{"id":42487,"date":"2023-04-02T07:41:01","date_gmt":"2023-04-02T02:11:01","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=42487"},"modified":"2023-04-03T10:01:22","modified_gmt":"2023-04-03T04:31:22","slug":"ml-aggarwal-class-7-solutions-for-icse-maths-chapter-6-ex-6-1","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-7-solutions-for-icse-maths-chapter-6-ex-6-1\/","title":{"rendered":"ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 6 Ratio and Proportion Ex 6.1"},"content":{"rendered":"

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 6 Ratio and Proportion Ex 6.1<\/h2>\n

Question 1.
\nExpress the following ratios in simplest form:
\n\"ML
\nSolution:
\n\"ML
\n\"ML<\/p>\n

Question 2.
\nFind the ratio of each of the following in simplest form:
\n(i) \u20b9 5 to 50 paise
\n(ii) 3 km to 300 m
\n(iii) 9 m to 27 cm
\n(iv) 15 kg to 210 g
\n(v) 25 minutes to 1.5 hours
\n(vi) 30 days to 36 hours
\nSolution:
\n(i) \u20b9 5 to 50 paise
\n= 500 paise : 50 paise
\n= 10 : 1 (Dividing by 50)
\n(ii) 3 km to 300 m
\n= 3000 m to 300 m
\n= 10 : 1 (Dividing by 300)
\n(iii) 9 m to 27 cm
\n= 9 \u00d7 100 cm : 27 cm
\n= 100 : 3 (Dividing by 9)
\n(iv) 15 kg to 210 g
\n= 15 \u00d7 1000 g : 210 g
\n= 15000 : 210
\n= 500 : 7 (Dividing by 30)
\n(v) 25 minutes to 1.5 hours
\n= 25 minutes to \\(\\frac { 3 }{ 2 }\\) \u00d7 60
\n= 25 : 90
\n= 5 : 18
\n(vi) 30 days to 36 hours
\n= 30 \u00d7 24 hours to 36 hours
\n= 720 : 36
\n= 20 : 1 (Dividing by 36)<\/p>\n

Question 3.
\nIf A : B = 3 : 4 and B : C = 8 : 9, then find A : C.
\nSolution:
\nA : B = 3 : 4 and B : C = 8 : 9
\n\\(\\frac { A }{ B }\\) = \\(\\frac { 3 }{ 4 }\\)
\n\"ML<\/p>\n

Question 4.
\nIf A : B = 5 : 8 and B : C = 18 : 25, then find A : B : C.
\nSolution:
\nA : B = 5 : 8 and B : C = 18 : 25
\nHere, In A : B, B = 8
\nand In B : C, B = 18
\nLCM of 8, 18 is 72
\n\"ML<\/p>\n

Question 5.
\nIf 3A = 2B = 5C, then find A : B : C.
\nSolution:
\nLet 3A = 2B = 5C = 1
\n\"ML<\/p>\n

Question 6.
\nOut of daily income of \u20b9 120, a labourer spends \u20b9 90 on food and shelter and saves the rest. Find the ratio of his
\n(i) spending to income
\n(ii) saving to income
\n(iii) saving to spending.
\nSolution:
\nDaily income = \u20b9 120
\nExpenditure = \u20b9 90
\nSavings = \u20b9 120 – \u20b9 90 = \u20b9 30
\n(i) Ratio between spending to income
\n= 90 : 120
\n= 3 : 4 (Dividing by 30)
\n(ii) Ratio between saving to income
\n= 30 : 120
\n= 1 : 4 (Dividing by 30)
\n(iii) Ratio between saving to spending
\n= 30 : 90
\n= 1 : 3 (Dividing by 30)<\/p>\n

Question 7.
\n5 grams of an alloy contains 3\\(\\frac { 3 }{ 4 }\\) grams copper and the rest is nickel. Find the ratio by weight of nickel to copper.
\nSolution:
\nTotal weight of an alloy = 5 gms
\n3 15
\nWeight of copper = 3\\(\\frac { 3 }{ 4 }\\) gms = \\(\\frac { 15 }{ 4 }\\) gms
\nWeight of nickel = Total weight of alloy – weight of copper
\n\"ML
\n\"ML<\/p>\n

Question 8.
\nA pole of height 3 metres is struck by a speeding car and breaks into two pieces such that the first piece is \\(\\frac { 1 }{ 2 }\\) of the second. Find the length of both pieces.
\nSolution:
\nTotal height of pole = 3 metres
\nLet length of 2nd piece = x
\nLength of 1st piece = \\(\\frac { 1 }{ 2 }\\) x
\nRatio of lengths of two parts = \\(\\frac { 1 }{ 2 }\\) x : 1x
\n\"ML
\nLength of 1st part = 1 m
\nLength of 2nd part = 2m<\/p>\n

Question 9.
\nHeights of Anshul and Dhruv are 1.04 m and 78 cm respectively. Divide 35 sweets between them in the ratio of their heights.
\nSolution:
\nHeight of Anshul : Height of Dhruv
\n1.4 m : 78 cm
\n(1.04 \u00d7 100) cm : 78 cm
\n= 104 : 78
\n= \\(\\frac { 104 }{ 78 }\\) (Dividing both by 2)
\n= \\(\\frac { 52 }{ 39 }\\) (Dividing boht by 13)
\n= \\(\\frac { 4 }{ 3 }\\)
\n= 4 : 3
\nRatio of heights of Anshul and Dhruv is 4 : 3
\nThus, we are to divide 35 sweets in the ratio 4 : 3
\nSum of the terms of the ratios = 4 + 3 = 7
\nShare of Anshul = \\(\\frac { 4 }{ 7 }\\) of 35 sweets
\n= \\(\\frac { 4 }{ 7 }\\) \u00d7 35
\n= 20 sweets
\nShare of Dhruv = \\(\\frac { 3 }{ 7 }\\) of 35 sweets
\n= \\(\\frac { 3 }{ 7 }\\) \u00d7 35
\n= 15 sweets<\/p>\n

Question 10.
\n\u20b9 180 are to be divided among three children in the ratio \\(\\frac { 1 }{ 3 } :\\frac { 1 }{ 4 } :\\frac { 1 }{ 6 }\\) Find the share of each child.
\nSolution:
\nFirst we will simplify the given ratio
\nGiven ratio \\(\\frac { 1 }{ 3 } :\\frac { 1 }{ 4 } :\\frac { 1 }{ 6 }\\)
\nTaking L.C.M. of 3, 4 and 6
\nL.C.M. of 3, 4 and 6 = 12
\n\"ML<\/p>\n

Question 11.
\nA natural number has been divided into two parts in the ratio 7 : 11. If the difference between the two parts is 20, find the number and the two parts.
\nSolution:
\nLet the first part = 7x
\nSecond part = 11x
\nAccording to given statement,
\n11x – 7x = 20
\n\u21d2 4x = 20
\n\u21d2 x = 5
\nFirst part = 7x = 7 \u00d7 5 = 35
\nSecond part = 11x = 11 \u00d7 5 = 55
\nand number will be 35 + 55 = 90<\/p>\n

Question 12.
\nA certain sum of money has been divided into two parts in the ratio 9 : 13. If the second part is \u20b9 260, find the total amount.
\nSolution:
\nLet the total amount = \u20b9 x
\nThe amount has been divided into two parts in the ratio 9 : 13.
\nSum of the terms of the ratio = 9 + 13 = 22
\nFirst part = \\(\\frac { 9 }{ 22 }\\) of total amount
\nSecond part = \\(\\frac { 13 }{ 22 }\\) of total amount
\nAccording to given statement
\n\"ML<\/p>\n

Question 13.
\nThe ratio of the present ages of Anjali and Ashu is 2 : 3. Five years hence, the ratio of theif ages will be 3 : 4. Find their present ages.
\nSolution:
\nRatio of present ages of Anjali and Ashu = 2 : 3
\nLet age of Anjali = 2x
\nand age of Ashu = 3x
\n5 years hence,
\nAge of Anjali = 2x + 5
\nand age of Ashu = 3x + 5
\n\\(\\frac { 2x+5 }{ 3x+5 }\\) = \\(\\frac { 3 }{ 4 }\\)
\n9x + 15 = 8x + 20
\n9x – 8x = 20 – 15
\nx = 5
\nPresent age of Anjali = 2x = 2 \u00d7 5 = 10 years
\nand age of Ashu = 3x = 3 \u00d7 5 = 15 years<\/p>\n

Question 14.
\nThe present ages of A and B are in the ratio 5 : 6. Three years ago, their ages were in the ratio 4 : 5. find their present ages.
\nSolution:
\nRatio of the present age of A and B = 5 : 6
\nLet age of A = 5x
\nand age of b = 6x
\n3 years ago,
\nAge of A was = 5x – 3
\nand age of B was = 6x – 3
\n\\(\\frac { 5x-3 }{ 6x-3 }\\) = \\(\\frac { 4 }{ 5 }\\)
\n\u21d2 25x – 15 = 24x – 12
\n\u21d2 25x – 24x = -12 + 15
\n\u21d2 x = 3
\nPresent age of A = 5x = 5 \u00d7 3 = 15 years
\nand age of B = 6x = 6 \u00d7 3 = 18 years<\/p>\n

Question 15.
\nTwo numbers are in the ratio 5 : 6. When 2 is added to first and 3 is added to second, they are in the ratio 4 : 5. Find the numbers.
\nSolution:
\nRatio in two numbers = 5 : 6
\nLet first number = 5x
\nThen second number = 6x
\nAdding 2 in the first and 3 in the second
\nA = 5x + 2
\nB = 6x + 3
\n\\(\\frac { 5x+2 }{ 6x+3 }\\) = \\(\\frac { 4 }{ 5 }\\)
\n25x + 10 = 24x + 12
\n25x – 24x = 12- 10
\nx = 2
\nFirst number = 5x = 5 \u00d7 2 = 10
\nand second = 6x = 6 \u00d7 2 = 12<\/p>\n

Question 16.
\nThe ratio of number of boys to the number of girls in a school of 1430 students is 7 : 6. If 26 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of number of boys to the number of girls may change to 8 : 7.
\nSolution:
\nNumber of students = 1430
\nRatio in number of boys and girls = 7 : 6
\nLet number of boys = 7x and of girls = 6x
\n7x + 6x = 1430
\n\u21d2 13x = 1430
\n\u21d2 x = 110
\nNumber of boys = 7x = 7 \u00d7 110 = 770
\nand number of girls = 6x = 6 \u00d7 110 = 660
\nNow adding 26 new girls, the number of girls will be = 660 + 26 = 686
\nLet new boys be added = y
\nThe number of boys = 770 + y
\nNow new ratio = 8 : 7
\n\\(\\frac { 770+y }{ 686 }\\) = \\(\\frac { 8 }{ 7 }\\)
\n5390 + 7y = 5488
\n7y = 5488 – 5390 = 98
\ny = 14
\nNumber of new boys admitted = 14<\/p>\n

Question 17.
\nWhich ratio is greater?
\n(i) 5 : 6 or 6 : 7
\n(ii) 13 : 24 or 17 : 32
\nSolution:
\n(i) 5 : 6 or 6 : 7
\n5 : 6 = \\(\\frac { 5 }{ 6 }\\) and 6 : 7 = \\(\\frac { 6 }{ 7 }\\)
\nConverting them into equivalent fraction by taking L.C.M. of 6 and 7 = 42
\n\"ML
\n\"ML<\/p>\n

ML Aggarwal Class 7 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 6 Ratio and Proportion Ex 6.1 Question 1. Express the following ratios in simplest form: Solution: Question 2. Find the ratio of each of the following in simplest form: (i) \u20b9 5 to 50 paise (ii) 3 km to 300 m (iii) 9 m to 27 …<\/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\/42487"}],"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=42487"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42487\/revisions"}],"predecessor-version":[{"id":158963,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42487\/revisions\/158963"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=42487"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=42487"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=42487"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}