{"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 Concise Mathematics Class 10 ICSE Solutions Ratio and Proportion (Including Properties and Uses)<\/h2>\n

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses)<\/strong><\/p>\n

Ratio and Proportion Exercise 7A – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 2.<\/strong><\/span>
\nIf x: y = 4: 7, find the value of (3x + 2y): (5x + y).
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 3.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 4.<\/strong><\/span>
\nIf (a – b): (a + b) = 1: 11, find the ratio (5a + 4b + 15): (5a – 4b + 3).
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 5.<\/strong><\/span>
\n\"selina-icse-solutions-class-10-maths-ratio-proportion-7a-q5-i\"
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 6.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 7.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 8.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 9.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 10.<\/strong><\/span>
\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\"selina-icse-solutions-class-10-maths-ratio-proportion-7a-q10\"<\/p>\n

Question 11.<\/strong><\/span>
\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\"Selina
\nThus, the required number which should be subtracted is 5.<\/p>\n

Question 12.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 13.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 14.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 15.<\/strong><\/span>
\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

Question 16.<\/strong><\/span>
\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\"Selina
\nThus, the original number of oranges and apples are 7 \u00d7\u00a05 = 35 and 13 \u00d7\u00a05 = 65 respectively.<\/p>\n

Question 17.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 18.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 19(i).<\/strong><\/span>
\nIf 3A = 4B = 6C; find A: B: C.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 19(ii).<\/strong><\/span>
\nIf 2a = 3b and 4b = 5c, find: a : c.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 20.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 21.<\/strong><\/span>
\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

Question 22.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 23.<\/strong><\/span>
\nFind sub-duplicate ratio of:
\n(i) 9: 16 (ii) (x – y)4<\/sup>: (x + y)6<\/sup>
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 24.<\/strong><\/span>
\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

Question 25.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 26.<\/strong><\/span>
\nIf (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 27.<\/strong><\/span>
\nIf m: n is the duplicate ratio of m + x: n + x; show that x2<\/sup>\u00a0= mn.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 28.<\/strong><\/span>
\nIf (3x – 9) : (5x + 4) is the triplicate ratio of 3 : 4, find the value of x.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 29.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 30(a).<\/strong><\/span>
\nIf r2\u00a0<\/sup>=<\/sub>pq, show that p : q is the duplicate ratio of (p + r) : (q + r).
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 30(b).<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Ratio and Proportion Exercise 7B – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong><\/span>
\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

Question 2.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 3.<\/strong><\/span>
\n\"Selina
\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

Question 4.<\/strong><\/span>
\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

Question 5.<\/strong><\/span>
\nIf x2<\/sup>, 4 and 9 are in continued proportion, find x.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 6.<\/strong><\/span>
\nWhat least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 7(i).<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 7(ii).<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 7(iii).<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 8.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 9.<\/strong><\/span>
\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>.
\nSolution:<\/strong><\/span>
\nSince y is the mean proportion between x and z
\nTherefore, y2 <\/sup>= xz
\nNow, we have to prove that xy+yz is the mean proportional between x2<\/sup>+y2<\/sup> and y2<\/sup>+z2<\/sup>, i.e.,
\n\"selina-icse-solutions-class-10-maths-ratio-proportion-ex-10b-q9\"
\nLHS = RHS
\nHence, proved.<\/p>\n

Question 10.<\/strong><\/span>
\nIf q is the mean proportional between p and r, show that:
\npqr (p + q + r)3<\/sup>\u00a0= (pq + qr + rp)3<\/sup>.
\nSolution:<\/strong><\/span>
\nGiven, q is the mean proportional between p and r.
\n\u21d2 q2<\/sup> = pr
\n\"Selina<\/p>\n

Question 11.<\/strong><\/span>
\nIf three quantities are in continued proportion; show that the ratio of the first to the third is the duplicate ratio of the first to the second.
\nSolution:<\/strong><\/span>
\nLet x, y and z be the three quantities which are in continued proportion.
\nThen, x : y :: y : z \u21d2\u00a0y2<\/sup> = xz ….(1)
\nNow, we have to prove that
\nx : z = x2 <\/sup>: y2<\/sup>
\nThat is we need to prove that
\nxy2 <\/sup>= x2<\/sup>z
\nLHS = xy2<\/sup> = x(xz) = x2<\/sup>z = RHS [Using (1)]
\nHence, proved.<\/p>\n

Question 12.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\nGiven, y is the mean proportional between x and z.
\n\u21d2 y2<\/sup> = xz
\n\"Selina<\/p>\n

Question 13.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"selina-icse-solutions-class-10-maths-ratio-proportion-ex-10b-q13\"<\/p>\n

Question 14.<\/strong><\/span>
\nFind two numbers such that the mean mean proportional between them is 12 and the third proportional to them is 96.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 15.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 16.<\/strong><\/span>
\nIf p: q = r: s; then show that:
\nmp + nq : q = mr + ns : s.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 17.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 18.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 19.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 20.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Ratio and Proportion Exercise 7C – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong><\/span>
\nIf a : b = c : d, prove that:
\n(i) 5a + 7b : 5a – 7b = 5c + 7d : 5c – 7d.
\n(ii) (9a + 13b) (9c – 13d) = (9c + 13d) (9a – 13b).
\n(iii) xa + yb : xc + yd = b : d.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 2.<\/strong><\/span>
\nIf a : b = c : d, prove that:
\n(6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b).
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 3.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 4.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 5.<\/strong><\/span>
\nIf (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d), prove that a: b = c: d.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 6.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina
\n\"Selina<\/p>\n

Question 7.<\/strong><\/span>
\nIf (a + b + c + d) (a – b – c + d) = (a + b – c – d) (a – b + c – d), prove that a: b = c: d.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 8.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 9.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 10.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\nGiven, a, b and c are in continued proportion.
\n\"Selina
\n\"Selina<\/p>\n

Question 11.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina
\n\"Selina<\/p>\n

Question 12.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 13.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 14.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 15.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Ratio and Proportion Exercise 7D – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong><\/span>
\nIf a: b = 3: 5, find:
\n(10a + 3b): (5a + 2b)
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 2.<\/strong><\/span>
\nIf 5x + 6y: 8x + 5y = 8: 9, find x: y.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 3.<\/strong><\/span>
\nIf (3x – 4y): (2x – 3y) = (5x – 6y): (4x – 5y), find x: y.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 4.<\/strong><\/span>
\nFind the:
\n(i) duplicate ratio of 2\u221a2 : 3\u221a5
\n(ii) triplicate ratio of 2a: 3b
\n(iii) sub-duplicate ratio of 9x2<\/sup>a4\u00a0<\/sup>: 25y6<\/sup>b2<\/sup>
\n(iv) sub-triplicate ratio of 216: 343
\n(v) reciprocal ratio of 3: 5
\n(vi) ratio compounded of the duplicate ratio of 5: 6, the reciprocal ratio of 25: 42 and the sub-duplicate ratio of 36: 49.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 5.<\/strong><\/span>
\nFind the value of x, if:
\n(i) (2x + 3): (5x – 38) is the duplicate ratio of \u221a5 : \u221a6
\n(ii) (2x + 1): (3x + 13) is the sub-duplicate ratio of 9: 25.
\n(iii) (3x – 7): (4x + 3) is the sub-triplicate ratio of 8: 27.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 6.<\/strong><\/span>
\nWhat quantity must be added to each term of the ratio x: y so that it may become equal to c: d?
\nSolution:<\/strong><\/span>
\nLet the required quantity which is to be added be p.
\nThen, we have:
\n\"Selina<\/p>\n

Question 7.<\/strong><\/span>
\nA woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 84 kg?
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 8.<\/strong><\/span>
\nIf 15(2x2<\/sup>\u00a0– y2<\/sup>) = 7xy, find x: y; if x and y both are positive.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 9.<\/strong><\/span>
\nFind the:
\n(i) fourth proportional to 2xy, x2<\/sup>\u00a0and y2<\/sup>.
\n(ii) third proportional to a2<\/sup>\u00a0– b2<\/sup>\u00a0and a + b.
\n(iii) mean proportional to (x – y) and (x3<\/sup>\u00a0– x2<\/sup>y).
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 10.<\/strong><\/span>
\nFind two numbers such that the mean proportional between them is 14 and third proportional to them is 112.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 11.<\/strong><\/span>
\nIf x and y be unequal and x: y is the duplicate ratio of x + z and y + z, prove that z is mean proportional between x and y.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 12.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 13.<\/strong><\/span>
\nIf (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove that:
\na: b = c: d.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 14.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 15.<\/strong><\/span>
\nThere are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3: 1. How any more girls should be added to the council so that the ratio of the number of boys to the number of girls may be 9: 5?
\nSolution:<\/strong><\/span>
\nRatio of number of boys to the number of girls = 3: 1
\nLet the number of boys be 3x and number of girls be x.
\n3x + x = 36
\n4x = 36
\nx = 9
\n\u2234 Number of boys = 27
\nNumber of girls = 9
\nLe n number of girls be added to the council.
\nFrom given information, we have:
\n\"Selina
\nThus, 6 girls are added to the council.<\/p>\n

Question 16.<\/strong><\/span>
\nIf 7x – 15y = 4x + y, find the value of x: y. Hence, use componendo and dividend to find the values of:
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 17.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 18.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 19.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 20.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 21.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 22.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 23.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 24.
\n\"Selina
\n<\/strong><\/span><\/p>\n

Question 25.
\n\"Selina
\n<\/strong><\/span><\/p>\n

Question 26.
\n\"Selina
\n<\/strong><\/span><\/p>\n

Question 27.
\n\"Selina
\n\"Selina
\n<\/strong><\/span><\/p>\n

Question 28.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 29.<\/strong><\/span>
\n\"Selina
\nSolution:<\/strong><\/span>
\nSince, q is the mean proportional between p and r,
\nq2<\/sup> = pr
\n\"Selina<\/p>\n

Question 30.<\/strong><\/span>
\nIf a, b and c are in continued proportion, prove that:
\na: c = (a2<\/sup>\u00a0+ b2<\/sup>) : (b2<\/sup>\u00a0+ c2<\/sup>)
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

More Resources for Selina Concise Class 10 ICSE Solutions<\/strong><\/p>\n