{"id":15517,"date":"2024-02-14T16:23:34","date_gmt":"2024-02-14T10:53:34","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=15517"},"modified":"2024-02-15T15:18:18","modified_gmt":"2024-02-15T09:48:18","slug":"selina-icse-solutions-class-10-maths-ratio-proportion","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/selina-icse-solutions-class-10-maths-ratio-proportion\/","title":{"rendered":"Selina Concise Mathematics Class 10 ICSE Solutions Ratio and Proportion (Including Properties and Uses)"},"content":{"rendered":"
Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses)<\/strong><\/p>\n Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> Question 13.<\/strong><\/span> Question 14.<\/strong><\/span> Question 15.<\/strong><\/span> Question 16.<\/strong><\/span> Question 17.<\/strong><\/span> Question 18.<\/strong><\/span> Question 19(i).<\/strong><\/span> Question 19(ii).<\/strong><\/span> Question 20.<\/strong><\/span> Question 21.<\/strong><\/span> Question 22.<\/strong><\/span> Question 23.<\/strong><\/span> Question 24.<\/strong><\/span> Question 25.<\/strong><\/span> Question 26.<\/strong><\/span> Question 27.<\/strong><\/span> Question 28.<\/strong><\/span> Question 29.<\/strong><\/span> Question 30(a).<\/strong><\/span> Question 30(b).<\/strong><\/span> Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7(i).<\/strong><\/span> Question 7(ii).<\/strong><\/span> Question 7(iii).<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span>Ratio and Proportion Exercise 7A – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf x: y = 4: 7, find the value of (3x + 2y): (5x + y).
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf (a – b): (a + b) = 1: 11, find the ratio (5a + 4b + 15): (5a – 4b + 3).
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nA school has 630 students. The ratio of the number of boys to the number of girls is 3 : 2. This ratio changes to 7 : 5 after the admission of 90 new students. Find the number of newly admitted boys.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nWhat quantity must be subtracted from each term of the ratio 9: 17 to make it equal to 1: 3?
\nSolution:<\/strong><\/span>
\nLet x be subtracted from each term of the ratio 9: 17.
\n
\nThus, the required number which should be subtracted is 5.<\/p>\n
\nThe monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each saves Rs. 80 every month, find their monthly pocket money.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nThe work done by (x – 2) men in (4x + 1) days and the work done by (4x + 1) men in (2x – 3) days are in the ratio 3: 8. Find the value of x.
\nSolution:<\/strong><\/span>
\nAssuming that all the men do the same amount of work in one day and one day work of each man = 1 units, we have,
\nAmount of work done by (x – 2) men in (4x + 1) days
\n= Amount of work done by (x – 2)(4x + 1) men in one day
\n= (x – 2)(4x + 1) units of work
\nSimilarly,
\nAmount of work done by (4x + 1) men in (2x – 3) days
\n= (4x + 1)(2x – 3) units of work
\nAccording to the given information,
\n<\/p>\n
\nThe bus fare between two cities is increased in the ratio 7: 9. Find the increase in the fare, if:
\n(i) the original fare is Rs 245;
\n(ii) the increased fare is Rs 207.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nBy increasing the cost of entry ticket to a fair in the ratio 10: 13, the number of visitors to the fair has decreased in the ratio 6: 5. In what ratio has the total collection increased or decreased?
\nSolution:<\/strong><\/span>
\nLet the cost of the entry ticket initially and at present be 10 x and 13x respectively.
\nLet the number of visitors initially and at present be 6y and 5y respectively.
\nInitially, total collection = 10x \u00d7 6y = 60 xy
\nAt present, total collection = 13x \u00d7\u00a05y = 65 xy
\nRatio of total collection = 60 xy: 65 xy = 12: 13
\nThus, the total collection has increased in the ratio 12: 13.<\/p>\n
\nIn a basket, the ratio between the number of oranges and the number of apples is 7: 13. If 8 oranges and 11 apples are eaten, the ratio between the number of oranges and the number of apples becomes 1: 2. Find the original number of oranges and the original number of apples in the basket.
\nSolution:<\/strong><\/span>
\nLet the original number of oranges and apples be 7x and 13x.
\nAccording to the given information,
\n
\nThus, the original number of oranges and apples are 7 \u00d7\u00a05 = 35 and 13 \u00d7\u00a05 = 65 respectively.<\/p>\n
\nIn a mixture of 126 kg of milk and water, milk and water are in ratio 5 : 2. How much water must be added to the mixture to make this ratio 3 : 2?
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n(A) If A: B = 3: 4 and B: C = 6: 7, find:
\n(i) A: B: C
\n(ii) A: C
\n(B) If A : B = 2 : 5 and A : C = 3 : 4, find
\n(i) A : B : C
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf 3A = 4B = 6C; find A: B: C.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf 2a = 3b and 4b = 5c, find: a : c.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nFind duplicate ratio of:
\n(i) 3: 4 (ii) 3\u221a3 : 2\u221a5
\nSolution:<\/strong><\/span>
\n(i) Duplicate ratio of 3 : 4 = 32 : 42 = 9 : 16
\n(ii) Duplicate ratio of 3\u221a3 : 2\u221a5 = (3\u221a3)\u00b2\u00a0: (2\u221a5)\u00b2 = 27 : 20<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nFind sub-duplicate ratio of:
\n(i) 9: 16 (ii) (x – y)4<\/sup>: (x + y)6<\/sup>
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nFind the sub-triplicate ratio of:
\n(i) 64: 27 (ii) x3<\/sup>: 125y3<\/sup>
\nSolution:<\/strong><\/span>
\n(i) Sub-triplicate ratio of 64 : 27 = \u221b64 : \u221b27 = 4 : 3
\n(ii) Sub-triplicate ratio of x\u00b3 : 125y\u00b3 =\u00a0\u221bx\u00b3 : \u221b125y\u00b3 = x : 5y<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf m: n is the duplicate ratio of m + x: n + x; show that x2<\/sup>\u00a0= mn.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf (3x – 9) : (5x + 4) is the triplicate ratio of 3 : 4, find the value of x.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nFind the ratio compounded of the reciprocal ratio of 15: 28, the sub-duplicate ratio of 36: 49 and the triplicate ratio of 5: 4.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf r2\u00a0<\/sup>=<\/sub>pq, show that p : q is the duplicate ratio of (p + r) : (q + r).
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\nRatio and Proportion Exercise 7B – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n
\nFind the fourth proportional to:
\n(i) 1.5, 4.5 and 3.5 (ii) 3a, 6a2<\/sup>\u00a0and 2ab2<\/sup>
\nSolution:<\/strong><\/span>
\n(i) Let the fourth proportional to 1.5, 4.5 and 3.5 be x.
\n\u21d2 1.5 : 4.5 = 3.5 : x
\n\u21d2 1.5 \u00d7\u00a0x = 3.5 4.5
\n\u21d2 x = 10.5
\n(ii) Let the fourth proportional to 3a, 6a2<\/sup> and 2ab2<\/sup> be x.
\n\u21d2 3a : 6a2<\/sup> = 2ab2<\/sup> : x
\n\u21d2 3a \u00d7\u00a0x = 2ab2<\/sup> 6a2<\/sup>
\n\u21d2 3a \u00d7\u00a0x = 12a3<\/sup>b2<\/sup>
\n\u21d2 x = 4a2<\/sup>b2<\/sup><\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n(i) Let the mean proportional between 6 + 3\u221a3 and 8 – 4\u221a3 be x.
\n\u21d2 6 + 3\u221a3, x and 8 – 4\u221a3 are in continued proportion.
\n\u21d2 6 + 3\u221a3 : x = x : 8 – 4\u221a3
\n\u21d2 x \u00d7\u00a0x = (6 + 3\u221a3) (8 – 4\u221a3)
\n\u21d2 x2 <\/sup>= 48 + 24\u221a3- 24\u221a3 – 36
\n\u21d2 x2 <\/sup>= 12
\n\u21d2 x = 2\u221a3
\n(ii) Let the mean proportional between a – b and a3<\/sup> – a2<\/sup>b be x.
\n\u21d2 a – b, x, a3<\/sup> – a2<\/sup>b are in continued proportion.
\n\u21d2 a – b : x = x : a3<\/sup> – a2<\/sup>b
\n\u21d2 x \u00d7\u00a0x = (a – b) (a3<\/sup> – a2<\/sup>b)
\n\u21d2 x2<\/sup> = (a – b) a2<\/sup>(a – b) = [a(a – b)]2<\/sup>
\n\u21d2 x = a(a – b)<\/p>\n
\nIf x + 5 is the mean proportional between x + 2 and x + 9; find the value of x.
\nSolution:<\/strong><\/span>
\nGiven, x + 5 is the mean proportional between x + 2 and x + 9.
\n\u21d2 (x + 2), (x + 5) and (x + 9) are in continued proportion.
\n\u21d2 (x + 2) : (x + 5) = (x + 5) : (x + 9)
\n\u21d2 (x + 5)2<\/sup> = (x + 2)(x + 9)
\n\u21d2 x2<\/sup> + 25 + 10x = x2<\/sup> + 2x + 9x + 18
\n\u21d2 25 – 18 = 11x – 10x
\n\u21d2 x = 7<\/p>\n
\nIf x2<\/sup>, 4 and 9 are in continued proportion, find x.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nWhat least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nWhat least number must be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf y is the mean proportional between x and z; show that xy + yz is the mean proportional between x2<\/sup>+y2<\/sup>\u00a0and y2<\/sup>+z2<\/sup>.