{"id":4698,"date":"2021-03-01T18:17:39","date_gmt":"2021-03-01T12:47:39","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=4698"},"modified":"2021-09-15T11:40:36","modified_gmt":"2021-09-15T06:10:36","slug":"ap-ssc-10th-class-maths-solutions-chapter-4-ex-4-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-4-ex-4-2\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.2<\/h2>\n

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers<\/h3>\n

Form a pair of linear equations for each of the following problems and find their solution.
\nQuestion 1.
\nThe ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly income.
\nAnswer:
\nGiven ratio of incomes of two persons = 9 : 7
\nSo let the incomes of each = Rs. 9x and Rs. 7x
\nand ratio of expenditures = 4 : 3
\nSo let the expenditures of each = 4y and 3y
\nthen earnings of each = (income – expenditure) of each
\n\u21d2 9x – 4y = Rs. 2000 and 7x – 3y = 200
\n\u2234 9x – 4y = 7x – 3y = 2000
\n\u21d2 9x – 7x = 4y – 3y
\n\u21d2 y = 2x
\nnow putting y = 2x in 9x – 4y = 2000 we get
\n9x – 4(2x) = 2000 \u21d2 x = 2000
\n\u2234 Income of each = 9x = 9(2000) = 18000
\nand 7x = 7(2000) = 14,000<\/p>\n

\"AP<\/p>\n

Question 2.
\nThe sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
\nAnswer:
\nLet the digit in units place be x
\nand the digit in tens place be y
\nthen the value of the number = 10y + x
\nNumber obtained by reversing the digits = 10x + y
\nBy problem,
\n(10y + x) + (10x + y) = 66
\nand x – y = 2
\n\u21d2 11x – 11y = 66 and x – y = 2
\n\u21d2 x + y = 6 and x – y = 2
\nSolving these two equations
\nx + y = 6
\nx – y = 2
\n(+) 2x = 8
\nx = \\(\\frac{8}{2}\\) = 4
\nSubstituting x = 4 in x + y = 6
\nwe get 4 + y = 6 \u21d2 y = 2
\nSubstituting x, y values in equations (10y + x) & (10x + y),
\nWe get 10y + x
\n= 10(2) + 4 = 20 + 4 = 24
\nand 10x + y = 10(4) + 2
\n= 40 + 2 = 42
\n\u2234 The number is 42 or 24
\nThus we have two such numbers.<\/p>\n

Question 3.
\nThe larger of two supplementary angles exceeds the smaller by 18\u00b0. Find the angles.
\nAnswer:
\nLet the pair of supplementary angles be x and y [and x > y]
\nthen we have x + y = 180\u00b0 …… (1)
\nBy problem, x = y + 18\u00b0
\n\u21d2 x – y = 18\u00b0 …… (2)
\nSolving the equations (1) and (2) we get
\n\"AP
\nand x = \\(\\frac{198}{2}\\) = 99\u00b0
\nSubstituting x = 99\u00b0 in equation (2) we get
\n99\u00b0 – y\u00b0 = 18\u00b0
\n\u21d2 y\u00b0 = 99\u00b0 – 18 = 81\u00b0
\n\u2234 The angles are 99\u00b0 and 81\u00b0.<\/p>\n

\"AP<\/p>\n

Question 4.
\nThe taxi charges in Hyderabad are fixed, along with the charge for the distance covered. For a distance of 10 km., the charge paid is Rs. 220. For a journey of 15 km. the charge paid is Rs. 310.
\ni) What are the fixed charges and charge per km?
\nii) How much does a person have to pay for travelling a distance of 25km?
\nAnswer:
\nLet the fixed charge be = Rs. x.
\nand the charge per one km = Rs. y.
\nBy problem, x + 10y = 220 x + 15y = 310
\nSolving (1) and (2) we get
\n\"AP
\n\u2234 y = \\(\\frac{-90}{-5}\\) = 18
\ni.e., charge per one km = Rs. 18
\nSubstituting y = 18 in equation (1) we get
\nx + 10 \u00d7 18 = 220
\n\u21d2 x = 220 – 180
\n\u21d2 x = Rs. 40
\n\u2234 Fixed charge = Rs. 40;
\nCharge per km = Rs. 18.<\/p>\n

ii) Now, the charge for travelling a distance of 25 km = 25 \u00d7 18
\n= Rs. 450 + 40
\n= Rs. 490<\/p>\n

Question 5.
\nA fraction becomes equal to \\(\\frac{4}{5}\\) if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes equal to \\(\\frac{1}{2}\\). What is the fraction?
\nAnswer:
\nLet the numerator of the fraction = x
\nand the denominator of the fraction = y
\nBy problem,
\n\\(\\frac{x+1}{y+1}\\) = \\(\\frac{4}{5}\\) and \\(\\frac{x-5}{y-5}\\) = \\(\\frac{1}{2}\\)
\n\u21d2 5(x + 1) = 4(y + 1) and 2(x – 5) = 1(y – 5)
\n5x + 5 = 4y + 4 and 2x – 10 = y – 5
\n\u21d2 5x – 4y = 4 – 5 and 2x – y = – 5 + 10
\n\u21d2 5x – 4y = – 1 …… (1)
\nand 2x – y = 5 …… (2)
\n\"AP
\n\u2234 y = \\(\\frac{-27}{-3}\\) = 9
\nSubstituting y = 9 in equation (2) we get
\n2x – 9 = 5
\n\u21d2 2x = 5 + 9
\n\u21d2 2x = 14 and
\nx = \\(\\frac{14}{2}\\) = 7
\nThus the fraction is \\(\\frac{x}{y}\\) = \\(\\frac{7}{9}\\)<\/p>\n

\"AP<\/p>\n

Question 6.
\nPlaces A and B are 100 km apart on a highway One car starts from A and another from B at the same time at different speeds. If the cars travel in the same direction, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
\nAnswer:
\n\"AP
\nLet the speed of the car which started from the place A = x kmph
\nand B = y kmph
\nDistance travelled by first car in 5h = 5x and in 1h = x
\nThe distance covered by second car in 5h = 5y and in 1h = y
\nBy problem when travelled in same direction,
\n5x – 5y = 100 \u21d2 x – y = 20 …… (1)
\nand when travelled towards each other
\nx + y = 100 ……. (2)
\nSolving (1) and (2),
\n\"AP
\n\u2234 x = \\(\\frac{120}{2}\\) = 60
\nSubstituting x = 60 in equation (1) we get
\n60 – y = 20
\n\u21d2 y = 60 – 20 = 40 kmph
\nThus the speed of the cars are 60 kmph and 40 kmph.<\/p>\n

Question 7.
\nTwo angles are complementary. The larger angle is 3\u00b0 less than twice the measure of the smaller angle. Find the measure of each angle.
\nAnswer:
\nLet the pair of complementary angles be x\u00b0 and y\u00b0 with x\u00b0 > y\u00b0
\nthen x\u00b0 + y\u00b0 = 90\u00b0 and
\nBy problem
\nx = 2y – 3\u00b0 \u21d2 x – 2y = – 3\u00b0
\nSolving these two equations we get,
\n\"AP
\n\u2234 y = \\(\\frac{93}{3}\\) = 31\u00b0
\nSubstituting y = 31\u00b0in x + y = 90\u00b0 we get
\nx + 31\u00b0 = 90\u00b0
\n\u21d2 x = 90\u00b0 – 31\u00b0 = 59\u00b0
\nThe angles are 59\u00b0 and 31\u00b0.<\/p>\n

\"AP<\/p>\n

Question 8.
\nAn algebra textbook has a total of 1382 pages. It is broken up into two parts. The second part of the book has 64 pages more than the first part. How many pages are in each part of the book?
\nAnswer:
\nLet the first part of the book contains x pages
\nand the second part of the book contains y pages By problem,
\nx + y = 1382 ….. (1)
\ny = x + 64 \u21d2 x – y = -64 …… (2)
\nSolving equations (1) and (2) we get
\n\"AP
\n\u2234 x = \\(\\frac{1318}{2}\\) = 659
\nSubstituting x = 659 in equation (1) we get
\n659 + y = 1382
\n\u21d2 y = 1382 – 659 = 723
\n\u2234 The number of pages in the first part = 659
\nSecond part = 723<\/p>\n

Question 9.
\nA chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the other is 80% solution. How much of each should be used to obtain 100 ml of a 68% solution?
\nAnswer:
\nLet the first solution contains 50% acid.
\nSecond solution contains 80% acid.
\nLet x ml of 1st solution and y ml of second solution are added.
\nThen x + y = 100
\nAcid content in the \u2018mix\u2019 is 50% of x + 80% of y = 68%
\n\"AP
\n\u2234 y = \\(\\frac{180}{3}\\) = 60
\nSubstituting y = 60 in equation (1) we get
\nx + 60 = 100
\n\u21d2 x = 100 – 60 = 40
\n\u2234 Quantity of first solution = 40 ml
\nQuantity of second solution = 60 ml<\/p>\n

\"AP<\/p>\n

Question 10.
\nSuppose you have Rs. 12000 to invest. You have to invest some amount at 10% and the rest at 15%. How much should be invested at each rate to yield 12% on the total amount invested ?
\nAnswer:
\nLet the amount to be invested @ 10% be Rs. x
\nand the amount to be invested @ 15% be Rs. y
\nBy problem x + y = 12000 ……. (1)
\nAlso 10% of x + 15% of y = 12% of 12000
\n\"AP
\n\u21d2 y = \\(\\frac{-24000}{-5}\\) = Rs. 4800
\nSubstituting y = 4800 in equation (1) we get
\nx + 4800 = 12000
\n\u21d2 x = 12000 – 4800 = 7200
\nThe invested @ 10% = Rs. 7200
\n@ 15% = Rs. 4800<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.2 10th Class Maths 4th Lesson Pair of Linear Equations in Two …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4698"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=4698"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4698\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=4698"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=4698"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=4698"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}