Question 1

Question 1(v)Which of the following is/are an Arithmetic Progression (A.P.) ?(i) 1, 4, 9, 16, ........(ii) 3,23,33,43,\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3},3​,23​,33​,43​, ............(iii) 8, 6, 4, 2, ...........only (i)only (ii)only (ii) and (iii)all (i), (ii) and (iii)

Accepted Answer

In first series :1, 4, 9, 16, ...........16 - 9 = 7, 9 - 4 = 5.Since, difference between consecutive terms are not equal.∴ It is not an A.P.In second series :3,23,33,43,\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3},3​,23​,33​,43​, ............43−33=33−23=23−3=34\sqrt{3} - 3\sqrt{3} = 3\sqrt{3} - 2\sqrt{3} = 2\sqrt{3} - \sqrt{3} = \sqrt{3}43​−33​=33​−23​=23​−3​=3​.Since, difference between consecutive terms are equal.∴ It is an A.P.In third series :8, 6, 4, 2, ..........2 - 4 = 4 - 6 = 6 - 8 = -2.Since, difference between consecutive terms are equal.∴ It is an A.P.Hence, Option 3 is the correct option.

Question 2

Question 1(vi)The table shows the values of x and y, where x is proportional to y. What are the values of M and N ?xy6M1218N6M = 4, N = 9M = 9, N = 3M = 9, N = 4M = 12, N = 0

Accepted Answer

Given, x is proportional to y.⇒6M=1218⇒M=6×1812⇒M=10812⇒M=9.1218=N6⇒N=12×618⇒N=7218⇒N=4.\Rightarrow \dfrac{6}{M} = \dfrac{12}{18} \[1em] \Rightarrow M = \dfrac{6 \times 18}{12} \[1em] \Rightarrow M = \dfrac{108}{12} \[1em] \Rightarrow M = 9. \[1.5em] \dfrac{12}{18} = \dfrac{N}{6} \[1em] \Rightarrow N = \dfrac{12 \times 6}{18} \[1em] \Rightarrow N = \dfrac{72}{18} \[1em] \Rightarrow N = 4.⇒M6​=1812​⇒M=126×18​⇒M=12108​⇒M=9.1812​=6N​⇒N=1812×6​⇒N=1872​⇒N=4.Hence, Option 3 is the correct option.

Question 3

Question 2(ii)In a recurring deposit account for 2 years, the total amount deposited by a person is ₹ 9600. If the interest earned by him is one-twelfth of his total deposit, then find :(a) the interest he earns(b) his monthly deposit(c) the rate of interest

Accepted Answer

(a) Given,Interest earned by man = One-twelfth of the total deposit= 112×9600\dfrac{1}{12} 	imes 9600121​×9600= ₹ 800.Hence, the interest earned = ₹ 800.(b) Given,In a recurring deposit account for 2 years (or 24 months), the total amount deposited by a person is ₹ 9600.Money deposited per month = 960024\dfrac{9600}{24}249600​ = ₹ 400.Hence, monthly deposit = ₹ 400.(c) By formula,Rate of interest = Interest earnedAmount invested×100\dfrac{	ext{Interest earned}}{	ext{Amount invested}} 	imes 100%Amount investedInterest earned​×100=8009600×100= \dfrac{800}{9600} 	imes 100% = \dfrac{100}{12} = 8\dfrac{1}{3}=9600800​×100 %.Hence, rate of interest = 8138\dfrac{1}{3}831​ %.

Question 4

Question 3(i)If a, b and c are in continued proportion, then prove that :3a2+5ab+7b23b2+5bc+7c2=ac\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}3b2+5bc+7c23a2+5ab+7b2​=ca​.

Accepted Answer

Since, a, b and c are in continued proportion.∴ab=bc⇒b2=ac.\therefore \dfrac{a}{b} = \dfrac{b}{c} \[1em] \Rightarrow b^2 = ac.∴ba​=cb​⇒b2=ac.Substituting, b2 = ac in L.H.S. of equation 3a2+5ab+7b23b2+5bc+7c2=ac\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}3b2+5bc+7c23a2+5ab+7b2​=ca​, we get :⇒3a2+5ab+7ac3ac+5bc+7c2⇒a(3a+5b+7c)c(3a+5b+7c)⇒ac.\Rightarrow \dfrac{3a^2 + 5ab + 7ac}{3ac + 5bc + 7c^2} \[1em] \Rightarrow \dfrac{a(3a + 5b + 7c)}{c(3a + 5b + 7c)} \[1em] \Rightarrow \dfrac{a}{c}.⇒3ac+5bc+7c23a2+5ab+7ac​⇒c(3a+5b+7c)a(3a+5b+7c)​⇒ca​.Hence, proved that 3a2+5ab+7b23b2+5bc+7c2=ac\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}3b2+5bc+7c23a2+5ab+7b2​=ca​.

Question 5

Question 3(ii)In the given diagram, O is the center of circle circumscribing the △ABC, CD is perpendicular to chord AB. ∠OAC = 32°. Find each of the unknown angles, x, y and z.

Accepted Answer

In △OAC,⇒ OA = OC (Radius of same circle)⇒ ∠OCA = ∠OAC = 32° (Angles opposite to equal sides are equal)By angle sum property of triangle,⇒ ∠OCA + ∠OAC + ∠AOC = 180°⇒ 32° + 32° + ∠AOC = 180°⇒ ∠AOC + 64° = 180°⇒ ∠AOC = 180° - 64°⇒ ∠AOC (x°) = 116°.We know that,Angle subtended by an arc at the center of the circle is twice the angle subtended at the remaining part of the circumference.⇒ ∠AOC = 2∠ABC⇒ x° = 2y°⇒ y° = x°2=116°2\dfrac{x°}{2} = \dfrac{116°}{2}2x°​=2116°​ = 58°.In △ BDC,⇒ ∠BDC + ∠BCD + ∠CBD = 180°⇒ 90° + z° + y° = 180°⇒ z° = 180° - 90° - y° = 90° - 58° = 32°.Hence, x° = 116°, y° = 58° and z° = 32°.

Question 6

Question 3(iii)Study the graph and answer each of the following :(a) Name the curve plotted(b) Total number of students(c) The median marks(d) Number of students scoring between 50 and 80 marks.

Accepted Answer

(a) From graph,The curve plotted is a cumulative frequency curve (ogive).(b) From graph,The total number of students = 40.(c) From graph,The median marks = 56.(d) From graph,No of students scoring below 80 = 37No of students scoring below 50 = 12∴ No. of students scoring between 50 and 80 = 37 - 12 = 25.Hence, the no. of students scoring between 50 and 80 = 25.

Question 7

Question 4(i)If A = [4−4−44]\begin{bmatrix*}[r] 4 & -4 \ -4 & 4 \end{bmatrix*}[4−4​−44​], find A2. If A2 = pA, then find the value of p.

Accepted Answer

Given,⇒ A2 = pA⇒[4−4−44][4−4−44]=p[4−4−44]⇒[4×4+(−4)×(−4)4×(−4)+(−4)×4−4×4+4×(−4)(−4)×(−4)+4×4]=p[4−4−44]⇒[16+16−16−16−16−1616+16]=p[4−4−44]⇒[32−32−3232]=p[4−4−44]⇒32[1−1−11]=4p[1−1−11]⇒32=4p⇒p=324=8.\Rightarrow \begin{bmatrix*}[r] 4 & -4 \ -4 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 4 & -4 \ -4 & 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \ -4 & 4 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times 4 + (-4) \times (-4) & 4 \times (-4) + (-4) \times 4 \ -4 \times 4 + 4 \times (-4) & (-4) \times (-4) + 4 \times 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \ -4 & 4 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] 16 + 16 & -16 - 16 \ -16 - 16 & 16 + 16 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \ -4 & 4 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] 32 & -32 \ -32 & 32 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \ -4 & 4 \end{bmatrix*} \[1em] \Rightarrow 32\begin{bmatrix*}[r] 1 & -1 \ -1 & 1 \end{bmatrix*} = 4p\begin{bmatrix*}[r] 1 & -1 \ -1 & 1 \end{bmatrix*} \[1em] \Rightarrow 32 = 4p \[1em] \Rightarrow p = \dfrac{32}{4} = 8.⇒[4−4​−44​][4−4​−44​]=p[4−4​−44​]⇒[4×4+(−4)×(−4)−4×4+4×(−4)​4×(−4)+(−4)×4(−4)×(−4)+4×4​]=p[4−4​−44​]⇒[16+16−16−16​−16−1616+16​]=p[4−4​−44​]⇒[32−32​−3232​]=p[4−4​−44​]⇒32[1−1​−11​]=4p[1−1​−11​]⇒32=4p⇒p=432​=8.Hence, p = 8.

Question 8

Question 4(ii)Solve the given equation x2 - 4x - 2 = 0 and express your answer correct to two places of decimal.

Accepted Answer

Given, equation : x2 - 4x - 2 = 0.By formula,x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}2a−b±b2−4ac​​Substituting values we get :⇒x=−(−4)±(−4)2−4×1×−22×1=4±16+82=4±242=4±262=2±6=2+6,2−6=2+2.449,2−2.449=4.449,−0.449\Rightarrow x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1\times -2}}{2 \times 1} \[1em] = \dfrac{4 \pm \sqrt{16 + 8}}{2} \[1em] = \dfrac{4 \pm \sqrt{24}}{2} \[1em] = \dfrac{4 \pm 2\sqrt{6}}{2} \[1em] = 2 \pm \sqrt{6} \[1em] = 2 + \sqrt{6}, 2 - \sqrt{6} \[1em] = 2 + 2.449, 2 - 2.449 \[1em] = 4.449, -0.449⇒x=2×1−(−4)±(−4)2−4×1×−2​​=24±16+8​​=24±24​​=24±26​​=2±6​=2+6​,2−6​=2+2.449,2−2.449=4.449,−0.449Hence, x = 4.449 or -0.449

Question 9

Question 4(iii)In the given diagram, △ ABC is right angled at ∠B. BDFE is a rectangle. AD = 6 cm, CE = 4 cm and BC = 12 cm.(a) prove that △ADF ~ △FEC.(b) prove that △ADF ~ △ABC.(c) find the length of FE(d) find area △ADF : area △ABC

Accepted Answer

(a) In △ADF,⇒ ∠ADF = 90°By angle sum property of triangle,⇒ ∠ADF + ∠AFD + ∠A = 180°⇒ 90° + ∠AFD + ∠A = 180°⇒ ∠AFD + ∠A = 180° - 90°⇒ ∠AFD + ∠A = 90° .........(1)In △ ABC,By angle sum property of triangle,⇒ ∠A + ∠B + ∠C = 180°⇒ ∠A + 90° + ∠C = 180°⇒ ∠A + ∠C = 180° - 90°⇒ ∠A + ∠C = 90° .........(2)From equation (1) and (2), we get :⇒ ∠AFD + ∠A = ∠A + ∠C⇒ ∠AFD = ∠CIn △ ADF and △ FEC,⇒ ∠ADF = ∠FEC (Both equal to 90°)⇒ ∠AFD = ∠C (Proved above)∴ △ ADF ~ △ FEC [By A.A. axiom]Hence, proved that △ ADF ~ △ FEC.(b) In △ ADF and △ ABC,⇒ ∠DAF = ∠BAC (Common angle)⇒ ∠ADF = ∠ABC (Both equal to 90°)∴ △ ADF ~ △ ABC [By A.A. axiom]Hence, proved that △ ADF ~ △ ABC.(c) From figure,⇒ DB = FE = x (let)⇒ AB = AD + DB = (6 + x) cm⇒ DF = BE = BC - CE = 12 - 4 = 8 cm.△ ADF ~ △ ABC [proved above]We know that,Corresponding sides of similar triangle are proportional.⇒ADAB=DFBC⇒66+x=812⇒8(6+x)=6×12⇒48+8x=72⇒8x=72−48⇒8x=24⇒x=248=3 cm.\Rightarrow \dfrac{AD}{AB} = \dfrac{DF}{BC} \[1em] \Rightarrow \dfrac{6}{6 + x} = \dfrac{8}{12} \[1em] \Rightarrow 8(6 + x) = 6 \times 12 \[1em] \Rightarrow 48 + 8x = 72 \[1em] \Rightarrow 8x = 72 - 48 \[1em] \Rightarrow 8x = 24 \[1em] \Rightarrow x = \dfrac{24}{8} = 3 \text{ cm}.⇒ABAD​=BCDF​⇒6+x6​=128​⇒8(6+x)=6×12⇒48+8x=72⇒8x=72−48⇒8x=24⇒x=824​=3 cm.Hence, FE = 3 cm.(d) We know that,The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.⇒Area of △ ADFArea of △ ABC=AD2AB2⇒Area of △ ADFArea of △ ABC=62(6+3)2⇒Area of △ ADFArea of △ ABC=6292⇒Area of △ ADFArea of △ ABC=3681=49.\Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{AD^2}{AB^2} \[1em] \Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{6^2}{(6 + 3)^2} \[1em] \Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{6^2}{9^2} \[1em] \Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{36}{81} = \dfrac{4}{9}.⇒Area of △ ABCArea of △ ADF​=AB2AD2​⇒Area of △ ABCArea of △ ADF​=(6+3)262​⇒Area of △ ABCArea of △ ADF​=9262​⇒Area of △ ABCArea of △ ADF​=8136​=94​.Hence, area △ ADF : area △ ABC = 4 : 9.

Question 10

Question 5(i)Shown below is a table illustrating the monthly income distribution in a company with 100 employees.Monthly income (in ₹ 10,000)Number of employees0-4554-8158-120612-160816-201220-244Using step-deviation method, find the mean monthly income of an employee.

Accepted Answer

In the given table,Class size (i) = 4.Monthly incomeNo.of employees (f)Class markd = x - Au = d/ifu0-4552-4-1-554-815A = 60008-120610410612-160814821616-2012181233620-2442216416TotalΣf = 100Σfu = 19By formula,Mean = A + ΣfuΣf×i\dfrac{Σfu}{Σf} 	imes iΣfΣfu​×i= 6 + 19100×4\dfrac{19}{100} 	imes 410019​×4= 6 + 76100\dfrac{76}{100}10076​= 6 + 0.76= 6.76Hence, mean = 6.76

Question 11

Question 6(i)There are three positive numbers in Geometric Progression (G.P.) such that :(a) their product is 3375(b) the result of the product of first and second number added to the product of second and third number is 750.Find the numbers.

Accepted Answer

(a) Let the three positive numbers be ar,a,ar\dfrac{a}{r}, a, arra​,a,ar.Given,Product of the numbers is 3375.⇒ar×a×ar=3375⇒a3=3375⇒a3=(15)3⇒a=15.\Rightarrow \dfrac{a}{r} \times a \times ar = 3375 \[1em] \Rightarrow a^3 = 3375 \[1em] \Rightarrow a^3 = (15)^3 \[1em] \Rightarrow a = 15.⇒ra​×a×ar=3375⇒a3=3375⇒a3=(15)3⇒a=15.(b) Given,Result of the product of first and second number added to the product of second and third number is 750.∴ar×a+a×ar=750⇒a2r+a2r=750⇒a2(1r+r)=750⇒152(1r+r)=750⇒(1+r2r)=750152⇒(1+r2r)=103⇒3(1+r2)=10r⇒3+3r2=10r⇒3r2−10r+3=0⇒3r2−9r−r+3=0⇒3r(r−3)−1(r−3)=0⇒(3r−1)(r−3)=0⇒3r−1=0 or r−3=0⇒3r=1 or r=3⇒r=13 or r=3.\therefore \dfrac{a}{r} \times a + a \times ar = 750 \[1em] \Rightarrow \dfrac{a^2}{r} + a^2r = 750 \[1em] \Rightarrow a^2\Big(\dfrac{1}{r} + r\Big) = 750 \[1em] \Rightarrow 15^2\Big(\dfrac{1}{r} + r\Big) = 750 \[1em] \Rightarrow \Big(\dfrac{1 + r^2}{r}\Big) = \dfrac{750}{15^2} \[1em] \Rightarrow \Big(\dfrac{1 + r^2}{r}\Big) = \dfrac{10}{3} \[1em] \Rightarrow 3(1 + r^2) = 10r \[1em] \Rightarrow 3 + 3r^2 = 10r \[1em] \Rightarrow 3r^2 - 10r + 3 = 0 \[1em] \Rightarrow 3r^2 - 9r - r + 3 = 0 \[1em] \Rightarrow 3r(r - 3) - 1(r - 3) = 0 \[1em] \Rightarrow (3r - 1)(r - 3) = 0 \[1em] \Rightarrow 3r - 1 = 0 \text{ or } r - 3 = 0 \[1em] \Rightarrow 3r = 1 \text{ or } r = 3 \[1em] \Rightarrow r = \dfrac{1}{3} \text{ or } r = 3.∴ra​×a+a×ar=750⇒ra2​+a2r=750⇒a2(r1​+r)=750⇒152(r1​+r)=750⇒(r1+r2​)=152750​⇒(r1+r2​)=310​⇒3(1+r2)=10r⇒3+3r2=10r⇒3r2−10r+3=0⇒3r2−9r−r+3=0⇒3r(r−3)−1(r−3)=0⇒(3r−1)(r−3)=0⇒3r−1=0 or r−3=0⇒3r=1 or r=3⇒r=31​ or r=3.Let r = 13\dfrac{1}{3}31​Numbers : ar,a,ar\dfrac{a}{r}, a, arra​,a,ar= 1513,15,15×13\dfrac{15}{\dfrac{1}{3}}, 15, 15 \times \dfrac{1}{3}31​15​,15,15×31​= 45, 15, 5.Let r = 3Numbers : ar,a,ar\dfrac{a}{r}, a, arra​,a,ar= 153,15,15×3\dfrac{15}{3}, 15, 15 \times 3315​,15,15×3= 5, 15, 45.Hence, numbers are 5, 15 and 45 or 45, 15 and 5.

Question 12

Question 6(ii)The table given below shows the ages of members of a society.Age (in years)Number of members of society25-350535-453245-556955-658065-756175-8513(a) Draw a histogram representing the above distribution.(b) Estimate the modal age of the members.

Accepted Answer

Steps of construction :Draw a histogram of the given distribution.Inside the highest rectangle, which represents the maximum frequency (or modal class), draw two lines AC and BD diagonally from the upper corners C and D of adjacent rectangles.Through the point K (the point of intersection of diagonals AC and BD), draw KL perpendicular to the horizontal axis.The value of point L on the horizontal axis represents the value of mode.From graph,L = 59 yearsHence, required mode = 59 years.

Question 13

Question 6(iii)A tent is in the shape of a cylinder surmounted by a conical top. If height and radius of the cylindrical part are 7 m each and the total height of the tent is 14 m. Find the :(a) quantity of air contained inside the tent.(b) radius of a sphere whose volume is equal to the quantity of air inside the tent.Use π=227\pi = \dfrac{22}{7}π=722​

Accepted Answer

(a) From figure,Radius of cylindrical part = Radius of conical part = r = 7 mHeight of cylindrical part (H) = 7 mHeight of conical part (h) = 14 - 7 = 7 m.Quantity of air inside the tent = Volume of cylindrical part + Volume of conical part=πr2H+13πr2h=227×72×7+13×227×72×7=22×49+13×22×49=1078+359.33=1437.33 m3.= πr^2H + \dfrac{1}{3}πr^2h \[1em] = \dfrac{22}{7} \times 7^2 \times 7 + \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 7 \[1em] = 22 \times 49 + \dfrac{1}{3} \times 22 \times 49 \[1em] = 1078 + 359.33 \[1em] = 1437.33 \text{ m}^3.=πr2H+31​πr2h=722​×72×7+31​×722​×72×7=22×49+31​×22×49=1078+359.33=1437.33 m3.Hence, the quantity of air inside the tent = 1437.33 m3.(b) Let radius of required sphere be R m.According to question,Volume of sphere = Quantity of air inside the tent∴43πR3=πr2H+13πr2h⇒43R3=r2H+13r2h⇒43R3=3r2H+r2h3⇒4R3=3r2H+r2h⇒R3=3×72×7+72×74⇒R3=3×49×7+49×74⇒R3=1029+3434⇒R3=13724⇒R3=343⇒R3=73⇒R=7 m.\therefore \dfrac{4}{3}πR^3 = πr^2H + \dfrac{1}{3}πr^2h \[1em] \Rightarrow \dfrac{4}{3}R^3 = r^2H + \dfrac{1}{3}r^2h \[1em] \Rightarrow \dfrac{4}{3}R^3 = \dfrac{3r^2H + r^2h}{3} \[1em] \Rightarrow 4R^3 = 3r^2H + r^2h \[1em] \Rightarrow R^3 = \dfrac{3 \times 7^2 \times 7 + 7^2 \times 7}{4} \[1em] \Rightarrow R^3 = \dfrac{3 \times 49 \times 7 + 49 \times 7}{4} \[1em] \Rightarrow R^3 = \dfrac{1029 + 343}{4} \[1em] \Rightarrow R^3 = \dfrac{1372}{4} \[1em] \Rightarrow R^3 = 343 \[1em] \Rightarrow R^3 = 7^3 \[1em] \Rightarrow R = 7 \text{ m}.∴34​πR3=πr2H+31​πr2h⇒34​R3=r2H+31​r2h⇒34​R3=33r2H+r2h​⇒4R3=3r2H+r2h⇒R3=43×72×7+72×7​⇒R3=43×49×7+49×7​⇒R3=41029+343​⇒R3=41372​⇒R3=343⇒R3=73⇒R=7 m.Hence, radius of sphere = 7 m.

Question 14

Question 7(i)The line segment joining A(2, -3) and B(-3, 2) is intercepted by the x-axis at the point M and the y-axis at the point N. PQ is perpendicular to AB at R and meets the y-axis at a distance of 6 units form the origin O, as shown in the diagram, at S. Find the :(a) coordinates of M and N.(b) coordinates of S(c) slope of AB.(d) equation of line PQ.

Accepted Answer

(a) From figure,Coordinates of M = (-1, 0) and coordinates of N = (0, -1).(b) From figure,Coordinates of S = (0, 6).(c) By formula,Slope = y2−y1x2−x1\dfrac{y_2 - y_1}{x_2 - x_1}x2​−x1​y2​−y1​​Slope of AB = 2−(−3)−3−2=2+3−5=5−5\dfrac{2 - (-3)}{-3 - 2} = \dfrac{2 + 3}{-5} = \dfrac{5}{-5}−3−22−(−3)​=−52+3​=−55​ = -1.Hence, slope of AB = -1.(d) We know that,Product of slope of perpendicular lines = -1.∴ Slope of AB × Slope of PQ = -1⇒ -1 × Slope of PQ = -1⇒ Slope of PQ = −1−1\dfrac{-1}{-1}−1−1​ = 1.Since, PQ passes through point S.By point-slope form,Equation of line : y - y1 = m(x - x1)⇒ y - 6 = 1(x - 0)⇒ y - 6 = x⇒ y = x + 6.Hence, equation of line PQ is y = x + 6.

Question 15

Question 7(ii)The angles of depression of two ships A and B on opposite sides of a light house of height 100 m are respectively 42° and 54°. The line joining the two ships passes through the foot of the light house.(a) Find the distance between the two ships A and B.(b) Give your final answer correct to the nearest whole number.(Use mathematical tables for this question)

Accepted Answer

Let ∠BCP = α and ∠ACP = βFrom figure,⇒ α + 54° = 90°⇒ α = 90° - 54° = 36°.⇒ β + 42° = 90°⇒ β = 90° - 42° = 48°.⇒ tan α = BPCP\dfrac{BP}{CP}CPBP​⇒ tan 36° = BP100\dfrac{BP}{100}100BP​⇒ 0.7265 = BP100\dfrac{BP}{100}100BP​⇒ BP = 0.7265 × 100 = 72.65 m⇒ tan β = APCP\dfrac{AP}{CP}CPAP​⇒ tan 48° = AP100\dfrac{AP}{100}100AP​⇒ 1.1106 = AP100\dfrac{AP}{100}100AP​⇒ AP = 1.1106 × 100 = 111.06 m(a) From figure,AB = AP + BP = 72.65 + 111.06 = 183.71 mHence, the distance between two ships = 183.71 m.(b) On rounding off,AB = 184 m.Hence, the distance between two ships = 184 m.

Question 16

Question 8(i)Solve the following inequation and write the solution and represent it on the real number line.3 - 2x ≥ x + 1−x3\dfrac{1 - x}{3}31−x​ > 2x5\dfrac{2x}{5}52x​, x ∈ R.

Accepted Answer

To prove:3 - 2x ≥ x + 1−x3\dfrac{1 - x}{3}31−x​ > 2x5\dfrac{2x}{5}52x​Solving L.H.S. of the above inequation, we get :⇒ 3 - 2x ≥ x + 1−x3\dfrac{1 - x}{3}31−x​⇒ 3 - 2x ≥ 3x+1−x3\dfrac{3x + 1 - x}{3}33x+1−x​⇒ 3(3 - 2x) ≥ 2x + 1⇒ 9 - 6x ≥ 2x + 1⇒ 2x + 6x ≤ 9 - 1⇒ 8x ≤ 8⇒ x ≤ 88\dfrac{8}{8}88​⇒ x ≤ 1 ............(1)Solving R.H.S. of the above equation, we get :⇒ x + 1−x3>2x5\dfrac{1 - x}{3} \gt \dfrac{2x}{5}31−x​>52x​⇒ 3x+1−x3>2x5\dfrac{3x + 1 - x}{3} \gt \dfrac{2x}{5}33x+1−x​>52x​⇒ 5(2x + 1) > 3 × 2x⇒ 10x + 5 > 6x⇒ 10x - 6x > -5⇒ 4x > -5⇒ x > −54-\dfrac{5}{4}−45​ ...........(2)From equation (1) and (2), we get :Solution set = {x : −54-\dfrac{5}{4}−45​ < x ≤ 1, x ∈ R}Representation of solution set on real number line is :

Question 17

Question 8(ii)ABCD is a cyclic quadrilateral in which BC = CD and EF is a tangent at A. ∠CBD = 43° and ∠ADB = 62°. Find :(a) ∠ADC(b) ∠ABD(c) ∠FAD

Accepted Answer

(a) In △ BDC,⇒ BC = CD (Equal sides)⇒ ∠BDC = ∠DBC = 43° (Angles opposite to equal sides are equal)From figure,⇒ ∠ADC = ∠ADB + ∠BDC = 62° + 43° = 105°.Hence, ∠ADC = 105°.(b) We know that,Opposite angles of a cyclic quadrilateral are supplementary.⇒ ∠ABC + ∠ADC = 180°⇒ ∠ABC + 105° = 180°⇒ ∠ABC = 180° - 105° = 75°.From figure,⇒ ∠ABD = ∠ABC - ∠DBC = 75° - 43° = 32°.Hence, ∠ABD = 32°.(c) From figure,⇒ ∠FAD = ∠ABD = 32°. (Angles in alternate segment are equal)Hence, ∠FAD = 32°.

Question 18

Question 8(iii)A(a, b), B(-4, 3) and C(8, -6) are the vertices of a △ ABC. Point D is on BC such that BD : DC is 2 : 1 and M(6, 0) is mid-point of AD. Find :(a) coordinates of point D.(b) coordinates of point A.(c) equation of a line parallel to line BC, through M.

Accepted Answer

(a) Given,BD : DC = 2 : 1.Let coordinates of D be (x, y)By section-formula,(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)(m1​+m2​m1​x2​+m2​x1​​,m1​+m2​m1​y2​+m2​y1​​)Substituting values, we get :⇒(x,y)=(2×8+1×(−4)2+1,2×−6+1×32+1)⇒(x,y)=(16−43,−12+33)⇒(x,y)=(123,−93)⇒(x,y)=(4,−3).\Rightarrow (x, y) = \Big(\dfrac{2 \times 8 + 1 \times (-4)}{2 + 1}, \dfrac{2 \times -6 + 1 \times 3}{2 + 1}\Big) \[1em] \Rightarrow (x, y) = \Big(\dfrac{16 - 4}{3}, \dfrac{-12 + 3}{3}\Big) \[1em] \Rightarrow (x, y) = \Big(\dfrac{12}{3}, \dfrac{-9}{3}\Big) \[1em] \Rightarrow (x, y) = (4, -3).⇒(x,y)=(2+12×8+1×(−4)​,2+12×−6+1×3​)⇒(x,y)=(316−4​,3−12+3​)⇒(x,y)=(312​,3−9​)⇒(x,y)=(4,−3).Hence, coordinates of D = (4, -3).(b) By mid-point formula,Mid-point = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)(2x1​+x2​​,2y1​+y2​​)Given,M(6, 0) is the mid-point of AD.∴(6,0)=(a+42,b+(−3)2)⇒(6,0)=(a+42,b−32)⇒a+42=6 and b−32=0⇒a+4=12 and b−3=0⇒a=12−4=8 and b=3.\therefore (6, 0) = \Big(\dfrac{a + 4}{2}, \dfrac{b + (-3)}{2}\Big) \[1em] \Rightarrow (6, 0) = \Big(\dfrac{a + 4}{2}, \dfrac{b - 3}{2}\Big) \[1em] \Rightarrow \dfrac{a + 4}{2} = 6 \text{ and } \dfrac{b - 3}{2} = 0 \[1em] \Rightarrow a + 4 = 12 \text{ and } b - 3 = 0 \[1em] \Rightarrow a = 12 - 4 = 8 \text{ and } b = 3.∴(6,0)=(2a+4​,2b+(−3)​)⇒(6,0)=(2a+4​,2b−3​)⇒2a+4​=6 and 2b−3​=0⇒a+4=12 and b−3=0⇒a=12−4=8 and b=3.A = (a, b) = (8, 3).Hence, coordinates of A = (8, 3).(c) By formula,Slope of line = y2−y1x2−x1\dfrac{y_2 - y_1}{x_2 - x_1}x2​−x1​y2​−y1​​Slope of line BC = −6−38−(−4)=−912=−34\dfrac{-6 - 3}{8 - (-4)} = \dfrac{-9}{12} = -\dfrac{3}{4}8−(−4)−6−3​=12−9​=−43​.We know that,Slope of parallel lines are equal.Slope of line parallel to BC = −34-\dfrac{3}{4}−43​.By point-slope form,Equation of line : y - y1 = m(x - x1)Equation of line parallel to BC and passing through M is :⇒ y - 0 = −34(x−6)-\dfrac{3}{4}(x - 6)−43​(x−6)⇒ y = −34(x−6)-\dfrac{3}{4}(x - 6)−43​(x−6)⇒ 4y = -3(x - 6)⇒ 4y = -3x + 18⇒ 3x + 4y = 18.Hence, equation of the required line is 3x + 4y = 18.

Question 19

Question 9(i)Use componendo and dividendo to find the value of x, when :x2+3x3x2+1=1413\dfrac{x^2 + 3x}{3x^2 + 1} = \dfrac{14}{13}3x2+1x2+3x​=1314​

Accepted Answer

Given,x2+3x3x2+1=1413\dfrac{x^2 + 3x}{3x^2 + 1} = \dfrac{14}{13}3x2+1x2+3x​=1314​Applying componendo and dividendo, we get :⇒x2+3x+3x2+1x2+3x−(3x2+1)=14+1314−13⇒x2+3x+3x2+1x2+3x−3x2−1=271⇒(x+1)3(x−1)3=3313⇒(x+1x−1)3=(31)3⇒x+1x−1=3⇒x+1=3(x−1)⇒x+1=3x−3⇒3x−x=1+3⇒2x=4⇒x=42=2.\Rightarrow \dfrac{x^2 + 3x + 3x^2 + 1}{x^2 + 3x - (3x^2 + 1)} = \dfrac{14 + 13}{14 - 13} \[1em] \Rightarrow \dfrac{x^2 + 3x + 3x^2 + 1}{x^2 + 3x - 3x^2 - 1}= \dfrac{27}{1} \[1em] \Rightarrow \dfrac{(x + 1)^3}{(x - 1)^3} = \dfrac{3^3}{1^3} \[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \Big(\dfrac{3}{1}\Big)^3 \[1em] \Rightarrow \dfrac{x + 1}{x - 1} = 3 \[1em] \Rightarrow x + 1 = 3(x - 1) \[1em] \Rightarrow x + 1 = 3x - 3 \[1em] \Rightarrow 3x - x = 1 + 3 \[1em] \Rightarrow 2x = 4 \[1em] \Rightarrow x = \dfrac{4}{2} = 2.⇒x2+3x−(3x2+1)x2+3x+3x2+1​=14−1314+13​⇒x2+3x−3x2−1x2+3x+3x2+1​=127​⇒(x−1)3(x+1)3​=1333​⇒(x−1x+1​)3=(13​)3⇒x−1x+1​=3⇒x+1=3(x−1)⇒x+1=3x−3⇒3x−x=1+3⇒2x=4⇒x=24​=2.Hence, x = 2.

Question 20

Question 9(ii)The total expenses of a trip for certain number of people is ₹ 18,000. If three more people join them, then the share of each reduces by ₹ 3,000. Take x to be the original number of people, form a quadratic equation in x and solve it to find the value of x.

Accepted Answer

Let no. of people be x.Total expense = ₹ 18,000Expense per person = ₹ 18000x\dfrac{18000}{x}x18000​Given,If three more people join them, then the share of each reduces by ₹ 3,000.No, of people now = x + 3Expense per person = ₹ 18000x+3\dfrac{18000}{x + 3}x+318000​According to question,⇒18000x−18000x+3=3000⇒18000(x+3)−18000xx(x+3)=3000⇒18000x+54000−18000xx2+3x=3000⇒54000x2+3x=3000⇒x2+3x=540003000⇒x2+3x=18⇒x2+3x−18=0⇒x2+6x−3x−18=0⇒x(x+6)−3(x+6)=0⇒(x−3)(x+6)=0⇒x−3=0 or x+6=0⇒x=3 or x=−6.\Rightarrow \dfrac{18000}{x} - \dfrac{18000}{x + 3} = 3000 \[1em] \Rightarrow \dfrac{18000(x + 3) - 18000x}{x(x + 3)} = 3000 \[1em] \Rightarrow \dfrac{18000x + 54000 - 18000x}{x^2 + 3x} = 3000 \[1em] \Rightarrow \dfrac{54000}{x^2 + 3x} = 3000 \[1em] \Rightarrow x^2 + 3x = \dfrac{54000}{3000} \[1em] \Rightarrow x^2 + 3x = 18 \[1em] \Rightarrow x^2 + 3x - 18 = 0 \[1em] \Rightarrow x^2 + 6x - 3x - 18 = 0 \[1em] \Rightarrow x(x + 6) - 3(x + 6) = 0 \[1em] \Rightarrow (x - 3)(x + 6) = 0 \[1em] \Rightarrow x - 3 = 0 \text{ or } x + 6 = 0 \[1em] \Rightarrow x = 3 \text{ or } x = -6.⇒x18000​−x+318000​=3000⇒x(x+3)18000(x+3)−18000x​=3000⇒x2+3x18000x+54000−18000x​=3000⇒x2+3x54000​=3000⇒x2+3x=300054000​⇒x2+3x=18⇒x2+3x−18=0⇒x2+6x−3x−18=0⇒x(x+6)−3(x+6)=0⇒(x−3)(x+6)=0⇒x−3=0 or x+6=0⇒x=3 or x=−6.Since, no. of people cannot be negative.∴ x = 3.Hence, original number of people = 3.

Solved 2024 Specimen Paper ICSE Class 10 Mathematics

Section A

Section B