{"id":22021,"date":"2022-01-13T17:13:00","date_gmt":"2022-01-13T11:43:00","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=22021"},"modified":"2022-01-28T10:05:53","modified_gmt":"2022-01-28T04:35:53","slug":"ap-board-7th-class-maths-solutions-chapter-3-ex-3-4","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-7th-class-maths-solutions-chapter-3-ex-3-4\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 3.4"},"content":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf<\/a> Chapter 3 Simple Equations Ex 3.4 Textbook Exercise Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.4<\/h2>\n

Question 1.
\nFind the height of the statue from the adjacent diagram.
\n\"AP
\nAnswer:
\nFrom the given diagram
\nHeight of the statue (BC) = x m
\nHeight of the Pedastal (AB) = 1.9 m
\nTotal Height (AC) = 3.6 m
\nBC + AB = 3.6 m
\n\u21d2 x + 1.9 m = 3.6 m
\n\u21d2 x + 1.9 – 1.9 = 3.6 – 1.9 (Subtract 1.9 on both sides)
\n\u21d2 x = 1.7 m
\n\u2234 Height of the statue = 1.7 m<\/p>\n

\"AP<\/p>\n

Question 2.
\nThe sum of twice a number and 4 is 80, find the number.
\nAnswer:
\nLet the number be x.
\nTwice the number = 2x
\nSum of twice a number and 4 = 80
\n2x + 4 = 80
\n\u21d2 2x + 4 – 4 = 80 – 4
\n(Subtract 4 on both sides)
\n\u21d2 2x = 76
\n\u21d2 \\(\\frac{2 x}{2}\\) = \\(\\frac{76}{2}\\) (Divide by 2 on both sides)
\n\u21d2 x = 38
\n\u2234 Number = 38<\/p>\n

Question 3.
\nThe difference between a number and one-fourth of itself is 24, find the number.
\nAnswer:
\nLet the number be a.
\nOne-fourth of a number = \\(\\frac{1}{4}\\) of a = \\(\\frac{a}{4}\\)
\nDifference of number and one-fourth of itself = 24
\n\"AP
\n\u21d2 a = 32
\n\u2234 The number a = 32<\/p>\n

Question 4.
\n\"AP
\nFind the value of x from the adjacent diagram.
\nAnswer:
\n\"AP
\nFrom the above diagram
\nAB = 12 cm; CD = x cm
\nEF = 5 cm; PQ = 24 cm
\nWe know,
\nAB + CD + EF = PQ
\n\u21d2 12 + x + 5 = 24
\n\u21d2 x + 17 = 24
\n\u21d2 x + 17 – 17 = 24 – 17
\n(Subtract 17 on both sides)
\n\u21d2 x = 7 cm<\/p>\n

\"AP<\/p>\n

Question 5.
\nTo convert temperature from Fahrenheit to Centigrade, we use the formula (F – 32) = \\(\\frac{9}{5}\\) \u00d7 C. If C = – 40\u00b0 C, then find F.
\nAnswer:
\nGiven formula (F – 32) = \\(\\frac{9}{5}\\) \u00d7 C and C = – 40\u00b0 C
\n\"AP
\n\u21d2 F – 32 = 9(- 8)
\n\u21d2 F – 32 = – 72
\n\u21d2 F – 32 + 32 = – 72 + 32 (Add 32 on both sides)
\n\u21d2 F = – 40\u00b0 F
\nTherefore we conclude that
\n– 40\u00b0 C = – 40\u00b0 F.<\/p>\n

Question 6.
\nRahim has \u20b9x, from which he spends \u20b96. If twice of the money left with him is \u20b986, then find x.
\nAnswer:
\nRahim has money = \u20b9x
\nMoney after spends \u20b96 = x – 6
\nTwice the money left with Rahim
\n= 2(x – 6) = \u20b986
\n\u21d2 – 2(x – 6) = 86
\n\u21d2 \\(\\frac{2(x-6)}{2}\\) = \\(\\frac{86}{2}\\) (Divide by 2 on both sides)
\n\u21d2 x – 6 = 43
\n\u21d2 x – 6 + 6 = 43 + 6 (Add 6 on both sides)
\n\u21d2 x = 49
\nTherefore money at Rahim = \u20b949.<\/p>\n

Question 7.
\nThe difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.
\nAnswer:
\nLet the smaller number is x.
\nLarger number = x + 7
\nSix times the smaller = 6.x
\nSix times smaller plus the larger = 77
\n\u21d2 6x + x + 7 = 77
\n\u21d2 7x + 7 = 77
\n\"AP
\n\u21d2 7x = 70
\n\u21d2 \\(\\frac{7 \\mathrm{x}}{7}\\) = \\(\\frac{70}{7}\\) (Divide by 7 on both sides) \u21d2x = 10
\n\u2234 Smaller number x = 10
\nLarger number = x + 7 = 10 + 7 = 17
\n\u2234 The numbers are 10 and 17.<\/p>\n

\"AP<\/p>\n

Question 8.
\nThe sum of three consecutive even numbers is 54. Find the numbers.
\nAnswer:
\nLet the three consecutive even numbers are 2x, 2(x +1), 2(x + 2).
\nSum of three consecutive even numbers = 54
\n\u21d2 2x + 2(x + 1) + 2(x + 2) = 54
\n\u21d2 2x + 2x + 2 + 2x + 4 = 54
\n\u21d2 6x + 6 = 54
\n\"AP
\n\u21d2 6x = 48
\n\u21d2 \\(\\frac{6 x}{6}\\) = \\(\\frac{48}{6}\\) (Divide by 6 on both sides)
\n\u21d2 x = 8
\nNumbers are, 2x, 2(x + 1), 2(x + 2)
\n2(8), 2(8 + 1), 2(8 + 2)
\n\u2234 The numbers are 16, 18, 20.<\/p>\n

Question 9.
\nIn a class of 48 students, the number of girls is one third the number of boys. Find the number of girls and boys in the class.
\nAnswer:
\nLet the number of boys = x
\nThe number of girls = \\(\\frac{1}{3}\\) of boys
\n= \\(\\frac{1}{3}\\) of x = \\(\\frac{x}{3}\\)
\nBoys + Girls = Total students
\n\u21d2 \\(\\frac{x}{1}+\\frac{x}{3}\\) = 48
\n\u21d2 \\(\\frac{3 x+x}{3}\\) = 48
\n\u21d2 \\(\\frac{4 x}{3}\\) = 48
\n\u21d2 \\(\\frac{4 x}{3}\\) \u00d7 3 = 48 \u00d7 3
\n(Multiply by 3 on both sides)
\n\u21d2 4x = 144
\n\u21d2 \\(\\frac{4 x}{4}\\) = \\(\\frac{144}{4}\\) (Divide by 4 on both sides)
\n\u21d2 x = 36
\n\u2234 Number of boys = 36
\nNumber of girls = \\(\\frac{x}{3}\\) = \\(\\frac{36}{3}\\) = 12
\nBoys = 36 and girls = 12<\/p>\n

\"AP<\/p>\n

Question 10.
\nThe present ages of Mary and Joseph are in the ratio 5 : 3. After 3 years sum of their ages will be 38. Find their present ages.
\nAnswer:
\nRatio of present ages of Mary and Joseph = 5:3 = 5x : 3x
\nAfter 3 years their ages
\n= 5x + 3 : 3x + 3
\nSum of their ages = 38
\n\u21d2 5x + 3 + 3x + 3 = 38
\n\u21d2 8x + 6 = 38
\n\u21d2 8x + 6 – 6 = 38 – 6 (Subtract 6 on both sides)
\n\u21d2 8x = 32
\n\u21d2 \\(\\frac{8 x}{8}\\) = \\(\\frac{32}{8}\\)
\n\u21d2 x = 4
\n\u21d2 Ratio of present ages = 5x : 3x
\n= 5(4) : 3(4) = 20 : 12
\nPresent ages of Mary and Joseph are 20 years and 12 years.<\/p>\n

Question 11.
\nA sum of \u20b9500 is in the form of notes of denominations of \u20b95 and \u20b910. If the total number of notes is 90, then find the number of notes of each type.
\n\"AP
\nAnswer:
\nLet, number of f5 notes = x
\nGiven sum of number of \u20b95 notes and \u20b910 notes is 90.
\nSo, number of \u20b910 notes = 90 – x
\nSum of \u20b95 notes = \u20b95 \u00d7 x = 5x
\nSum of \u20b9 10 notes = \u20b9 10 \u00d7 (90 – x)
\n= 900 – 10x
\n\u21d2 5x + 900 – 10x = 500
\n\u21d2 – 5x + 900 = 500
\n\u21d2 – 5x + 900 – 900 = 500 – 900 (Subtract 900 on both sides)
\n\u21d2 – 5x = – 400
\n\u21d2 \\(\\frac{-5 x}{-5}\\) = \\(\\frac{-400}{-5}\\)
\n(Divide by – 5 on both sides)
\n\u21d2 x = 80
\n\u2234 Number of \u20b95 notes x = 80
\nNumber of \u20b910 notes
\n= 90 – x = 90 – 80 = 10.<\/p>\n

\"AP<\/p>\n

Question 12.
\nJohn and Ismail donated some money to the Relief Fund. The amount paid by Ismail is \u20b985 more than twice that of John. If the total money paid by them is \u20b94000, then find the amount of money donated by John.
\nAnswer:
\nLet the amount donated by John = \u20b9x
\nThe amount donated by Ismail is \u20b985 more than twice of John.
\nSo, the amount donated by Ismail = \u20b9(2x + 85)
\nTotal amount = John amount + Ismail amount = \u20b94000
\n\u21d2 x + 2x + 85 = 4000
\n\u21d2 3x + 85 = 4000
\n\u21d2 3x + 85 – 85 = 4000 – 85 (Subtract 85 on both sides)
\n\u21d2 3x = 3915
\n\u21d2 \\(\\frac{3 x}{3}\\) = \\(\\frac{3915}{3}\\) (Divide by 3 on both sides)
\n\u21d2 x = 1305
\n\u2234 Money donated by John = x = \u20b9 1305
\n\u2234 Money donated by Ismail = 2x + 85
\n= 2(1305) + 85
\n= \u20b92695<\/p>\n

Question 13.
\nLength of a rectangle is 4 m less than 3 times its breadth. If the perimeter of rectangle is 32 m. Then find its length and breadth.
\nAnswer:
\nLet breadth of a rectangle (b) = a m
\nLength of rectangle (l) = 4 m less than 3 times of breadth
\n= (3 \u2219 a – 4) m
\nGiven perimeter = 2(l + b) = 32 m
\n\u21d2 2[3a – 4 + a] = 32
\n\u21d2 2[4a – 4] = 32
\n\u21d2 (8a – 8) = 32 (Distributive property)
\n\u21d2 8a – 8 + 8 = 32 + 8 (Add 8 on both sides)
\n\u21d2 8a = 40
\n\u21d2 \\(\\frac{8 \\mathrm{a}}{8}\\) = \\(\\frac{40}{8}\\) (Divide by 8 on both sides) o o
\n\u21d2 a = 5
\n\u2234 breadth (b) = 5 m
\nlength (l) = 3a – 4
\n= 3(5) – 4 = 15 – 4 = 11 m
\n\u2234 length = 11 m and breadth = 5 m.<\/p>\n

\"AP<\/p>\n

Question 14.
\nA bag contains some number of white balls, twice the number of white balls are blue balls, thrice the number of blue balls are the red balls. If the total number of balls in the bag are 27. Then calculate the number of balls of each colour present in the bag.
\nAnswer:
\nLet the number of white balls = x
\nNumber of blue balls = twice the number of white balls
\n= 2(x) = 2x
\nNumber of red balls
\n= thrice the number of blue balls = 3(2x) = 6x
\nTotal balls = White + Blue + Red = 27
\n\u21d2 x + 2x + 6x = 27
\n\u21d2 9x = 27
\n\u21d2 \\(\\frac{9 x}{9}\\) = \\(\\frac{27}{9}\\) (Divide by 9 on both sides) 9 9 .
\n\u21d2 x = 3
\nTherefore white balls x = 3
\nBlue balls = 2x = 2(3) = 6
\nRed balls = 6x = 6(3) = 18<\/p>\n

Question 15.
\n\"AP
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

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

SCERT\u00a0AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.4 Textbook Exercise Questions and Answers. AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.4 Question 1. Find the height of the statue from the adjacent diagram. Answer: From the given diagram Height of the statue (BC) = x m …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[9],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22021"}],"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\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=22021"}],"version-history":[{"count":3,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22021\/revisions"}],"predecessor-version":[{"id":23876,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22021\/revisions\/23876"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=22021"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=22021"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=22021"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}