{"id":13766,"date":"2020-09-09T16:43:51","date_gmt":"2020-09-09T11:13:51","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=13766"},"modified":"2021-07-02T15:01:54","modified_gmt":"2021-07-02T09:31:54","slug":"kseeb-solutions-for-class-7-maths-chapter-1-ex-1-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-1-ex-1-4\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4"},"content":{"rendered":"

Students can Download Chapter 1 Integers Ex 1.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 7 Maths Chapter 1 Integers Ex 1.4<\/h2>\n

Question 1.
\nEvaluate each of the following
\na) (-30) \u00f7 10
\nSolution:
\n\"KSEEB<\/p>\n

b) 50 \u00f7 (-5)
\nSolution:
\n\"KSEEB<\/p>\n

c) (-36) \u00f7 (-9)
\nSolution:
\n\"KSEEB
\n= 4<\/p>\n

d) (-49) \u00f7 (49)
\nSolution:
\n\"KSEEB
\n= -1<\/p>\n

e) 13 \u00f7 [(-2) + 1]
\nSolution:
\n= 13 \u00f7 (-2 + 10)
\n= 13 \u00f7 (-1)
\n\\(=\\frac{13}{-1}\\)
\n= -13<\/p>\n

\"KSEEB<\/p>\n

f) 0 \u00f7 (-12)
\n\"KSEEB
\n= 0<\/p>\n

g) (-31) \u00f7 [(-30) + (-1)]
\n= (-31) \u00f7 [-30 – 1]
\n= (-31) \u00f7 (-31)
\n\"KSEEB
\n= 1<\/p>\n

h) [(-36) \u00f7 12] \u00f7 3
\nSolution:
\n\"KSEEB
\n\"KSEEB
\n= -1<\/p>\n

i) [(-6) + 5] \u00f7 [(- 2) + 1]
\n=[-6 + 5] \u00f7 [-2 + 1]
\n= (-1) \u00f7 (-1)
\n\"KSEEB
\n= 1<\/p>\n

Question 2.
\nVerify that a \u00f7 (b + c) \u2260 (a \u00f7 b) + (a \u00f7 c) for each of the following values of a, b and c.
\na) a = 12, b = -4, c = 2
\nSolution:
\na) a = 12, b = -4, c = 2
\nTo verify a \u00f7 (b + c) \u2260 (a \u00f7 b) + (a \u00f7 c)
\n\"KSEEB
\n\u2234 LHS \u2260 RHS
\nhence verified<\/p>\n

\"KSEEB<\/p>\n

b) a = (-10), b = 1, c = 1
\n\"KSEEB
\n\u2234 LHS \u2260 RHS
\nhence verified<\/p>\n

Question 3.
\nFill in the blanks :<\/p>\n

a) 369 \u00f7 ___ = 369
\n369 \u00f7 1<\/u> = 369
\nLet number be x
\n\"KSEEB
\nx = 1<\/p>\n

b) (-75) \u00f7 ___ = -1
\n(-75) \u00f7 75<\/u> = -1
\nLet number be x
\n\"KSEEB
\nx = 75<\/p>\n

c) (-206) \u00f7 ___ = 1
\n(-206) \u00f7 -206<\/u> = 1
\nLet number be x
\n\"KSEEB
\nx = -206<\/p>\n

d) -87 \u00f7 ___ = 87
\n-87 \u00f7 -1<\/u> = 87
\nlet the number be x
\n\"KSEEB
\nx = -1<\/p>\n

\"KSEEB<\/p>\n

e) ___ \u00f7 1 = -87
\n87<\/u> \u00f7 1 = -87
\nLet the number be x
\n\"KSEEB
\n1 \u00d7 -87 = x
\n-87 = x
\nx = -87<\/p>\n

f) ___ \u00f7 48 = -1
\n48<\/u> \u00f7 48 = -1
\nLet number be x
\n\"KSEEB
\n– 1 \u00d7 48 = x
\n-48 = x
\n\u2234 x = -48<\/p>\n

g) 20 \u00f7 ___ = -2
\n20 \u00f7 -10<\/u> = (-2)
\n\"KSEEB
\nx = -10<\/p>\n

\"KSEEB<\/p>\n

h) __ \u00f7 (4) = -3
\n-12<\/u> \u00f7 (4) = -3
\nLet number be x
\n\"KSEEB
\nx = -3 \u00d7 4
\nx=-12<\/p>\n

Question 4.
\nWrite five pairs of integers (a, b) such that a – b = -3. One such pair is (6, -2) because 6 \u00f7 (-2) = (-3)
\nSolution:
\na \u00f7 b = -3
\n\"KSEEB
\n\u2234 The other five pairs are (-6, 2), (-3, 1), (-9, 3), (-12, 4), (-15, 5)<\/p>\n

Question 5.
\nThe temperature at 12 noon was 10\u00b0C above zero. If it decreases at the rate of 2\u00b0C per hour until midnight, at what time would the temperature be 8\u00b0C below zero? What would be the temperature at mid-night ?
\nSolution:
\nTemperature at 12pm = 10\u00b0C
\nTemperature decreases at 2\u00b0C per hour
\nwe need to find, time when temperature is 8\u00b0C below zero, ie., -8\u00b0C
\n\u2234 Total decrease = Final temperature – initial temperature = -8\u00b0C -10\u00b0C
\n= -8 – 10 = -18\u00b0C
\nNow,
\nTemperature decrease in 1 hour \u00d7 Total number of hours = Total decrease
\n\"KSEEB
\nSo, temperature becomes -8\u00b0C = 9 hours
\n\u2234 Temperature is -8\u00b0C at 9PM
\nNow,
\nWe need to find temperature at mid – night temperature at 12AM = Temperature at 12PM
\nTemperature change in 12 hours
\nTemperature change in 1 hour = -2\u00b0C
\nTemperature change in 12 hour = -2 \u00d7 12\u00b0C = -24\u00b0C
\nNow,
\nTemperature at 12AM – Temperature at 12PM = Temperature change in 12 hours. Temperature at 12AM – 10\u00b0C = -24\u00b0C
\nTemperature at 12AM = – 24\u00b0C + 10\u00b0C
\nTemperature at 12 AM = -14\u00b0C
\n\u2234 Temperature at Mid – night = -14\u00b0C<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIn a class test (+3) marks are given for every correct answer and (-2) marks are given for every incorrect answer and no marks for not attempting any question.
\nSolution:
\nMarks given for correct answer = +3
\nMarks given for incorrect answer = -2
\nMarks given for not attempting = 0
\ni) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly ?
\nSolution:
\nTotal marks scored = 20
\nNumber of correct answer = 12
\nMarks of for correct answer = 12 \u00d7 3 = 36
\nAlso
\nMarks for incorrect answer = (-2) \u00d7 Number of incorrect answer
\nMarks for not attempt answer = 0 \u00d7 Number of not attempt = 0
\nNow,
\nTotal marks = marks for correct answer + marks for incorrect answer + marks for not attempt
\n20 = 36 + (-2) \u00d7 No of incorrect answers + 0
\n20 – 36 = (-2) \u00d7 No of incorrect answers
\n-16 = (-2) \u00d7 No of incorrect answer
\n\\(\\frac{-16}{-2}\\) = No of incorrect answers
\n8 = No of incorrect answers
\nSo, Radhika attempted 8 questions incorrectly<\/p>\n

ii) Mohini scores -5 marks in this test, thought she has got 7 correct answers. How many questions has she attempted incorrectly?
\nSolution:
\nTotal marks = – 5
\nNo of correct answers = 7
\nMarks for correct answers = 7 \u00d7 3 = 21
\nAlso,
\nMarks for incorrect answer = (- 2) \u00d7 No. of incorrect answers
\nMarks for not attempt answer = 0 \u00d7 No of not attempt answers
\nNow,
\nTotal marks = marks for correct answer + marks for incorrect answer + marks for not attempt
\n– 5 = 21 + (- 2) \u00d7 No. of incorrect answers + 0
\n– 5 – 21 = (- 2) \u00d7 No. of incorrect answers
\n– 26 = – 2 \u00d7 No. of incorrect answers
\n\\(\\frac{-26}{-2}\\) = No. of incorrect answers
\n13 = No, of incorrect answers.
\n\u2234 Mohini attempted 13 questions incorrectly<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nAn elevator descends into a mine shaft at the rate of 6m\/min. If the descent starts from 10m above the ground level, how long will it take to reach -350 m.
\nSolution:
\nInitial position = 10m final position = -350 m
\nTotal distance elevator has to travel = initial position – final position
\n= 10 – (- 350)
\n= 10 + 350 = 360 m
\nNow,
\nElevator descends at rate 6m\/min
\n\u2234 Elevator travels 6m in = 1 minute
\nElevator travels 1 m in: = \\(\\frac{1}{6}\\)
\nElevator travels 360m in = \\(\\frac{1}{6}\\) \u00d7 360 minutes
\n= 60 minutes
\n= 1 hour
\n\u2234 It will take the elevator 1 hour.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 1 Integers Ex 1.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 7 Maths Chapter 1 Integers Ex 1.4 Question 1. Evaluate each of the following a) …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[84],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13766"}],"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=13766"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13766\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=13766"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=13766"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=13766"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}