text: ## RD Sharma Solutions For Class 7 Maths Chapter 9 Ratio And Proportion Free Online
  
  Exercise 9.1 Page No: 9.6
  1. If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y
  Solution:
  Given x: y = 3: 5
  We can write above equation as
  x/y = 3/5
  5x = 3y
  x = 3y/5
  By substituting the value of x in given equation 3x + 4y: 8x + 5y we get,
  3x + 4y: 8x + 5y = 3 (3y/5) + 4y: 8 (3y/5) + 5y
  = (9y + 20y)/5: (24y + 25y)/5
  = 29y/5: 49y/5
  = 29y: 49y
  = 29: 49
  2. If x: y = 8: 9, find the ratio (7x – 4y): 3x + 2y.
  Solution:
  Given x: y = 8: 9
  We can write above equation as
  x/y = 8/9
  9x = 8y
  x = 8y/9
  By substituting the value of x in the given equation (7x – 4y): 3x + 2y we get,
  (7x – 4y): 3x + 2y = 7 (8y/9) – 4y: 3 (8y/9): + 2y
  = (56y – 36y)/9: 42y/9
  = 20y/9: 42y/9
  = 20y: 42y
  = 20: 42
  = 10: 21
  3. If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers.
  Solution:
  Given two numbers are in the ratio 6: 13
  Let the required number be 6x and 13x
  The LCM of 6x and 13x is 78x
  = 78x = 312
  x = (312/78)
  x = 4
  Thus the numbers are 6x = 6 (4) = 24
  13x = 13 (4) = 52
  4. Two numbers are in the ratio 3: 5. If 8 is added to each number, the ratio becomes 2:3. Find the numbers.
  Solution:
  Let the required numbers be 3x and 5x
  Given that if 8 is added to each other then ratio becomes 2: 3
  That is 3x + 8: 5x + 8 = 2: 3
  (3x + 8)/ (5x + 8) = 2/3
  3 (3x + 8) = 2 (5x + 8)
  9x + 24 = 10x + 16
  By transposing
  24 – 16 = 10x – 9x
  x = 8
  Thus the numbers are 3x = 3 (8) = 24
  And 5x = 5 (8) = 40
  5. What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3
  Solution:
  Let the number to be added is x
  Then (7 + x) + (13 + x) = (2/3)
  (7 + x) 3 = 2 (13 + x)
  21 + 3x = 26 + 2x
  3x – 2x = 26 – 21
  x = 5
  Hence the required number is 5
  6. Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers
  Solution:
  Given that three numbers are in the ratio 2: 3: 5 and sum of them is 800
  Therefore sum of the terms of the ratio = 2 + 3 + 5 = 10
  First number = (2/10) × 800
  = 2 × 80
  = 160
  Second number = (3/10) × 800
  = 3 × 80
  = 240
  Third number = (5/10) × 800
  = 5 × 80
  = 400
  The three numbers are 160, 240 and 400
  7. The ages of two persons are in the ratio 5: 7. Eighteen years ago their ages were in the ratio 8: 13. Find their present ages.
  Solution:
  Let present ages of two persons be 5x and 7x
  Given ages of two persons are in the ratio 5: 7
  And also given that 18 years ago their ages were in the ratio 8: 13
  Therefore (5x – 18)/ (7x – 18) = (8/13)
  13 (5x – 18) = 8 (7x – 18)
  65x – 234 = 56x – 144
  65x – 56x = 234 – 144
  9x = 90
  x = 90/9
  x = 10
  Thus the ages are 5x = 5 (10) = 50 years
  And 7x = 7 (10) = 70 years
  8. Two numbers are in the ratio 7: 11. If 7 is added to each of the numbers, the ratio becomes 2: 3. Find the numbers.
  Solution:
  Let the required numbers be 7x and 11x
  If 7 is added to each of them then
  (7x + 7)/ (11x + 7) = (2/3)
  3 (7x + 7) = 2 (11x + 7)
  21x + 21 = 22x + 14
  22x – 21x = 21 – 14
  x = 21 – 14 = 7
  Thus the numbers are 7x = 7 (7) =49
  And 11x = 11 (7) = 77
  9. Two numbers are in the ratio 2: 7. 11 the sum of the numbers is 810. Find the numbers.
  Solution:
  Given two numbers are in the ratio 2: 7
  And their sum = 810
  Sum of terms in the ratio = 2 + 7 = 9
  First number = (2/9) × 810
  = 2 × 90
  = 180
  Second number = (7/9) × 810
  = 7 × 90
  = 630
  10. Divide Rs 1350 between Ravish and Shikha in the ratio 2: 3.
  Solution:
  Given total amount to be divided = 1350
  Sum of the terms of the ratio = 2 + 3 = 5
  Ravish share of money = (2/5) × 1350
  = 2 × 270
  = Rs. 540
  And Shikha’s share of money = (3/5) × 1350
  = 3 × 270
  = Rs. 810
  11. Divide Rs 2000 among P, Q, R in the ratio 2: 3: 5.
  Solution:
  Given total amount to be divided = 2000
  Sum of the terms of the ratio = 2 + 3 + 5 = 10
  P’s share of money = (2/10) × 2000
  = 2 × 200
  = Rs. 400
  And Q’s share of money = (3/10) × 2000
  = 3 × 200
  = Rs. 600
  And R’s share of money = (5/10) × 2000
  = 5 × 200
  = Rs. 1000
  12. The boys and the girls in a school are in the ratio 7:4. If total strength of the school be 550, find the number of boys and girls.
  Solution:
  Given that boys and the girls in a school are in the ratio 7:4
  Sum of the terms of the ratio = 7 + 4 = 11
  Total strength = 550
  Boys strength = (7/11) × 550
  = 7 × 50
  = 350
  Girls strength = (4/11) × 550
  = 4 × 50
  = 200
  13. The ratio of monthly income to the savings of a family is 7: 2. If the savings be of Rs. 500, find the income and expenditure.
  Solution:
  Given that the ratio of income and savings is 7: 2
  Let the savings be 2x
  2x = 500
  So, x = 250
  Therefore,
  Income = 7x
  Income = 7 × 250 = 1750
  Expenditure = Income – savings
  = 1750 – 500
  = Rs.1250
  14. The sides of a triangle are in the ratio 1: 2: 3. If the perimeter is 36 cm, find its sides.
  Solution:
  Given sides of a triangle are in the ratio 1: 2: 3
  Perimeter = 36cm
  Sum of the terms of the ratio = 1 + 2 + 3 = 6
  First side = (1/6) × 36
  = 6cm
  Second side = (2/6) × 36
  = 2 × 6
  = 12cm
  Third side = (3/6) × 36
  = 6 × 3
  = 18cm
  15. A sum of Rs 5500 is to be divided between Raman and Amen in the rate 2: 3. How much will each get?
  Solution:
  Given total amount to be divided = 5500
  Sum of the terms of the ratio = 2 + 3 = 5
  Raman’s share of money = (2/5) × 5500
  = 2 × 1100
  = Rs. 2200
  And Aman’s share of money = (3/5) × 5500
  = 3 × 1100
  = Rs. 3300
  16. The ratio of zinc and copper in an alloy is 7: 9. It the weight of the copper in the alloy is 11.7 kg, find the weight of the zinc in the alloy.
  Solution:
  Given that ratio of zinc and copper in an alloy is 7: 9
  Let their ratio = 7x: 9x
  Weight of copper = 11.7kg
  9x = 11.7
  x = 11.7/9
  x = 1.3
  Weight of the zinc in the alloy = 1.3 × 7
  = 9.10kg
  17. In the ratio 7: 8. If the consequent is 40, what a the antecedent
  Solution:
  Given ratio = 7: 8
  Let the ratio of consequent and antecedent 7x: 8x
  Consequent = 40
  8x = 40
  x = 40/8
  x = 5
  Antecedent = 7x = 7 × 5 = 35
  18. Divide Rs 351 into two parts such that one may be to the other as 2: 7.
  Solution:
  Given total amount is to be divided = 351
  Ratio 2: 7
  The sum of terms = 2 + 7
  = 9
  First ratio of amount = (2/9) × 351
  = 2 × 39
  = Rs. 78
  Second ratio of amount = (7/9) × 351
  = 7 × 39
  = Rs. 273
  19. Find the ratio of the price of pencil to that of ball pen, if pencil cost Rs.16 per score and ball pen cost Rs.8.40 per dozen.
  Solution:
  One score contains 20 pencils
  And cost per score = 16
  Therefore pencil cost = 16/20
  = Rs. 0.80
  Cost of one dozen ball pen = 8.40
  1 dozen = 12
  Therefore cost of pen = 8.40/12
  = Rs 0.70
  Ratio of the price of pencil to that of ball pen = 0.80/0.70
  = 8/7
  = 8: 7
  20. In a class, one out of every six students fails. If there are 42 students in the class, how many pass?
  Solution:
  Given, total number of students = 42
  One out of 6 student fails
  x out of 42 students
  16 = x/42
  x = 42/6
  x = 7
  Number of students who fail = 7 students
  No of students who pass =Total students – Number of students who fail
  = 42 – 7
  = 35 students.
  
  Exercise 9.2 Page No: 9.10
  1. Which ratio is larger in the following pairs?
  (i) 3: 4 or 9: 16
  (ii) 15: 16 or 24: 25
  (iii) 4: 7 or 5: 8
  (iv) 9: 20 or 8: 13
  (v) 1: 2 or 13: 27
  Solution:
  (i) Given 3: 4 or 9: 16
  LCM for 4 and 16 is 16
  3: 4 can be written as = 3/4
  3/4 × (4/4) = 12/16
  And we have 9/16
  Clearly 12 > 9
  Therefore 3: 4 > 9: 16
  (ii) Given 15: 16 or 24: 25
  LCM for 16 and 25 is 400
  15: 16 can be written as = 15/16
  15/16 × (25/25) = 375/400
  And we have 24/25
  24/25 × (16/16) = 384/400
  Clearly 384 > 375
  Therefore 15: 16 < 24: 25
  (iii) Given 4: 7 or 5: 8
  LCM for 7 and 8 is 56
  4: 7 can be written as = 4/7
  4/7 × (8/8) = 32/56
  And we have 5/8
  5/8 × (7/7) = 35/56
  Clearly 35 > 32
  Therefore 4: 7 < 5: 8
  (iv) Given 9: 20 or 8: 13
  LCM for 20 and 13 is 260
  9: 20 can be written as = 9/20
  9/20 × (13/13) = 117/260
  And we have 8/13
  8/13 × (20/20) = 160/260
  Clearly 160 > 117
  Therefore 9: 20 < 8: 13
  (v) Given 1: 2 or 13: 27
  LCM for 2 and 27 is 54
  1: 2 can be written as = 1/2
  1/2 × (27/27) = 27/54
  And we have 13/27
  13/27 × (2/2) = 26/54
  Clearly 27 > 26
  Therefore 1: 2 > 13: 27
  2. Give the equivalent ratios of 6: 8.
  Solution:
  Given 6: 8
  By multiplying both numerator and denominator by 2 we equivalent ratios
  6/8 × (2/2) = 12/16
  And also by dividing both numerator and denominator by 2 we equivalent ratios
  (6/2)/ (8/2) = 3/4
  Two equivalent ratios are 3: 4 = 12: 16
  3. Fill in the following blanks:
  12/20 = …. /5 = 9/….
  Solution:
  12/20 = 3/5 = 9/15
  Explanation:
  Consider 12/20 = …. /5
  Let unknown value be x
  Therefore 12/20 = x/5
  On cross multiplying
  x = 60/20
  x = 3
  Consider 12/20 = 9/….
  Let the unknown value be y
  Therefore 12/20 = 9/y
  On cross multiplying we get
  y = 180/12
  y = 15
  
  Exercise 9.3 Page No: 9.13
  1. Find which of the following are in proportion?
  (i) 33, 44, 66, 88
  (ii) 46, 69, 69, 46
  (iii) 72, 84, 186, 217
  Solution:
  (i) Given 33, 44, 66, 88
  Product of extremes = 33 × 88 = 2904
  Product of means = 44 × 66 = 2904
  Therefore product of extremes = product of means
  Hence given numbers are in proportion.
  (ii) Given 46, 69, 69, 46
  Product of extremes = 46 × 46 = 2116
  Product of means = 69 × 69 = 4761
  Therefore product of extremes is not equal to product of means
  Hence given numbers are not in proportion.
  (iii) Given 72, 84, 186, 217
  Product of extremes = 72 × 217 = 15624
  Product of means = 84 × 186 = 15624
  Therefore product of extremes = product of means
  Hence given numbers are in proportion.
  2. Find x in the following proportions:
  (i) 16: 18 = x: 96
  (ii) x: 92 = 87: 116
  Solution:
  (i) Given 16: 18 = x: 96
  In proportion we know that product of extremes = product of means
  16/18 = x/96
  On cross multiplying
  x = (18 × 96)/ 16
  x = 256/3
  (ii) Given x: 92 = 87: 116
  In proportion we know that product of extremes = product of means
  x/ 92 = 87/116
  On cross multiplying
  x = (87 × 92)/ 116
  x = 69
  3. The ratio of income to the expenditure of a family is 7: 6. Find the savings if the income is Rs.1400.
  Solution:
  Given that income = 1400
  Given the ratio of income and expenditure = 7: 6
  7x = 1400
  Therefore x = 200
  Expenditure = 6x = 6 × 200 = Rs.1200
  Savings = Income – Expenditure
  = 1400 -1200
  = Rs.200
  4. The scale of a map is 1: 4000000. What is the actual distance between the two towns if they are 5cm apart on the map?
  Solution:
  Given that the scale of map = 1: 4000000
  Let us assume the actual distance between towns is x cm
  1: 4000000 =5: x
  x = 5 × 4000000
  x = 20000000 cm
  We know that 1km = 1000 m
  1m = 100 cm
  Therefore
  x = 200 km
  5. The ratio of income of a person to his savings is 10: 1. If his savings for one year is Rs.6000, what is his income per month?
  Solution:
  Given that the ratio of income of a person to his savings is 10: 1
  Savings per year = 6000
  Savings per month = 6000/12
  = Rs.500
  Then let income per month be x
  x: 500 = 10:1
  x = 500 × 10
  x = 5000
  Income per month is Rs. 5000
  6. An electric pole casts a shadow of length 20 meters at a time when a tree 6 meters high casts a shadow of length 8 meters. Find the height of the pole.
  Solution:
  Given that length electric pole shadow is 20m
  Height of the tree: Length of the shadow of tree
  Height of the pole: Length of the shadow of pole
  x: 20 = 6: 8
  x = 120/8
  x = 15
  Therefore height of the pole is 15 meters
  Courtesy : CBSE