{"id":42577,"date":"2024-02-29T05:09:12","date_gmt":"2024-02-28T23:39:12","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=42577"},"modified":"2024-02-29T15:04:27","modified_gmt":"2024-02-29T09:34:27","slug":"selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/","title":{"rendered":"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B"},"content":{"rendered":"

Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B<\/h2>\n

Question 1.
\nA retailer buys an article, from a wholesaler, at a discount of 15% on the printed price. He marks up its printed price by 10%. Due to competition in the market, he allows a discount of 5% to the buyer. If the buyer pays \u20b9 6,771.60 for the article inclusive of 8% VAT, find :
\n(i) the printed price of the article,
\n(ii) the proft percent of the retailer.
\n(iii) VAT paid by the retailer.
\nSolution:
\nLet the marked price of an article = Rs. x
\nDiscount = 15%
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina<\/p>\n

Question 2.
\nThe monthly instalment of a recuring deposit account is \u20b9 2.400. If the account is held for 1 year 6 months and its maturity value is \u20b9 47,304, fnd the rate of interest.
\nSolution:
\nMonthly deposit = \u20b9 2400
\nPeriod (n) = 1 year 6 months
\n\u2234 Total Principal for 1 month
\n\"Selina
\n= 2400 \u00d7 9 \u00d7 19
\n= \u20b9 410400
\n\u2234 Maturity value = \u20b9 47304
\nand actual principal = 2400 \u00d7 18 = \u20b9 43200
\n\u2234 Interest = \u20b9 (47304 – 43200)
\n= \u20b9 4104
\nLet rate of interest = r% p.a.
\n\"Selina<\/p>\n

Question 3.
\nThe maturity value of a recurring deposit account is \u20b9 42,400. If the account is held for 2 years and rate of interest is 10% per annum, find the amount of each monthly instalment.
\nSolution:
\nMaturity value = \u20b9 42400
\nPeriod (n) = 2 years = 24 months
\nRate of interest (r) = 10% p.a.
\nLet monthly instalment of recurring deposit = \u20b9 v
\n\u2234 Interest = \u20b9 (42400 – x \u00d7 24)
\n= \u20b9 42400 – 24x ……………(i)
\n\"Selina
\n\u21d2 5x = 84800 – 48x
\n48x + 5x = 84800
\n53x = 84800
\nx = \\(\\frac{84800}{53}\\)
\nx = \u20b9 1600
\n\u2234 Monthly instalment = \u20b9 1600<\/p>\n

Question 4.
\nA man invests equal amount of money in two companies A and B. Company A pays a dividend of 15% and its T 100 shares are available at 20% discount. The shares of company B has a nominal value of \u20b9 25 and are available at 20% premium. If at the end of one year, the man gets equal dividends from both the companies, find the rate of dividend paid by company B.
\nSolution:
\nLet in each type of investment = \u20b9 100
\nIn the company,
\nFace value of each share = \u20b9 100
\n\"Selina
\n\u2234 Rate of dividend = 22.5%<\/p>\n

Question 5.
\nA sum of \u20b9 54,000 is invested partly in shares paying 6% dividend at 40% premium and partly in 5% shares at 25% premium. If the nomial value of one share in each company is \u20b9 100 and the total income of the man is \u20b9 2,240, find the money invested in the second company.
\nSolution:
\nTotal investment = \u20b9 54000
\nLet amount invested in first kind of shares = x
\n\u2234 Amount invested in second kind of shares = 5400 – x
\nIn first kind of shares
\nRate of dividend = 6%
\nM. V. = 100 + 40 = \u20b9 140
\nN. V. = \u20b9 100
\n\u2234 Income = \\(\\frac{x \\times 6}{140}=\\frac{3 x}{70}\\)
\nIn second kind of shares
\nRate of dividend = 5%
\nM. V. = 100 + 25 = 125
\nN. V. = \u20b9 100
\n\"Selina<\/p>\n

Question 6.
\nSolve and graph the solution set of :
\n(i) 2x – 9 < 7 and 3x + 9 \u2264 25; x \u03f5 R.
\n(ii) 3x – 2 > 19 or 3 – 2x \u2265 – 7; x \u03f5 R.
\nSolution:
\n(i) 2x – 9 < 7 and 3x + 9 \u2264 25; x \u03f5 R.
\n2x < 7 + 9 and 3x \u2264 25 – 9 \u21d2 2x < 16 and 3x \u2264 16
\n\"Selina<\/p>\n

(ii) 3x – 2 > 19 or 3 – 2x \u2265 – 7
\n\u21d2 3x > 19 + 2 or -2x \u2265 -7 – 3
\n\u21d2 3x > 21 or -2x \u2265 -10
\n\u2234 x > 7 or – x \u2265 – 5
\n\u2234 x > 7 or x \u2264 5
\n\"Selina<\/p>\n

Question 7.
\nUse formula to solve the quadratic equation x2<\/sup> + x – (a + 1) (a + 2) = 0.
\nSolution:
\nx2<\/sup> + x – (a + 1) (a + 2) = 0
\nHere a = 1, b = 1 and c = -(a + 1) (a + 2) = -(a2<\/sup> + 3a + 2)
\n\u2234 D = b2<\/sup> – 4ac
\n= (1)2<\/sup> – 4 \u00d7 {-(a2<\/sup> + 3a + 2)}
\n= 1 + 4(a2<\/sup> + 3a + 2)
\n= 1 + 4a2<\/sup> + 12a + 8
\n= 4a2<\/sup> + 12a + 9 = (2a + 3)2<\/sup>
\n\"Selina<\/p>\n

Question 8.
\nBy selling an article for \u20b9 96, a man gains as much percent as its cost price. Find the cost price of the article.
\nSolution:
\nS.P. of an article = \u20b9 96
\nLet C.P. of the article = \u20b9 x
\n\u2234 Gain = x%
\n\\(\\therefore \\text { S.P. }=\\frac{\\mathrm{CP}(100+g a i n \\%)}{100}\\)
\n\\(96=\\frac{x(100+x)}{100}\\)
\n\u21d2 .x(100 + x) = 9600
\n\u21d2 100x + x2<\/sup> -9600 = 0
\n\u21d2 x2<\/sup> +100x – 9600 = 0
\n\u21d2 x2<\/sup> + 160x – 60x – 9600 = 0
\n\u21d2 x (x + 160) – 60 (x + 160) = 0
\n\u21d2 (x + 160) (x – 60) = 0
\nEither x + 160 = 0, then x = -160 which is not possible
\ni.e. x – 60 = 0, then x = 60
\n\u2234 cost price = \u20b9 60<\/p>\n

Question 9.
\nA trader bought a number of articles for \u20b9 900, five were damaged and he sold each of the rest at \u20b9 2 more than what he paid for it. It on the whole he gains \u20b980, find the number of articles bought.
\nSolution:
\nLet number of articles = x
\nC.P. of x articles = \u20b9 900
\n\"Selina
\n\u2234 S.P. = \u20b9 900 + 80 = \u20b9 980
\n\\(\\therefore \\frac{(x-5)(900+2 x)}{x}=980\\)
\n\u21d2 900x + 2x2<\/sup> – 4500 – 10x = 980x
\n\u21d2 2x2<\/sup> + 890x – 4500 – 980x = 0
\n\u21d2 2x2<\/sup> – 90x – 4500 = 0
\n\u21d2 x2<\/sup> – 45x – 2250 = 0
\n\u21d2 x2<\/sup> – 75x + 30x – 2250 = 0
\n\u21d2 x(x – 75) + 30(x – 75) = 0
\n\u21d2 (x – 75) (x + 30) = 0
\nEither x – 75 = 0, then x = 75 or x + 30 = 0, then x = -30 which is not possible.
\n\u2234 Number of articles purchased = 75<\/p>\n

Question 10.
\n1077 boxes of oranges were loaded in three trucks. While unloading them, 7, 12 and 8 boxes were found rotten in the trucks respectively. If the number of remaining boxes in the three trucks are in the ratio 4 : 6 : 5, find the number of boxes loaded originally in each truck.
\nSolution:
\nTotal boxes in original =1077
\nRatio in the remaining boxes = 4 : 6 : 5
\nLet boxes unloaded in first truck = 4x
\nin second truck = 6x
\nand in third truck = 5x
\n\u2234 Number of boxes loaded in the first trucks = 4x + 7
\nin second truck = 6x + 12
\nand in third truck = 5x + 8
\nNow 4x + 7 + 6x + 12 + 5x + 8 = 1077
\n\u21d2 15x + 21 = 1077
\n15x = 1077 -27
\n\u21d2 15x = 1050
\n\u21d2 \\(x=\\frac{1050}{15}=70\\)
\n\u2234 Number of boxes loaded in the first truck
\n= 70 \u00d7 4 + 7 = 280 + 7 = 287
\nNumber of boxes in the second truck
\n= 70 \u00d7 6 + 12
\n= 420 + 12 = 432
\nand in the third truck
\n= 70 \u00d7 5 + 8
\n= 350 + 8 = 358<\/p>\n

Question 11.
\nIf a \u2260 b and a : b is the duplicate ratio of (a + c) and (b + c), show that a, c and b are in continued proportion.
\nSolution:
\n\u2235 a : b is the duplicate ratio of (a + c) and (b + c)
\n\u2234 \\(\\frac{(a+c)^{2}}{(b+c)^{2}}=\\frac{a}{b}\\)
\n\u21d2 b(a + c)2<\/sup> = a(b + c)2<\/sup>
\n\u21d2 b(a2<\/sup> + c2<\/sup> + 2ac) = a(b2<\/sup> + c2<\/sup>+ 2bc)
\n\u21d2 a2<\/sup>b + bc2<\/sup> + 2abc = ab2<\/sup> + c2<\/sup>a + 2ab2<\/sup>
\n\u21d2 a2<\/sup>b + bc2<\/sup> + 2abc – 2abc = ab2<\/sup> + c2<\/sup>a
\n\u21d2 a2<\/sup>b + bc2<\/sup> = ab2<\/sup> + c2<\/sup>a
\n\u21d2 c2<\/sup>a – bc2<\/sup> = a2<\/sup>b – ab2<\/sup>
\n\u21d2 c2<\/sup>(a – b) – ab (a – b)
\n\u21d2 c2<\/sup> = ab (\u2235 a \u2260 b)
\n\\(\\frac{a}{c}=\\frac{c}{b}\\)
\nHence a, ec and b are in continued proportion.<\/p>\n

Question 12.
\nIf \\(16\\left(\\frac{a-x}{a+x}\\right)^{3}=\\left(\\frac{a+x}{a-x}\\right)\\) show that a = 3x
\nSolution:
\n\\(16\\left(\\frac{a-x}{a+x}\\right)^{3}=\\left(\\frac{a+x}{a-x}\\right)\\)
\n\u21d2 16 [(a – x)3<\/sup> (a – x)] = (a + x) (a – x)3<\/sup>
\n\u21d2 16 [(a – x)4<\/sup>] = [(a + x)4<\/sup>]
\n\u21d2 (2)4<\/sup> (a – x)4<\/sup> = (a + x)4<\/sup>
\nWe get,
\n\u21d2 2 (a – x) = a + x
\n\u21d2 2a – 2x = a + x
\n\u21d2 2a – a = x + 2x
\n\u2234 a = 3x
\nHence proved.<\/p>\n

Question 13.
\nSolve for x, using the properties of proportionality \\(\\frac{1+x+x^{2}}{1-x+x^{2}}=\\frac{62(1+x)}{63(1-x)}\\)
\nSolution:
\n\"Selina<\/p>\n

Question 14.
\nShows that (2x + 7) is a factor of 2x3<\/sup> + 7x2<\/sup> – 4x – 14. Hence, solve the equation : 2x3<\/sup> + 7x2<\/sup> – 4x – 14 = 0.
\nSolution:
\nLet f(x) = 2x3<\/sup> + 7x2<\/sup> – 4x – 14 and 2x + 7 = 0
\n\u21d2 2x = -7
\n\"Selina
\n\u2234 Remainder is zero
\nHence 2x + 7 is a factor f(x)
\nNow dividing f(x) by 2x + 7, we get
\nQuotient = x2<\/sup> – 2
\n\"Selina<\/p>\n

Question 15.
\n(i) What number should be subtracted from 2x3<\/sup>\u00a0– 5x2<\/sup> + 5x + 8 so that the resulting polynomial has a factor 2x – 3 ?
\n(ii) The expression 4x32<\/sup> – bx2<\/sup> + x – c leaves remainders 0 and 30 when divided by (x + 1) and (2x – 3) respectively. Calculate the values of b and c. Hence, factorise the expression completely.
\nSolution:
\nLet f(x) = 2x3<\/sup> – 5x2<\/sup> + 5x + 8
\nand 2x – 3 = 0,
\n\u2234 2x = 3
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina<\/p>\n

Question 16.
\nIf for two matrices M and N, N = \\(\\left[\\begin{array}{cc}{3} & {2} \\\\ {2} & {-1}\\end{array}\\right]\\) and product M \u00d7 N = [-1 4]; find matrix M.
\nSolution:
\nLet the order of matrix M be a \u00d7 b
\n\"Selina
\nSince, the product of matrices is possible only when the number of columns in the first matrix is equal to the number of rows in the second
\n\u2234 b = 2
\nAlso, the number of rows of product (resulting matrix) is equal to the number of rows of first matrix
\n\u2234 a = 1 \u21d2 order of matrix
\nM = a \u00d7 b = 1 \u00d7 2
\nLet M = [r y]
\n\\([x y]\\left[\\begin{array}{cc}{3} & {2} \\\\ {2} & {-1}\\end{array}\\right]=\\left[\\begin{array}{ll}{-1} & {4}\\end{array}\\right]\\)
\n\u21d2 [3x + 2y – 2x – y] = [-1 4]
\n\u21d2 3x + 2y = -1 ….(i)
\nand 2x – y = 4 ……(ii)
\nFrom eq. (i)
\n\\(x=\\frac{-1-2 y}{3}\\) …(iii)
\nPut the value of equation (iii) in equation (ii) We get,
\n\"Selina<\/p>\n

Question 17.
\nIf the sum of first 20 terms of an A.P. is same as the sum of its first 28 terms. Find the sum of its 48 terms.
\nSolution:
\nIn an A.P.
\nSum of first 20 terms = Sum of first 28 terms
\nLet a be the first term and d be the common difference, therefore
\n\\(\\frac{20}{2}[2 a-(20-1) d]=\\frac{28}{2}[2 a-(28-1) d]\\)
\n10(2a – 19d) = 14(2a – 27d)
\n20a – 190d= 28a- 318d
\n378d – 190d = 28a – 20a
\n\u21d2 8a= 188d
\n2a = 47d
\nNow, sum of 48 terms
\n\\(\\frac{48}{2}[2 a-(48-1) d]\\)
\n= 24[2a – 47 d]
\n= 24[47d – 41d] = 0
\n\u2234 Sum of first 48 terms = 0<\/p>\n

Question 18.
\nIf a, b, c are in A.P., show that : (b + c), (c + a) and (a + b) are also in A.P.
\nSol.
\na, b, c are in A.P.
\n\u2234 a – (a + b + c), b – (a + b + c), c – (a + b + c) are in A.P.
\n\u21d2 -(b + c), -(a + a), -(a + b) are in A.P.
\n\u21d2 (b + c), (c + a), (a + b) are in A.P.<\/p>\n

Question 19.
\nIf a, b, c are in GP.; a, x, b are in A.P. and b, y, c are also in A.P.
\nProve that: \\(\\frac{1}{x}+\\frac{1}{y}=\\frac{2}{b}\\).
\nSolution:
\na, b, c are in G.P.
\n\u2234 B2<\/sup> = ac
\na, x, b are in A.P.
\n\"Selina<\/p>\n

Question 20.
\nEvaluate: 9 + 99 + 999 + ………. upto n terms.
\nSolution:
\n9 + 99 + 999 + ……… n terms
\n= (10 – 1) + (100 – 1) + (1000 – 1) + …… n terms
\n= 10 + 100 + 1000 + ………. n terms -(1 + 1 + 1 + ……… n terms)
\n\"Selina<\/p>\n

Question 21.
\nFind the point on the y-axis whose distance from the points (3, 2) and (-1, 1.5) are in the ratio 2:1.
\nSolution:
\nLet P he the point on the linesegment joining the two points A (3, 2) and B (-1, 1.5) and lies on y-axis.
\nLet co-ordinates of P be (0, y)
\n\u2235 P divides the line segment AB in the ratio of 2 : 1
\n\"Selina
\n\"Selina<\/p>\n

Question 22.
\nIn what ratio does the point P (a, 2) divide the line segment joining the points A (5, -3) and B (-9, 4) ? Also, find the value of ‘a’.
\nSolution:
\nLet the point P (a, 2) divides the line segment joining the points A (5, -3) and B (-9, 4) in the ratio m1<\/sub> : m2<\/sub>
\n\"Selina
\nHence Ratio is 5 : 2
\nand a = -5<\/p>\n

Question 23.
\nA straight line makes on the co-ordinate axes positive interceps whose sum is 5. If the line passes through the point P(-3, 4), find its equation.
\nSolution:
\nLet a, b he the positive intercepts made by the line which passes through P(-3, 4) and a + b = 5
\nNow equation of the line
\n\"Selina
\n\u2235 a + b = 5 \u21d2 b = 5 – a
\n\u2234 -3 (5 – a) + 4a = a (5 – a)
\n\u21d2 -15 + 3a + 4a = 5a – a2<\/sup>
\n\u21d2 -15 + 7a = 5a – a2<\/sup>
\n\u21d2 a2<\/sup> + 7a – 5a – 15 = 0
\n\u21d2 a2<\/sup> + 2a – 15 = 0
\n\u21d2 a2<\/sup> + 5a – 3a – 15 = 0
\n\u21d2 a(a + 5) – 3(a + 5) = 0
\n\u21d2 (a + 5) (a – 3) = 0
\nEither a + 5 = 0 then a = -5, which is not possible as a and b are positive or a – 3 = 0, then a = 3
\n\u2234 b = 5 – a = 5- 3 = 2
\n\u2234 Equation of the line will be
\n\\(\\frac{x}{a}+\\frac{y}{b}=1 \\Rightarrow \\frac{x}{3}+\\frac{y}{2}=1\\)
\nor 2x + 3y = 6<\/p>\n

Question 24.
\nThe line 3x – 4y + 12 = 0 meets x-axis at points A and y-axis at point B, Find :
\n(i) the co-ordinates of A and B.
\n(ii) equation of perpendicular bisector of line segment AB
\nSolution:
\nEquation of line is 3x – 4y + 12 = 0, which intersects x-axis at A and y-axis at B
\nLet co-ordinates of A be (x, 0) and of B be (0, y)
\n\"Selina
\n\u2235 These lie on the line
\n\u2234 3x – 4 \u00d7 0 + 12 = 0
\n\u21d2 3x + 12 = 0 \u21d2 3x = -12
\n\\(x=\\frac{-12}{3}=-4\\)
\n\u2234 Co-ordinates of A will be (-4, 0)
\nAgain 3 \u00d7 0 – 4y +12 = 0
\n0 – 4y = -12
\n\u21d2 – 4y = -12
\n\u21d2 \\(y=\\frac{-12}{-4}=3\\)
\n\u2234 Co-ordinates of B will be (0, 3)<\/p>\n

(ii) Let PQ be the perpendicular bisector of line segment AB which intersects it at M
\n\u2234 Co-ordinates of M will be \\(\\left(\\frac{-4+0}{2}, \\frac{0+3}{2}\\right)\\) or \\(\\left(-2, \\frac{3}{2}\\right)\\)
\nSlope of line AB = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{3-0}{0-(-4)}=\\frac{3}{4}\\)
\n\"Selina<\/p>\n

Question 25.
\nIn a triangle PQR, L and M are two points on the base QR such that \u2220LPQ = \u2220QRP and \u2220RPM = \u2220RQB Prove that :
\n(i) \u2206 PQL ~ \u2206 RPM
\n(ii) QL \u00d7 RM = PL \u00d7 PM
\n(iii) PQ2<\/sup> = QR \u00d7 QL
\nSolution\u2014
\nGiven : In \u2206 PQR, L and M are two points on base QR such that \u2220LPQ = \u2220QRP and \u2220RPM = \u2220RQP
\n\"Selina
\nTo prove :
\n(i) \u2206PQL ~ \u2206 RPM
\n(ii) QL \u00d7 RM = PL x PM
\n(iii) PQ2<\/sup> = QR \u00d7 QL
\n\"Selina<\/p>\n

Question 26.
\nIn a rectahgle ABCD, its diagonal AC = 15 cm and \u2220ACD = \u03b1. If cot \u03b1 = \\(\\frac{3}{2}\\), find the perimeter and the area of the rectangle.
\nSolution:
\n\"Selina
\n\"Selina
\n\"Selina<\/p>\n

Question 27.
\nIn the given figure : AB is a diameter of the circle. Chords AC and AD produced to meet the tangent to the circle at a point B in points P and Q respectively. Prove that :
\n\"Selina
\nAB2<\/sup> = AC \u00d7 AP
\n(ii) C, B, Q and P are concyclic.
\nSolution:
\nGiven : AB is a diameter of a circle – chords ‘ AC and AD produced meet the tangent to the circle at B in points P and Q, respectively.
\n\"Selina
\nTo prove :
\n(i) AB2<\/sup> = AC \u00d7 AP
\n(ii) C,D,Q and P are concyclic.
\nProof : (i) \u2235 AB is diameter and PQ is tangent.
\n\u2234 AB \u22a5 PQ.
\nNow, in right \u2206 APB,
\nPB2<\/sup> = AP2<\/sup> – AB2<\/sup> …..(i)
\n\u2235 PB is tangent and PCA is secant
\n\u2234 PB2<\/sup> = PA.PC = AP.PC …..(ii)
\nFrom (i) and (ii),
\nAP.PC – AP2<\/sup> – AB2<\/sup>
\n\u21d2 AB2<\/sup> = AP – AP . PC
\n= AP (AP – PC)
\n= AP \u00d7 AC –<\/p>\n

(iii) \u2235 AOB is diameter and PQ is tangent and CD is chord,
\n\u2234 AOB \u22a5 PQ + AE \u22a5 CD
\n\u2234 AP = AQ and AC = AD
\n\u2234 CD || PQ and \u2220P = \u2220Q, \u2220ACE = vADE
\nBut \u2220P = \u2220ACE and \u2220Q = \u2220ADE (Corresponding angles)
\nBut \u2220ACE + \u2220ECP = 180\u00b0 (Linear pair)
\n\u2234 \u2220P + \u2220ECP = 180\u00b0
\n\u21d2 \u2220Q+ \u2220ECP = 180\u00b0
\nBut there are the opposite angles of quadrilateral CDQP.
\n\u2234 CDQP is a cyclic quadrilateral Hence C,D,Q,P are concyclic.
\nHence proved.<\/p>\n

Question 28.
\nUse ruler and compasses for this question.
\n(i) Construct an isosceles triangle ABC in which AB = AC = 7.5 cm and BC = 6 cm
\n(ii) Draw AD, the perpendicular from vertex A to side BC.
\n(iii) Draw a circle with centre A and radius 2.8 cm, cutting AD at E.
\n(iv) Construct another circle to circumscribe the triangle BCE.
\nSolution:
\nSteps of construction :
\n(i) Draw a line segment BC = 6cm
\n(ii) with centres B and C, radius 75 cm draw to arcs intersecting each other at A
\n(iii) Join AB and AC
\n(iv) Draw AD perpendicular bisector of BC
\n(v) with centre A and radius 2.8 cm draw a circle which intersects AD at E
\n(vi) Join EB and draw its perpendicular bisector intersecting AD at O.
\n(vii) With centre O and radius OE, draw a circle which passes through B and C and touches the first circle at E externally. This is the required circle.
\n\"Selina<\/p>\n

Question 29.
\nIn triangle ABC, \u2220BAC = 90\u00b0, AB = 6 cm and BC = 10 cm. A circle is drawn inside the triangle which touches all the sides of the triangle (i.e., an incircle of A ABC is drawn). Find the area of the triangle excluding the circle.
\nSolution:
\nIn \u2206 ABC, \u2220BAC = 90\u00b0,
\nAB = 6cm, BC = 10cm
\n\"Selina
\n\u2234 AC2<\/sup> = BC2<\/sup> – AB2<\/sup> (By Pythagoras Theorem)
\n= 102<\/sup> – 62<\/sup>
\n= 100 – 36 = 64 = 82<\/sup>
\n\u2234 AC = 8 cm.
\n\u2234 Area \u2206 ABC = \\(\\frac{1}{2}\\) \u00d7 AB \u00d7 AC
\n= \\(\\frac{1}{2}\\) \u00d7 6 \u00d7 8 = 24 cm2<\/sup>
\nA circle is drawn which touches the sides of the \u2206 ABC at D, E and F respectively Join OD, OF which are the radius of the circle.
\n\"Selina
\n\"Selina<\/p>\n

Question 30.
\nA conical vessel of radius 6 cm and height 8 cm is completely filled with water. A shpere is lowered into the water and its size is such that when it touches the sides it is just immersed. What fraction of water overflows?
\nSolution:
\nRadius of conical vessel = 6 cm
\nand height = 8cm
\n\"Selina
\n\"Selina<\/p>\n

Question 31.
\nProve that:
\n\"Selina
\nSolution:
\n\"Selina
\n\"Selina<\/p>\n

Question 32.
\nSolve x \u03f5 W, o\u00b0 \u2264 x \u2264 90\u00b0.
\n(i) 3 tan2<\/sup> 2x = 1
\n(ii) tan2<\/sup> x = 3 (secx – 1)
\nSolution:
\n\"Selina
\ntan2<\/sup> x = 3 (sec x – 1)
\n\u21d2 sec2<\/sup> x – 1 = 3 (sec x – 1)
\n\u21d2 (sec x – 1) (sec x + 1) -3 (sec x – 1) = 0
\n\u21d2 (sec x – 1) [sec x + 1 – 3] = 0
\n\u21d2 (sec x – 1) (sec x – 2) = 0
\nEither sec x – 1 = 0, then sec x = 1 = sec 0\u00b0
\nComparing, x = 0\u00b0
\nor sec x – 2 = 0, then sec x = 2 = sec 60\u00b0
\n\u2234 x = 60\u00b0
\n\u2234 x = 0\u00b0, 60\u00b0<\/p>\n

Question 33.
\nThe angle of elevation of the top of a tower as observed from a point on the ground is \u2018a\u2019 and on moving \u2018a’ metre towards the tower, the angle of elevation is \u2018P\u2019.
\nProve that the height of the tower is : \\(\\frac{\\alpha \\tan \\alpha \\tan \\beta}{\\tan \\beta-\\tan \\alpha}\\).
\nSolution:
\nLet AB be tower and two points C and D are two points making angles of elevation with top Of tower A area and P respectively and CD = a,
\nLet AB = h and BD = x
\n\"Selina
\n\"Selina<\/p>\n

Question 34.
\nThe mean of the following frequency distribution is 50, but the frequencies f1<\/sub> and f2<\/sub> in class 20 – 40 and 60 – 80 respectively are not known. Find these frequencies.<\/p>\n\n\n\n\n
Class<\/strong><\/td>\n0-20<\/td>\n2040<\/td>\n40-60<\/td>\n60-80<\/td>\n80-100<\/td>\n<\/tr>\n
Frequencies<\/strong><\/td>\n17<\/td>\nf1<\/sub><\/td>\n32<\/td>\nf2<\/sub><\/td>\n19<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Given that the sum of frequencies is 120.
\nSolution:
\nMeans = 50<\/p>\n\n\n\n\n\n\n\n\n\n
Class<\/strong><\/td>\nFrequencies (f)<\/strong><\/td>\nClass mark x<\/strong><\/td>\nf \u00d7 x<\/strong><\/td>\n<\/tr>\n
0-20<\/td>\n17<\/td>\n10<\/td>\n170<\/td>\n<\/tr>\n
2040<\/td>\nf1<\/sub><\/td>\n30<\/td>\n30f1<\/sub><\/td>\n<\/tr>\n
40-60<\/td>\n32<\/td>\n50<\/td>\n1600<\/td>\n<\/tr>\n
60-80<\/td>\nf2<\/sub><\/td>\n70<\/td>\n70f2<\/sub><\/td>\n<\/tr>\n
80-100<\/td>\n19<\/td>\n90<\/td>\n1710<\/td>\n<\/tr>\n
Total<\/td>\n68+ f1\u00a0<\/sub>+ f2<\/sub> = 120<\/td>\n<\/td>\n3480 + 30f1<\/sub> + 70f2<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"Selina<\/p>\n

Question 35.
\nA card is drawn at random from a well- shuffled deck of 52 playing cards. Find the probability that it is :
\n(i) an ace
\n(ii) a jack of hearts
\n(iii) a three of clubs or a six of diamonds
\n(iv) a heart
\n(v) any suit except heart
\n(vi) a ten or a spade
\n(vii) neither a four nor a club
\n(viii) a picture card
\n(ix) a spade or a picture card.
\nSolution:
\nNumber of possible outcome = 52
\n(i) Number of ace in the deck = 4
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina<\/p>\n

Selina Concise Mathematics Class 10 ICSE Solutions<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B Question 1. A retailer buys an article, from a wholesaler, at a discount of 15% on the printed price. He marks up its printed price by 10%. Due to competition in the market, he allows a discount of 5% to the buyer. If the […]<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[3034],"tags":[],"yoast_head":"\nSelina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B\" \/>\n<meta property=\"og:description\" content=\"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B Question 1. A retailer buys an article, from a wholesaler, at a discount of 15% on the printed price. He marks up its printed price by 10%. Due to competition in the market, he allows a discount of 5% to the buyer. If the […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/\" \/>\n<meta property=\"og:site_name\" content=\"A Plus Topper\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/aplustopper\/\" \/>\n<meta property=\"article:published_time\" content=\"2024-02-28T23:39:12+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-02-29T09:34:27+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Mathematics-Class-10-ICSE-Solutions-Mixed-Practice-Set-B-image-1.png\" \/>\n<meta name=\"twitter:card\" content=\"summary\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Prasanna\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"15 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/www.aplustopper.com\/#organization\",\"name\":\"Aplus Topper\",\"url\":\"https:\/\/www.aplustopper.com\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/aplustopper\/\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.aplustopper.com\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"contentUrl\":\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"width\":1585,\"height\":375,\"caption\":\"Aplus Topper\"},\"image\":{\"@id\":\"https:\/\/www.aplustopper.com\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/www.aplustopper.com\/#website\",\"url\":\"https:\/\/www.aplustopper.com\/\",\"name\":\"A Plus Topper\",\"description\":\"Improve your Grades\",\"publisher\":{\"@id\":\"https:\/\/www.aplustopper.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/www.aplustopper.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Mathematics-Class-10-ICSE-Solutions-Mixed-Practice-Set-B-image-1.png\",\"contentUrl\":\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Mathematics-Class-10-ICSE-Solutions-Mixed-Practice-Set-B-image-1.png\",\"width\":328,\"height\":563,\"caption\":\"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B image - 1\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#webpage\",\"url\":\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/\",\"name\":\"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B - A Plus Topper\",\"isPartOf\":{\"@id\":\"https:\/\/www.aplustopper.com\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#primaryimage\"},\"datePublished\":\"2024-02-28T23:39:12+00:00\",\"dateModified\":\"2024-02-29T09:34:27+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/www.aplustopper.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B\"}]},{\"@type\":\"Article\",\"@id\":\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#webpage\"},\"author\":{\"@id\":\"https:\/\/www.aplustopper.com\/#\/schema\/person\/2533e4338ba14fc0e4001efcca2f8794\"},\"headline\":\"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B\",\"datePublished\":\"2024-02-28T23:39:12+00:00\",\"dateModified\":\"2024-02-29T09:34:27+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#webpage\"},\"wordCount\":2934,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/www.aplustopper.com\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Mathematics-Class-10-ICSE-Solutions-Mixed-Practice-Set-B-image-1.png\",\"articleSection\":[\"ICSE\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#respond\"]}]},{\"@type\":\"Person\",\"@id\":\"https:\/\/www.aplustopper.com\/#\/schema\/person\/2533e4338ba14fc0e4001efcca2f8794\",\"name\":\"Prasanna\",\"image\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.aplustopper.com\/#personlogo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"caption\":\"Prasanna\"},\"url\":\"https:\/\/www.aplustopper.com\/author\/prasanna\/\"}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B - A Plus Topper","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/","og_locale":"en_US","og_type":"article","og_title":"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B","og_description":"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B Question 1. A retailer buys an article, from a wholesaler, at a discount of 15% on the printed price. He marks up its printed price by 10%. Due to competition in the market, he allows a discount of 5% to the buyer. If the […]","og_url":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/","og_site_name":"A Plus Topper","article_publisher":"https:\/\/www.facebook.com\/aplustopper\/","article_published_time":"2024-02-28T23:39:12+00:00","article_modified_time":"2024-02-29T09:34:27+00:00","og_image":[{"url":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Mathematics-Class-10-ICSE-Solutions-Mixed-Practice-Set-B-image-1.png"}],"twitter_card":"summary","twitter_misc":{"Written by":"Prasanna","Est. reading time":"15 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/www.aplustopper.com\/#organization","name":"Aplus Topper","url":"https:\/\/www.aplustopper.com\/","sameAs":["https:\/\/www.facebook.com\/aplustopper\/"],"logo":{"@type":"ImageObject","@id":"https:\/\/www.aplustopper.com\/#logo","inLanguage":"en-US","url":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","contentUrl":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","width":1585,"height":375,"caption":"Aplus Topper"},"image":{"@id":"https:\/\/www.aplustopper.com\/#logo"}},{"@type":"WebSite","@id":"https:\/\/www.aplustopper.com\/#website","url":"https:\/\/www.aplustopper.com\/","name":"A Plus Topper","description":"Improve your Grades","publisher":{"@id":"https:\/\/www.aplustopper.com\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/www.aplustopper.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#primaryimage","inLanguage":"en-US","url":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Mathematics-Class-10-ICSE-Solutions-Mixed-Practice-Set-B-image-1.png","contentUrl":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Mathematics-Class-10-ICSE-Solutions-Mixed-Practice-Set-B-image-1.png","width":328,"height":563,"caption":"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B image - 1"},{"@type":"WebPage","@id":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#webpage","url":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/","name":"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B - A Plus Topper","isPartOf":{"@id":"https:\/\/www.aplustopper.com\/#website"},"primaryImageOfPage":{"@id":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#primaryimage"},"datePublished":"2024-02-28T23:39:12+00:00","dateModified":"2024-02-29T09:34:27+00:00","breadcrumb":{"@id":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/www.aplustopper.com\/"},{"@type":"ListItem","position":2,"name":"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B"}]},{"@type":"Article","@id":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#article","isPartOf":{"@id":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#webpage"},"author":{"@id":"https:\/\/www.aplustopper.com\/#\/schema\/person\/2533e4338ba14fc0e4001efcca2f8794"},"headline":"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set B","datePublished":"2024-02-28T23:39:12+00:00","dateModified":"2024-02-29T09:34:27+00:00","mainEntityOfPage":{"@id":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#webpage"},"wordCount":2934,"commentCount":0,"publisher":{"@id":"https:\/\/www.aplustopper.com\/#organization"},"image":{"@id":"https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#primaryimage"},"thumbnailUrl":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Mathematics-Class-10-ICSE-Solutions-Mixed-Practice-Set-B-image-1.png","articleSection":["ICSE"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/www.aplustopper.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-b\/#respond"]}]},{"@type":"Person","@id":"https:\/\/www.aplustopper.com\/#\/schema\/person\/2533e4338ba14fc0e4001efcca2f8794","name":"Prasanna","image":{"@type":"ImageObject","@id":"https:\/\/www.aplustopper.com\/#personlogo","inLanguage":"en-US","url":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","caption":"Prasanna"},"url":"https:\/\/www.aplustopper.com\/author\/prasanna\/"}]}},"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/posts\/42577"}],"collection":[{"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/comments?post=42577"}],"version-history":[{"count":1,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/posts\/42577\/revisions"}],"predecessor-version":[{"id":167575,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/posts\/42577\/revisions\/167575"}],"wp:attachment":[{"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/media?parent=42577"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/categories?post=42577"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/tags?post=42577"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}