{"id":19750,"date":"2020-01-14T11:47:59","date_gmt":"2020-01-14T06:17:59","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19750"},"modified":"2021-07-02T15:29:38","modified_gmt":"2021-07-02T09:59:38","slug":"2nd-puc-basic-maths-question-bank-chapter-7-ex-7-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-7-ex-7-1\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1"},"content":{"rendered":"

Students can Download Basic Maths Exercise 7.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Ratios and Proportions Ex 7.1 One Mark Questions and Answers ( 1 \u00d7 10 = 10)<\/strong><\/p>\n

Question 1.
\nIf 3:5 is a ratio, find the antecedent and consequent.
\nAnswer:
\nAntecedent = 3, Consequent = 5.<\/p>\n

Question 2.
\nx gets a salary of7 20,000, y gets a salary of 5,000. Find the ratio of their salaries.
\nAnswer:
\nRatio of their salaries is 20000:50000 = 2:5.<\/p>\n

Question 3.
\nFind the inverse ratio of 4:5.
\nAnswer:
\nInverse ratio of 4 : 5 is 5 : 4<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nA house consumes 30 kgs of wheat and 4kg of sugar compare the consumption of wheat and sugar in the form of ratio.
\nAnswer:
\nWheat: Sugar = 30 kg: 4 kg = 15:2. 6.<\/p>\n

Question 5.
\nMr. X completes a job is 3 hours and Mr. Y completes the same job is 45 minutes compare their performances.
\nAnswer:
\n3 hrs.: 45 minutes = 3 \u00d7 60 : 45 = 180 : 45
\n= 4:1.<\/p>\n

Question 6.
\nFind the compound ratio of 3 : 4 and 4: 7.
\nAnswer:
\nCompound ratio = \\(\\frac{3}{4} \\cdot \\frac{4}{7}=\\frac{3}{7}=3: 7\\)<\/p>\n

Question 7.
\nFind the compound ratio of 1:2, 2:3 and 3:5.
\nAnswer:
\nCompound ratio = \\(\\frac{1}{2} \\times \\frac{2}{3} \\times \\frac{3}{5}=\\frac{1}{5}=1: 5\\)<\/p>\n

Question 8.
\nFind the duplicate ratio of 5 : 4.
\nAnswer:
\nDuplicate ratio of 5 : 4 is 52<\/sup>: 42<\/sup> = 25: 16.<\/p>\n

Question 9.
\nFind the triplicate ratio of 3 : 5.
\nAnswer:
\nTriplicate ratio of 3 : 5 is 33<\/sup>: 53<\/sup> = 27 : 125.<\/p>\n

Question 10.
\nFind the subduplicate ratio of 9:49.
\nAnswer:
\nSubduplicate ratio of 9 : 49 is \u221a9:\u221a49 = 3:7<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nFind the subtriplicate ratio of 125: 64.
\nAnswer:
\nSubtriplicate ratio of 125 : 64
\n= 3\u221a125:3\u221a64 = 5:4<\/p>\n

Question 12.
\nFind the value of x of 5 : 20 = 3 :x.
\nAnswer:
\n5x = 20 x 3\u21d2 x = \\(\\frac{20 \\times 3}{5}=\\) = 12 \u21d2 X = 12<\/p>\n

Part – B<\/span><\/p>\n

2nd PUC Basic Maths Ratios and Proportions Ex 7.1 Two Marks Questions and Answers (2 \u00d7 10 = 20)<\/strong><\/p>\n

Question 1.
\nFind the ratio between two numbers such that their sum is 40 and their difference is 8.
\nAnswer:
\nLet the two numbers be x & y
\nGiven x + y = 40 ……. (1)
\n& x – y = 8 … (2)
\nSolving we get
\nAdding 2x = 48 \u21d2 x = 24 & y = 40 \u2013 24 = 16
\n\u2234Ratio of the numbers is 24 : 16 \u21d2 3:2.<\/p>\n

Question 2.
\nA ratio is the lowest term is 3:8. If the difference between the quantities is 25. Find the quantities.
\nAnswer:
\nLet the numbers are 3x and 8x
\nGiven 8x – 3x = 25
\n5x = 25 \u21d2 X = 5
\n\u2234 Two numbers are 3(5) & 8(5)
\ni.e., 15 and 40.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nTwo numbers are is the ratio 3:5. If 5 is added to each , they are is the ratio. 2:3 find the numbers.
\nAnswer:
\nLet two numbers be 3x & 5x
\nIf 5 is added to each then \\(\\frac{3 x+5}{5 x+5}=\\frac{2}{3}\\)
\n9x + 15 = 10x + 10
\nx = 5
\n\u2234 The two numbers are 3(5) & 5(5) = 15 & 25.<\/p>\n

Question 4.
\nWhat must be added to each term is the ratio 2 : 3 so that is becomes 5:6.
\nAnswer:
\nLet x must be added to each term of 2 : 3 then
\n\\(\\frac{2+x}{3+x}=\\frac{5}{6}\\)
\n\u21d2 12 + 6x = 15 + 5x
\nx = 3.<\/p>\n

Question 5.
\nWhat must be added to each term is the ratio 4:5 so that is becomes 7:8.
\nAnswer:
\nLet x must be added to each of 4:5 then
\n\\(\\frac{7-x}{4-x}=\\frac{5}{2}\\)
\nx = 3<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nWhat must be subtracted from each term is the ratio 7:4 so that it becomes 5:2.
\nAnswer:
\nLet x must be subtracted from each of 7 : 4 then \\(\\frac{7-x}{4-x}=\\frac{5}{2}\\)
\n\u21d2 14 – 2x = 20 – 5x 5x \u2013 2x = 20 – 14
\n3x = 6 \u21d2 x= 2.<\/p>\n

Question 7.
\nWhat must be subtracted from each term is the ratio 8:7 so that it becomes 4:3.
\nAnswer:
\nLet x must be subtracted from each of 8:7 then
\n\\(\\frac{8-x}{7-x}=\\frac{4}{3}\\)
\n24 – 3x = 28 – 4x
\n4x – 3x = 28 – 24
\nx = 4<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nIf a: b = 2:3, b:c=3:5 and c:d= 5:7 find a:d.
\nAnswer:
\n\"2nd\"2nd<\/p>\n

Question 9.
\nIf a: b = 2:3 and b:c= 6:13 find a:b:c.
\nAnswer:
\nGiven
\n\\(\\frac{a}{b}=\\frac{2}{3}, \\frac{b}{c}=\\frac{6}{13} \\)
\n\\(\\frac{a}{b}=\\frac{2 \\times 2}{2 \\times 2}=\\frac{4}{6} \\)
\n\u2234 a:b:c = 4 : 6 : 13<\/p>\n

Question 10.
\nIf a : b = 3:4, b:c=8 = 15 find a:b:c.
\nAnswer:
\n\\(\\frac{a}{b}=\\frac{3}{4}, \\frac{b}{c}=\\frac{8}{15}\\)
\n\\(\\frac{a}{b}=\\frac{3 \\times 2}{4 \\times 2}=\\frac{6}{8} \\frac{b}{c}=\\frac{8}{15}\\)
\n\u2234 a : b : c = 6 : 8 : 15<\/p>\n

Part-C<\/span><\/p>\n

2nd PUC Basic Maths Ratios and Proportions Ex 7.1 Three Marks Questions and Answers (3 \u00d7 16 = 48)<\/strong><\/p>\n

Question 1.
\nDivide 1800 is the ratio 3 : 4 : 5.
\nAnswer:
\nSum of the ratios 3 + 4 + 5 = 12
\n\u2234\u00a0 \\(\\frac{3}{12}\\) \u00d7 1800 = 450
\n\\(\\frac{4}{12}\\)1800 = 600
\n\\(\\frac{4}{12}\\)1800 = 750<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nIf x : y = 3; 4 find \\(\\frac{2 x^{2}+3 y^{2}}{x^{2}+y^{2}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\nIf a:b= 2:3 and x:y = 4:7.
\nFind \\(\\frac{5 a x+4 b y}{8 a x+3 b y}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\nThe angles of a triangle are in the ratio 3:4:5. Find the angles.
\nAnswer:
\nSum of the ratios 3 + 4 + 5 = 12
\nW. K. T sum of the angles of a triangle = 75\u00b0
\n\u2234\u00a0 \\(\\frac{3}{12}\\) \u00d7 180 = 45\u00b0, 12
\n\\(\\frac{3}{12}\\) \u00d7 180 = 60\u00b0,
\n\\(\\frac{3}{12}\\) \u00d7 180 = 75\u00b0
\n\u2234 The three angles of a triangle are 45\u00b0, 60\u00b0, 75\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nAn article is sold at 40% gain an the cost price. Find the ratio of the selling price and cost price.
\nAnswer:
\nAn article is sold at 40% gain
\n100 + 40 = 40
\nIf the cost price is 100 then the selling price 100 + expenditure is 4x – 1000 40 = 140
\n\u2234 \\(\\frac{S P}{C P}=\\frac{140}{100}=\\frac{7}{5}\\)
\n\u2234 SP : CP = 7 : 5<\/p>\n

Question 6.
\nIf the monthly incomes of A and B are is the ratio 3:4 and their expenditures are is the ratio 1 :2. If each saves 1000 find the monthly incomes.
\nAnswer:
\nLet monthly income of A is 3x and expenditure
\nis 3x – 1000 and monthly income of B is 4x and
\nGiven \\(\\frac{3 x-1000}{4 x-1000}=\\frac{1}{2}\\)
\n6x – 4000 = 4x – 2000
\n\u21d2 2x = 2000 \u21d2 x = 1000
\n\u2234 The monthly income of A = 3x = 3 \u00d7 1000 = 3000
\n\u2234 The monthly income of B is 3(500) = 2000.<\/p>\n

Question 7.
\nIf the monthly incomes of A and Bare in the ratio 3:4 and their expenditure are in the ratio 1:2.If each saves 2000. Find their monthly incomes.
\nAnswer:
\nLet their monthly incomes are 3x and 4x and their expenditure are 3x \u2013 2000 and 4x – 2000
\nGiven difference of their squares is \\(\\frac{3 x-2000}{4 x-2000}=\\frac{1}{2}\\)
\n6x – 4000 = 4x – 2000
\n\u21d2 2x = 2000 \u21d2 x = 1000
\nThe monthly income of A = 3x = 3 \u00d7 1000 = 3000
\nThe monthly income of B = 4x = 4 x 1000 = 4000<\/p>\n

Question 8.
\nTwo numbers are is the ratio 6:7. If the difference of their squares is 117 find the two numbers.
\nAnswer:
\nLet the two numbers are 6x and 7x
\ni.e., (7x)2<\/sup> – (6x)2<\/sup>= 117
\n49x2<\/sup> – 36x2<\/sup>= 117
\n13x2<\/sup> = 117
\n\u21d2 x = 6
\n\u2234 The numbers are 3(6) and 46) = 18 and 21.<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nTwo numbers are is the ratio 3:4. If the sum of their squares is 900 find the two numbers.
\nAnswer:
\nGiven \\(\\frac{x}{y}=\\frac{3}{4}, \\frac{y}{z}=\\frac{7}{9}\\)
\nGiven sum of their squares = 900
\n(3x)2<\/sup> + (4x)2<\/sup>= 900
\n9x2<\/sup> + 16x2<\/sup> = 900
\n25x2<\/sup> = 900
\n\u21d2 x2<\/sup> = \\(\\frac{900}{25}\\) = 36 \u21d2 x = 6
\n\u2234 The numbers are 3(6) and 4 (6) = 18 and 24<\/p>\n

Question 10.
\nIf x:y= 3:4 and y:z= 7:9. Find x:y:z.
\nAnswer:
\n\"2nd<\/p>\n

Part – D<\/span><\/p>\n

2nd PUC Basic Maths Ratios and Proportions Ex 7.1 Five Marks Questions and Answers (5 \u00d7 16 = 80)<\/strong><\/p>\n

Question 1.
\nDivide \u20b9 800 is three parts such that three times of the first, five times of the second and six times the third are equal.
\nAnswer:
\nLet the 3 parts be x, y and z
\nGiven 3x = 5y = 62
\nLet 3x = 6z and 5y = =\\(\\frac{6}{5} \\)z
\nx = 2z, y = 2
\n\u2234 x : y : z = 2z : \\(\\frac{6}{5}\\)z : z= 10:6:5
\nSum of the ratios = 10 + 6 + 5 = 21
\n\"2nd<\/p>\n

Question 2.
\nDivide \u20b9 3262 among x, y and z such that if\u00a0 \u20b935, \u20b915 and \u20b912 are deducted from their respective shares, the remainders are in the ratio 3:5:8.
\nAnswer:
\nx + y + z = 3262 and x – 35:y – 15: Z – 12 = 3:5:8 x-35 3 y-155 y -15 5
\n\"2nd
\n5x – 175 = 3y – 45,
\n5x \u2013 3y = 130,
\n5x = 130 + 3y,
\nx= 26+\\(\\frac{3}{5} \\)y
\n8y – 120 = 52 – 60
\n8y – 5z = 60
\n5z = 8y – 60
\nz = \\(\\frac{8}{5} \\)y – 12
\nSubstitute for x and z
\nButy x + y + z = 3262
\n\"2nd
\n= 26 + 609
\nx = 635
\nx = 635, y = 1015 and z = 1612
\nz = 1624 – 12
\nz = 1612<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nIf x:y= 2:3. Find the value of \\(\\frac{2 x^{3}+3 y^{3}}{x^{3}+y^{3}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\nDivide 5880 is to three parts, such that ‘B’ receive thrice as ‘A’ and C receives \\(\\frac{5^{t h}}{6}\\) of what B receives.
\nAnswer:
\nLet A receive \u20b9x and B receives \u20b9 3x and C receive \\(\\frac{5}{6}\\)(3x) = \\(\\frac{5}{2}\\)x
\n\u2234 A:B:C= 1:3:\\(\\frac{5}{2}\\)= 2:6:5
\n\u2234 A’s share = \\(\\frac{2}{13}\\) \u00d7 5880 = 904.6
\nB’s share = \\(\\frac{6}{13}\\) x 5880 = 2713.8 :
\nC’s share = \\(\\frac{5}{13}\\)x 5880 = 2261.5<\/p>\n

Question 5.
\nIf \\(\\frac{2 x^{2}+3 y^{2}}{x^{3}+y^{2}}=\\frac{2}{41}\\) Find x:y.
\nAnswer:
\n\"2nd<\/p>\n

Question 6.
\nDivide 6000 into three parts is the ratio \\(\\frac{1}{2}: \\frac{1}{3}: \\frac{1}{6}\\)
\nAnswer:
\nGiven \\(\\frac{1}{2}: \\frac{1}{3}: \\frac{1}{6}\\) multiply by 6, we get 3:2:1
\n1st part = \\(\\frac{3}{6}\\) \u00d7 6000 = 3000
\n2nd part = \\(\\frac{3}{6}\\) \u00d7 6000 = 2000
\n3rd part = \\(\\frac{3}{6}\\) \u00d7 6000= 1000<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nDivide 17,640 a many P, Q, R and S such that O gets \\(7^{\\text {th }}5\\) of P, R gets \\(5^{\\text {th }}8\\) of Q and S gets \\(\\frac{2^{t h}}{13}\\) of the sum of Q and R.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nThe Rajdhani express takes 18 hours to reach Delhi from Bhubaneshwar while Nilachala Express takes
\n24 hours for the same of Delhi is 2880 kms from Bhubaneshwar find the ratio between the average speeds of the two trains.
\nAnswer:
\nDistance Speed = \\frac{\\text { Distance }}{\\text { Time taken }}
\nSpeed of Ragdhari express =\\(=\\frac{2880}{18}\\)
\nSpeed of Nilachala express = \\(\\frac{2880}{24}\\)
\nRatio of average speed = \\(\\frac{2880}{18}: \\frac{2880}{24}\\) = 24:18 = 4:3<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 7.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 Part – A 2nd …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19750"}],"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=19750"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19750\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19750"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19750"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19750"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}