Q: Question 1(vii)The printed price of an article is ₹ 3080. If the rate of GST is 10% then the GST charged is :₹ 154₹ 308₹ 30.80₹ 15.40

GST charged = GST (%) × M.P.= 10100×3080\dfrac{10}{100} \times 308010010×3080= ₹ 308.Hence, Option 2 is the correct option.

Q: Question 4(i)If A=[1324],B=[1224],C=[4115] and I=[1001]A = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*}, C = \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} \text{ and } I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}A=[1234],B=[1224],C=[4115] and I=[1001]. Find A(B + C) - 14I.

B + C = [1224]+[4115]=[5339]\begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*}[1224]+[4115]=[5339]Substituting values we get :A(B+C)−14I=[1324][5339]−14[1001]=[1×5+3×31×3+3×92×5+4×32×3+4×9]−[140014]=[5+93+2710+126+36]−[140014]=[14302242]−[140014]=[14−1430−022−042−14]=[0302228].A(B + C) - 14I = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \\ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 + 9 & 3 + 27 \\ 10 + 12 & 6 + 36 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 & 30 \\ 22 & 42 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.A(B+C)−14I=[1234][5339]−14[1001]=[1×5+3×32×5+4×31×3+3×92×3+4×9]−[140014]=[5+910+123+276+36]−[140014]=[14223042]−[140014]=[14−1422−030−042−14]=[0223028].Hence, A(B + C) - 14I = [0302228].\begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.[0223028].

Question 1

Question 1(vii)The printed price of an article is ₹ 3080. If the rate of GST is 10% then the GST charged is :₹ 154₹ 308₹ 30.80₹ 15.40

Accepted Answer

GST charged = GST (%) × M.P.= 10100×3080\dfrac{10}{100} 	imes 308010010​×3080= ₹ 308.Hence, Option 2 is the correct option.

Question 2

Question 1(xiii)Volume of a cylinder of height 3 cm is 48π. Radius of the cylinder is :48 cm16 cm4 cm24 cm

Accepted Answer

Let radius of cylinder be r cm.Given,Height of cylinder (h) = 3 cmBy formula,⇒ Volume of cylinder = πr2h⇒ 3πr2 = 48π⇒ 3r2 = 48⇒ r2 = 483\dfrac{48}{3}348​⇒ r2 = 483\dfrac{48}{3}348​⇒ r2 = 16⇒ r = 16\sqrt{16}16​ = 4 cm.Hence, Option 3 is the correct option.

Question 3

Question 1(xv)The solution set for the inequation 2x + 4 ≤ 14, x ∈ W is :{1, 2, 3, 4, 5}{0, 1, 2, 3, 4, 5}{1, 2, 3, 4}{0, 1, 2, 3, 4}

Accepted Answer

Solving,⇒ 2x + 4 ≤ 14⇒ 2x ≤ 14 - 4⇒ 2x ≤ 10⇒ x ≤ 102\dfrac{10}{2}210​⇒ x ≤ 5.Since, x ∈ W,∴ x = {0, 1, 2, 3, 4, 5}.Hence, Option 2 is the correct option.

Question 4

Question 2(ii)Salman deposits ₹ 1,000 every month in a recurring deposit account for 2 years. If he receives ₹ 26,000 on maturity, find :(a) total interest Salman earns(b) the rate of interest.

Accepted Answer

(a) Given,Salman deposits ₹ 1000 every month in a recurring deposit account for 2 years.Total deposit = ₹ 1,000 × 2 × 12 = ₹ 24,000.By formula,Total interest earned = Maturity value - Total deposit= ₹ 26,000 - ₹ 24,000= ₹ 2,000.Hence, interest earned = ₹ 2,000.(b) Let rate of interest be r%. Time (n) = 24 monthsBy formula,Interest = P×n×(n+1)2×12×r100\dfrac{P \times n \times (n + 1)}{2 \times 12} \times \dfrac{r}{100}2×12P×n×(n+1)​×100r​Substituting values we get :⇒2000=1000×24×252×12×r100⇒2000=1000×25×r100⇒2000=250r⇒r=2000250⇒r=8\Rightarrow 2000 = 1000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \[1em] \Rightarrow 2000 = 1000 \times 25 \times \dfrac{r}{100} \[1em] \Rightarrow 2000 = 250r \[1em] \Rightarrow r = \dfrac{2000}{250} \[1em] \Rightarrow r = 8%.⇒2000=1000×2×1224×25​×100r​⇒2000=1000×25×100r​⇒2000=250r⇒r=2502000​⇒r=8Hence, rate of interest earned = 8%.

Question 5

Question 3(iii)Use graph sheet to answer this question. Take 2 cm = 1 unit along both the axes.(a) Plot A, B, C where A(0, 4), B(1, 1) and C(4, 0).(b) Reflect A and B on the x-axis and name them as E and D respectively.(c) Reflect B through the origin and name it F. Write down the coordinates of F.(d) Reflect B and C on the y-axis and name them as H and G respectively.(e) Join points A, B, C, D, E, F, G, H and A in order and name the closed figure formed.

Accepted Answer

The points are shown in the graph below:(c) From graph,F = (-1, -1)Hence, co-ordinates of F = (-1, -1).(e) From graph,The closed figure formed is a star.

Question 6

Question 4(i)If A=[1324],B=[1224],C=[4115] and I=[1001]A = \begin{bmatrix*}[r] 1 & 3 \ 2 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & 2 \ 2 & 4 \end{bmatrix*}, C = \begin{bmatrix*}[r] 4 & 1 \ 1 & 5 \end{bmatrix*} 	ext{ and } I = \begin{bmatrix*}[r] 1 & 0 \ 0 & 1 \end{bmatrix*}A=[12​34​],B=[12​24​],C=[41​15​] and I=[10​01​]. Find A(B + C) - 14I.

Accepted Answer

B + C = [1224]+[4115]=[5339]\begin{bmatrix*}[r] 1 & 2 \ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 1 \ 1 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 3 \ 3 & 9 \end{bmatrix*}[12​24​]+[41​15​]=[53​39​]Substituting values we get :A(B+C)−14I=[1324][5339]−14[1001]=[1×5+3×31×3+3×92×5+4×32×3+4×9]−[140014]=[5+93+2710+126+36]−[140014]=[14302242]−[140014]=[14−1430−022−042−14]=[0302228].A(B + C) - 14I = \begin{bmatrix*}[r] 1 & 3 \ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 5 & 3 \ 3 & 9 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \ 0 & 1 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \ 0 & 14 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 5 + 9 & 3 + 27 \ 10 + 12 & 6 + 36 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \ 0 & 14 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 14 & 30 \ 22 & 42 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \ 0 & 14 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 14 - 14 & 30 - 0 \ 22 - 0 & 42 - 14 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 0 & 30 \ 22 & 28 \end{bmatrix*}.A(B+C)−14I=[12​34​][53​39​]−14[10​01​]=[1×5+3×32×5+4×3​1×3+3×92×3+4×9​]−[140​014​]=[5+910+12​3+276+36​]−[140​014​]=[1422​3042​]−[140​014​]=[14−1422−0​30−042−14​]=[022​3028​].Hence, A(B + C) - 14I = [0302228].\begin{bmatrix*}[r] 0 & 30 \ 22 & 28 \end{bmatrix*}.[022​3028​].

Question 7

Question 4(ii)ABC is a triangle whose vertices are A(1, -1), B(0, 4) and C(-6, 4), D is the mid-point of BC. Find the :(a) coordinates of D.(b) equation of the median AD.

Accepted Answer

(a) By formula,Mid-point (M) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)(2x1​+x2​​,2y1​+y2​​)Given,D is the mid-point of BC.∴Co-ordinates of D=(0+(−6)2,4+42)=(−62,82)=(−3,4).\therefore \text{Co-ordinates of D} = \Big(\dfrac{0 + (-6)}{2}, \dfrac{4 + 4}{2}\Big) \[1em] = \Big(\dfrac{-6}{2}, \dfrac{8}{2}\Big) \[1em] = (-3, 4).∴Co-ordinates of D=(20+(−6)​,24+4​)=(2−6​,28​)=(−3,4).Hence, co-ordinates of D = (-3, 4).(b) By two-point form :Equation of a 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​)Substituting values we get :Equation of AD :⇒ y - (-1) = 4−(−1)(−3)−1(x−1)\dfrac{4 - (-1)}{(-3) - 1}(x - 1)(−3)−14−(−1)​(x−1)⇒ y + 1 = 5−4(x−1)\dfrac{5}{-4}(x - 1)−45​(x−1)⇒ -4(y + 1) = 5(x - 1)⇒ -4y - 4 = 5x - 5⇒ 5x + 4y = -4 + 5⇒ 5x + 4y = 1.Hence, equation of median AD is 5x + 4y = 1.

Question 8

Question 4(iii)In the given figure, O is the center of the circle. PQ is a tangent to the circle at T. Chord AB produced meets the tangent at P. AB = 9 cm, BP = 16 cm, ∠PTB = 50°, ∠OBA = 45°. Find :(a) length of PT(b) ∠BAT(c) ∠BOT(d) ∠ABT

Accepted Answer

(a) We know that,If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.Since, PT is tangent from point P and PAB is secant.∴ PT2 = PA × PB⇒ PT2 = (9 + 16) × 16⇒ PT2 = 25 × 16⇒ PT2 = 400⇒ PT = 400\sqrt{400}400​ = 20 cm.Hence, PT = 20 cm.(b) Since, angle between tangent and chord at point of contact is equal to angle of alternative segment.∴ ∠BAT = ∠PTB = 50°.Hence, ∠BAT = 50°.(c) We know that,Angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at some point on the remaining circumference.∴ ∠BOT = 2 × ∠BAT = 2 × 50° = 100°.Hence, ∠BOT = 100°.(d) From figure,⇒ OB = OT (Radii of same circle)⇒ ∠OBT = ∠OTB = x (let) (Angles opposite to equal sides are equal)In △ BOT,By angle sum property of triangle,⇒ ∠OBT + ∠OTB + ∠BOT = 180°⇒ x + x + 100° = 180°⇒ 2x = 180° - 100°⇒ 2x = 80°⇒ x = 80°2\dfrac{80°}{2}280°​ = 40°.∴ ∠OBT = 40°From figure,⇒ ∠ABT = ∠ABO + ∠OBT = 45° + 40° = 85°.Hence, ∠ABT = 85°.

Question 9

Question 5(i)Mrs. Arora bought the following articles from a departmental store :S.No.ItemPriceRate of GSTDiscount1Hair oil₹ 120018%₹ 1002Cashew nuts₹ 60012%-Find the :(a) Total GST paid.(b) Total bill amount including GST.

Accepted Answer

(a) By formula,GST on an article = (Price of item - Discount) × Rate of GSTGST on hair oil = (1200 - 100) × 18%= 1100 × 18100\dfrac{18}{100}10018​= 11 × 18= ₹ 198.GST on cashew nuts = (600 - 0) × 12%= 600 × 12100\dfrac{12}{100}10012​= 6 × 12= ₹ 72.Total GST paid = ₹ 198 + ₹ 72 = ₹ 270.Hence, total GST paid = ₹ 270.(b) Total bill amount including GST = Total cost price + Total GST paid= [(₹ 1200 - ₹ 100) + ₹ 600] + ₹ 270= ₹ 1100 + ₹ 600 + ₹ 270= ₹ 1970.Hence, total bill amount = ₹ 1970.

Question 10

Question 5(ii)Solve the following inequation. Write down the solution set and represent it on the real number line.-5(x - 9) ≥ 17 - 9x > x + 2, x ∈ R.

Accepted Answer

Given, equation :-5(x - 9) ≥ 17 - 9x > x + 2Solving L.H.S. of the given equation :⇒ -5(x - 9) ≥ 17 - 9x⇒ -5x + 45 ≥ 17 - 9x⇒ -5x + 9x ≥ 17 - 45⇒ 4x ≥ -28⇒ x ≥ −284-\dfrac{28}{4}−428⇒ x ≥ -7 ............(1)Solving L.H.S. of the given equation :⇒ 17 - 9x > x + 2⇒ x + 9x < 17 - 2⇒ 10x < 15⇒ x < 1510\dfrac{15}{10}1015⇒ x < 32\dfrac{3}{2}23⇒ x < 1.5 .............(2)From equation (1) and (2), we get :Solution set : {x : -7 ≤ x < 1.5}Solution set on the number line :

Question 11

Question 5(iii)In the given figure, AC // DE // BF. If AC = 24 cm, EG = 8 cm, GB = 16 cm, BF = 30 cm.(a) Prove △ GED ~ △ GBF(b) Find DE(c) Find DB : AB.

Accepted Answer

(a) Given,DE // BF∴ ∠DEG = ∠GBF (Alternate angles are equal)∠EDG = ∠GFB (Alternate angles are equal)⇒ △GED ~ △ GBF (By A.A. axiom)Hence, proved that △GED ~ △ GBF.(b) We know that,Corresponding sides of similar triangle are proportional.Since, △ GED ~ △ GBF⇒DEBF=EGBG⇒DE30=816⇒DE=30×816⇒DE=15 cm.\Rightarrow \dfrac{DE}{BF} = \dfrac{EG}{BG} \[1em] \Rightarrow \dfrac{DE}{30} = \dfrac{8}{16} \[1em] \Rightarrow DE = 30 \times \dfrac{8}{16} \[1em] \Rightarrow DE = 15 \text{ cm}.⇒BFDE​=BGEG​⇒30DE​=168​⇒DE=30×168​⇒DE=15 cm.Hence, DE = 15 cm.(c) In △ BAC and △ BDE,⇒ ∠BAC = ∠BDE (Corresponding angles are equal)⇒ ∠ABC = ∠DBE (Common angle)⇒ △ BAC ~ △ BDE (By A.A. axiom)We know that,Corresponding sides of similar triangle are proportional.Since, △ BAC ~ △ BDE⇒ABBD=ACDE⇒BDAB=DEAC⇒BDAB=1524⇒BDAB=58\Rightarrow \dfrac{AB}{BD} = \dfrac{AC}{DE} \[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{DE}{AC} \[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{15}{24} \[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{5}{8}⇒BDAB​=DEAC​⇒ABBD​=ACDE​⇒ABBD​=2415​⇒ABBD​=85​⇒ BD : AB = 5 : 8.Hence, BD : AB = 5 : 8.

Question 12

Question 6(ii)The 5th and 9th term of an Arithmetic Progression are 4 and -12 respectively. Find :(a) the first term(b) common difference(c) sum of 16 terms of the A.P.

Accepted Answer

If first term is a and common difference is d of the A.P.By formula,nth term = an = a + (n - 1)dGiven,⇒ 5th term = 4⇒ a5 = 4⇒ a + (5 - 1)d = 4⇒ a + 4d = 4 .........(1)⇒ 9th term = -12⇒ a9 = -12⇒ a + (9 - 1)d = -12⇒ a + 8d = -12 .........(2)Subtracting equation (1) from (2), we get :⇒ (a + 8d) - (a + 4d) = -12 - 4⇒ a - a + 8d - 4d = -16⇒ 4d = -16⇒ d = −164-\dfrac{16}{4}−416​⇒ d = -4.Substituting value of d in equation (1), we get :⇒ a + 4d = 4⇒ a + 4(-4) = 4⇒ a - 16 = 4⇒ a = 4 + 16 = 20.(a) Hence, first term of A.P. = 20.(b) Common difference of A.P. = -4.(c) By formula,Sum of n terms of A.P. = n2(a+l)\dfrac{n}{2}(a + l)2n​(a+l)Sum of 16 terms of A.P. = n2(a+a16)\dfrac{n}{2}(a + a_{16})2n​(a+a16​)=162[a+a+(16−1)d]=8(2a+15d)=8×(2×20+15×−4)=8×(40−60)=8×−20=−160.= \dfrac{16}{2}[a + a + (16 - 1)d] \[1em] = 8(2a + 15d) \[1em] = 8 \times (2 \times 20 + 15 \times -4) \[1em] = 8 \times (40 - 60) \[1em] = 8 \times -20 \[1em] = -160.=216​[a+a+(16−1)d]=8(2a+15d)=8×(2×20+15×−4)=8×(40−60)=8×−20=−160.Hence, sum of 16 terms of A.P. = -160.

Question 13

Question 6(iii)A and B are two points on the x-axis and y-axis respectively.(a) Write down the co-ordinates of A and B.(b) P is a point on AB such that AP : PB = 3 : 1. Using section formula find the coordinates of point P.(c) Find the equation of a line passing through P and perpendicular to AB.

Accepted Answer

(a) From figure,A = (4, 0) and B = (0, 4).(b) Let coordinates of P 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)=(3×0+1×43+1,3×4+1×03+1)=(0+44,12+04)=(44,124)=(1,3).\Rightarrow (x, y) = \Big(\dfrac{3 \times 0 + 1 \times 4}{3 + 1}, \dfrac{3 \times 4 + 1 \times 0}{3 + 1}\Big) \[1em] = \Big(\dfrac{0 + 4}{4}, \dfrac{12 + 0}{4}\Big) \[1em] = \Big(\dfrac{4}{4}, \dfrac{12}{4}\Big) \[1em] = (1, 3).⇒(x,y)=(3+13×0+1×4​,3+13×4+1×0​)=(40+4​,412+0​)=(44​,412​)=(1,3).Hence, coordinates of P = (1, 3).(c) By formula,Slope = y2−y1x2−x1\dfrac{y_2 - y_1}{x_2 - x_1}x2​−x1​y2​−y1​​Substituting values we get :Slope of AB = 4−00−4=4−4\dfrac{4 - 0}{0 - 4} = \dfrac{4}{-4}0−44−0​=−44​ = -1.We know that,Product of slope of perpendicular lines = -1.∴ Slope of AB × Slope of line perpendicular to AB = -1⇒ -1 × Slope of line perpendicular to AB = -1⇒ Slope of line perpendicular to AB = −1−1\dfrac{-1}{-1}−1−1​ = 1.Line passing through P and perpendicular to AB :⇒ y - y1 = m(x - x1)⇒ y - 3 = 1(x - 1)⇒ y - 3 = x - 1⇒ y = x - 1 + 3⇒ y = x + 2.Hence, required equation is y = x + 2.

Question 14

Question 7(i)A bag contains 25 cards, numbered through 1 to 25. A card is drawn at random. What is the probability that the number on the card drawn is :(a) a multiple of 5(b) a perfect square(c) a prime number ?

Accepted Answer

Total number of outcomes = 25(a) Number of multiples of 5 between 1 to 25 = 5 (5, 10, 15, 20, 25).P(that number on card is a multiples of 5)= No. of multiples of 5Total number of outcomes=525=15\dfrac{	ext{No. of multiples of 5}}{	ext{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}Total number of outcomesNo. of multiples of 5​=255​=51​.Hence, the required probability is 15\dfrac{1}{5}51​.(b) Number of perfect squares between 1 to 25 = 5 (1, 4, 9, 16, 25).P(that number on card is a perfect square)= No. of perfect squaresTotal number of outcomes=525=15\dfrac{	ext{No. of perfect squares}}{	ext{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}Total number of outcomesNo. of perfect squares​=255​=51​.Hence, the required probability is 15\dfrac{1}{5}51​.(c) Number of prime numbers between 1 to 25 = 9 (2, 3, 5, 7, 11, 13, 17, 19, 23).P(that number is a prime number)= No. of prime numbersTotal number of outcomes=925\dfrac{	ext{No. of prime numbers}}{	ext{Total number of outcomes}} = \dfrac{9}{25}Total number of outcomesNo. of prime numbers​=259​.Hence, the required probability is 925\dfrac{9}{25}259​.

Question 15

Question 7(ii)A man covers a distance of 100 km, travelling with a uniform speed of x km/hr. Had the speed been 5 km/hr more it would have taken 1 hour less. Find x the original speed.

Accepted Answer

In first case :Distance = 100 kmSpeed = x km/hrTime = DistanceSpeed=100x\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{100}{x}SpeedDistance​=x100​In second case :Distance = 100 kmSpeed = (x + 5) km/hrTime = DistanceSpeed=100x+5\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{100}{x + 5}SpeedDistance​=x+5100​According to question,The difference between the time taken between first and second case is 1 hour.∴100x−100x+5=1⇒100(x+5)−100xx(x+5)=1⇒100x+500−100xx2+5x=1⇒500x2+5x=1⇒500=x2+5x⇒x2+5x−500=0⇒x2+25x−20x−500=0⇒x(x+25)−20(x+25)=0⇒(x−20)(x+25)=0⇒x−20=0 or x+25=0⇒x=20 or x=−25.\therefore \dfrac{100}{x} - \dfrac{100}{x + 5} = 1 \[1em] \Rightarrow \dfrac{100(x + 5) - 100x}{x(x + 5)} = 1 \[1em] \Rightarrow \dfrac{100x + 500 - 100x}{x^2 + 5x} = 1 \[1em] \Rightarrow \dfrac{500}{x^2 + 5x} = 1 \[1em] \Rightarrow 500 = x^2 + 5x \[1em] \Rightarrow x^2 + 5x - 500 = 0 \[1em] \Rightarrow x^2 + 25x - 20x - 500 = 0 \[1em] \Rightarrow x(x + 25) - 20(x + 25) = 0 \[1em] \Rightarrow (x - 20)(x + 25) = 0 \[1em] \Rightarrow x - 20 = 0 \text{ or } x + 25 = 0 \[1em] \Rightarrow x = 20 \text{ or } x = -25.∴x100​−x+5100​=1⇒x(x+5)100(x+5)−100x​=1⇒x2+5x100x+500−100x​=1⇒x2+5x500​=1⇒500=x2+5x⇒x2+5x−500=0⇒x2+25x−20x−500=0⇒x(x+25)−20(x+25)=0⇒(x−20)(x+25)=0⇒x−20=0 or x+25=0⇒x=20 or x=−25.Since, speed cannot be negative.∴ x = 20 km/hr.Hence, original speed = 20 km/hr.

Question 16

Question 7(iii)A solid is in the shape of a hemisphere of radius 7 cm, surmounted by a cone of height 4 cm. The solid is immersed completely in a cylindrical container filled with water to a certain height so that the solid is completely submerged in water. If the radius of the cylinder is 14 cm, find the rise in the water level.

Accepted Answer

Let the rise in water level be x cm.∴ Volume of water that rises by x cm in the cylindrical container = Volume of hemisphere submerged + Volume of cone submerged⇒ π × (14)2 × x = 23×π×73+13×π×72×4\dfrac{2}{3} 	imes π 	imes 7^3 + \dfrac{1}{3} 	imes π 	imes 7^2 	imes 432​×π×73+31​×π×72×4⇒ 196πx = 23×π×343+13×π×196\dfrac{2}{3} 	imes π 	imes 343 + \dfrac{1}{3} 	imes π 	imes 19632​×π×343+31​×π×196⇒ 196x = 23×343+13×196\dfrac{2}{3} 	imes 343 + \dfrac{1}{3} 	imes 19632​×343+31​×196⇒ 196x = 686+1963\dfrac{686 + 196}{3}3686+196​⇒ 196x = 8823\dfrac{882}{3}3882​⇒ x = 8823×196=882588\dfrac{882}{3 	imes 196} = \dfrac{882}{588}3×196882​=588882​ = 1.5 cm.Hence, rise in water level = 1.5 cm

Question 17

Question 8(ii)What number must be added to each of the number 4, 6, 8, 11 in order to get the four numbers in proportion ?

Accepted Answer

Let x be added to each number.So, 4 + x, 6 + x, 8 + x and 11 + x must be in proportion.∴4+x6+x=8+x11+x⇒(4+x)×(11+x)=(8+x)×(6+x)⇒44+4x+11x+x2=48+8x+6x+x2⇒44+15x+x2=48+14x+x2⇒15x−14x=48−44⇒x=4.\therefore \dfrac{4 + x}{6 + x} = \dfrac{8 + x}{11 + x} \[1em] \Rightarrow (4 + x) \times (11 + x) = (8 + x) \times (6 + x) \[1em] \Rightarrow 44 + 4x + 11x + x^2 = 48 + 8x + 6x + x^2\[1em] \Rightarrow 44 + 15x + x^2 = 48 + 14x + x^2 \[1em] \Rightarrow 15x - 14x = 48 - 44 \[1em] \Rightarrow x = 4.∴6+x4+x​=11+x8+x​⇒(4+x)×(11+x)=(8+x)×(6+x)⇒44+4x+11x+x2=48+8x+6x+x2⇒44+15x+x2=48+14x+x2⇒15x−14x=48−44⇒x=4.Hence, required number is 4.

Question 18

Question 9(i)Using componendo and dividendo solve for x : 2x+2+2x−12x+2−2x−1=3\dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 32x+2​−2x−1​2x+2​+2x−1​​=3.

Accepted Answer

Given : 2x+2+2x−12x+2−2x−1=3\dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 32x+2​−2x−1​2x+2​+2x−1​​=3Applying componendo and dividendo, we get :⇒2x+2+2x−1+2x+2−2x−12x+2+2x−1−(2x+2−2x−1)=3+13−1⇒22x+222x−1=42⇒2x+22x−1=2⇒2x+2=22x−1\Rightarrow \dfrac{\sqrt{2x + 2} + \sqrt{2x - 1} + \sqrt{2x + 2} - \sqrt{2x - 1}}{\sqrt{2x + 2} + \sqrt{2x - 1} - (\sqrt{2x + 2} - \sqrt{2x - 1})} = \dfrac{3 + 1}{3 - 1} \[1em] \Rightarrow \dfrac{2\sqrt{2x + 2}}{2\sqrt{2x - 1}} = \dfrac{4}{2} \[1em] \Rightarrow \dfrac{\sqrt{2x + 2}}{\sqrt{2x - 1}} = 2 \[1em] \Rightarrow \sqrt{2x + 2} = 2\sqrt{2x - 1}⇒2x+2​+2x−1​−(2x+2​−2x−1​)2x+2​+2x−1​+2x+2​−2x−1​​=3−13+1​⇒22x−1​22x+2​​=24​⇒2x−1​2x+2​​=2⇒2x+2​=22x−1​Squaring both sides we get :⇒ 2x + 2 = 4(2x - 1)⇒ 2x + 2 = 8x - 4⇒ 8x - 2x = 2 + 4⇒ 6x = 6⇒ x = 66\dfrac{6}{6}66​ = 1.Hence, x = 1.

Question 19

Question 9(iii)From the top of a tower 100 m high, a man observes the angle of depression of two ships A and B, on opposite sides of the tower as 45° and 38° respectively. If the foot of tower and the ships are in the same horizontal line, find the distance between two ships A and B to the nearest metre.

Accepted Answer

Let CD be the tower.From figure,⇒ ∠A = ∠EDA = 45° (Alternate angles are equal)⇒ ∠B = ∠FDB = 38° (Alternate angles are equal)In △ ACD,⇒ tan A = CDAC\dfrac{CD}{AC}ACCD​⇒ tan 45° = 100AC\dfrac{100}{AC}AC100​⇒ AC = 100tan 45°=1001\dfrac{100}{	ext{tan 45°}} = \dfrac{100}{1}tan 45°100​=1100​ = 100 m.In △ BCD,⇒ tan B = CDBC\dfrac{CD}{BC}BCCD​⇒ tan 38° = 100BC\dfrac{100}{BC}BC100​⇒ BC = 100tan 38°=1000.7813\dfrac{100}{	ext{tan 38°}} = \dfrac{100}{0.7813}tan 38°100​=0.7813100​ = 127.99 m.From figure,The distance between ships A and B = AC + BC= 100 + 127.99= 227.99 m ≈ 228 m.Hence, the distance between the two ships A and B = 228 m.

Question 20

Question 10(i)Factorize completely using factor theorem : 2x3 - x2 - 13x - 6.

Accepted Answer

Substituting x = -2 in 2x3 - x2 - 13x - 6, we get :⇒ 2(-2)3 - (-2)2 - 13(-2) - 6⇒ 2(-8) - 4 + 26 - 6⇒ -16 - 4 + 20⇒ -20 + 20⇒ 0.∴ x + 2 is a factor of the polynomial 2x3 - x2 - 13x - 6.Dividing, 2x3 - x2 - 13x - 6 by x + 2, we get :x+2)2x2−5x−3x+2)2x3−x2−13x−6‾x+2))+−2x3+−4x2‾x+2x3−2−5x2−13xx+2)x3−2−+5x2−+10x‾x+2)x3−2x2(3)−3x−6x+2)x3−2x2(31)−+3x−+6‾x+2)x3−2x2(31)−2x×\begin{array}{l} \phantom{x + 2)}{\quad 2x^2 -5x - 3} \ x + 2\overline{\smash{\big)}\quad 2x^3 - x^2 - 13x - 6} \ \phantom{x + 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}4x^2} \ \phantom{{x + 2}x^3-2}-5x^2 - 13x \ \phantom{{x + 2)}x^3-2}\underline{\underset{+}{-}5x^2 \underset{+}{-} 10x} \ \phantom{{x + 2)}{x^3-2x^{2}(3)}}-3x - 6 \ \phantom{{x + 2)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}3x \underset{+}{-} 6} \ \phantom{{x + 2)}{x^3-2x^{2}(31)}{-2x}}	imes \end{array}x+2)2x2−5x−3x+2)2x3−x2−13x−6​x+2))−+​2x3−+​4x2​x+2x3−2−5x2−13xx+2)x3−2+−​5x2+−​10x​x+2)x3−2x2(3)−3x−6x+2)x3−2x2(31)+−​3x+−​6​x+2)x3−2x2(31)−2x×​∴ 2x3 - x2 - 13x - 6 = (x + 2)(2x2 - 5x - 3)= (x + 2)(2x2 - 6x + x - 3)= (x + 2)[2x(x - 3) + 1(x - 3)]= (x + 2)(2x + 1)(x - 3).Hence, 2x3 - x2 - 13x - 6 = (x + 2)(2x + 1)(x - 3).

Solved 2023 Question Paper ICSE Class 10 Mathematics

Section A

Section B