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Download RATIO AND PROPORTION Ratio : The ratio of two qualities a and b
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RATIO AND PROPORTION .c om Ratio : The ratio of two qualities a and b in the same units, is the fraction a/b and we write it as a:b. In the ratio a:b, A is known as the antecedent and b is known as the consequent. Ex: The ratio 5:9 represents 5/9 with antecedent=5 , consequent=9 ak s Proportion: The equality of two ratios is called proportion. If a:b=c:d, we write a:b::c:d and we say that a,b,c and d are in proportion. Here a and b are called extremes, while b and c are called mean terms. Product of means=product of extremes Thus, a:b::c:d or (ππ × ππ) = (ππ × ππ) xa m fre Compounded Ratio Ratios are compounded by multiplying together the antecedents for a new antecedent, and the consequents for a new consequent. Ex- Find the compounded ratio of the following ratios3:4, 2:3, 4:5 and 3:8 Sol: The required ratio = 3 20 . Important terms: Fourth proportional: If a:b::c:d, then d is called the fourth proportional to a,b and c. Third proportional: If a:b::b:c, then c is called third proportional to a and b. Mean proportional: Mean proportional between a and b is SQRT(a*b). Compounded ratio: The compounded ratio of the ratios (a:b), (c:d),(e:f) is (ace:bdf). Duplicate Ratio: If the given ratio is (a:b) then its duplicate ratio is (ππ2 : ππ 2 ). w 6. Sub-duplicate ratio of (a:b) is οΏ½βππ: βπποΏ½. 7. Triplicate ratio of (a:b) is (ππ3 : ππ 3 ) 3 3 8. Sub-triplicate ratio of (a:b) is οΏ½ βππ: βπποΏ½. 9. Reciprocal ratio of a : b is b : a 10. Inverse Ratio of a:b is b:a. w w = .e 1. 2. 3. 4. 5. 3×2×4×3 4×3×5×8 Important Properties: 1. Invertendo. If a : b :: c : d then b : a :: d : c Alternendo. If a : b :: c : d then a : c :: b : d Componendo. If a : b :: c : d then (a +b) : b :: (c +d) : d Dividendo. If a : b :: c : d then (a -b) : b :: (c -d) : d Componendo and Dividendo. If a : b :: c : d then (a +b) : (a -b) :: (c +d) : (c -d) i.e., a/b = c/d => (a +b)/(a - b) = (c +d)/(c +d) 6. If a/b = c/d = e/f = ..., then each ratio = (a +c +e +...)/(b +d +f +...) Which is same as ( pa + qc + re +. . . )/( pb + qd + rf + . . . ) i.e a/b = c/d = e/f =(a +c +e +...)/(b +d +f +...) = ( pa + qc + re +. . . )/( pb + qd + rf + . . . ) = ( pna + qnc + rne +. . . )1/n/( pnb + qnd + rnf + . . . )1/n ak s .c om 2. 3. 4. 5. where p, q, r, etc. are constants such that all of them are simultaneously not equal to zero. xa m fre Some important terms used: 1. Ratio of equality : If the antecedent is equal to the consequent in a ratio then the ratio is known as the ratio of equality. Eg β 6:6 2. Ratio of greater inequality : If the antecedent is greater than the consequent in a ratio then the ratio is known as the ratio of greater inequality. Eg β 11:6 3. Ratio of less inequality : If the antecedent is less than the consequent in a ratio then the ratio is known as the ratio of less inequality. Eg β 6:11 .e Proportional Division The process by which a quantity may be divided into parts which bear a given ratio to one another, is called proportional division and the parts are known as proportional parts. For Example : Divide the quantity βaβ in the ratio x:y:z then First Part = π₯π₯ π₯π₯+π¦π¦+π§π§ w Second Part = w w Third Part = π¦π¦ × ππ π₯π₯+π¦π¦+π§π§ π§π§ π₯π₯+π¦π¦+π§π§ × ππ × ππ VARIATION: 1. We say that x is directly proportional to y, if x=ky for some constant k and we write π₯π₯ π¦π¦ 2. We say that x is inversely proportional to y, if xy=k for some constant and we write 1 π₯π₯ π¦π¦ .c om Solved examples: 1. Divide 121 into 3 parts such that the ratio of the three numbers is 2:4:5. Ist part = IInd part = 4 2 2+4+5 2+4+5 IIIrd part = × 121 = 22 × 121 = 44 5 2+4+5 × 121 = 55 Sol: xa m fre 2. If a:b = 4:5 and b:c = 3:7. Find a:b:c? ak s Sol: a:b = 4:5 b:c = 3:7 Now, we will equalize the value of b in both the ratios Thus, we get a : b = (4 : 5) x 3 b:c = (3 : 7) x 5 .e Thus we get, a : b = 12 : 15 b:c = 15 : 35 w And hence the answer is a: b: c = 12:15:35. w w 3. Find the fourth proportion to 3,5,6. Sol: Let is the fourth proportion to 3,5,6. Thus by definition we get 3 : 5 :: 6 : x On solving β ππ ππ = ππ ππ ππππ ππ = ππ×ππ ππ ππππ ππ = ππππ 4. Find third proportion to 4,16. Sol: Let the third proportion to 4, 16 be x. Then, according to the definition we get, 4 16 16 = π₯π₯ ππππ π₯π₯ = Hence, the third proportion to 4, 16 is 64. 16 ×16 4 = 64. ak s 4 βΆ 16 βΆ : 16 βΆ π₯π₯ ππππ .c om Hence, the fourth proportion to 3, 5, 6 is 10. 5. Find three numbers in the ratio of 1 : 2 : 3, so that the sum of their squares is equal to 350. xa m fre Sol: Let the numbers be x, 2x and 3x. Then we have, π₯π₯ 2 + (2π₯π₯)2 + (3π₯π₯)2 = 350 ππππ π₯π₯ 2 + 4π₯π₯ 2 + 9π₯π₯ 2 = 350 ππππ 14π₯π₯ 2 = 350 ππππ π₯π₯ 2 = 25 ππππ π₯π₯ = 5 Hence the required numbers are β Ist part = x = 5 IInd part = 2x = 2×5 = 10 .e IIIrd part = 3x = 3 × 5 = 15. w 6. Find mean proportion between 9 and 36. w w Sol: Let the mean proportion be x. Then according to the question we get β 9 βΆ π₯π₯ βΆ : π₯π₯ βΆ 36 ππππ 9 π₯π₯ = π₯π₯ 36 ππππ π₯π₯ 2 = 9 × 36 ππππ π₯π₯ = (18) ππππ (β18) Hence the mean proportion between 9 and 36 is 18 or -18. 7. 3A=4B=5C then A:B:C? Sol: Let 3A=4B=5C=k then we get ππ ππ π΄π΄ = , π΅π΅ = , πΆπΆ = 3 4 ππ ππ ππ ππ 5 π΄π΄: π΅π΅: πΆπΆ = : : = 20: 15: 12 .c om 3 4 5 8. If 0.75 : x :: 5 : 8, then x is equal to: Sol: 0.75 π₯π₯ = 5 8 ππππ 0.75 × 8 = 5π₯π₯ ππππ π₯π₯ = 6 5 ππππ π₯π₯ = 1.20 ak s 9. The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries? Sol: Let A = 2k, B = 3k and C = 5k. 115 A's new salary = 100 110 C's new salary = 120 23ππ 10 33ππ xa m fre B's new salary = ππππ 2ππ = Therefore, οΏ½ 100 100 23ππ 33ππ 10 : 10 ππππ 3ππ = 10 ππππ 5ππ = 6ππ : 6πποΏ½ = 23: 33: 60 1 10. If Rs. 782 be divided into three parts, proportional to βΆ 1 .e Sol: We are given the proportion as βΆ 2 2 3 βΆ 3 4 1 2 = 12 οΏ½ βΆ 2 2 3 w Now according to the question the first part will be = 3 βΆ 3 3 4 , then the first part is: βΆ οΏ½= 4 6 6+8+9 12 2 βΆ 24 3 × 782 = = π π π π . 204 βΆ 36 4 6 23 =6βΆ8βΆ9 × 782 11. Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is? w w 2 Sol: Let the numbers be 3x and 5x. Then, 3π₯π₯β9 5π₯π₯β9 = 12 23 ππππ 23(3π₯π₯ β 9) = 12(5π₯π₯ β 9 ππππ 9π₯π₯ = 99 ππππ π₯π₯ = 11 The smaller number = (3 × 11) = 33 .c om 12. In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there? Sol: Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively. 25π₯π₯ Then, sum of their values = π π π π . οΏ½ Now, according to the question- 100 60π₯π₯ 100 + 10×2π₯π₯ 100 + 5×3π₯π₯ 100 = 30 ππππ π₯π₯ = 30 ×100 60 60 100 = 50 π₯π₯ ak s Hence, the number of 5 p coins = (3 x 50) = 150. οΏ½ = π π π π . 13. In a partnership, two men invest $2000 and $3000. If the net profit of $13500 at the end of the year is divided in accordance with the amount each partner invested, then the man who invested more gets? 2000 3000 2 = . 3 xa m fre Sol: The ratio in which each of them invested is Now the second man invests more. The amount of profit who invested more = 3 2+3 × 13500 = 3 5 × 13500 = $8100 14. 24-carat gold is pure gold. 18-carat gold is 3/4 gold and 20-carat gold is 5/6 gold. The ratio of pure gold in 20-carat gold to pure gold in 18-carat gold is? Sol: The amount of pure gold in 20 carat gold is 5/6 gold. .e The amount of pure gold in 18 carat gold is 3/4 gold. Now the ratio of pure gold in 20 carat gold and 18 carat gold is w 5 5×4 20 10 = = =6= 3 6×3 18 9 4 w w 15. The ratio of Science and Arts students in a college is 4:3.If 14 Science students shift to Arts then the ratio becomes 1:1.Find the total strength of Science and Arts students. Sol: The ratio of Science and Arts students is 4:3. Now when 14 Science students shift to Arts then the ratio becomes= 4π₯π₯β14 3π₯π₯+14 Now according to the question: 4π₯π₯β14 = 1 ππππ 4π₯π₯ β 14 = 3π₯π₯ + 14 ππππ π₯π₯ = 28 3π₯π₯+14 .c om Now, the total strength of Science and Arts students = 4π₯π₯ + 3π₯π₯ = 7π₯π₯ = 7 × 28 = 196 16. A certain amount of money is to be divided between A and B in the ratio 3:1.But because of a miscalculation A received 1/14 of the total money additionally. Find the ratio in which they divided the money. Sol: Let the money to be divided be Rs. 28. Now, according to the question- ak s Now, the original shares for A and B should be Rs. 21 and Rs. 7 respectively. A received 1/14th of the total adiitionally. So he got π π π π . 1 14 × 28 = π π π π . 2 extra. xa m fre Now the changes ratio in which they divided the money for A:B = 21+2 7β2 = 23 5 . 17. The value of a diamond is proportional to the square of its weight. It is broken into three pieces whose weights are in the ratio 3:2:5. Find the loss incurred due to breakage if the value of the original diamond is Rs.20,000. .e Sol: Let the original weight of the diamond be 10x. Now, when it is the three parts will have the weights 3x, 2x and 5x. According to the question20000 ππππ π₯π₯ 2 ππ = 200 ππ(10π₯π₯)2 = 20000 ππππ 100π₯π₯ 2 ππ = 20000 ππππ π₯π₯ 2 ππ = 100 Now when the diamond breaks- w w w The price will be = ππ(3π₯π₯)2 + ππ(2π₯π₯)2 + ππ(5π₯π₯)2 = 9ππππ 2 + 4ππππ 2 + 25ππππ 2 = 38πππ₯π₯ 2 On putting value of kx2 from above we get the final price as = 38 × 200 = π π π π . 7600 Thus the loss incurred = Rs. (20000 β 7600) = Rs. 12400. 18. Two numbers are in the ratio 5:6.If 22 is added to the first number and 22 is subtracted from the second number then the ratio of the two numbers becomes 6:5.Find the sum of the two numbers. Sol: Let the numbers be 5x and 6x. Now according to the question1st number becomes 5x+22 and the 2nd number becomes 6x-22. 5π₯π₯+22 .c om Now the new ratio = 6π₯π₯β22 According to the question β 5π₯π₯ + 22 6 = or 5(5π₯π₯ + 22) = 6(6π₯π₯ β 22) ππππ 25π₯π₯ + 110 = 36π₯π₯ β 132 6π₯π₯ β 22 5 ak s ππππ 11π₯π₯ = 110 + 132 ππππ 11π₯π₯ = 242 ππππ π₯π₯ = 22 Now the sum of the two numbers = 5π₯π₯ + 6π₯π₯ = 11π₯π₯ = 11 × 22 = 242. 19. The ratio of two numbers is 5:6.If a number is added to both the numbers the ratio becomes 7:8.If the larger number exceeds the smaller number by 10, find the number added. xa m fre Sol: Let the two numbers be 5x and 6x. Now according to the question β 6x-5x=10 or x=10. Now when a number βnβ is added is added to both then the new ratio is given by β 5π₯π₯ + ππ 7 = ππππ 8(5π₯π₯ + ππ) = 7(6π₯π₯ + ππ) ππππ 40π₯π₯ + 8ππ = 42π₯π₯ + 7ππ 6π₯π₯ + ππ 8 .e ππππ 8ππ β 7ππ = 42π₯π₯ β 40π₯π₯ ππππ ππ = 2π₯π₯ ππππ ππ = 2 × 10 = 20 Thus, we get that the number added to both the numbers was 20. w 20. Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit's present salary? w w Sol: Let the salaries of Ravi and Sumit be 2x and 3x respectively. Now according to the question β 2π₯π₯ + 4000 40 = ππππ 57(2π₯π₯ + 4000) = 40(3π₯π₯ + 4000) 3π₯π₯ + 4000 57 ππππ 114π₯π₯ + 228000 = 120π₯π₯ + 160000 ππππ 6π₯π₯ = 68000 ππππ 3π₯π₯ = 34000 Now, Sumitβs preseny salary = Rs. (3x+4000) = Rs. (34000 + 4000) = Rs. 38000 Sol: Let the three number be a, b and c. 2 : 3 b:c = 5:8 Now, a : b : c = 10 : 15 : 24 Now, the second number = ( Refer to Q2. Above) 15 10+15+24 22. A and B together have Rs. 1210. If of much amount does B have? 4 15 × 98 = 15 49 × 98 = π π π π . 30 ak s Now, a : b = .c om 21. The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is: 2 A's amount is equal to of B's amount, how 5 xa m fre Sol: Let A and B have the amounts x and y. Now according to the question β 4 2 π₯π₯ = π¦π¦ ππππππ π₯π₯ + π¦π¦ = 1210 15 5 So we get, 3 π₯π₯ = π¦π¦ and 2 3 2 π¦π¦ + π¦π¦ = 1210 ππππ 5 2 π¦π¦ = 1210 ππππ π¦π¦ = π π π π . 484 .e 23. Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is: w Sol: Let the third number be x. w w Then, first number = 120% of x = Second number = 150% of x = 120π₯π₯ 100 150π₯π₯ 100 = = 3π₯π₯ 2 6π₯π₯ 5 Therefore, the ratio of first two numbers = οΏ½ 6π₯π₯ 3π₯π₯ 5 : 2 οΏ½ = 12π₯π₯: 15π₯π₯ = 4: 5 24. Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats? .c om Sol: Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively. Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x). Therefore the required ratio = ( 140% ππππ 5π₯π₯ βΆ 150% ππππ 7π₯π₯ βΆ 175% ππππ 8π₯π₯ =2βΆ3βΆ4 ak s = 140 × 5 βΆ 150 × 7 βΆ 175 × 8 25. The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio? xa m fre Sol: Originally, let the number of boys and girls in the college be 7x and 8x respectively. Their increased number is (120% of 7x) and (110% of 8x). Therefore the required ratio = 120% ππππ 7π₯π₯ βΆ 110% ππππ 8π₯π₯ = 120 × 7: 110 × 8 = 21 βΆ 22 26. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B's share? .e Sol: Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively. Then, 4x - 3x = 1000 Or x = 1000. w B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000. w w 27. If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number? Sol: πΏπΏπΏπΏπΏπΏ 40% ππππ π΄π΄ = ππβππππ, 2 40 100 2 3 π΄π΄ = 2 π΅π΅ 2 3 π΅π΅ ππππ π΄π΄ = π΅π΅ ππππ π΄π΄ βΆ π΅π΅ = 5 βΆ 3 5 3 2 Sol: Quantity of milk = 60 × litres = 40 litres 3 Quantity of water in it = (60- 40) litres = 20 litres. New ratio = 1 : 2 Then, Now, ππππππππ π€π€π€π€π€π€π€π€π€π€ 40 20+π₯π₯ = = 1 40 20+π₯π₯ 2 xa m fre Or 20 + x = 80 or x = 60 ak s Let quantity of water to be added further be x litres. w w .e Therefore, Quantity of water to be added = 60 litres. w .c om 28. In a mixture 60 litres, the ratio of milk and water 2 : 1. If the this ratio is to be 1 : 2, then the quantity of water to be further added is:
