\nQuestion 1.
\nFill in the blanks:
\n(i) In the ratio 3 : 5, the first term is …………. and second term is ………..
\n(ii) In a ratio, the first term is also called ……….. and second term is also called …….
\n(iii) If two terms of a ratio have no common factor (except 1), then the ratio is said to be in …….
\n(iv) To simplify a ratio, we divide the two terms by their …….
\n(v) The simplest form of the ratio 8 : 12 is ……
\n(vi) 90 cm : 1.5 m = ……….
\n(vii) Method of comparison of two quantities of the same kind (in same units) by division is known as …………
\n(viii) When two ratios are equal, they are said to be in ………
\n(ix) When four quantities are in proportion, then the product of ………… is equal to product of middle terms.
\n(x) 4.5 omo is equal to ………
\nSolution:
\n(i) In the ratio 3 : 5, the first term is 3 and second term is 5.
\n(ii) In a ratio, the first term is also called antecedent and second term is also called consequent.
\n(iii) If two terms of a ratio have no common factor (except 1), then the ratio is said to be in simplest form.
\n(iv) To simplify a ratio, we divide the two terms by their H.C.F.
\n(v) The simplest form of the ratio 8 : 12 is 2 : 3.
\n(vi) 90 cm : 1.5 m = 3 : 5.
\n(vii) Method of comparison of two quantities of the same kind (in same units) by division is known as ratio.
\n(viii)When two ratios are equal, they are said to be in proportion.
\n(ix) When four quantities are in proportion, then the product of extreme terms is equal to product of middle terms.
\n(x) 4.5 of \u20b940 is equal to \u20b91.80.<\/p>\n
Question 2.
\nState whether the following statemtns are true (T) or false (F):
\n(i) Ratio exists only between two quantities of the same kind.
\n(ii) Ratio has no units.
\n(iii) If a \u2260 b, then the ratio a: bis different from the ratio b : a.
\n(iv) If we multiply or divide both terms of a ratio by the same non-zero number, then the ratio remains the same.
\n(v) The ratio a: b is said to be in simplest form if HCF of a and b is 1.
\n(vi) In some situations, comparison of two quantities (of same kind) by difference does not make much sense.
\nSolution:
\n(i) Ratio exists only between two quantities of the same kind. True
\n(ii) Ratio has no units. True
\n(iii) If a \u2260 b, then the ratio a : b is different from the ratio b : a. True
\n(iv) If we multiply or divide both terms of a ratio by the same non-zero number, then the ratio remains the same. True
\n(v) The ratio a: bis said to be in simplest form if HCF of a and b is 1. True
\n(vi) In some situations, comparison of two quantities (of same kind) by difference does not make much sense. True<\/p>\n
Multiple Choice Questions<\/strong> Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. Value Based Questions<\/strong> Higher Order Thinking Skills (Hots)<\/strong> Question 2. Question 3.
\nChoose the correct answer from the given four options (3 to 18):<\/strong>
\nQuestion 3.
\nA ratio equivalent to 5 : 7 is
\n(a) 10:21
\n(b) 15 : 14
\n(c) 20 : 28
\n(d) 25 : 49
\nSolution:
\n5 : 7
\n\\(\\Rightarrow \\frac{5}{7} \\times \\frac{4}{4}=\\frac{20}{28}=20 : 28\\) (c)<\/p>\n
\nThe ratio 384 : 480 in the simplest form is
\n(a) 2 : 5
\n(b) 3 : 5
\n(c) 5 : 4
\n(d) 4 : 5
\nSolution:
\n384 : 480
\nDividing by 96, we get
\n\\(=\\frac{384}{96} : \\frac{480}{96} \\Rightarrow 4 : 5\\) (d)<\/p>\n
\nThe ratio of 20 minutes to 1 hour is
\n(a) 20 : 1
\n(b) 1 : 3
\n(c) 1 : 4
\n(d) 2 : 5
\nSolution:
\n20 min : 1 hour
\n20 min : 60 minutes
\n= 20 : 60
\nDivide both terms by
\n\\(=\\frac{20}{20} : \\frac{60}{20}=1 : 3\\)
\n\u21d2 1 :3 (b)<\/p>\n
\nThe ratio of 150 g to 2 kg is
\n(a) 75 : 1
\n(b) 40 : 3
\n(c) 3 : 40
\n(d) 3 : 200
\nSolution:
\nWe have, 150 g to 2 kg
\n= 150 g : 2 \u00d7 1000 g
\n= 150 g : 2000 g
\nDivide both terms by 50
\n\\(=\\frac{150}{50} : \\frac{2000}{50}\\)
\n= 3 : 40 (c)<\/p>\n
\nIn a class of 40 students, 25 students play cricket and the remaining play tennis. The ratio of number of students playing crickets to the number of students playing tennis is
\n(a) 5 : 8
\n(b) 5 : 3
\n(c) 3 : 5
\n(d) 8 : 5
\nSolution:
\nTotal number of students = 40
\nStudent play cricket = 25
\nStudent play tennis = 40 – 25 = 15
\nNumber of students : Number of students
\nplay cricket play tennis
\n= 25 : 15
\nDivide both terms by 5
\n= \\(\\frac{25}{5} : \\frac{15}{5}\\)
\n= 5:3 (b)<\/p>\n
\nTwo numbers are in the ratio 3 : 5. If the sum of numbers is 144, then the smaller number is
\n(a) 54
\n(b) 72
\n(c) 90
\n(d) 48
\nSolution:
\nLet any number = x
\nFirst number : Second number
\n3 : 5
\nSum of the numbers = 144
\n\u21d2 3x + 5x = 144
\n\u21d2 8x = 144
\n\\(x=\\frac{144}{8}=18\\)
\nFirst number = 3 \u00d7 18 = 54
\nSecond number = 5 \u00d7 18 = 90
\n\u2234 The smallest number = 54 (a)<\/p>\n
\nThe ratio of number of girls to the number of boys in a class is 5 : 4. If there are 25 girls in the class, then the number of boys in the class is
\n(a) 15
\n(b) 20
\n(c) 30
\n(d) 40
\nSolution:
\nLet the number of boys in the class = x
\nAccording to question,
\nGirls : Boys = 5 : 4
\n25 : x = 5 : 4
\n\\(\\frac{25}{x}=\\frac{5}{4}\\)
\n\\(x=\\frac{25 \\times 4}{5}=20\\)
\nHence number of boys = 20 (b)<\/p>\n
\nThe ratio of the number of sides of a square and the number of edges of a cube is
\n(a) 1 : 2
\n(b) 1 : 3
\n(c) 1 : 4
\n(d) 2 : 3
\nSolution:
\nNumber of sides of square = 4
\nEdges of cube = 12
\n\u2234 Ratio = 4 : 12
\n\u21d21 : 3 (b)<\/p>\n
\nIn shelf, the books with green cover and that with brown cover are in the ratio 2:3. If there are 18 books with green cover, then the number of books with brown cover is
\n(a) 12
\n(b) 24
\n(c) 27
\n(d) 36
\nSolution:
\nLet the brown covered books = x
\nand green covered books =18
\nGreen covered books: Brown covered books
\n= 2 : 3
\n18 : x = 2 : 3
\n\u21d2 \\(\\frac{18}{x}=\\frac{2}{3}\\)
\n\u21d2 \\(x=\\frac{18 \\times 3}{2}=9 \\times 3 \\Rightarrow x=27\\) (c)<\/p>\n
\nIn a box, the ratio of the number of red marbles to that of blue marbles is 4 :7. Which of the following could be the total number of marbles in the box?
\n(a) 14
\n(b) 21
\n(c) 22
\n(d) 28
\nSolution:
\nThe ratio of red marbles to blue marbles = 4:7
\n\u21d2 So total marbles can be
\n4x + 7x = y
\n11 x=y
\ny should be a multiple of 11
\n\u2234 Total number of marble in the box are 22 (c)<\/p>\n
\nIf a, b, c and d are in proportion, then
\n(a) ab = cd
\n(b) ad = be
\n(c) ac = bd
\n(d) none of these
\nSolution:
\na, b, c and d are in proportion, then
\n\u21d2 \\(\\frac{a}{b}=\\frac{c}{d}\\)
\n\u2234 ad = be (b)<\/p>\n
\nIf the weight of 5 bags of rice is 272 kg, then the weight of 1 bag of rice is
\n(a) 50.4 kg
\n(b) 54.4 kg
\n(c) 54.004 kg
\n(d) 54.04 kg
\nSolution:
\nWeight of 5 bags of rice = 272 kg
\nWeight of 1 bag of rice = \\(\\frac{272}{5}\\) kg
\n= 54.4 kg (b)<\/p>\n
\nIf 7 pencils cost \u20b935, then the cost of one dozen pencils is
\n(a) \u20b960
\n(b) \u20b970
\n(c) \u20b930
\n(d) \u20b95
\nSolution:
\n7 pencils cost = \u20b935
\n1 dozen = 12 pencils
\nCost of 1 pencil = \u20b9\\(\\frac{35}{7}\\)
\n\u2234 Cost of 12 pencils (1 dozen) = \\(\\frac{35}{7} \\times 12\\)
\n= \u20b960 (a)<\/p>\n
\nThe ratio 2 : 3 expressed as percentage is
\n(a) 40%
\n(b) 60%
\n(c) \\(66 \\frac{2}{3} \\%\\)
\n(d) \\(33 \\frac{1}{3} \\%\\)
\nSolution:
\nGiven, 2 : 3 = \\(\\frac{2}{3}\\)
\n= \\(\\left(\\frac{2}{3} \\times 100\\right) \\%=\\frac{200}{3}=66 \\frac{2}{3} \\%\\) (c)<\/p>\n
\n0.025 when expressed as percentage is
\n(a) 250%
\n(b) 25%
\n(c) 4%
\n(d) 2.5%
\nSolution:
\n0.025 =\\(\\frac{25}{1000}\\) \u00d7 100 = 2.5% (d)<\/p>\n
\nIn a class, 45% of the students are girls. If there are 18 girls in the class, then the total number of students in the class is
\n(a) 44
\n(b) 40
\n(c) 36
\n(d) 30
\nSolution:
\n% of girls in class = 45%
\nTotal number of girls in class = 18
\nLet total students = x
\nAs per question,
\n45% of x = 18
\n\\(\\frac{45}{100} x=18\\)
\n\\(x=\\frac{18 \\times 100}{45}\\)
\n\u2234Total students = 40 students (b)<\/p>\n
\nQuestion 1.
\nStudents of a colony decided to go to an old age home in their vicinity to wish Happy New year and get blessings from old people.
\nThey carried the following items with them:
\nBouquets 63, New Year Cards 70 and chocolates bars 140. Answer the following questions:
\n(i) What is the ratio of number of bouquets to the number of chocolate bars?
\n(ii) What is the ratio of number of cards to the number of sum of all items?
\nSolution:
\n(i) Number of bouquets = 63
\nNumber of chocolates =140
\n\u2234 Ratio of bouquets to number of chocolate bars.
\n63 : 140 = 9 : 20
\n(ii) Total number of cards = 70
\nSum of all items = 63 + 70 + 140 = 273
\n\u2234 Ratio = 70 : 273 = 10: 39<\/p>\n
\nQuestion 1.
\nDivide \u20b96000 among Irfan, Nagma and Ishan in the raito 3 : 5 : 7.
\nSolution:
\nTotal amount = \u20b96000
\nRatio in Irfan, Nagma and Ishan = 3 : 5 : 7
\nSum of ratios = 3 + 5 + 7= 15
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-8-Ratio-and-Proportion-Objective-Type-Questions-1.png\")
<\/p>\n
\nSapna weighs 54 kg on earth and 9 kg on moon. If a monkey weighs 3.5 kg on moon, then how much will it weigh on the earth?
\nSolution:
\nSapna weight on earth : Sapna weight on moon = Monkey weight on earth : Monkey weight on moon
\n54 : 9 = x : 3.5
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-8-Ratio-and-Proportion-Objective-Type-Questions-2.png\")
<\/p>\n
\nIf 5 men can do a certain construction work in 14 days, then how long will 7 men take to complete the same construction work?
\nSolution:
\n5 men can do construction on work in = 14 days
\n1 man can do construction work in = 14 \u00d7 5 days
\n7 men can do construction work in
\n=\\(\\frac{14 \\times 5}{7}\\) = 10 days<\/p>\nML Aggarwal Class 6 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"