Worked out problems on ratio and proportion are explained here in detailed description using step-by-step procedure. Solved examples involving different questions related to comparison of ratios in ascending order or descending order, simplification of ratios and also word problems on ratio proportion. Sample questions and answers are given below in the worked out problems on ratio and proportion to get the basic concepts of solving ratio proportion. 1. Arrange the following ratios in descending order. 2 : 3, 3 : 4, 5 : 6, 1 : 5 Solution: Given ratios are 2/3, 3/4, 5/6, 1/5 The L.C.M. of 3, 4, 6, 5 is 2 × 2 × 3 × 5 = 60 Now, 2/3 = (2 × 20)/(3 × 20) = 40/60 3/4 = (3 × 15)/(4 × 15) = 45/60 5/6 = (5 × 10)/(6 × 10) = 50/60 1/5 = (1 × 12)/(5 × 12) = 12/60 Clearly, 50/60 > 45/60 > 40/60 > 12/60 Therefore, 5/6 > 3/4 > 2/3 > 1/5 So, 5 : 6 > 3 : 4 > 2 : 3 > 1 : 5
Solution: Sum of the terms of the ratio = 3 + 4 = 7 Sum of numbers = 63 Therefore, first number = 3/7 × 63 = 27 Second number = 4/7 × 63 = 36 Therefore, the two numbers are 27 and 36.3. If x : y = 1 : 2, find the value of (2x + 3y) : (x + 4y) Solution: x : y = 1 : 2 means x/y = 1/2Now, (2x + 3y) : (x + 4y) = (2x + 3y)/(x + 4y) [Divide numerator and denominator by y.] = [(2x + 3y)/y]/[(x + 4y)/2] = [2(x/y) + 3]/[(x/y) + 4], put x/y = 1/2 We get = [2 (1/2) + 3)/(1/2 + 4) = (1 + 3)/[(1 + 8)/2] = 4/(9/2) = 4/1 × 2/9 = 8/9 Therefore the value of (2x + 3y) : (x + 4y) = 8 : 9
4. A bag contains $510 in the form of 50 p, 25 p and 20 p coins in the ratio 2 : 3 : 4. Find the number of coins of each type.
5. If 2A = 3B = 4C, find A : B : C Solution: Let 2A = 3B = 4C = x So, A = x/2 B = x/3 C = x/4 The L.C.M of 2, 3 and 4 is 12 Therefore, A : B : C = x/2 × 12 : x/3 × 12 : x/4 = 12 = 6x : 4x : 3x = 6 : 4 : 3 Therefore, A : B : C = 6 : 4 : 36. What must be added to each term of the ratio 2 : 3, so that it may become equal to 4 : 5? Solution: Let the number to be added be x, then (2 + x) : (3 + x) = 4 : 5 ⇒ (2 + x)/(5 + x) = 4/5 5(2 + x) = 4(3 + x) 10 + 5x = 12 + 4x 5x - 4x = 12 - 10 x = 27. The length of the ribbon was originally 30 cm. It was reduced in the ratio 5 : 3. What is its length now? Solution: Length of ribbon originally = 30 cm Let the original length be 5x and reduced length be 3x. But 5x = 30 cm x = 30/5 cm = 6 cm Therefore, reduced length = 3 cm = 3 × 6 cm = 18 cmMore worked out problems on ratio and proportion are explained here step-by-step. 8. Mother divided the money among Ron, Sam and Maria in the ratio 2 : 3 : 5. If Maria got $150, find the total amount and the money received by Ron and Sam. Solution: Let the money received by Ron, Sam and Maria be 2x, 3x, 5x respectively. Given that Maria has got $ 150. Therefore, 5x = 150 or, x = 150/5 or, x = 30 So, Ron got = 2x = $ 2 × 30 = $60 Sam got = 3x= 3 × 60 = $90 Therefore, the total amount $(60 + 90 + 150) = $300 9. Divide $370 into three parts such that second part is 1/4 of the third part and the ratio between the first and the third part is 3 : 5. Find each part. Solution: Let the first and the third parts be 3x and 5x. Second part = 1/4 of third part. = (1/4) × 5x = 5x/4 Therefore, 3x + (5x/4) + 5x = 370 (12x + 5x + 20x)/4 = 370 37x/4 = 370 x = (370 × 4)/37 x = 10 × 4 x = 40 Therefore, first part = 3x = 3 × 40 = $120 Second part = 5x/4 = 5 × 40/4 = $50 Third part = 5x = 5 × 40 = $ 20010. The first, second and third terms of the proportion are 42, 36, 35. Find the fourth term. Solution: Let the fourth term be x. Thus 42, 36, 35, x are in proportion. Product of extreme terms = 42 ×x Product of mean terms = 36 X 35 Since, the numbers make up a proportion Therefore, 42 × x = 36 × 35 or, x = (36 × 35)/42 or, x = 30 Therefore, the fourth term of the proportion is 30.
11. Set up all possible proportions from the numbers 8, 12, 20, 30. Solution: We note that 8 × 30 = 240 and 12 × 20 = 240 Thus, 8 × 30 = 12 × 20 ………..(I) Hence, 8 : 12 = 20 : 30 ……….. (i) We also note that, 8 × 30 = 20 × 12 Hence, 8 : 20 = 12 : 30 ……….. (ii) (I) can also be written as 12 × 20 = 8 × 30 Hence, 12 : 8 = 30 : 20 ……….. (iii) Last (I) can also be written as 12 : 30 = 8 : 20 ……….. (iv) Thus, the required proportions are 8 : 12 = 20 : 30 8 : 20 = 12 : 30 12 : 8 = 30 : 20 12 : 30 = 8 : 20
Solution: Number of girls in the class = 18 Ratio of boys and girls = 4 : 3 According to the question, Boys/Girls = 4/5 Boys/18 = 4/5 Boys = (4 × 18)/3 = 24 Therefore, total number of students = 24 + 18 = 42.13. Find the third proportional of 16 and 20. Solution: Let the third proportional of 16 and 20 be x. Then 16, 20, x are in proportion. This means 16 : 20 = 20 : x So, 16 × x = 20 × 20 x = (20 × 20)/16 = 25 Therefore, the third proportional of 16 and 20 is 25.● Ratio and Proportion What is Ratio and Proportion? Worked out Problems on Ratio and Proportion Practice Test on Ratio and Proportion ● Ratio and Proportion - Worksheets Worksheet on Ratio and Proportion 8th Grade Math Practice From Worked out Problems on Ratio and Proportion to HOME PAGE
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