{"id":43344,"date":"2023-04-11T07:20:55","date_gmt":"2023-04-11T01:50:55","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=43344"},"modified":"2023-04-12T12:33:17","modified_gmt":"2023-04-12T07:03:17","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-1-ex-1-6","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-1-ex-1-6\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 1 Rational Numbers Ex 1.6"},"content":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 1 Rational Numbers Ex 1.6<\/h2>\n

Question 1.
\nIn a bag, there are 20 kg of fruits. If 7\\(\\frac { 1 }{ 6 }\\) kg of these fruits be oranges and 8\\(\\frac { 2 }{ 3 }\\) kg of thee are apples and rest are grapes. Find the mass of the grapes in the bag.
\nSolution:
\nTotal fruits in a bag = 20 kg
\n\"ML
\n\"ML<\/p>\n

Question 2.
\nThe population of a city is 6,63,432. If \\(\\frac { 1 }{ 2 }\\) of the population are adult males and \\(\\frac { 1 }{ 3 }\\) of the population are adult females, then find the number of children in the city.
\nSolution:
\nPopulation of a city = 6,63,432
\nNumber of adult males = \\(\\frac { 1 }{ 2 }\\) of 6,63,432 = 3,31,716
\nNumber of adult females = \\(\\frac { 1 }{ 3 }\\) of 6,63,432 = 2,21,144
\nRemaining population = 6,63,432 – (3,31,716 + 2,21,144)
\n= 6,63,432 – 5,52,860
\n= 1,10,572
\nNumber of children = 1,10,572<\/p>\n

Question 3.
\nIn an election of housing society, there are 30 voters. Each of them gives the vote. Three persons X, Y and Z are standing for the post of Secretary. If Mr X got \\(\\frac { 2 }{ 5 }\\) of the total votes and Mr Z got \\(\\frac { 1 }{ 3 }\\) of the total votes, then find the number of votes which Mr Y got.
\nSolution:
\nNumber of voters = 30
\nand number of person for election = X, Y, Z
\nX get \\(\\frac { 2 }{ 5 }\\) of total votes = \\(\\frac { 2 }{ 5 }\\) of 30 = 12
\nZ get \\(\\frac { 1 }{ 3 }\\) of total votes = \\(\\frac { 1 }{ 3 }\\) of 30 = 10
\nRemaining votes = 30 – (12 + 10) = 30 – 22 = 8
\nY gets votes = 8<\/p>\n

Question 4.
\nA person earns \u20b9 100 in a day. If he spent \u20b9 14\\(\\frac { 2 }{ 7 }\\) on food and \u20b9 30\\(\\frac { 2 }{ 3 }\\) on petrol. How much did he save on that day?
\nSolution:
\nA man’s earnings per day = \u20b9 100
\n\"ML<\/p>\n

Question 5.
\nIn an examination, 400 students appeared. If \\(\\frac { 2 }{ 3 }\\) of the boys and all 130 girls passed in examination, then find how many boys failed in examination?
\nSolution:
\nNumber of students appears = 400
\nNumber of boys = \\(\\frac { 2 }{ 3 }\\) of total boys passed the examination
\nand all girls 130 passed
\nNumber of total boys = 400 – 130 = 270
\nNumber of students passed = \\(\\frac { 2 }{ 3 }\\) of 270 = 180
\nand number of boys failed = 270 – 180 = 90<\/p>\n

Question 6.
\nA car is moving at the speed of 40\\(\\frac { 2 }{ 3 }\\) km\/h. Find how much distance will it cover in \\(\\frac { 9 }{ 10 }\\) hrs?
\nSolution:
\n\"ML<\/p>\n

Question 7.
\nFind the area of a square lawn whose one side is 5\\(\\frac { 7 }{ 9 }\\) m long.
\nSolution:
\n\"ML<\/p>\n

Question 8.
\nPerimeter of a rectangle is 15\\(\\frac { 3 }{ 7 }\\) m. If the length is 4\\(\\frac { 2 }{ 7 }\\) m, find its breadth.
\nSolution:
\n\"ML
\n\"ML<\/p>\n

Question 9.
\nRahul had a rope of 325\\(\\frac { 4 }{ 5 }\\) m long. He cut off a 150\\(\\frac { 3 }{ 5 }\\) m long piece, then he divided the rest of the rope into 3 parts of equal length. Find the length of each part.
\nSolution:
\n\"ML<\/p>\n

Question 10.
\nIf 3\\(\\frac { 1 }{ 2 }\\) litre of petrol costs \u20b9 270\\(\\frac { 3 }{ 8 }\\), then find the cost of 4 litre of petrol.
\nSolution:
\n\"ML
\n\"ML<\/p>\n

Question 11.
\nRamesh earns \u20b9 40,000 per month. He spends \\(\\frac { 3 }{ 8 }\\) of the income on food, \\(\\frac { 1 }{ 5 }\\) of the remaining on LIC premium and then \\(\\frac { 1 }{ 2 }\\) of the remaining on other expenses. Find how much money is left with him?
\nSolution:
\nMonthly earnings of Ramesh = \u20b9 40000
\nExpenditure on food = \\(\\frac { 3 }{ 8 }\\) of \u20b9 40000 = \u20b9 15000
\nRemaining income = 40000 – 15000 = \u20b9 25000
\nExpenditure on LIC premium = \\(\\frac { 1 }{ 5 }\\) of \u20b9 25000 = \u20b9 5000
\nRemaining amount = \u20b9 25000 – \u20b9 5000 = \u20b9 20000
\nExpenditure on other expenses = \\(\\frac { 1 }{ 2 }\\) of \u20b9 20000 = \u20b9 10000
\nRemaining amount left = \u20b9 20000 – \u20b9 10000 = \u20b9 10000<\/p>\n

Question 12.
\nA, B, C, D and E went to a restaurant for dinner. A paid \\(\\frac { 1 }{ 2 }\\) of the bill, B paid \\(\\frac { 1 }{ 5 }\\) of the bill and rest of the bill was shared equally by C, D and E. What fraction of the bill was paid by each?
\nSolution:
\nLet total bill of the restaurant = 1
\nA pays the bill = \\(\\frac { 1 }{ 2 }\\)
\nB pays the bill = \\(\\frac { 1 }{ 5 }\\)
\n\"ML<\/p>\n

Question 13.
\n\\(\\frac { 2 }{ 5 }\\) of total number of students of a school come by car while \\(\\frac { 1 }{ 4 }\\) of students come by bus to school. All the other students walk to school of which \\(\\frac { 1 }{ 3 }\\) walk on their own and the rest are escorted by their parents. If 224 students come to school walking on their own, how many students study in the school? Solution:
\nLet total number of students = 1
\nStudents who come by car = \\(\\frac { 2 }{ 5 }\\)
\nWho come by bus = \\(\\frac { 1 }{ 4 }\\)
\nWho come walking = \\(\\frac { 1 }{ 3 }\\) of remaining
\nRest student = 1 – (\\(\\frac { 2 }{ 5 }\\) + \\(\\frac { 1 }{ 4 }\\))
\n\"ML<\/p>\n

Question 14.
\nA mother and her two sons got a room constructed for \u20b9 60,000. The elder son contributes \\(\\frac { 3 }{ 8 }\\) of his mother’s contribution while the younger son contributes \\(\\frac { 1 }{ 2 }\\) of his mother’s share. How much do the three contribute individually?
\nSolution:
\nCost of a room = \u20b9 60000
\nContribution of elder son = \\(\\frac { 3 }{ 8 }\\) of her mother’s contribution
\nand contribution of younger son = \\(\\frac { 1 }{ 2 }\\) of his mother’s share
\nLet mother contribution = 1
\nThen elder son’s contribution = \\(\\frac { 3 }{ 8 }\\)
\nand younger’s = \\(\\frac { 1 }{ 2 }\\)
\nNow ratios in their share = 1 : \\(\\frac { 3 }{ 8 }\\) : \\(\\frac { 1 }{ 2 }\\)
\n= 8 : 3 : 4
\nSum of ratios = 8 + 3 + 4 = 15
\n\"ML<\/p>\n

Question 15.
\nIn a class of 56 students, the number of boys is \\(\\frac { 2 }{ 5 }\\) th of the number of girls. Find the number of boys and girls.
\nSolution:
\nNumber of students in a class = 56
\nLet number of girls = 1
\n\"ML<\/p>\n

Question 16.
\nA man donated \\(\\frac { 1 }{ 10 }\\) of his money to a school, \\(\\frac { 1 }{ 6 }\\) th of the remaining to a church and the remaining money he distributed equally among his three children. If each child gets \u20b9 50000, how much money did the man originally have?
\nSolution:
\nLet total money of a man = 1
\nAmount donated to a school = \\(\\frac { 1 }{ 10 }\\)
\nRemaining money = 1 – \\(\\frac { 1 }{ 10 }\\) = \\(\\frac { 9 }{ 10 }\\)
\n\"ML<\/p>\n

Question 17.
\nIf \\(\\frac { 1 }{ 4 }\\) of a number is added to \\(\\frac { 1 }{ 3 }\\) of that number, the result is 15 greater than half of that number. Find the number.
\nSolution:
\nLet a number be = x
\nThen according to the condition,
\n\"ML<\/p>\n

Question 18.
\nA student was asked to multiply a given number by \\(\\frac { 4 }{ 5 }\\). By mistake, he divided the given number by \\(\\frac { 4 }{ 5 }\\). His answer was 36 more than the correct answer. What was the given number?
\nSolution:
\nLet given number = x
\nAccording to the condition,
\n\"ML<\/p>\n

ML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 1 Rational Numbers Ex 1.6 Question 1. In a bag, there are 20 kg of fruits. If 7 kg of these fruits be oranges and 8 kg of thee are apples and rest are grapes. Find the mass of the grapes in the bag. Solution: Total …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/43344"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=43344"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/43344\/revisions"}],"predecessor-version":[{"id":159053,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/43344\/revisions\/159053"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=43344"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=43344"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=43344"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}