2 = 50 + 2 = 52 m.<\/p>\n
\nThe base of an isosceles triangle is \\(\\frac { 4 }{ 3 }\\) cm. The perimeter of the triangle is 4\\(\\frac { 2 }{ 15 }\\) cm. What is the length of either of the remaining equal sides?
\nSolution:
\nLet the length of each of equal sides of the triangle be x cm.
\nPerimeter of the triangle = sum of the three sides
\n<\/p>\n
\nSum of two numbers be 95. If one exceeds the other by 15, find the numbers.
\nSolution:
\nLet one number be x
\nOther number = x + 15
\nAs per the condition of the question, we get
\nx + (x + 15) = 95
\n\u21d2 x + x + 15 = 95
\n\u21d2 2x + 15 = 95
\n\u21d2 2x = 95 – 15 [transposing 15 from (+) to (-)]
\n\u21d2 2x = 80
\n\u21d2 x = \\(\\frac { 80 }{ 2 }\\) [transposing 2 from (\u00d7) to (\u00f7)]
\n\u21d2 x = 40
\nOther number = 95 – 40 = 55
\nThus, the required numbers are 40 and 55.<\/p>\n
\nTwo numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
\nSolution:
\nLet the two numbers be 5x and 3x.
\nAs per the conditions, we get
\n5x – 3x = 18
\n\u21d2 2x = 18
\n\u21d2 x = 18 \u00f7 2 [Transposing 2 from (\u00d7) to (\u00f7)]
\n\u21d2 x = 9.
\nThus, the required numbers are 5 \u00d7 9 = 45 and 3 \u00d7 9 = 27<\/p>\n
\nThree consecutive integers add up to 51. What are these integers?
\nSolution:
\nLet the three consecutive integers be x, x + 1 and x + 2.
\nAs per the condition, we get
\nx + (x + 1) + (x + 2) = 51
\n\u21d2 x + x + 1 + x + 2 = 51
\n\u21d2 3x + 3 = 51
\n\u21d2 3x = 51 – 3 [transposing 3 to RHS]
\n\u21d2 3x = 48
\n\u21d2 x = 48 \u00f7 3 [transposing 3 to RHS]
\n\u21d2 x = 16
\nThus, the required integers are 16, 16 + 1 = 17 and 16 + 2 = 18, i.e., 16, 17 and 18.<\/p>\n
\nThe sum of three consecutive multiples of 8 is 888. Find the multiples.
\nSolution:
\nLet the three consecutive multiples of 8 be 8x, 8x + 8 and 8x + 16.
\nAs per the conditions, we get
\n8x + (8x + 8) + (8x + 16) = 888
\n\u21d2 8x + 8x + 8 + 8x + 16 = 888
\n\u21d2 24x + 24 = 888
\n\u21d2 24x = 888 – 24 (transposing 24 to RHS)
\n\u21d2 24x = 864
\n\u21d2 x = 864 \u00f7 24 (transposing 24 to RHS)
\n
\n\u21d2 x = 36
\nThus, the required multiples are
\n36 \u00d7 8 = 288, 36 \u00d7 8 + 8 = 296 and 36 \u00d7 8 + 16 = 304,
\ni.e., 288, 296 and 304.<\/p>\n
\nThree consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3, and 4 respectively, they add up to 74. Find these numbers.
\nSolution:
\nLet the three consecutive integers be x, x + 1 and x + 2.
\nAs per the condition, we have
\n2x + 3(x + 1) + 4(x + 2) = 74
\n\u21d2 2x + 3x + 3 + 4x + 8 = 74
\n\u21d2 9x + 11 = 74
\n\u21d2 9x = 74 – 11 (transposing 11 to RHS)
\n\u21d2 9x = 63
\n\u21d2 x = 63 \u00f7 9
\n\u21d2 x = 7 (transposing 7 to RHS)
\nThus, the required numbers are 7, 7 + 1 = 8 and 7 + 2 = 9, i.e., 7, 8 and 9.<\/p>\n
\nThe ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?
\nSolution:
\nLet the present ages of Rahul and Haroon he 5x years and 7x years respectively.
\n4 years later, the age of Rahul will be (5x + 4) years.
\n4 years later, the age of Haroon will be (7x + 4) years.
\nAs per the conditions, we get
\n(5x + 4) + (7x + 4) = 56
\n\u21d2 5x + 4 + 7x + 4 = 56
\n\u21d2 12x + 8 = 56
\n\u21d2 12x = 56 – 8 (transposing 8 to RHS)
\n\u21d2 12x = 48
\n\u21d2 x = 48 \u00f7 12 = 4 (transposing 12 to RHS)
\nHence, the required age of Rahul = 5 \u00d7 4 = 20 years.
\nand the required age of Haroon = 7 \u00d7 4 = 28 years.<\/p>\n
\nThe number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the numbers of girls. What is the total class strength?
\nSolution:
\nLet the number of boys be 7x
\nand the number of girls be 5x
\nAs per the conditions, we get
\n7x – 5x = 8
\n\u21d2 2x = 8
\n\u21d2 x = 8 \u00f7 2 = 4 (transposing 2 to RHS)
\nthe required number of boys = 7 \u00d7 4 = 28
\nand the number of girls = 5 \u00d7 4 = 20
\nHence, total class strength = 28 + 20 = 48<\/p>\n
\nBaichung\u2019s father is 26 years younger than Baichung\u2019s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
\nSolution:
\nLet the age of Baichung be x years.
\nThe age of his father = x + 29 years,
\nand the age of his grandfather = x + 29 + 26 = (x + 55) years.
\nAs per the conditions, we get
\nx + x + 29 + x + 55 = 135
\n\u21d2 3x + 84 = 135
\n\u21d2 3x = 135 – 84 (transposing 84 to RHS)
\n\u21d2 3x = 51
\n\u21d2 x = 51 \u00f7 3 (transposing 3 to RHS)
\n\u21d2 x = 17
\nHence Baichung\u2019s age = 17 years
\nBaichung\u2019s father\u2019s age = 17 + 29 = 46 years,
\nand grand father\u2019s age = 46 + 26 = 72 years.<\/p>\n
\nFifteen years from now Ravi\u2019s age will be four times his present age. What is Ravi\u2019s present age?
\nSolution:
\nLet the present age of Ravi be x years.
\nAfter 15 years, his age will be = (x + 15) years
\nAs per the conditions, we get
\n\u21d2 x + 15 = 4x
\n\u21d2 15 = 4x – x (transposing x to RHS)
\n\u21d2 15 = 3x
\n\u21d2 15 \u00f7 3 = x (transposing 3 to LHS)
\n\u21d2 x = 5
\nHence, the present age of Ravi = 5 years.<\/p>\n
\nA rational number is such that when you multiply it by \\(\\frac { 5 }{ 2 }\\) and add \\(\\frac { 2 }{ 3 }\\) to the product, you get \\(\\frac { -7 }{ 12 }\\). What is the number?
\nSolution:
\nLet the required rational number be x.
\nAs per the condition, we get
\n
\nHence, the required rational number is \\(\\frac { -1 }{ 2 }\\)<\/p>\n
\nLakshmi is a cashier in a bank. She has currency notes of denominations \u20b9 100, \u20b9 50 and \u20b9 10, respectively. The ratio of these notes is 2 : 3 : 5. The total cash with Lakshmi is \u20b9 4,00,000. How many notes of each denomination does she have?
\nSolution:
\nLet the number of \u20b9 100, \u20b9 50 and \u20b9 10 notes be 2x, 3x and 5x respectively.
\nConverting all the denominations into rupees, we have
\n2x \u00d7 100, 3x \u00d7 50 and 5x \u00d7 10 i.e. 200x, 150x and 50x
\nAs per the conditions, we have
\n200x + 150x + 50x = 4,00,000
\n\u21d2 400x = 4,00,000
\n\u21d2 x = 4,00,000 \u00f7 400 (transposing 400 to RHS)
\n\u21d2 x = 1,000
\nHence, the required number of notes of
\n\u20b9 100 notes = 2 \u00d7 1000 = 2000
\n\u20b9 50 notes = 3 \u00d7 1000 = 3000
\nand \u20b9 10 notes = 5 \u00d7 1000 = 5000<\/p>\n
\nI have a total of \u20b9 300 in coins of denomination \u20b9 1, \u20b9 2 and \u20b9 5. The number of \u20b9 2 coins is 3 times the number of \u20b9 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
\nSolution:
\nLet the number of \u20b9 5 coins be x.
\nNumber of \u20b9 2 coins = 3x
\nTotal number of coins = 160
\nNumber of \u20b9 1 coin = 160 – (x + 3x) = 160 – 4x
\nConverting the number of coins into rupees, we have
\nx coins of \u20b9 5 amount to \u20b9 5x
\n3x coins of \u20b9 2 amount to \u20b9 3x \u00d7 2 = \u20b9 6x
\nand (160 – 4x) coins of \u20b9 1 amount to \u20b9 1 \u00d7 (160 – 4x) = \u20b9 (160 – 4x)
\nAs per the conditions, we have
\n5x + 6x + 160 – 4x = 300
\n\u21d2 7x + 160 = 300
\n\u21d2 7x = 300 – 160 (transposing 160 to RHS)
\n\u21d2 7x = 140
\n\u21d2 x = 140 \u00f7 7 (transposing 7 to RHS)
\n\u21d2 x = 20
\nThus, number of \u20b9 5 coins = 20
\nNumber of \u20b9 2 coins = 3 \u00d7 20 = 60
\nand Number of \u20b9 1 coins = 160 – 4 \u00d7 20 = 160 – 80 = 80<\/p>\n
\nThe organisers of an essay competition decide that a winner in the competition gets a prize of \u20b9 100 and a participant who does not win gets a prize of \u20b9 25. The total prize money distributed is \u20b9 3,000. Find the number of winners, if the total number of participants is 63.
\nSolution:
\nLet the number of winners = x
\nNumber of participants who does not win the prize = (63 – x)
\nAmount got by winners = \u20b9 100 \u00d7 x = \u20b9 100x
\nAmount got by loosers = \u20b9 (63 – x) \u00d7 25 = \u20b9 (1575 – 25x)
\nAs per the conditions, we get
\n100x + 1575 – 25x = 3000
\n\u21d2 75x + 1575 = 3000
\n\u21d2 75x = 3000 – 1575 (transposing 1575 to RHS)
\n\u21d2 75x = 1425
\n\u21d2 x = 1425 \u00f7 75 (Transposing 75 to RHS)
\n
\n\u21d2 x = 19
\nThus, the number of winners = 19<\/p>\nMore CBSE Class 8 Study Material<\/h4>\n
\n