{"id":427,"date":"2025-01-09T10:17:09","date_gmt":"2025-01-09T04:47:09","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=427"},"modified":"2025-01-10T15:54:04","modified_gmt":"2025-01-10T10:24:04","slug":"samacheer-kalvi-9th-maths-guide-chapter-3-ex-3-14","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-9th-maths-guide-chapter-3-ex-3-14\/","title":{"rendered":"Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.14"},"content":{"rendered":"

Students can download Maths Chapter 3 Algebra Ex 3.14 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.14<\/h2>\n

Question 1.
\nThe sum of a two digit number and the number formed by interchanging the digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number. Find the first number.
\nSolution:
\nLet the ten’s digit be x and the unit digit be y.
\nThe number is 10x + y
\nIf the digits are interchanged
\nThe new number is 10y + x
\nBy the given first condition
\n10x + y + 10y + x = 110
\n11x + 11y = 110
\nx + y = 10 \u2192 (1) (Divided by 11)
\nAgain by the given second condition
\n10x + y – 10 = 5(x + y ) + 4
\n10x + y – 10 = 5x + 5y + 4
\n5x – 4y = 14 \u2192 (2)
\n(1) \u00d7 5 \u21d2 5x + 5y = 50 \u2192 (3)
\n(2) \u00d7 1 \u21d2 5x – 4y = 14 \u2192 (2)
\n(3) – (2) \u21d2 9y = 36
\ny = 36\/9
\n= 4
\nSubstitute the value of y = 4 in (1)
\nx + y = 10
\nx + 4 = 10
\nx = 10 – 4
\n= 6
\n\u2234 The number is (10 \u00d7 6 + 4) = 64<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nThe sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \\(\\frac{1}{2}\\). Find the fraction.
\nSolution:
\nLet the numerator be “x” and the denominator be “y”
\n\u2234 The fraction is \\(\\frac{x}{y}\\)
\nBy the given first condition
\nx + y = 12 \u2192 (1)
\nAgain by the second condition
\n\\(\\frac{x}{y+3}\\) = \\(\\frac{1}{2}\\)
\n2x = y + 3
\n2x – y = 3 \u2192 (2)
\n(1) + (2) \u21d2 3x = 15
\nx = \\(\\frac{15}{3}\\) = 5
\nSubstitute the value of x = 5 in (1)
\n5 + y = 12
\ny = 12 – 5
\n= 7
\n\u2234 The fraction is \\(\\frac{5}{7}\\)<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nABCD is a cyclic quadrilateral such that \u2220A = (4y + 20)\u00b0, \u2220B = (3y -5)\u00b0, \u2220C = (4x)\u00b0 and \u2220D = (7x + 5)\u00b0. Find the four angles.
\nSolution:
\n\"Samacheer
\nABCD is a cyclic quadrilateral \u2220A + \u2220C = 180\u00b0
\n(Sum of the opposite angles of a cyclic quadrilateral is 180\u00b0)
\n(4y + 20)\u00b0 + (4x)\u00b0 = 180\u00b0
\n4y + 20 + 4x = 180
\n4x + 4y = 180 – 20
\n4x + 4y = 160
\nx + y = 40 \u2192 (1) (divided by 4)
\n\u2220B + \u2220D = 180\u00b0 (Sum of the opposite angles of a cyclic quadrilateral)
\n(3y – 5)\u00b0 + (7x + 5)\u00b0 = 180\u00b0
\n3y – 5 + 7x + 5 = 180
\n7x + 3y = 180 \u2192 (2)
\n(1) \u00d7 3 \u21d2 3x + 3y = 120 \u2192 (3)
\n(3) – (2) \u21d2 -4x = – 60
\n4x = 60
\nx = \\(\\frac{60}{4}\\)
\nSubstitute the value of x = 15 in (1)
\n15 + y = 40
\ny = 40 – 15 = 25
\n\u2220A = 4y + 20 = 4(25) + 20 = 100 + 20 = 120\u00b0
\n\u2234 \u2220A = 120\u00b0
\n\u2220B = 3y – 5 = 3(25) – 5 = 75 – 5 = 70
\n\u2234 \u2220B = 70\u00b0
\n\u2220C = 4x = 4(15) = 60
\n\u2234 \u2220C = 60\u00b0
\n\u2220D = 7x + 5 = 7(15) + 5
\n\u2220D = 105 + 5 = 110\u00b0
\n\u2234 \u2220A= 120\u00b0, \u2220B = 70\u00b0, \u2220C = 60\u00b0 and \u2220D = 110\u00b0<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nOn selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains Rs 2000. But if he sells the T.V. at 10% gain and the fridge at 5% loss, he gains Rs.1500 on the transaction. Find the actual price of the T.V. and the fridge.
\nSolution:
\nLet the cost price of the TV be Rs “x” and the cost price of the fridge be Rs “y”.
\nBy the given condition
\n\"Samacheer
\nMultiply by 20
\nx + 2y = 40000 \u2192 (1)
\nAgain by the given second condition
\n\"Samacheer
\nMultiply by 20
\n2x – y = 30000 \u2192 (2)
\n(2) \u00d7 2 \u21d2 4x – 2y = 60000 \u2192 (3)
\n(1) + (3) \u21d2 5x + 0 = 100000
\nx = \\(\\frac{100000}{5}\\)
\n= 20000
\nSubstitute the value of x = 20000 in (1)
\n20000 + 2y = 40000
\n2y = 40000 – 20000
\n= 20000
\ny = \\(\\frac{20000}{2}\\)
\n= 10000
\nCost price of a TV = Rs 20,000
\nCost price of a fridge = Rs 10,000<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nTwo numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.
\nSolution:
\nLet the two numbers be x and y.
\nBy the given first condition
\nx : y = 5 : 6
\n6x = 5y (Product of the extreme is equal to the product of the means)
\n6x – 5y = 0 \u2192 (1)
\nAgain by the given second condition
\nx – 8 : y – 8 = 4 : 5
\n5(x – 8) = 4(y – 8)
\n5x – 40 = 4y – 32
\n5x – 4y = – 32 + 40
\n5x – 4y = 8 \u2192 (2)
\n(1) \u00d7 4 \u21d2 24x – 20y = 0 \u2192 (3)
\n(2) \u00d7 5 \u21d2 25x – 20y = 40 \u2192 (4)
\n(3) – (4) \u21d2 – x + 0 = -40
\n\u2234 x = 40
\nSubstitute the value of x = 40 in (1)
\n6(40) – 5y = 0
\n240 – 5y = 0 \u21d2 – 5y = -240
\n5y = 240
\ny = \\(\\frac{240}{5}\\)
\n= 48
\nThe two numbers are 40 and 48 [\u2234 The ratio of the number = 40 : 48 are 5 : 6]<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\n4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indian and 5 Chinese can finish it in 4 days. How long would it take for 1 Indian to do it? How long would it ‘ take for 1 Chinese to do it?
\nSolution:
\nLet the time taken by a Indian be “x”
\nTime taken by a Chinese be “y”
\nWork done by a Indian in one day = \\(\\frac{1}{x}\\)
\nWork done by a Chinese in one day = \\(\\frac{1}{y}\\)
\nBy the given first condition
\n(4 Indian + 4 Chinese) finish the work in 3 days
\n\\(\\frac{4}{x}\\) + \\(\\frac{4}{y}\\) = \\(\\frac{1}{3}\\) \u2192 (1)
\nAgain by the given second condition
\n(2 Indian + 5 Chinese) finish the work in 4 days
\n\\(\\frac{2}{x}\\) + \\(\\frac{5}{y}\\) = \\(\\frac{1}{4}\\) \u2192 (2)
\nSolve the equation (1) and (2)
\nLet \\(\\frac{1}{x}\\) = a; \\(\\frac{1}{y}\\) = b
\n4a + 4b = \\(\\frac{1}{3}\\)
\n12a + 12b = 1 \u2192 (3) (Multiply by 3)
\n2a + 5b = \\(\\frac{1}{4}\\)
\n8a + 20b = 1 \u2192 (4) (Multiply by 4)
\n(3) \u00d7 (2) \u21d2 24a + 24b = 2 \u2192 (5)
\n(4) \u00d7 (3) \u21d2 24a + 60b = 3 \u2192 (6)
\n(5) – (6) \u21d2 -36b = -1
\nb = \\(\\frac{1}{36}\\)
\nSubstitute the value of b = \\(\\frac{1}{36}\\) in (3)
\n12a + 12(\\(\\frac{1}{36}\\)) = 1
\n12a + \\(\\frac{1}{3}\\) = 1
\n36a + 1 = 3
\n36a = 2
\na = \\(\\frac{2}{36}\\) = \\(\\frac{1}{18}\\)
\nBut \\(\\frac{1}{x}\\) = a \u21d2 \\(\\frac{1}{x}\\) = \\(\\frac{1}{18}\\)
\nx = 18
\n\\(\\frac{1}{y}\\) = b \u21d2 \\(\\frac{1}{y}\\) = \\(\\frac{1}{36}\\)
\ny = 36
\n\u2234 Time taken by a 1 Indian is 18 days
\nTime taken by a 1 Chinese is 36 days<\/p>\n

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

Students can download Maths Chapter 3 Algebra Ex 3.14 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.14 Question …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[3],"tags":[],"class_list":["post-427","post","type-post","status-publish","format-standard","hentry","category-class-9"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/427"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=427"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/427\/revisions"}],"predecessor-version":[{"id":39826,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/427\/revisions\/39826"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=427"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=427"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=427"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}