{"id":11378,"date":"2023-06-08T14:57:09","date_gmt":"2023-06-08T09:27:09","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=11378"},"modified":"2023-06-09T17:32:49","modified_gmt":"2023-06-09T12:02:49","slug":"ap-board-6th-class-maths-solutions-chapter-6-intext-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-6th-class-maths-solutions-chapter-6-intext-questions\/","title":{"rendered":"AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic InText Questions"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 6th Class Maths Solutions<\/a> Chapter 6 Basic Arithmetic InText Questions and Answers.<\/p>\n

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic InText Questions<\/h2>\n

\"AP<\/p>\n

Check Your Progress <\/span>(Page No. 84)<\/p>\n

Question 1.
\nExpress the terms 45 and 70 by using ratio symbol.
\nSolution:
\nGiven terms are 45 and 70
\nRatio = 45 : 70
\nIt can be read as 45 is to 70.<\/p>\n

Question 2.
\nWrite antecedent in the ratio 7:15.
\nSolution:
\nGiven ratio 7 : 15
\nIn the ratio first term is called antecedent.
\nIn 7 : 15 antecedent is 7.<\/p>\n

Question 3.
\nWrite the consequent in the ratio 8 : 13.
\nSolution:
\nGiven ratio 8 : 13
\nIn the ratio second term is called consequent.
\nIn 8 : 13 consequent is 13.<\/p>\n

\"AP<\/p>\n

Question 4.
\nExpress the ratio 35 : 55 in the simplest form.
\nSolution:
\nGiven ratio 35 : 55 (or)
\nTo write the ratio in the simplest form we have to divide by the common factor of two terms 35 and 55.
\nCommon factor is 5.
\nNow divide by 5,
\n\\(\\frac{35}{55}=\\frac{35 \\div 5}{55 \\div 5}=\\frac{7}{11}\\)
\nSimplest form of \\(\\) is \\(\\frac{7}{11}\\)<\/p>\n

Question 5.
\nIn the given figure, find the ratio of
\ni) Shaded part to unshaded parts.
\nii) Shaded part to total parts,
\niii) Unshaded parts to total parts.
\n\"AP
\nSolution:
\ni) In the given figur.e,
\nNumber of shaded parts = 1
\nNumber of unshaded parts = 3
\nRatio = shaded parts : unshaded parts = 1:3<\/p>\n

ii) Number of shaded parts = 1
\nTotal parts = 4
\nRatio = shaded parts : total parts = 1:4<\/p>\n

iii) Number of unshaded parts = 3
\nTotal parts = 4
\nRatio = unshaded parts : total parts = 3:4<\/p>\n

Question 6.
\nExpress the following in the form of ratio.
\na) The length of a rectangle is triple its breadth. ‘
\nb) In a school, the workload of teaching 19 sections has been assigned to 38 teachers.
\nSolution:
\na) Let breadth of rectangle = x or one part = 1 part
\nlength of rectangle = triple the breadth
\n= 3 x x = 3x = 3 parts
\nRatio = l : b = x : 3x =\\(\\frac{1 x}{3 x}=\\frac{1}{3}\\) = 1:3<\/p>\n

b) Given number of sections = 19
\nNumber of teachers = 38
\nRatio = 19 : 38 = \\(\\frac{19}{38}=\\frac{1}{2}\\) = 1 : 2<\/p>\n

\"AP<\/p>\n

(Page No. 88)<\/p>\n

Question 1.
\nWhich ratio is larger in the following pairs ?
\n(a) 5 : 4 or 9 : 8
\n(b) 12 : 14 or 16 : 18
\n(c) 8: 20 or 12: 15
\n(d)4:7 or 7:11
\nSolution:
\na) 5 : 4 or 9 : 8
\nWrite the given ratios as fractions, we have 5 : 4 = \\(\\frac{5}{4}\\) and 9 : 8 = \\(\\frac{9}{8}\\)
\nNow find the LCM of the denominators of 4 and 8 is 8.
\nMake the denominator of the each fraction equal to 8.
\nWe have \\(\\frac{5}{4} \\times \\frac{2}{2}=\\frac{10}{8}\\) and \\(\\frac{9}{8} \\times \\frac{1}{1}=\\frac{9}{8}\\)
\nClearly we know that 10 > 9
\n\u2234 \\(\\frac{10}{8}>\\frac{9}{8}\\) (or) 10 : 8 > 9 : 8
\n10 : 8 is equal to 5 : 4
\nTherefore the larger ratio is 5 : 4.<\/p>\n

b) 12 : 14 or 16:18
\n12 : 14 = \\(\\frac{12}{14}=\\frac{6}{7}\\) and 16 : 18 = \\(\\frac{16}{18}=\\frac{8}{9}\\)
\nNow find the LCM of the denominators of 7 and 9 is 63.
\nMake the denominator of the each fraction equal to 63.
\nwe have \\(\\frac{6}{7} \\times \\frac{9}{9}=\\frac{54}{63}\\) and \\(\\frac{8}{9} \\times \\frac{7}{7}=\\frac{56}{63}\\)
\nClearly, we know that 54 < 56
\n\u2234 \\(\\frac{54}{63}<\\frac{56}{63}\\) (or) 54:63 < 56:63
\n56 : 63 is equal to 16 : 18 (or) 8 : 9
\n\u2234 The larger ratio is 16 : 18.<\/p>\n

c) 8 : 20 or 12 : 15
\nWrite the given ratios as fractions we have
\n8:20 = \\(\\frac{8}{20}=\\frac{2}{5}\\) and 12:15 = \\(\\frac{12}{15}=\\frac{4}{5}\\)
\n\\(\\frac{2}{5}\\) and \\(\\frac{4}{5}\\)<\/p>\n

Clearly \\(\\frac{2}{5}\\) < \\(\\frac{4}{5}\\)
\ni.e., 2:5 < 4 : 5 (or) 8: 20 < 12: 15
\nTherefore the larger ratio is 12 : 15.<\/p>\n

\"AP<\/p>\n

d) 4: 7 or 7: 11
\nWrite the given ratios as fractions, we have 4 7
\n4 : 7 = \\(\\frac{4}{7}\\) and 7:11 = \\(\\frac{7}{11}\\) .
\nNow find the LCM of the denominators of 7 and 11 is 77.
\nMake the denominators of the each fraction equal to 77.
\nWe have \\(\\frac{4}{7} \\times \\frac{11}{11}=\\frac{44}{77}\\) and \\(\\frac{7}{11} \\times \\frac{7}{7}=\\frac{49}{77}\\)
\n\\(\\frac{44}{7}\\) and \\(\\frac{49}{77}\\)
\nClearly we know that 44 < 49
\n\u2234 \\(\\frac{44}{77}<\\frac{49}{77}\\) (or) 44 : 77 < 49 : 77
\ni.e.,4: 7 < 7 : 11
\nTherefore the larger ratio is 7 : 11<\/p>\n

Question 2.
\nFind three equivalent ratios of 12 : 16.
\nSolution:
\nGiven ratio is 12 : 16
\nWrite the given ratio as fraction we have 12:16= \\(\\frac{12}{16}=\\frac{3}{4}\\)
\nNow, write equivalent fractions of \\(\\frac{3}{4}\\)
\n\"AP
\ni. e., 6 : 8 = 9 : 12 = 12 : 16 = 15 : 20 = 18 : 24
\n\u2234 Equivalent ratios of 12 : 16 are 6 : 8, 9 : 12, 12 : 16, 15 : 20 and 18 : 24.<\/p>\n

\"AP<\/p>\n

(Page No. 90)<\/p>\n

Question 1.
\nCheck whether the following terms are in proportion ?
\n1) 5,6,7,8
\n2) 3,5,6,10
\n3) 4,8,7,14
\n4) 2,12,3,18
\nSolution:
\n1) Given, 5, 6, 7, 8
\nIf a, b, c, d are in proportion i.e., a : b :: c : d
\nIf 5, 6, 7, 8 are in proportion i.e., 5 : 6 : : 7 : 8
\nWe know that, product of extremes = Product of means [a x d : b x c]
\n5 x 8 = 6 x 7
\n40 \u2260 42
\nSo, 5, 6, 7, 8 are not in proportion.<\/p>\n

2) Given, 3, 5, 6, 10
\nIf a, b, c, d are in proportion i.e., a : b :: c : d
\nIf 3, 5, 6, 10 are in proportion i.e., 3 : 5 :: 6 : 10
\nWe know that, product of extremes = Product of means a x d = b x c
\n3 x 10 = 5 x 6
\n30 = 30
\nSo, 3, 5, 6, 10 are in proportion.<\/p>\n

3) Given, 4, 8, 7, 14.
\nIf a, b, c, d are in proportion i.e., a : b : : c : d
\nIf 4, 8, 7, 14 are in proportion i.e., 4 : 8 : : 7 : 14
\nWe know that, product of extremes = Product of means a x d = b x c
\n4 x 14 = 8 x 7
\n56 = 56
\nSo, 4, 8, 7, 14 are in proportion.<\/p>\n

4) Given, 2, 12, 3, 18
\nIf a, b, c, d are in proportion i.e., a : b :: c : d
\nIf 2, 12, 3, 18 are in proportion i.e., 2 : 12 : : 3 : 18
\nWe know that, product of extremes = Product of means [ a x d = b x c ]
\n2 x 18 = 12 x 3
\n36 = 36
\nSo, 2, 12, 3, 18 are in proportion.<\/p>\n

\"AP<\/p>\n

Let’s Explore<\/span> (Page No. 92)<\/p>\n

Question 1.
\nRead the table and fill in the boxes.
\n\"AP
\nPrepare two similar problems and ask your friend to solve them
\nSolution:<\/p>\n\n\n\n\n\n\n
Weight<\/td>\nCost of Tomato<\/td>\nCost of Potato<\/td>\n<\/tr>\n
5 kg<\/td>\n\u20b9 75<\/td>\n\u20b9 60<\/td>\n<\/tr>\n
1 kg<\/td>\n\u20b915<\/td>\n\u20b9 12<\/td>\n<\/tr>\n
3 kg<\/td>\n\u20b9 45<\/td>\n\u20b9 36<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(Page. No. 94)<\/p>\n

Question 1.
\nRepresent the following in other forms.
\n\"AP
\nSolution:
\n\"AP<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic InText Questions and Answers. AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic InText Questions Check Your Progress (Page No. 84) Question 1. Express the terms 45 and 70 by using ratio symbol. Solution: Given terms are 45 and 70 Ratio … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[10],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/11378"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=11378"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/11378\/revisions"}],"predecessor-version":[{"id":41099,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/11378\/revisions\/41099"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=11378"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=11378"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=11378"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}