Question 1

Question 1(iv)If matrix 𝐴 = [2202] and A2=[4x04]\begin{bmatrix*}[r] 2 & 2 \ 0 & 2 \end{bmatrix*}	ext{ and } A^2 = \begin{bmatrix*}[r] 4 & x \ 0 & 4 \end{bmatrix*}[20​22​] and A2=[40​x4​], then the value of x is :24810

Accepted Answer

⇒A2=[2202][2202]⇒A2=[2×2+2×02×2+2×20×2+2×00×2+2×2]⇒[4x04]=[4+04+40+00+4]⇒[4x04]=[4804]⇒x=8.\phantom{\Rightarrow} A^2 = \begin{bmatrix*}[r] 2 & 2 \ 0 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 2 \ 0 & 2 \end{bmatrix*} \[1em] \Rightarrow A^2 = \begin{bmatrix*}[r] 2 \times 2 + 2\times 0 & 2 \times 2 + 2 \times 2 \ 0 \times 2 + 2 \times 0 & 0 \times 2 + 2 \times 2 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 + 0 & 4 + 4 \ 0 + 0 & 0 + 4 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 8 \ 0 & 4 \end{bmatrix*} \[1em] \Rightarrow x = 8.⇒A2=[20​22​][20​22​]⇒A2=[2×2+2×00×2+2×0​2×2+2×20×2+2×2​]⇒[40​x4​]=[4+00+0​4+40+4​]⇒[40​x4​]=[40​84​]⇒x=8.Hence, Option 3 is the correct option.

Question 2

Question 1(xv)In the given diagram, ∆ABC ∼ ∆PQR. If AD and PS are bisectors of ∠BAC and ∠QPR respectively then:∆ABC ∼ ∆PQS∆ABD ∼ ∆PQS∆ABD ∼ ∆PSR∆ABC ∼ ∆PSR

Accepted Answer

Given,∆ABC ∼ ∆PQR∴ ∠A = ∠P⇒ ∠A2=∠P2\dfrac{∠A}{2} = \dfrac{∠P}{2}2∠A​=2∠P​⇒ ∠BAD = ∠QPS∠B = ∠Q [∵ ∆ABC ∼ ∆PQR]In ∆ABD ∼ ∆PQS,⇒ ∠BAD = ∠QPS⇒ ∠B = ∠Q∴ ∆ABD ∼ ∆PQS (By A.A. axiom)Hence, Option 2 is the correct option.

Question 3

Question 2(i)A = [x011],B=[40y1] and C=[40x1]\begin{bmatrix*}[r] x & 0 \ 1 & 1 \end{bmatrix*}, B = \begin{bmatrix*}[r] 4 & 0 \ y & 1 \end{bmatrix*}	ext{ and } C = \begin{bmatrix*}[r] 4 & 0 \ x & 1 \end{bmatrix*}[x1​01​],B=[4y​01​] and C=[4x​01​].Find the values of x and y, if AB = C.

Accepted Answer

⇒AB=C⇒[x011][40y1]=[40x1]⇒[x×4+0×yx×0+0×11×4+1×y1×0+1×1]=[40x1]⇒[4x+00+04+y0+1]=[40x1]⇒[4x04+y1]=[40x1]⇒4x=4 and 4+y=x⇒x=44 and 4+y=x⇒x=1 and 4+y=1⇒x=1 and y=1−4=−3.\phantom{\Rightarrow} AB = C \[1em] \Rightarrow \begin{bmatrix*}[r] x & 0 \ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 0 \ y & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \ x & 1 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] x \times 4 + 0 \times y & x \times 0 + 0 \times 1 \ 1 \times 4 + 1 \times y & 1 \times 0 + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \ x & 1 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] 4x + 0 & 0 + 0 \ 4 + y & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \ x & 1 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] 4x & 0 \ 4 + y & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \ x & 1 \end{bmatrix*} \[1em] \Rightarrow 4x = 4 \text{ and } 4 + y = x \[1em] \Rightarrow x = \dfrac{4}{4} \text{ and } 4 + y = x \[1em] \Rightarrow x = 1 \text{ and } 4 + y = 1 \[1em] \Rightarrow x = 1 \text{ and } y = 1 - 4 = -3.⇒AB=C⇒[x1​01​][4y​01​]=[4x​01​]⇒[x×4+0×y1×4+1×y​x×0+0×11×0+1×1​]=[4x​01​]⇒[4x+04+y​0+00+1​]=[4x​01​]⇒[4x4+y​01​]=[4x​01​]⇒4x=4 and 4+y=x⇒x=44​ and 4+y=x⇒x=1 and 4+y=1⇒x=1 and y=1−4=−3.Hence, x = 1 and y = -3.

Question 4

Question 2(ii)A solid metallic cylinder is cut into two identical halves along its height. The diameter of the cylinder is 7 cm and the height is 10 cm. Find :(a) The total surface area (both the halves).(b) The total cost of painting the two halves at the rate of ₹ 30 per cm2.(use π=227)\Big(	ext{use } \pi = \dfrac{22}{7}\Big)(use π=722​)

Accepted Answer

(a) Given,Diameter of cylinder (d) = 7 cmRadius of cylinder (r) = d2=72\dfrac{d}{2} = \dfrac{7}{2}2d​=27​ = 3.5 cmHeight of cylinder (h) = 10 cmTotal surface area (both the halves) = Total surface area of cylinder + Area of two rectangles= [2πr(h + r)] + [2 × (l × b)]= [2πr(h + r)] + [2 × (h × d)]= [2×227×3.5×(3.5+10)]+[2×10×7]\Big[2 	imes \dfrac{22}{7} 	imes 3.5 	imes (3.5 + 10) \Big] + [2 	imes 10 	imes 7][2×722​×3.5×(3.5+10)]+[2×10×7]= (2 × 22 × 0.5 × 13.5) + 140= 297 + 140= 437 cm2.Hence, total surface area of both the halves = 437 cm2.(b) Total cost of painting the two halves = Total surface area × Rate= 437 × 30= ₹ 13,110.Hence, total cost of painting the two halves = ₹ 13,110.

Question 5

Question 2(iii)15, 30, 60, 120 ...... are in G.P. (Geometric Progression).(a) Find the nth term of this G.P. in terms of n.(b) How many terms of the above G.P. will give the sum 945 ?

Accepted Answer

Given,G.P. : 15, 30, 60, 120 ......First term (a) = 15Common ratio (r) = 3015\dfrac{30}{15}1530​ = 2(a) nth term of G.P. = arn = 15 x 2nHence, nth term of the given G.P. is 15 x 2n(b) Let sum of n terms of G.P. is 945.By formula,Sum of n terms of G.P. = a(rn−1)(r−1)\dfrac{a(r^n - 1)}{(r - 1)}(r−1)a(rn−1)​Substituting values we get :⇒945=15.(2n−1)2−1⇒945=15.(2n−1)1⇒2n−1=94515⇒2n−1=63⇒2n=63+1⇒2n=64⇒2n=26⇒n=6.\Rightarrow 945 = \dfrac{15.(2^n - 1)}{2 - 1} \[1em] \Rightarrow 945 = \dfrac{15.(2^n - 1)}{1} \[1em] \Rightarrow 2^n - 1 = \dfrac{945}{15} \[1em] \Rightarrow 2^n - 1 = 63 \[1em] \Rightarrow 2^n = 63 + 1 \[1em] \Rightarrow 2^n = 64 \[1em] \Rightarrow 2^n = 2^6 \[1em] \Rightarrow n = 6.⇒945=2−115.(2n−1)​⇒945=115.(2n−1)​⇒2n−1=15945​⇒2n−1=63⇒2n=63+1⇒2n=64⇒2n=26⇒n=6.Hence, sum of 6 terms of G.P. = 945.

Question 6

Question 3(i)Factorize: sin3 θ + cos3 θHence, prove the following identity :sin3θ+cos3θsin θ + cos θ\dfrac{	ext{sin}^3 θ + 	ext{cos}^3 θ}{	ext{sin θ + cos θ}}sin θ + cos θsin3θ+cos3θ​ + sin θ cos θ = 1

Accepted Answer

Factorizing,⇒ sin3 θ + cos3 θ = (sin θ + cos θ)(sin2 θ + cos2 θ - sin θ cos θ)⇒ sin3 θ + cos3 θ = (sin θ + cos θ)(1 - sin θ cos θ) ...........(1)To prove,sin3θ+cos3θsin θ + cos θ\dfrac{	ext{sin}^3 θ + 	ext{cos}^3 θ}{	ext{sin θ + cos θ}}sin θ + cos θsin3θ+cos3θ​ + sin θ cos θ = 1Substituting value of sin3 θ + cos3 θ from equation (1) in L.H.S. of above equation :⇒ (sin θ + cos θ)(1 - sin θ cos θ)sin θ + cos θ+sin θ cos θ\dfrac{	ext{(sin θ + cos θ)(1 - sin θ cos θ)}}{	ext{sin θ + cos θ}} + 	ext{sin θ cos θ}sin θ + cos θ(sin θ + cos θ)(1 - sin θ cos θ)​+sin θ cos θ⇒ 1 - sin θ cos θ + sin θ cos θ⇒ 1.Since, L.H.S. = R.H.S.Hence, proved that sin3θ+cos3θsin θ + cos θ+sin θ cos θ=1\dfrac{	ext{sin}^3 θ + 	ext{cos}^3 θ}{	ext{sin θ + cos θ}} + 	ext{sin θ cos θ} = 1sin θ + cos θsin3θ+cos3θ​+sin θ cos θ=1.

Question 7

Question 3(iii)Use graph sheet for this question.(a) Plot A(0, 3), B(2, 1) and C(4, -1).(b) Reflect point B and C in y-axis and name their images as B' and C' respectively. Plot and write coordinates of the points B' and C'.(c) Reflect point A in the line BB' and name its images as A'.(d) Plot and write coordinates of point A'.(e) Join the points ABA'B' and give the geometrical name of the closed figure so formed.

Accepted Answer

From figure,Coordinates of point A' = (0, -1), B' = (-2, 1) and C' = (-4, -1).The closed figure ABA'B' is a square.

Question 8

Question 4(i)Suresh has a recurring deposit account in a bank. He deposits ₹2000 per month and the bank pays interest at the rate of 8% per annum. If he gets ₹1040 as interest at the time of maturity, find in years total time for which the account was held.

Accepted Answer

Let time be n months.By formula,I = P×n(n+1)2×12×r100P \times \dfrac{n(n + 1)}{2\times 12} \times \dfrac{r}{100}P×2×12n(n+1)​×100r​Substituting values we get :⇒1040=2000×n(n+1)24×8100⇒1040=2000×n(n+1)300⇒n(n+1)=1040×3002000⇒n(n+1)=156⇒n2+n−156=0⇒n2+13n−12n−156=0⇒n(n+13)−12(n+13)=0⇒(n−12)(n+13)=0⇒n−12=0 or n+13=0⇒n=12 or n=−13.\Rightarrow 1040 = 2000 \times \dfrac{n(n + 1)}{24} \times \dfrac{8}{100} \[1em] \Rightarrow 1040 = 2000 \times \dfrac{n(n + 1)}{300} \[1em] \Rightarrow n(n + 1) = \dfrac{1040 \times 300}{2000} \[1em] \Rightarrow n(n + 1) = 156 \[1em] \Rightarrow n^2 + n - 156 = 0 \[1em] \Rightarrow n^2 + 13n - 12n - 156 = 0 \[1em] \Rightarrow n(n + 13) - 12(n + 13) = 0 \[1em] \Rightarrow (n - 12)(n + 13) = 0 \[1em] \Rightarrow n - 12 = 0 \text{ or } n + 13 = 0 \[1em] \Rightarrow n = 12 \text{ or } n = -13.⇒1040=2000×24n(n+1)​×1008​⇒1040=2000×300n(n+1)​⇒n(n+1)=20001040×300​⇒n(n+1)=156⇒n2+n−156=0⇒n2+13n−12n−156=0⇒n(n+13)−12(n+13)=0⇒(n−12)(n+13)=0⇒n−12=0 or n+13=0⇒n=12 or n=−13.Since, no. of months cannot be negative.∴ n = 12.Hence, total time for which the account was held = 12 months.

Question 9

Question 4(ii)The following table gives the duration of movies in minutes.Duration (in minutes)No. of movies100-1105110-12010120-13017130-1408140-1506150-1604Using step–deviation method, find the mean duration of the movies.

Accepted Answer

In the given table i is the class interval which is equal to 10.ClassClass mark (x)d = (x - A)u = d/iFrequency (f)fu100-110105-30-35-15110-120115-20-210-20120-130125-10-117-17130-140A = 1350080140-15014510166150-16015520248TotalΣf = 50Σfu = -38Mean = A + ΣfuΣf×i=135+−3850×10\dfrac{Σfu}{Σf} 	imes i = 135 + \dfrac{-38}{50} 	imes 10ΣfΣfu​×i=135+50−38​×10= 135+−38050135 + \dfrac{-380}{50}135+50−380​= 135 - 7.6= 127.4Hence, mean = 127.4

Question 10

Question 4(iii)If (a+b)3(a−b)3=6427\dfrac{(a + b)^3}{(a - b)^3} = \dfrac{64}{27}(a−b)3(a+b)3​=2764​(a) Find a+ba−b\dfrac{a + b}{a - b}a−ba+b​(b) Hence using properties of proportion, find a : b.

Accepted Answer

(a) Solving,⇒(a+b)3(a−b)3=6427⇒(a+b)3(a−b)3=4333⇒(a+ba−b)3=(43)3⇒a+ba−b=43\Rightarrow \dfrac{(a + b)^3}{(a - b)^3} = \dfrac{64}{27} \[1em] \Rightarrow \dfrac{(a + b)^3}{(a - b)^3} = \dfrac{4^3}{3^3} \[1em] \Rightarrow \Big(\dfrac{a + b}{a - b}\Big)^3 = \Big(\dfrac{4}{3}\Big)^3 \[1em] \Rightarrow \dfrac{a + b}{a - b} = \dfrac{4}{3} \[1em]⇒(a−b)3(a+b)3​=2764​⇒(a−b)3(a+b)3​=3343​⇒(a−ba+b​)3=(34​)3⇒a−ba+b​=34​Hence, a+ba−b=43.\dfrac{a + b}{a - b} = \dfrac{4}{3}.a−ba+b​=34​.(b) Solving further,⇒3(a+b)=4(a−b)⇒3a+3b=4a−4b⇒4a−3a=3b+4b⇒a=7b⇒ab=71⇒a:b=7:1.\Rightarrow 3(a + b) = 4(a - b) \[1em] \Rightarrow 3a + 3b = 4a - 4b \[1em] \Rightarrow 4a - 3a = 3b + 4b \[1em] \Rightarrow a = 7b \[1em] \Rightarrow \dfrac{a}{b} = \dfrac{7}{1} \[1em] \Rightarrow a : b = 7 : 1.⇒3(a+b)=4(a−b)⇒3a+3b=4a−4b⇒4a−3a=3b+4b⇒a=7b⇒ba​=17​⇒a:b=7:1.Hence, a : b = 7 : 1.

Question 11

Question 5(ii)The angle of elevation of the top of a 100 m high tree from two points A and B on the opposite side of the tree are 52° and 45° respectively. Find the distance AB, to the nearest metre.

Accepted Answer

From figure,⇒ tan 52° = CDAC\dfrac{CD}{AC}ACCD​⇒ 1.28 = 100AC\dfrac{100}{AC}AC100​⇒ AC = 1001.28\dfrac{100}{1.28}1.28100​ = 78.125 m⇒ tan 45° = CDBC\dfrac{CD}{BC}BCCD​⇒ 1 = 100BC\dfrac{100}{BC}BC100​⇒ BC = 100 m⇒ AB = AC + BC = 78.125 + 100 = 178.125 mHence, AB = 178.125 m

Question 12

Question 6(i)Solve the following quadratic equation for x and give your answer correct to three significant figures :2x2 - 10x + 5 = 0

Accepted Answer

Comparing equation 2x2 - 10x + 5 = 0 with ax2 + bx + c = 0, we get :a = 2, b = -10 and c = 5By formula,⇒ x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}2a−b±b2−4ac​​Substituting values we get :⇒x=−(−10)±(−10)2−4×2×52×2=10±100−404=10±604=10±2154=2(5±15)4=5±152=5±3.872=5+3.872,5−3.872=8.872,1.132=4.44,0.565\Rightarrow x = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 2 \times 5}}{2 \times 2} \[1em] = \dfrac{10 \pm \sqrt{100 - 40}}{4} \[1em] = \dfrac{10 \pm \sqrt{60}}{4} \[1em] = \dfrac{10 \pm 2\sqrt{15}}{4} \[1em] = \dfrac{2(5 \pm \sqrt{15})}{4} \[1em] = \dfrac{5 \pm \sqrt{15}}{2} \[1em] = \dfrac{5 \pm 3.87}{2} \[1em] = \dfrac{5 + 3.87}{2}, \dfrac{5 - 3.87}{2} \[1em] = \dfrac{8.87}{2}, \dfrac{1.13}{2} \[1em] = 4.44, 0.565⇒x=2×2−(−10)±(−10)2−4×2×5​​=410±100−40​​=410±60​​=410±215​​=42(5±15​)​=25±15​​=25±3.87​=25+3.87​,25−3.87​=28.87​,21.13​=4.44,0.565Hence, x = 4.44, 0.565

Question 13

Question 6(ii)The nth term of an Arithmetic Progression (A.P.) is given by the relation Tn = 6(7 - n). Find ∶(a) its first term and common difference(b) sum of its first 25 terms

Accepted Answer

(a) Tn = 6(7 - n)Substituting n = 1, we get :T1 = 6(7 - 1) = 6 × 6 = 36.Substituting n = 2, we get :T2 = 6(7 - 2) = 6 × 5 = 30.Common difference (d) = T2 - T1= 30 - 36 = -6.Hence, common difference (d) = -6 and first term = 36.(b) By formula,Sum upto n terms = n2[2a+(n−1)d]\dfrac{n}{2}[2a + (n - 1)d]2n​[2a+(n−1)d]Sum upto 25 terms=252[2×36+(25−1)×−6]=12.5×[72+24×−6]=12.5×[72−144]=12.5×−72=−900.\text{Sum upto 25 terms} = \dfrac{25}{2}[2 \times 36 + (25 - 1) \times -6] \[1em] = 12.5 \times [72 + 24 \times -6] \[1em] = 12.5 \times [72 - 144] \[1em] = 12.5 \times -72 \[1em] = -900.Sum upto 25 terms=225​[2×36+(25−1)×−6]=12.5×[72+24×−6]=12.5×[72−144]=12.5×−72=−900.Hence, sum of its first 25 terms = -900.

Question 14

Question 6(iii)In the given diagram ∆ADB and ∆ACB are two right angled triangles with ∠ADB = ∠BCA = 90°. If AB = 10 cm, AD = 6 cm, BC = 2.4 cm and DP = 4.5 cm(a) Prove that ∆APD ∼ ∆BPC.(b) Find the length of BD and PB(c) Hence, find the length of PA(d) Find area ∆APD : area ∆BPC

Accepted Answer

(a) In ∆APD and ∆BPC,⇒ ∠APD = ∠BPC (Vertically opposite angles are equal)⇒ ∠ADP = ∠BCP (Both equal to 90°)Hence, proved that ∆APD ∼ ∆BPC.(b) In ∆ADB,By pythagoras theorem,⇒ AB2 = AD2 + BD2⇒ 102 = 62 + BD2⇒ BD2 = 100 - 36⇒ BD2 = 64⇒ BD = 64\sqrt{64}64​ = 8 cm.⇒ PB = BD - PD = 8 - 4.5 = 3.5 cmHence, BD = 8 cm and PB = 3.5 cm.(c) In ∆APD,By pythagoras theorem,⇒ AP2 = AD2 + DP2⇒ AP2 = 62 + (4.5)2⇒ AP2 = 36 + 20.25⇒ AP2 = 56.25⇒ AP = 56.25\sqrt{56.25}56.25​ = 7.5 cmHence, length of AP = 7.5 cm.(d) We know that,Ratio of area of similar triangles is equal to the square of the corresponding sides.∴Area of △APDArea of △BPC=AD2BC2=62(2.4)2=6×62.4×2.4=1×10.4×0.4=10×104×4=10016=254=25:4.\therefore \dfrac{\text{Area of △APD}}{\text{Area of △BPC}} = \dfrac{AD^2}{BC^2} \[1em] = \dfrac{6^2}{(2.4)^2} \[1em] = \dfrac{6 \times 6}{2.4 \times 2.4} \[1em] = \dfrac{1 \times 1}{0.4 \times 0.4} \[1em] = \dfrac{10 \times 10}{4 \times 4} \[1em] = \dfrac{100}{16} \[1em] = \dfrac{25}{4} \[1em] = 25 : 4.∴Area of △BPCArea of △APD​=BC2AD2​=(2.4)262​=2.4×2.46×6​=0.4×0.41×1​=4×410×10​=16100​=425​=25:4.Hence, area ∆APD : area ∆BPC = 25 : 4.

Question 15

Question 7(i)In the given diagram, an isosceles ∆ABC is inscribed in a circle with centre O. PQ is a tangent to the circle at C. OM is perpendicular to chord AC and ∠COM = 65°. Find :(a) ∠ABC(b) ∠BAC(c) ∠BCQ

Accepted Answer

(a) From figure,∠AOC = ∠AOM + ∠COM = 65° + 65° = 130°.We know that,Angle at the center is twice the angle formed by the same arc at any other point of the circle.⇒ ∠AOC = 2∠ABC⇒ ∠ABC = 12×∠AOC=12×130°\dfrac{1}{2} 	imes ∠AOC = \dfrac{1}{2} 	imes 130°21​×∠AOC=21​×130° = 65°.Hence, ∠ABC = 65°.(b) In △ABC,⇒ AB = AC (Given)⇒ ∠ACB = ∠ABC = 65° (Opposite angles of equal sides are equal)By angle sum property of triangle,⇒ ∠ACB + ∠ABC + ∠BAC = 180°⇒ 65° + 65° + ∠BAC = 180°⇒ ∠BAC = 180° - 65° - 65° = 50°.Hence, ∠BAC = 50°.(c) We know that,The angle formed between the tangent and the chord through the point of contact of the tangent is equal to the angle formed by the chord in the alternate segment.∴ ∠BCQ = ∠BAC = 50°.Hence, ∠BCQ = 50°.

Question 16

Question 7(ii)Solve the following inequation, write down the solution set and represent it on the real number line.-3 + x ≤ 7x2+2\dfrac{7x}{2} + 227x​+2 < 8 + 2x, x ∈ I.

Accepted Answer

Given,-3 + x ≤ 7x2+2\dfrac{7x}{2} + 227x​+2 < 8 + 2xSolving L.H.S. of the above inequation :⇒−3+x≤7x2+2⇒7x2−x≥−3−2⇒7x−2x2≥−5⇒5x2≥−5⇒x≥−5×25⇒x≥−2 ..........(1)\Rightarrow -3 + x \le \dfrac{7x}{2} + 2 \[1em] \Rightarrow \dfrac{7x}{2} -x \ge -3 - 2 \[1em] \Rightarrow \dfrac{7x - 2x}{2} \ge -5 \[1em] \Rightarrow \dfrac{5x}{2} \ge -5 \[1em] \Rightarrow x \ge \dfrac{-5 \times 2}{5} \[1em] \Rightarrow x \ge -2 \text{ ..........(1)}⇒−3+x≤27x​+2⇒27x​−x≥−3−2⇒27x−2x​≥−5⇒25x​≥−5⇒x≥5−5×2​⇒x≥−2 ..........(1)Solving R.H.S. of the above inequation :⇒7x2+2<8+2x⇒7x2−2x<8−2⇒7x−4x2<6⇒3x2<6⇒x<6×23⇒x<4 ..........(2)\Rightarrow \dfrac{7x}{2} + 2 \lt 8 + 2x \[1em] \Rightarrow \dfrac{7x}{2} - 2x \lt 8 - 2 \[1em] \Rightarrow \dfrac{7x - 4x}{2} \lt 6 \[1em] \Rightarrow \dfrac{3x}{2} \lt 6 \[1em] \Rightarrow x \lt \dfrac{6 \times 2}{3} \[1em] \Rightarrow x \lt 4 \text{ ..........(2)}⇒27x​+2<8+2x⇒27x​−2x<8−2⇒27x−4x​<6⇒23x​<6⇒x<36×2​⇒x<4 ..........(2)From inequation (1) and (2), we get :-2 ≤ x < 4.Since, x ∈ I.x = {-2, -1, 0, 1, 2, 3}.Hence, solution set = {-2, -1, 0, 1, 2, 3}.

Question 17

Question 7(iii)In the given diagram, ABC is a triangle, where B(4, -4) and C(-4, -2). D is a point on AC.(a) Write down the coordinates of A and D.(b) Find the coordinates of the centroid of ∆ABC.(c) If D divides AC in the ratio k : 1, find the value of k.(d) Find the equation of the line BD.

Accepted Answer

(a) From graph,Co-ordinates of A = (0, 6) and D = (-3, 0)(b) By formula,Co-ordinates of centroid = (x1+x2+x33,y1+y2+y33)\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)(3x1​+x2​+x3​​,3y1​+y2​+y3​​)=(0+4+(−4)3,6+(−4)+(−2)3)=(03,03)=(0,0).= \Big(\dfrac{0 + 4 + (-4)}{3}, \dfrac{6 + (-4) + (-2)}{3}\Big) \[1em] = \Big(\dfrac{0}{3}, \dfrac{0}{3}\Big) \[1em] = (0, 0).=(30+4+(−4)​,36+(−4)+(−2)​)=(30​,30​)=(0,0).Hence, co-ordinates of centroid of ∆ABC = (0, 0).(c) 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​​)Given,D divides AC in the ratio k : 1.∴(−3,0)=(k×−4+1×0k+1,k×−2+1×6k+1)⇒(−3,0)=(−4k+0k+1,−2k+6k+1)⇒(−3,0)=(−4kk+1,−2k+6k+1)⇒−3=−4kk+1 and 0=−2k+6k+1⇒−3(k+1)=−4k and 0=−2k+6⇒3(k+1)=4k and 2k=6⇒3k+3=4k and k=62⇒4k−3k=3 and k=3⇒k=3.\therefore (-3, 0) = \Big(\dfrac{k \times -4 + 1 \times 0}{k + 1}, \dfrac{k \times -2 + 1 \times 6}{k + 1}\Big) \[1em] \Rightarrow (-3, 0) = \Big(\dfrac{-4k + 0}{k + 1}, \dfrac{-2k + 6}{k + 1}\Big) \[1em] \Rightarrow (-3, 0) = \Big(\dfrac{-4k}{k + 1}, \dfrac{-2k + 6}{k + 1}\Big) \[1em] \Rightarrow -3 = -\dfrac{4k}{k + 1} \text{ and } 0 = \dfrac{-2k + 6}{k + 1} \[1em] \Rightarrow -3(k + 1) = -4k \text{ and } 0 = -2k + 6 \[1em] \Rightarrow 3(k + 1) = 4k \text{ and } 2k = 6 \[1em] \Rightarrow 3k + 3 = 4k \text{ and } k = \dfrac{6}{2} \[1em] \Rightarrow 4k - 3k = 3 \text{ and } k = 3 \[1em] \Rightarrow k = 3.∴(−3,0)=(k+1k×−4+1×0​,k+1k×−2+1×6​)⇒(−3,0)=(k+1−4k+0​,k+1−2k+6​)⇒(−3,0)=(k+1−4k​,k+1−2k+6​)⇒−3=−k+14k​ and 0=k+1−2k+6​⇒−3(k+1)=−4k and 0=−2k+6⇒3(k+1)=4k and 2k=6⇒3k+3=4k and k=26​⇒4k−3k=3 and k=3⇒k=3.Hence, k = 3.(d) By two point form,Equation of line :y - y1 = y2−y1x2−x1(x−x1)\dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)x2​−x1​y2​−y1​​(x−x1​)Equation of BD :⇒ y - (-4) = 0−(−4)−3−4(x−4)\dfrac{0 - (-4)}{-3 - 4}(x - 4)−3−40−(−4)​(x−4)⇒ y + 4 = 4−7(x−4)\dfrac{4}{-7}(x - 4)−74​(x−4)⇒ -7(y + 4) = 4(x - 4)⇒ -7y - 28 = 4x - 16⇒ 4x + 7y - 16 + 28 = 0⇒ 4x + 7y + 12 = 0.Hence, equation of BD is 4x + 7y + 12 = 0.

Question 18

Question 8(i)The polynomial 3x3 + 8x2 - 15x + k has (x - 1) as a factor. Find the value of k. Hence factorize the resulting polynomial completely.

Accepted Answer

⇒ x - 1 = 0⇒ x = 1.Given, (x - 1) is a factor of 3x3 + 8x2 - 15x + k.Thus, on substituting x = 1 in 3x3 + 8x2 - 15x + k, the remainder will be zero.⇒ 3.(1)3 + 8.(1)2 - 15(1) + k = 0⇒ 3.1 + 8.1 - 15 + k = 0⇒ 3 + 8 - 15 + k = 0⇒ 11 - 15 + k = 0⇒ k - 4 = 0⇒ k = 4.Polynomial = 3x3 + 8x2 - 15x + 4On dividing (3x3 + 8x2 - 15x + 4) by (x - 1), we get :x−1)3x2+11x−4x−1)3x3+8x2−15x+4‾x−1))+−3x3−+3x2‾x−131x3−211x2−15xx−1)x3−2+−11x2−+11x‾x−1)31x3−2+1−4x+4x−1)31x3−2+11−+4x+−4‾x−1)31x3−2+11+1×\begin{array}{l} \phantom{x - 1)}{\quad 3x^2 + 11x - 4} \ x - 1\overline{\smash{\big)}\quad 3x^3 + 8x^2 - 15x + 4} \ \phantom{x - 1)}\phantom{)}\underline{\underset{-}{+}3x^3 \underset{+}{-}3x^2} \ \phantom{{x - 1}31x^3-2}11x^2 - 15x \ \phantom{{x - 1)}x^3-2}\underline{\underset{-}{+}11x^2 \underset{+}{-} 11x} \ \phantom{{x - 1)}31x^3-2+1}-4x + 4 \ \phantom{{x - 1)}31x^3-2+11}\underline{\underset{+}{-}4x \underset{-}{+} 4} \ \phantom{{x - 1)}31x^3-2+11+1}	imes \end{array}x−1)3x2+11x−4x−1)3x3+8x2−15x+4​x−1))−+​3x3+−​3x2​x−131x3−211x2−15xx−1)x3−2−+​11x2+−​11x​x−1)31x3−2+1−4x+4x−1)31x3−2+11+−​4x−+​4​x−1)31x3−2+11+1×​⇒ 3x3 + 8x2 - 15x + 4 = (x - 1)(3x2 + 11x - 4)= (x - 1)[3x2 + 12x - x - 4]= (x - 1)[3x(x + 4) - 1(x + 4)]= (x - 1)(3x - 1)(x + 4).Hence, 3x3 + 8x2 - 15x + 4 = (x - 1)(3x - 1)(x + 4).

Question 19

Question 8(ii)The following letters A, D, M, N, O, S, U, Y of the English alphabet are written on separate cards and put in a box. The cards are well shuffled and one card is drawn at random. What is the probability that the card drawn is a letter of the word,(a) MONDAY?(b) which does not appear in MONDAY?(c) which appears both in SUNDAY and MONDAY?

Accepted Answer

Letters written on cards = {'A', 'D', 'M', 'N', 'O', 'S', 'U', 'Y'}No. of cards = 8(a) Letters of the word MONDAY present in the cards = {'M', 'O', 'N', 'D', 'A', 'Y'}Probability that the card drawn is a letter of the word MONDAY= Letters of MONDAY presentNo. of cards=68=34\dfrac{	ext{Letters of MONDAY present}}{	ext{No. of cards}} = \dfrac{6}{8} = \dfrac{3}{4}No. of cardsLetters of MONDAY present​=86​=43​.Hence, required probability = 34\dfrac{3}{4}43​.(b) Letters of the word not present in MONDAY = {'S', 'U'}Probability that the card drawn is not a letter of the word MONDAY= Letters not present in MONDAYNo. of cards=28=14\dfrac{	ext{Letters not present in MONDAY}}{	ext{No. of cards}} = \dfrac{2}{8} = \dfrac{1}{4}No. of cardsLetters not present in MONDAY​=82​=41​.Hence, required probability = 14\dfrac{1}{4}41​.(c) Letters of the word present in SUNDAY and MONDAY are {'N', 'D', 'A', 'Y'}.Probability that the card drawn has a letter which appears both in SUNDAY and MONDAY= Letters present in both MONDAY and SUNDAYNo. of cards=48=12\dfrac{	ext{Letters present in both MONDAY and SUNDAY}}{	ext{No. of cards}} = \dfrac{4}{8} = \dfrac{1}{2}No. of cardsLetters present in both MONDAY and SUNDAY​=84​=21​.Hence, required probability = 12\dfrac{1}{2}21​.

Question 20

Question 8(iii)Oil is stored in a spherical vessel occupying 34\dfrac{3}{4}43​ of its full capacity. Radius of this spherical vessel is 28 cm. This oil is then poured into a cylindrical vessel with a radius of 21 cm. Find the height of the oil in the cylindrical vessel (correct to the nearest cm).Take π=227\pi = \dfrac{22}{7}π=722​

Accepted Answer

Given,Radius of spherical vessel (r) = 28 cmVolume of spherical vessel (v) = 43πr3\dfrac{4}{3}πr^334​πr3Volume of oil in vessel = 34v\dfrac{3}{4}v43​vSubstituting values we get :⇒v=43×227×283⇒34v=34×43×227×283⇒34v=227×283.\Rightarrow v = \dfrac{4}{3} \times \dfrac{22}{7} \times 28^3 \[1em] \Rightarrow \dfrac{3}{4}v = \dfrac{3}{4} \times \dfrac{4}{3} \times \dfrac{22}{7} \times 28^3 \[1em] \Rightarrow \dfrac{3}{4}v = \dfrac{22}{7} \times 28^3.⇒v=34​×722​×283⇒43​v=43​×34​×722​×283⇒43​v=722​×283.Radius of cylindrical vessel (R) = 21 cmLet height of oil in cylindrical vessel be h cm.Volume of oil = Volume of cylinder upto which oil is filled (πR2h)⇒227×283=227×212×h⇒283=212×h⇒h=283212⇒h=21952441⇒h=49.77≈50 cm.\Rightarrow \dfrac{22}{7} \times 28^3 = \dfrac{22}{7} \times 21^2 \times h \[1em] \Rightarrow 28^3 = 21^2 \times h \[1em] \Rightarrow h = \dfrac{28^3}{21^2}\[1em] \Rightarrow h = \dfrac{21952}{441} \[1em] \Rightarrow h = 49.77 ≈ 50 \text{ cm}.⇒722​×283=722​×212×h⇒283=212×h⇒h=212283​⇒h=44121952​⇒h=49.77≈50 cm.Hence, height of the oil in the cylindrical vessel = 50 cm.

Solved 2024 Question Paper ICSE Class 10 Mathematics

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