{"id":42556,"date":"2024-02-08T16:10:26","date_gmt":"2024-02-08T10:40:26","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=42556"},"modified":"2024-02-09T10:14:08","modified_gmt":"2024-02-09T04:44:08","slug":"selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-a","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/selina-concise-mathematics-class-10-icse-solutions-mixed-practice-set-a\/","title":{"rendered":"Selina Concise Mathematics Class 10 ICSE Solutions Mixed Practice Set A"},"content":{"rendered":"

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

Question 1.
\nA shokeeper buys an article for \u20b9 10,800 at 10% discount on the marked price. The shopkeeper increases the marked price of the article by 15% and then sells it for \u20b9 14,628 including sales tax. If the sales tax at each stage is the same, find :
\n(i) the rate of sales tax.
\n(ii) the profit percent made by the shopkeeper.
\n(iii) the amount of VAT paid by the shopkeeper.
\nSolution:
\nIn the first case,
\nCost price for the shopkeeper or sale price of the whole seller = \u20b9 10800
\nRate of discount = 10%
\n\"Selina
\nNow increase in M.P. by the shopkeeper = 15%
\n\u2234 Increase marked price
\n= \u20b9 \\(\\frac{12000 \\times(100+15)}{100}\\) = \u20b9 13800<\/p>\n

(i) Selling price = \u20b9 14628
\n\u2234 Amount of sales tax = \u20b9 14628 – 13800 = \u20b9 828
\n\u2234 Rate of sales tax = \\(\\frac{828 \\times 100}{13800}=6 \\%\\)<\/p>\n

(ii) Now S. P. = \u20b9 13800
\nand C. P. = \u20b9 10800
\n\u2234 Total gain = S. P. – C. P.
\n= \u20b9 13800 – 10800 = \u20b9 3000
\nGain%= \\(\\frac{\\text { Total gain } \\times 100}{\\mathrm{CP}}\\)
\n\"Selina<\/p>\n

(iii) Since rate of sales tax are same
\n\u2234 Amount of VAT paid by the shopkeeper
\n= 10800 \u00d7 6% = \u20b9 648
\nAlso amount of VAT received by the shopkeeper = \u20b9 828
\nTotal amount of VAT paid by shopkeeper to sale tax department = \u20b9 828 – \u20b9 648 = \u20b9 180<\/p>\n

Question 2.
\nA manufacturer of furniture sells a sofa set to a retailer for Rs. 23,000. The retailer, in turn, sells the sofa set to a customer for Rs. 29,500. Find the VAT rate as percent, if the retailer pays Rs. 520 as VAT.
\nSolution:
\nS.P. of manufacturer = Rs. 23000
\nor and selling price of the set to the retailer
\n= Rs. 29500
\nor cost price of the set to the customer = Rs. 29500
\nVAT paid by the retailer = Rs. 520
\nVAT paid by the manufacture = Rs. (29500 – 23000) = Rs. 6500
\n\u2234 Rate of VAT
\n\"Selina<\/p>\n

Question 3.
\nMr Kumar has recurring deposit account in a bank for 4 years at 10% p.a. rate of interest. If he gets \u20b9 21,560 as interest at the time of maturity, find :
\n(i) the monthly instalment paid by Mr Kumar.
\n(ii) the amount of maturity of this recurring deposit account.
\nSolution:
\nInterest received on recurring deposit = \u20b9 21560
\nRate of interest (r) = 10% p.a.
\nPeriod (n) = 4 years or 48 months
\n(i) Let monthly instalment (P) = \u20b9 x
\n\"Selina
\n\u2234 Monthly instalment = \u20b9 2200<\/p>\n

(ii) Amount of maturity = P \u00d7 n + Interest
\n= \u20b9 2200 \u00d7 48 + 21560
\n= \u20b9 105600 + 21560
\n= \u20b9 127160<\/p>\n

Question 4.
\nThe maturity value of a cumulative deposit account is Rs. 1,20,400. If each monthly instalment for this account is Rs. 1,600 and the rate of interest is 10% per year, find the time for which the account was held.
\nSolution:
\nMaturity value of a cumulative deposit = Rs. 1,20,400
\nDeposit per month = Rs. 1600
\nRate if interest (r) = 10 %
\nLet period = n months
\n\u2234 Principal for 1 month
\n\"Selina
\n20n2<\/sup> + 20n = 3 (120400 – 1600n).
\n\u21d2 20n2<\/sup> + 20n = 361200 – 4800n
\n\u21d2 20n2<\/sup> + 20n + 4800n – 361200 = 0
\n\u21d2 20n2<\/sup> + 4820n – 561200 = 0
\nn2<\/sup> + 241 n – 18060 = 0 (Dividing by 20)
\n\u21d2 n2<\/sup> + 301 n \u2014 60n – 18060 = 0
\n\u21d2 n (n + 301) – 60 (n + 301) = 0
\n\u21d2 (n + 301) (n – 60) = 0
\nEither n + 301 = 0, then n = -301 which is not possible or n – 60 = 0 then n = 60
\n\u2234 Period = 60 months or 5 years<\/p>\n

Question 5.
\nRajat invested Rs. 24,000 in 7% hundred rupee shares at 20% discount After one year, he sold these shares at \u20b9 75 each and invested the proceeds in 18% twenty five rupee shares at 64% premium. Find :
\n(i) his gain or loss after one year.
\n(ii) his annual income from the second investment.
\n(iii) the percentage of increase in return on his original investment.
\nSolution:
\nIn first case,
\nInvestment = \u20b9 24000,
\nHe invested in 7% hundred rupees shares at 20% discount
\n\u2234 Market value = 100 – 20 = \u20b9 80
\n\u2234 Number of shares purchased = \\(\\frac{24000 \\times 1}{80}\\)
\n= 300
\nand amount of dividend = \\(\\frac{24000 \\times 7}{80}\\)
\n= \u20b9 2100
\nIn second case, by selling 300 shares at the rate of \u20b9 75
\n= 300 \u00d7 \u20b9 75 = \u20b9 22500
\nTotal investment = 22500 + 2100 = \u20b9 24600
\nBy investing \u20b9 24600 in 18% share at 64% per annum
\n= \\(\\frac{24600 \\times 18}{164}\\) = \u20b9 2700
\n(i) \u2234 Gain = \u20b9 2700 – \u20b9 2100 = \u20b9 600
\n(ii) Income from second investment = \u20b9 2700
\n(iii) Percentage of increase = \\(\\frac{600 \\times 100}{24000}=\\frac{5}{2} \\%\\)
\n= 2.5%<\/p>\n

Question 6.
\nA man sold some? 20 shares, paying 8% dividend at 10% discount and invested the proceeds in ? 10 shares paying 12% dividend, at 50% premium. If the change in his annual income is \u20b9 600, find the number of shares sold by the man.
\nSolution:
\nChange in income = \u20b9 600
\nLet number of shares = x
\nIn first case,
\nFace value = \u20b9 20
\nMarket value at 10% discount
\n= \\(20 \\times \\frac{90}{100}\\) = \u20b9 18
\nRate of dividend = 8%
\n.\u2019. Total dividend = x \u00d7 \\(\\frac{8 \\times 20}{100}\\)
\n= \u20b9 \\(\\frac{8}{5} x\\)
\nIn second case,
\nInvestment = x \u00d7 \u20b9 18 = \u20b9 18x
\nFace value = \u20b9 10
\nM. value = \u20b9 50% premium
\n\"Selina
\n\u2234 Number of shares = 3750<\/p>\n

Question 7.
\nFind the values of x, which satisfy the inequation:
\n\\(-2 \\leq \\frac{1}{2}-\\frac{2}{3} x \\leq 1 \\frac{5}{6}, x \\in \\mathrm{W}\\)
\nGraph the solution on the number line.
\nSolution:
\nHere in \\(-2 \\leq \\frac{1}{2}-\\frac{2}{3} x \\leq 1 \\frac{5}{6}\\)
\n\"Selina<\/p>\n

Question 8.
\nSolve (using formula) the equation
\n\"Selina
\nSolution:
\n\"Selina
\n\"Selina<\/p>\n

Question 9.
\nBy selling an article for \u20b9 24, a man gains as much percent as its cost price. Find the cost price of the article.
\nSolution:
\nLet CP = \u20b9 x
\n\u2234 Gain = x%
\n\"Selina
\n\u21d2 100x + x2<\/sup> = 2400
\n\u21d2 x2<\/sup> + 100x – 2400 = 0
\n\u21d2 x2<\/sup> + 120x – 20 – 2400 = 0
\n\u21d2 x(x + 120) – 20(x + 120) = 0
\n\u21d2 (x – 20) (x + 120) = 0
\nEither x – 20 = 0, then x = 20 or x + 120 = 0, then x \u2260 -120
\n\u2235 C.P. cannot be negative
\n\u2234 Cost price = \u20b9 20<\/p>\n

Question 10.
\nB takes 16 days less than A to do a certain piece of work. If both woring together can complete the work in 15 days. In how many days will B alone complete the work?
\nSolution:
\nLet A can do a piece of work in = x days
\n\u2234 B will do the same work in = (x – 16) days But Both A and B can do the work in = 15 days.
\nNow A\u2019s 1 day\u2019s work = \\(\\frac{1}{x}\\)
\nB\u2019s 1 day\u2019s work = \\(\\frac{1}{x-16}\\)
\nBoth will do the work in one day =\\(\\frac{1}{5}\\)
\n\"Selina
\n\u21d2 x(x – 16) = 15(2x – 16)
\n\u21d2 x2<\/sup> – 16x = 30x – 240
\n\u21d2 x2<\/sup> – 16x – 30x + 240 = 0
\n\u21d2 x2<\/sup> – 46x + 240 = 0
\n\u21d2 x2<\/sup> – 6x – 40 + 240 = 0
\n\u21d2 x (x – 6) – 40 (x – 6) = 0
\n\u21d2 (x – 6) (x – 40) = 0
\nEither x – 6 = 0, then x = 6 which is not possible
\nor x – 40 = 0, then x = 40 B will do the whole work in = 40 – 16 = 24 days<\/p>\n

Question 11.
\nDivide \u20b9 1,870 into three parts in such a way that half of the first part, one-third of the second part and one-sixth of the third part are all equal.
\nSolution:
\nTotal amount = \u20b9 1870
\nLet in each case, the result is x
\nThen \\(\\frac{1}{2}\\) of first share = \\(\\frac{1}{3}\\) of second share = \\(\\frac{1}{6}\\) of third share = x.
\n\u2234 First share = 2x
\nSecond share = 3x
\nand third share = 6x
\n\u2234 2x + 3x + 6x = 1870
\n\u21d2 11x = 1870
\n\u21d2 \\(x=\\frac{1870}{11}=170\\)
\n\u2234 First share = 2x = 2 \u00d7 170 = \u20b9 340
\nSecond share = 3x = 3 \u00d7 170 = \u20b9 510
\nThird share = 6x = 6 \u00d7 170 = \u20b9 1020<\/p>\n

Question 12.
\nIf a + c = bm and \\(\\frac{1}{b}+\\frac{1}{d}=\\frac{m}{c}\\), prove that a b c and d are in proportion
\nSolution:
\n\"Selina<\/p>\n

Question 13.
\nIf \\(\\frac{4 x+3 y}{4 x-3 y}=\\frac{7}{4}\\), use properties to find the value of \\(\\frac{2 x^{2}-11 y^{2}}{2 x^{2}+11 y^{2}}\\).
\nSolution:
\n\"Selina
\n\"Selina<\/p>\n

Question 14.
\nUse the properties of proportionality to solve
\n\"Selina
\nSolution:
\n\"Selina
\n\"Selina<\/p>\n

Question 15.
\nWhat number should be added to 2x3<\/sup> – 3x2<\/sup> – 8x + 3 so that the resulting polynomial leaves the remainder 10 when divided by 2x + 1 ?
\nSolution:
\nLet p{n) = 2x3<\/sup> – 3x2<\/sup> – 8x + 3
\nand 2x + 1 = 0, then 2x = – 1
\n\"Selina
\nBut remainder is given = 10
\n\u2234 10 – 6 = 4 is to be added
\nHence 4 is to be added<\/p>\n

Question 16.
\nIf \\(A=\\left[\\begin{array}{ll}{1} & {-1} \\\\ {z} & {-1}\\end{array}\\right], B=\\left[\\begin{array}{cc}{x} & {1} \\\\ {4} & {-1}\\end{array}\\right]\\) = A2<\/sup> + B2<\/sup> = (A2<\/sup> + B2<\/sup>) find the value of x. State, whether A2<\/sup> – B22<\/sup>and (A + B)2<\/sup> are always equal or not.
\nSolution:
\n\"Selina
\n\"Selina
\n\"Selina
\nHence x = 1<\/p>\n

Question 17.
\nFind the 10th term of the sequence 10, 8, 6, …..
\nSolution:
\nSequence is 10, 8, 6, ……
\nHere inA.P.
\na= 10 and d= 8 – 10 = -2
\n\u2234 T10<\/sub> = a + (n – 1)d = 10 + (10 – 1) \u00d7 (-2)
\n= 10 + 9 \u00d7 (-2) = 10 -18 = -8<\/p>\n

Question 18.
\nIf the 5th<\/sup> and 11th<\/sup> terms of an A.P. are 16 and 34 respectively. Find the A.P., …..
\nSolution:
\nIn an A.P.
\n5th<\/sup> term = 16
\n11th<\/sup> term=34
\nLet a be the first term and d be the common difference, then
\n\"Selina
\na + 4 \u00d7 3 = 16 \u21d2 a + 12 = 16
\na = 16 – 12 = 4
\n\u2234 A.P. is 4, 7, 10, 13, 16,<\/p>\n

Question 19.
\nIf pth<\/sup> term of an A.P. is q and its qth<\/sup> term is p. Show that its rth<\/sup> term is (p + q – r).
\nSolution:
\npth<\/sup> term = q
\nqth<\/sup> term = p
\nShow that (p + q – r) is rth<\/sup> term
\nLet a be the first term and d be the common difference
\n\u2234 pth<\/sup> term = a + (p – 1)d=q
\nqth<\/sup> term = a + (q – 1 )d=p
\nSubtracting,
\n(p – 1 – q + 1 )d = q – p
\n(p – q)d = q – p
\n\\(d=\\frac{-(p-q)}{p-q}=-1\\)
\n\u2234 a + (p – 1)(-1) = q
\na + (-p) + 1 = q \u21d2 a = p + q – 1
\nNow Tr<\/sub> = rth<\/sup> term
\na + (r – 1)d
\n= (p + q – 1) + (r – 1)(-1)
\n= p + q – 1 – r + 1
\n= p + q – r<\/p>\n

Question 20.
\nIf nth<\/sup> term of an A.P. is (2 n -1). Find its 7th<\/sup> term.
\nSolution:
\nnth<\/sup> term of an A.P. = 2n – 1
\n\u2234 7th<\/sup> term = 2n – 1
\n= 2 \u00d7 7 – 1 = 14 – 1 = 13<\/p>\n

Question 21.
\nIf the sum of first n terms of an A.P. is 3n2<\/sup> + 2n, find its rth<\/sup> term.
\nSolution:
\nSum of first n terms = 3n2<\/sup> + 2n
\n\u2234 1st\u00a0<\/sup>term(a) = 3(1)2<\/sup> + 2(1) = 3 + 2 = 5
\n2nd<\/sup> term = S2<\/sub> – S1<\/sub> = 3(2)2<\/sup> + 2 \u00d7 2 – 5
\n= 12 + 4 – 5
\n= 16 – 5 = 11
\n\u2234 d = T2<\/sub> -T1<\/sub> = 11 – 5 = 6
\n\u2234 Tr<\/sub> or rth<\/sup> term = a + (r – 1 )d
\n= 5 + (r – 1) x 6
\n= 5 + 6r – 6 = 6r – 1<\/p>\n

Question 22.
\nFor an A.P., the sum of its terms is 60, common difference is 2 and last term is 18. Find the number of terms in the A.P.
\nSolution:
\nLet number of terms in the A.P. = n
\nThen Sn<\/sub> = 60, d = 2, l = 18
\nl = a + (n – 1 )d
\n18 = a + (n – 1) \u00d7 2 = a + 2n – 2
\na + 2n = 18 + 2 = 20 \u21d2 a = 20 – 2n …(i)
\n\\(\\frac{n}{2}\\)[2a + (n – 1)d] = 60
\n\u21d2 \\(\\frac{n}{2}\\)[a + a + (n – 1)d] = 60
\n\u21d2 \\(\\frac{n}{2}\\)[a + 18] = 60
\n\u21d2 n(a+18) = 120
\nn(20 – 2n + 18)= 120 \u21d2 n(38 – 2n) = 120
\n38n -2n2<\/sup> – 120 = 0
\n-2n2<\/sup> + 38n – 120 = 0
\nn2<\/sup> – 19n + 60 = 0
\n\u21d2 n2<\/sup> = 4n – 15n + 60 = 0
\n\u21d2 n(n – 4) – 15(n – 4) = 0
\n\u21d2 (n – 4) (n – 15) = 0
\nEither n – 4 = 0 then n = 4 or n – 15 = 0, then n= 15
\n\u2234 n = 4 or 15<\/p>\n

Question 23.
\nFind the geometric progression whose 5th<\/sup> term is 48 and 8th<\/sup> term is 384.
\nSolution:
\nIn G.P.
\nT5<\/sub> = 48, T8<\/sub> = 384
\nLet a be the first term and r be the common ratio
\n\"Selina
\n\u2234 GP. will be 3, 6, 12, 24,<\/p>\n

Question 24.
\nSum of how many terms of the G.P. \\(\\frac{2}{9}-\\frac{1}{3}+\\frac{1}{2}\\) – …….. is \\(\\frac{55}{72}\\) ?
\nSolution:
\nG.P. is \\(\\frac{2}{9}-\\frac{1}{3}+\\frac{1}{2}\\)
\nSum = \\(\\frac{55}{72}\\)
\nLet a be the first term and r be the common difference.
\n\"Selina
\n\"Selina<\/p>\n

Question 25.
\nFind the sum of n terms of the sequence: 5 + 55 + 555 + ………
\nSolution:
\n5 + 55 + 555 + …… n term
\n5[1 + 11 + 111 + ………n terms]
\n\"Selina
\n\"Selina<\/p>\n

Question 26.
\nWhat point on x-axis is equidistant from the points (6, 7) and (4, -3) ?
\nSolution:
\n\u2235 The point is on x-axis
\n\u2234 its y = 0
\nLet the point be (x, 0)
\nThen distance between (x, 0) and (6, 7) is equal to the distance between (x, 0) and (4, -3)
\n\"Selina<\/p>\n

Question 27.
\nCalculate the ratio in which the line segment A(6, 5) and B(4, -3) is dividend by the line y = 2.
\nSolution:
\n\u2235 The line y = 2 is parallel to x-axis
\nLet the point P be (x, 2) which divided The line segment joining the points A(6, 5), B(4, -3)
\nLet the ratio be m1<\/sub> : m2<\/sub>
\n\"Selina
\n\u21d2 2m1<\/sub> + 2m2<\/sub> = -3m1<\/sub> + 5m2<\/sub>
\n\u21d2 2m1<\/sub> + 3m1<\/sub> = 5m2<\/sub> – 2m2<\/sub>
\n\u21d2 5m1<\/sub> = 3m2<\/sub>
\n\\(\\frac{m_{1}}{m_{2}}=\\frac{3}{5}\\)
\n\u2234 Ratio is 3 : 5<\/p>\n

Question 28.
\nFind the equations of the diagonals of a rectangle whose sides are x + 1 = 0, x – 4 = 0, y + 1 = 0 and y – 2 = 0.
\nSolution:
\n\u2235 Lines x + 1 = 0 and y – 4 = 0 are parallel to y-axis and lines y + 1 = 0 and y – 2 = 0 are parallel to x-axis which form a rectangle ABCD.
\n\"Selina
\n\u2234 Co-ordinates of A will be (-1, -1), of B will be (4, -1) of C will be (4, 2) and of D will be (-1, 2)
\n\u2235 AC and BD are the diagonals of the given rectangle ABCD.
\n\u2234 Equation of AC
\n\"Selina
\n\"Selina
\nHence equations of diagonals are 3x – 5y – 2 and 3x + 5y = 1<\/p>\n

Question 29.
\nThe line 4x + 5y + 20 = 0 meets x-axis at point A and y-axis at point B. Find :
\n(i) The co-ordinate of points A and B
\n(ii) The co-ordinates of point P in AB such that AB : BP = 5 : 3.
\n(iii) The equation of the line through P and perpendiculars to AB.
\nSolution:
\n(i) The equation of given line is
\n4x + 5y + 20 = 0
\n\u2235 If meets x-axis at A
\n\u2234 its y = 0
\nThen 4x + 5 \u00d7 0 + 20 = 0
\n\u21d2 4x + 20 = 0
\n\u21d2 4x = -20
\n\u21d2 \\(x=\\frac{-20}{4}=-5\\)
\n\u2234 Co-ordinates of A will be (-5, 0)
\n\u2235 It meets y-axis at B.
\n\u2234 its x = 0
\nThen 4 \u00d7 0 + 5y + 20 = 0
\n\u21d2 5y + 20 = 0 \u21d2 5y = -20
\n\u21d2 \\(y=\\frac{-20}{5}=-4\\)
\n\u2234 Co-ordinates of B will be (0, -4)<\/p>\n

(ii) Let point P(x, y) divides the line AB in such way that AB : BP = 5 : 3
\nAB = AP + PB
\n\u21d2 5 = AP + 3
\n\u21d2 AP = 5 – 3 = 2
\n\u2234\u00a0 AP : PB = 2 : 3
\n\"Selina
\n\"Selina<\/p>\n

Question 30.
\nIn the given figure, DE || BC and AE : EC = 5:4, find :
\n(i) DE : BC
\n(ii) DO : DC
\n(iii) are of \u2206DOE : area of \u2206DCE
\n\"Selina
\nSolution:
\n\"Selina
\n\"Selina
\n\u2234 DO : DC = 5 : 14
\n(iii) \u2235 \u2206DOE and \u2206DCE are on the same line and their vertex is common.
\n\u2234 area of (\u2206 DOE): area of (\u2206 DCE) = DO : DC
\n= 5 : 14 {Proved in (ii)}<\/p>\n

Question 31.
\nIf chords AB and CD of a circle intersect each other at a point inside the circle, prove that PA \u00d7 PB = PC \u00d7 PD.
\nSolution:
\nGiven : Two chords AB and CD of a circle such that they intersect each other at a point P lying inside the circle
\nTo prove : PA \u00d7 PB = PC \u00d7 PD
\nConstruction : AC and BD are joined at P
\n\"Selina
\nProof : In the above figure, P lies inside the circle
\nIn \u2206s PCA and \u2206s PBD, we have
\n\u2220 PCA = \u2220 PBD [Angles in the same segment]
\n\u2220 APC = \u2220 BPD [Vertically opposite angles]
\n\u2234 \u2206PCA ~ \u2206PBD [AA similarity]
\nHence, \\(\\frac{P A}{P D}=\\frac{P C}{P B}\\) or PA \u00d7 PB = PC \u00d7 PD
\nHence proved.<\/p>\n

Question 32.
\nP and Q are centres of circles of radii 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of a circle of radius x cm which touches the above circles externally. Given that \u2220PRQ = 90\u00b0, write an equation in x and solve it.
\nSolution:
\nP and Q are the centres of radii 9 cm and 2 cm respectively where PQ = 12 cm. R is the centre of another circle which touches the other circles such that \u2220PRQ = 90\u00b0
\n\"Selina
\nNow in right \u2206 PQR,
\nPQ2<\/sup> = PR2<\/sup> + QR2<\/sup>
\n\u21d2 (17)2<\/sup> = (9 + x)2<\/sup> + (2 + x)2<\/sup> (Pythagoras theorem)
\n\u21d2 289 = 81 + 18x + x2<\/sup> + 4 + 4x + x2<\/sup>
\n\u21d2 2x2<\/sup> + 22x + 85 – 289 = 0
\n\u21d2 2x2<\/sup> + 22x – 204 = 0<\/p>\n

(i) \u2234 x2<\/sup> + 11x – 102 = 0 (Dividing by 2)
\n\u21d2 x2<\/sup> – 6x + 17x – 102 = 0
\n\u21d2 x(x – 6) + 17 (x – 6) = 0
\n\u21d2 (x – 6) (x + 17) = 0
\nEither x – 6 = 0 then x = 6 or x + 17 = 0 then x = -17 But it is not possible.<\/p>\n

(ii) \u2234 x = 6<\/p>\n

Question 33.
\nUse ruler and a pair of compasses only in this question :
\n(i) Draw circle on AB = 6.4 cm as diameter.
\n(ii) Construct another circle of radius 2.5 cm to touch the circle in (i) above and the diameter AB produced.
\nSolution:
\nSteps of construction :
\n(i) Draw a line segment AB = 6.4 cm.
\n(ii) Bisect AB at D.
\n(iii) With centre D and diameter AB, draw a circle
\n(iv) Draw a line LM parallel to AB at a distance of 2.5 cm.
\n(v) With centre D and radius 3.2 + 2.5 = 5.7 cm, draw an arc intersecting the line LM at 0.
\n(v) With centre 0 and radius 2.5 cm, draw a circle which touches the given circle at E and AB produced at C.
\nThis is the required circle.
\n\"Selina
\nP.Q. The side of a square is 10 cm. Find the . area of the circumscribed and inscribed circles.
\nSolution:
\nSide of the square (a) = 10 cm
\n\u2234 Radius of the inscribed circle (r) = \\(\\frac{1}{2}\\) a = \\(\\frac{10}{2}\\) = 5cm
\n\"Selina
\n\"Selina<\/p>\n

Question 34.
\nThe internal and external diameters of a hollow hemispherical vessel are 14 cm and 21 cm respectively. The cost of silver plating of 1 cm2<\/sup> of its surface is \u20b9 32. Find the total cost of silver plating the vessel all over.
\nSolution:
\nIn a hollow hemispherical vessel,
\nThe external diameter = 21 cm
\nand internal diameter = 14 cm
\n\u2234 Outer radius (R) = \\(\\frac{21}{2}\\) cm
\nand Inner radius = \\(\\frac{14}{2}\\) = 7 cm
\nTotal surface area of the bowl
\n= 2\u03c0R2<\/sup> + 2\u03c0r2<\/sup> + (\u03c0R2<\/sup> – \u03c0r2<\/sup>.)
\n= 2\u03c0R2<\/sup> + 2\u03c0r2<\/sup> + \u03c0R2<\/sup> – \u03c0r2<\/sup>2
\n= 3\u03c0R2<\/sup> + \u03c0r2<\/sup>
\n\"Selina<\/p>\n

Question 35.
\nProve that:
\n\"Selina
\nSolution:
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina<\/p>\n

Question 36.
\nSolve for x, 0\u00b0 \u2264 x \u2264 90\u00b0
\n(i) 4 cos2<\/sup> 2x – 3 = 0
\n(ii) 2 sin2<\/sup> x – sin x = 0
\nSolution:
\n\"Selina<\/p>\n

Question 37.
\nA ladder rests against a wall at an angle a to the horizontal. Its foot is pulled away from the wall through a distance a, so that it slides is distance b down the wall making an angle \u03b2 with the horizontal. Show that :
\n\"Selina
\nSolution:
\nIn the figure AB and CD represent the same ladder:
\n\"Selina
\n\"Selina<\/p>\n

Question 38.
\nFor the following frequency distirubtion, draw an ogive and then use it to estimate median.<\/p>\n\n\n\n\n
C.I.<\/strong><\/td>\n450-550<\/td>\n550-650<\/td>\n650-750<\/td>\n750-850<\/td>\n850-950<\/td>\n950-1050<\/td>\n1050-1150<\/td>\n<\/tr>\n
f<\/strong><\/td>\n40<\/td>\n68<\/td>\n86<\/td>\n120<\/td>\n90<\/td>\n40<\/td>\n6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

For the same distribution, as given above, draw a histogram and then use it to estimate the mode.
\nSolution:<\/p>\n\n\n\n\n\n\n\n\n\n\n
C.I.<\/strong><\/td>\nf<\/strong><\/td>\nCf<\/strong><\/td>\n<\/tr>\n
450-550<\/td>\n40<\/td>\n40<\/td>\n<\/tr>\n
550-650<\/td>\n68<\/td>\n108<\/td>\n<\/tr>\n
650-750<\/td>\n86<\/td>\n194<\/td>\n<\/tr>\n
750-850<\/td>\n120<\/td>\n314<\/td>\n<\/tr>\n
850-950<\/td>\n90<\/td>\n404<\/td>\n<\/tr>\n
950-1050<\/td>\n40<\/td>\n444<\/td>\n<\/tr>\n
1050-1150<\/td>\n6<\/td>\n450<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Taking C.l. on x-axis and C.I. on y-axis.
\nPlot the points (550, 40), (650, 108) (750, 194), (850, 314), (950, 404), (1050, 444) and (1150, 450) on the graph and join them with free hand to obtain an ogive which is shown.
\n\"Selina
\n\"Selina
\non y-axis, mark A = 225 and from A, draw a line parallel to x-axis meeting it at M. From M, draw a perpendicular on x-axis meeting it at N.
\n\u2234 M is the median which is 780 approximately Hence median = 780
\nHistogram :
\n(i) Take C.I. along x-axis and\/along y-axis. Draw the histogram as shown in the figure.
\n(ii) In side the highest rectangle which represented the maximum frequency Join AC and BD from upper corners C and D of the adjacent rectangle which intersect each other at P.
\n\"Selina
\nFrom P, draw a perpendicular PM on x-axis M is the required mode which is 803 (approx)
\nHence mode = 803<\/p>\n

Question 39.
\n(i) Find the probability of drawing an ace or a jack from a pack of 52 cards.
\n(ii) A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag.
\nSolution:
\n(i) Total number of ace = 4
\nTotal number of jack = 4
\nTotal number of cards = 52
\nP (Ace or.Jack) = P (Ace) T P (Jack)
\n\\(=\\frac{4}{52}+\\frac{4}{52}=\\frac{2}{13}\\)
\n(ii) Let the number of blue balls be x Number of red balls = 5 Total number of balls = Number of blue balls + Number of red balls
\n= x + 5
\n\"Selina<\/p>\n

Question 40.
\nIn a single throw of two dice, what is the probability of getting
\n(i) a total of 9
\n(ii) two aces (ones)
\n(iii) at least one ones
\n(iv) doublet
\n(v) five on one die and six on the other
\n(vi) a multiple of 2 on one and a multiple of 3 on the other ?
\nSolution:
\nTwo dice are thrown having 1 to 6 each on their faces
\n\u2234 Number of possible outcome = 6 \u00d7 6 = 36
\n(ii) A sum of them is 9 are (3, 6), (4, 5), (5, 4), (6, 3)
\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 A Question 1. A shokeeper buys an article for \u20b9 10,800 at 10% discount on the marked price. The shopkeeper increases the marked price of the article by 15% and then sells it for \u20b9 14,628 including sales tax. If the sales tax at each …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42556"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=42556"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42556\/revisions"}],"predecessor-version":[{"id":167286,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42556\/revisions\/167286"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=42556"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=42556"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=42556"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}