{"id":9587,"date":"2024-02-23T10:10:15","date_gmt":"2024-02-23T04:40:15","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9587"},"modified":"2024-02-24T17:55:04","modified_gmt":"2024-02-24T12:25:04","slug":"ts-10th-class-maths-solutions-chapter-4-ex-4-2","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-maths-solutions-chapter-4-ex-4-2\/","title":{"rendered":"TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2"},"content":{"rendered":"

Students can practice 10th Class Maths Solutions Telangana<\/a> Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 to get the best methods of solving problems.<\/p>\n

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Exercise 4.2<\/h2>\n

Question 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
\n\u20b9 2,000 per month, find their monthly income.
\nSolution:
\nLet the income of the first person be \u20b9 9x and that of the second person be \u20b9 7x.
\nFurther, let the expenditure of the first person and the second person be \u20b9 4y and 3y respectively.
\nThen, the saving of first person = 9x – 4y
\nBy problem, 9x – 4y = 2,000 —— (1)
\nThe saving of second person = 7x – 3y
\nSolving (1) and (2), we get
\nWe equate the coefficients of ‘y’ in the equations.
\n\"TS
\n\u2234 x = 2,000
\nSubstitute x = 2,000 in equation (1), we get
\n9 \u00d7 2,000 – 4y = 2,000
\n18,000 – 4y = 2,000
\n-4y = 2,000 – 18,000 = -16,000
\n\u2234 y = \\(\\frac{-16000}{-4}\\) = 4,000
\nTherefore, monthly income of the first person = ?9x
\n= \u20b9 9 \u00d7 2,000
\n= \u20b9 18,000
\nMonthly income of the second person = \u20b9 7x
\n= \u20b9 7 \u00d7 2,000
\n= \u20b9 14,000<\/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 ?
\nSolution:
\nLet the digit at units place be x.
\nAnd the digit at tens place be y.
\nThe number will be yx.
\nThe value of the number = y \u00d7 10 + x \u00d7 1
\n= 10y + x
\n\"TS
\nNumber obtained by reversing the digits = xy
\nThen the value of the reversed number = 10 \u00d7 x + y \u00d7 1
\n= 10x + y
\nBy problem, we have
\n(10y + x) + (10x + y) = 66
\n\u21d2 11x + 11y = 66
\n\u21d2 x + y = 6 —– (1)
\nIt is given that the digits differ by 2.
\nSo, x – y = 2 (or) y – x = 2 —- (2)
\nSolving (1) and (2), we get
\n\"TS
\nSubstitute x = 4 in (1), we get
\nx + y = 6 \u21d2 y = 6 – 4 = 2
\nSubstitute y = 4 in (1),
\nx + y = 6 \u21d2 x = 6 – 4 = 2
\n\u2234 The required number is 24.
\n\u2234 The required number is 42.
\nThere are two numbers (i.e.,) 24 and 42.<\/p>\n

\"TS<\/p>\n

Question 3.
\nThe larger of two supplementary angles exceeds the smaller by 18\u00b0. Find the angles.
\nSolution:
\nTwo angles are said to be supplementary if their sum is 180\u00b0.
\nLet the smaller supplementary angle be x\u00b0. and the larger supplementary angle be y\u00b0.
\nWe know that sum of these two angles is 180\u00b0.
\n\u2234 x + y = 180\u00b0 —- (1)
\nThe larger angle exceeds the smaller by 18\u00b0. Then, y = x + 18
\n\u21d2 -x + y = 18 —- (2)
\nSolving (1) & (2), we get
\n\"TS
\nSubstitute y = 99 in (1), we get
\nx + 99 = 180
\n\u21d2 x = 180 – 99 = 81
\nTherefore, the required angles are 81\u00b0, 99\u00b0.<\/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 \u20b9 220. For a journey of 15 km, the charge paid is \u20b9 310.
\ni) What are the fixed charges and charge per km ?
\nii) How much does a penon have to pay for travelling a distance of 25 km?
\nSolution:
\nLet the fixed charge be \u20b9 x.
\nAnd charge per km be \u20b9y.
\nFor a distance of 10km, the charge paid is \u20b9 220.
\nThen x + 10y = 220 —– (1)
\nFor a distance of 15 km, the charge paid is \u20b9 310.
\nThen, x + 15y = 310 —- (2) ;
\nSolving (1) & (2), we get
\n\"TS
\nSubstitute y = 18 in equation (1), we get
\nx + (10 \u00d7 18) = 220
\nx + 180 = 220
\n\u2234 x = 220 – 180 = 40
\nFixed charge = \u20b9 40
\nCharge per km = \u20b9 18
\n\u2234 Charge for 25 km = \u20b9 450<\/p>\n

Question 5.
\nA fraction becomes if \\(\\frac{4}{5}\\) is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes \\(\\frac{1}{2}\\). What is the fraction ? (AP Mar. 15)
\nSolution:
\nLet the fraction be; \\(\\frac{x}{y}\\)
\nIf 1 is added to both numerator and denominator then the fraction = \\(\\frac{x+1}{y+1}\\)
\nBy problem, \\(\\frac{x+1}{y+1}\\) = \\(\\frac{4}{5}\\)
\n\u21d2 5(x + 1)= 4(y + 1)
\n\u21d2 5x + 5 = 4y +4
\n\u21d2 5x – 4y = 4 – 5 = -1
\n\u21d2 5x – 4y = -1 —- (1)
\nIf 5 is subtracted from both numerator and denominator, then
\nthe fraction = \\(\\frac{x-5}{y-5}\\)
\nBy problem, \\(\\frac{x-5}{y-5}\\) = \\(\\frac{1}{2}\\)
\n\u21d2 2(x – 5) = 1(y – 5)
\n\u21d2 2x – 10 = y – 5
\n\u21d2 2x – y = -5 + 10 = 5
\n\u21d2 2x – y = 5 —- (2)
\nSolving (1) & (2), we get
\n\"TS
\nSubstitute x = 7 in (2),
\n\u21d2 2 \u00d7 7 – y = 5
\n\u21d2 14 – y = 5
\n\u21d2 -y = 5 – 14 = -9
\n\u2234 y = 9
\n\u2234 The required fraction is \\(\\frac{7}{9}\\).<\/p>\n

\"TS<\/p>\n

Question 6.
\nPlaces A and B are 1oo 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?
\nSolution:
\nLet the speed of the car that starts from A be x kmph.
\nLet the speed of the car that starts from B be y kmph.
\nDistance between A and B is 1oo km.
\n\"TS
\n\u2234 Speed \u00d7 Time = Distance travelled
\n(x – y) \u00d7 5 = 100
\n5x – 5y = 100
\n\u21d2 x – y = 20 —- (1)
\nIf the two cars travel in the opposite direction, their relative speed will be (x + y) kmph. It is also given that the cars will meet in 1 hour.
\n\u2234 Speed \u00d7 Time = Distance travelled
\n(x + y) \u00d7 1 = 100
\n\u21d2 x + y = 100 —- (2)
\nSolving (1) & (2), we get
\n\"TS
\nSubstitute x = 60 in equation (2),
\n60 + y = 100
\n\u2234 y = 100 – 60 = 40
\n\u2234 The speeds 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
\nmeasure of each angle.
\nSolution:
\nTwo angles are said to be complementary if their sum is 90\u00b0.
\nLet the smaller angle be x\u00b0 and the larger angle y\u00b0.
\n\u2234 x + y = 90\u00b0 —- (1)
\nGiven that the larger angle is 30 less than twice the measure of the smaller angle.
\n\u2234 y = 2x – 3
\n\u21d2 2x – y = 3 — (2)
\nSolving (1) & (2), we get
\n\"TS
\nSubstitute x = 31 in (1),
\nThen, 31 + y = 90
\n\u2234 y = 90 – 31 = 59
\nThe two angles are 31\u00b0 and 59\u00b0<\/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
\nmore than the first part. How many pages are in each part of the book ?
\nSolution:
\nLet the number of pages in the first part of algebra textbook be ‘x’ and that of second part be ‘y’. The textbook contains 1382 pages.
\n\u2234 x + y = 1382 —- (1)
\nIt is given that the second part of the book has 64 pages more than the first part.
\n\u2234 y = x + 64
\n\u21d2 x – y = – 64 —- (2)
\nSolving (1) & (2), we get
\n\"TS
\nSubstitute x = 659 in (1), we get
\n659 + y = 1382
\n\u2234 y = 1382 – 659 = 723
\nFirst part of the book has 659 pages whereas the second part has 723 pages.<\/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 ?
\nSolution:
\nLet 50% solution to be used be ‘x’ ml.
\nAnd 80% solution to be used be ‘y’ ml.
\nQuantity of solution to be obtained = 100 ml
\n\u2234 x + y = 100 —– (1)
\n50% of ‘x’ ml = \\(\\frac{x \\times 50}{100}\\) = \\(\\frac{1}{2} \\mathrm{x}\\)
\n80% of ‘y’ ml = \\(\\frac{\\mathrm{y} \\times 80}{100}\\) = \\(\\frac{4}{5} \\mathrm{y}\\)
\nBy problem, \\(\\frac{1}{2}\\)x + \\(\\frac{4}{5}\\)y = 68
\nMultiplying each terms by 10, we get
\n[\\(\\frac{1}{2}\\)x \u00d7 10] + [\\(\\frac{4}{5}\\)y \u00d7 10] = 68 \u00d7 10
\n5x + 8y = 680 —– (2)
\nSolving equation (1) & (2), we have
\n\"TS
\n\u2234 y = \\(\\frac{180}{3}\\) = 60
\nSubstitute y = 60 in (1), we have
\nx + y = 100
\nx + 60 = 100
\n\u2234 x = 100 – 60 = 40
\nTherefore, 50% solution to be used = 40 ml.
\n80% solution to be used 60 ml<\/p>\n

\"TS<\/p>\n

Question 10.
\nSuppose you have \u20b9 12,000 to invest. You have to invest some amount at 10% and the rest at 15%. How much should be
\ninvested at each rate to yield 12% on the total amount invested?
\nSolution:
\nTotal amount to be invested = \u20b9 12,000\/-
\nLet the amount invested at 10% be \u20b9 x.
\nLet the amount invested at 15% be \u20b9 y.
\n\u2234 x + y = \u20b9 12,000 —- (1)
\nAmount yielded at 10% of
\nx = \\(\\frac{x \\times 10}{100}\\) = \\(\\frac{x}{10}\\)
\nAmount yielded at 15% of
\ny = \\(\\frac{y \\times 15}{100}\\) = \\(\\frac{3 \\mathrm{y}}{20}\\)
\nyielded at 12% of \u20b9 12,000
\n= \\(\\frac{12000 \\times 12}{100}\\) = 1440
\n\u2234 \\(\\frac{x}{10}\\) + \\(\\frac{3 y}{20}\\) = \\(\\frac{1440}{1}\\)
\n\u2234 2x + 3y = 28800 —- (2)
\n\"TS
\nSubstitute y = 4800 in (1), we get
\nx + 4800 = 12000
\n\u2234 x = 12000 – 4800 = 7200
\nThe amounts to be invested at 10% and 15% are \u20b9 7,200 and \u20b9 4,800.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can practice 10th Class Maths Solutions Telangana Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 to get the best methods of solving problems. TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Exercise 4.2 Question 1. The ratio of incomes of two persons is 9: 7 … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[16],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9587"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=9587"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9587\/revisions"}],"predecessor-version":[{"id":9669,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9587\/revisions\/9669"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9587"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9587"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9587"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}