Q: Question 1(xi)An article which is marked at ₹ 1200 is available at a discount of 20% and the rate of GST is 18%. The amount of SGST is :₹ 216.00₹ 172.80₹ 108.00₹ 86.40

M.P. = ₹ 1200Discount = 20% = 20100×1200\dfrac{20}{100} \times 120010020×1200 = ₹ 240Price after discount = ₹ 1200 - ₹ 240 = ₹ 960.GST = 18%SGST = GST2=182\dfrac{GST}{2} = \dfrac{18}{2}2GST=218 % = 9%= 9100×960\dfrac{9}{100} \times 9601009×960= 9 × 9.6= ₹ 86.40Hence, Option 4 is the correct option.

Q: Question 4(ii)Solve the following inequation, write the solution set and represent it on the real number line.5x - 21 < 5x7−6≤−337+x\dfrac{5x}{7} - 6 \le -3\dfrac{3}{7} + x75x−6≤−373+x, x ∈ R.

Given, inequation : 5x - 21 < 5x7−6≤−337+x\dfrac{5x}{7} - 6 \le -3\dfrac{3}{7} + x75x−6≤−373+xSolving L.H.S. of the inequation :⇒5x−21<5x7−6⇒5x−5x7<21−6⇒35x−5x7<15⇒30x7<15⇒x<7×1530⇒x<72 ..........(1)\Rightarrow 5x - 21 \lt \dfrac{5x}{7} - 6\\[1em] \Rightarrow 5x - \dfrac{5x}{7} \lt 21 - 6 \\[1em] \Rightarrow \dfrac{35x - 5x}{7} \lt 15 \\[1em] \Rightarrow \dfrac{30x}{7} \lt 15 \\[1em] \Rightarrow x \lt \dfrac{7 \times 15}{30} \\[1em] \Rightarrow x \lt \dfrac{7}{2} \text{ ..........(1)}⇒5x−21<75x−6⇒5x−75x<21−6⇒735x−5x<15⇒730x<15⇒x<307×15⇒x<27 ..........(1)Solving R.H.S. of the inequation :⇒5x7−6≤−337+x⇒5x7−6≤−247+x⇒x−5x7≥−6+247⇒7x−5x7≥−42+247⇒2x7≥−187⇒2x≥−18⇒x≥−182⇒x≥−9 ..........(2)\Rightarrow \dfrac{5x}{7} - 6 \le -3\dfrac{3}{7} + x \\[1em] \Rightarrow \dfrac{5x}{7} - 6 \le -\dfrac{24}{7} + x \\[1em] \Rightarrow x - \dfrac{5x}{7} \ge -6 + \dfrac{24}{7} \\[1em] \Rightarrow \dfrac{7x - 5x}{7} \ge \dfrac{-42 + 24}{7} \\[1em] \Rightarrow \dfrac{2x}{7} \ge \dfrac{-18}{7} \\[1em] \Rightarrow 2x \ge -18 \\[1em] \Rightarrow x \ge \dfrac{-18}{2} \\[1em] \Rightarrow x \ge -9 \text{ ..........(2)}⇒75x−6≤−373+x⇒75x−6≤−724+x⇒x−75x≥−6+724⇒77x−5x≥7−42+24⇒72x≥7−18⇒2x≥−18⇒x≥2−18⇒x≥−9 ..........(2)From equation (1) and (2),Solution set = {x : -9 ≤ x < 72\dfrac{7}{2}27, x ∈ R}Hence, solution set = {x : -9 ≤ x < 72\dfrac{7}{2}27, x ∈ R}.

Question 1

Question 1(iii)In the adjoining figure, AC is a diameter of the circle, AP = 3 cm and PB = 4 cm and QP ⊥ AB. If the area of △ APQ is 18 cm2, then the area of shaded portion QPBC is :32 cm249 cm280 cm298 cm2

Accepted Answer

We know that,Angle in a semi-circle is a right angle.∴ ∠ABC = 90°.In △ APQ and △ ABC,⇒ ∠APQ = ∠ABC (Both equal to 90°)⇒ ∠PAQ = ∠BAC (Common angles)∴ △ APQ ~ △ ABC (By A.A. axiom)We know that,The ratio of area of similar triangles is equal to the ratio of the square of the corresponding sides.∴Area of △ APQArea of △ ABC=AP2AB2⇒18Area of △ ABC=AP2AB2⇒Area of △ ABC=AB2AP2×18⇒Area of △ ABC=(3+4)232×18⇒Area of △ ABC=7232×18⇒Area of △ ABC=499×18=98 cm2.\therefore \dfrac{\text{Area of △ APQ}}{\text{Area of △ ABC}} = \dfrac{AP^2}{AB^2} \[1em] \Rightarrow \dfrac{18}{\text{Area of △ ABC}} = \dfrac{AP^2}{AB^2} \[1em] \Rightarrow \text{Area of △ ABC} = \dfrac{AB^2}{AP^2} \times 18 \[1em] \Rightarrow \text{Area of △ ABC} = \dfrac{(3 + 4)^2}{3^2} \times 18 \[1em] \Rightarrow \text{Area of △ ABC} = \dfrac{7^2}{3^2} \times 18 \[1em] \Rightarrow \text{Area of △ ABC} = \dfrac{49}{9} \times 18 = 98 \text{ cm}^2.∴Area of △ ABCArea of △ APQ​=AB2AP2​⇒Area of △ ABC18​=AB2AP2​⇒Area of △ ABC=AP2AB2​×18⇒Area of △ ABC=32(3+4)2​×18⇒Area of △ ABC=3272​×18⇒Area of △ ABC=949​×18=98 cm2.From figure,Area of QPBC = Area of △ ABC - Area of △ APQ = 98 - 18 = 80 cm2.Hence, Option 3 is the correct option.

Question 2

Question 1(vi)A solid sphere is cut into identical hemispheres.Statement 1 : The total volume of two hemispheres is equal to the volume of the original sphere.Statement 2 : The total surface area of two hemispheres together is equal to the surface area of the original sphere.Which of the following is valid ?Both the statements are trueBoth the statements are falseStatement 1 is true, and statement 2 is falseStatement 1 is false, and statement 2 is true

Accepted Answer

When a solid sphere is cut into identical hemispheres.Let radius of sphere be r.Volume of a sphere = 43πr3\dfrac{4}{3}πr^334​πr3Volume of hemisphere = 23πr3\dfrac{2}{3}πr^332​πr3Volume of 2 hemisphere = 2×23πr3=43πr32 	imes \dfrac{2}{3}πr^3 = \dfrac{4}{3}πr^32×32​πr3=34​πr3.The total volume of two hemispheres is equal to the volume of the original sphere.Surface area of sphere = 4πr2Surface area of hemisphere = 3πr2Surface area of 2 hemisphere = 2 × 3πr2 = 6πr2.The surface area of two hemispheres is not equal to the surface area of the original sphere.Hence, Option 3 is the correct option.

Question 3

Question 1(x)A mixture of paint is prepared by mixing 2 parts of red pigments with 5 parts of the base. Using the given information in the following table, find the values of a, b and c to get the required mixture of paint.Parts of red pigmentParts of base254ab12.56ca = 10, b = 10, c = 10a = 5, b = 2, c = 5a = 10, b = 5, c = 10a = 10, b = 5, c = 15

Accepted Answer

Given,2 parts of red pigments is mixed with 5 parts of the base.∴4a=25⇒a=4×52=202=10.∴b12.5=25⇒b=25×12.5=2×2.5=5∴6c=25⇒c=6×52=3×5=15.\therefore \dfrac{4}{a} = \dfrac{2}{5} \[1em] \Rightarrow a = \dfrac{4 \times 5}{2} = \dfrac{20}{2} = 10. \[1em] \therefore \dfrac{b}{12.5} = \dfrac{2}{5} \[1em] \Rightarrow b = \dfrac{2}{5} \times 12.5 = 2 \times 2.5 = 5 \[1em] \therefore \dfrac{6}{c} = \dfrac{2}{5} \[1em] \Rightarrow c = \dfrac{6 \times 5}{2} = 3 \times 5 = 15.∴a4​=52​⇒a=24×5​=220​=10.∴12.5b​=52​⇒b=52​×12.5=2×2.5=5∴c6​=52​⇒c=26×5​=3×5=15.Hence, Option 4 is the correct option.

Question 4

Question 1(xi)An article which is marked at ₹ 1200 is available at a discount of 20% and the rate of GST is 18%. The amount of SGST is :₹ 216.00₹ 172.80₹ 108.00₹ 86.40

Accepted Answer

M.P. = ₹ 1200Discount = 20% = 20100×1200\dfrac{20}{100} 	imes 120010020​×1200 = ₹ 240Price after discount = ₹ 1200 - ₹ 240 = ₹ 960.GST = 18%SGST = GST2=182\dfrac{GST}{2} = \dfrac{18}{2}2GST​=218​ % = 9%= 9100×960\dfrac{9}{100} 	imes 9601009​×960= 9 × 9.6= ₹ 86.40Hence, Option 4 is the correct option.

Question 5

Question 1(xv)Given, x + 2 ≤ x3+3\dfrac{x}{3} + 33x​+3 and x is a prime number. The solution set for x is :∅{0}{1}{0, 1}

Accepted Answer

Solving the given equation :⇒x+2≤x3+3⇒x−x3≤3−2⇒3x−x3≤1⇒2x3≤1⇒x≤32⇒x≤1.5\Rightarrow x + 2 \le \dfrac{x}{3} + 3 \[1em] \Rightarrow x - \dfrac{x}{3} \le 3 - 2 \[1em] \Rightarrow \dfrac{3x - x}{3} \le 1 \[1em] \Rightarrow \dfrac{2x}{3} \le 1 \[1em] \Rightarrow x \le \dfrac{3}{2} \[1em] \Rightarrow x \le 1.5⇒x+2≤3x​+3⇒x−3x​≤3−2⇒33x−x​≤1⇒32x​≤1⇒x≤23​⇒x≤1.5Since, x is a prime number less than 1.5Solution set is empty.Hence, Option 1 is the correct option.

Question 6

Question 2(i)While factorizing a given polynomial, using remainder and factor theorem, a student finds that (2x + 1) is a factor of 2x3 + 7x2 + 2x - 3.(a) Is the student's solution correct stating that (2x + 1) is a factor of the given polynomial ?(b) Given a valid reason for your answer.Also, factorize the given polynomial completely.

Accepted Answer

⇒ 2x + 1 = 0⇒ 2x = -1⇒ x = −12-\dfrac{1}{2}−21​Substituting x = −12-\dfrac{1}{2}−21​ in 2x3 + 7x2 + 2x - 3, we get :⇒2×(−12)3+7×(−12)2+2×(−12)−3⇒2×−18+7×14+(−1)−3⇒−14+74−4⇒−1+74−4⇒64−4⇒6−164⇒−104⇒−52.\Rightarrow 2 \times \Big(-\dfrac{1}{2}\Big)^3 + 7 \times \Big(-\dfrac{1}{2}\Big)^2 + 2 \times \Big(-\dfrac{1}{2}\Big) - 3 \[1em] \Rightarrow 2 \times -\dfrac{1}{8} + 7 \times \dfrac{1}{4} + (-1) - 3 \[1em] \Rightarrow -\dfrac{1}{4} + \dfrac{7}{4} - 4 \[1em] \Rightarrow \dfrac{-1 + 7}{4} - 4 \[1em] \Rightarrow \dfrac{6}{4} - 4 \[1em] \Rightarrow \dfrac{6 - 16}{4} \[1em] \Rightarrow \dfrac{-10}{4} \[1em] \Rightarrow -\dfrac{5}{2}.⇒2×(−21​)3+7×(−21​)2+2×(−21​)−3⇒2×−81​+7×41​+(−1)−3⇒−41​+47​−4⇒4−1+7​−4⇒46​−4⇒46−16​⇒4−10​⇒−25​.Since, remainder is not equal to zero.Hence, (2x + 1) is not a factor of the given polynomial.Substituting x = 12\dfrac{1}{2}21​ in 2x3 + 7x2 + 2x - 3, we get :⇒2×(12)3+7×(12)2+2×12−3⇒2×18+7×14+1−3⇒14+74−2⇒84−2⇒2−2⇒0.\Rightarrow 2 \times \Big(\dfrac{1}{2}\Big)^3 + 7 \times \Big(\dfrac{1}{2}\Big)^2 + 2 \times \dfrac{1}{2} - 3 \[1em] \Rightarrow 2 \times \dfrac{1}{8} + 7 \times \dfrac{1}{4} + 1 - 3 \[1em] \Rightarrow \dfrac{1}{4} + \dfrac{7}{4} - 2 \[1em] \Rightarrow \dfrac{8}{4} - 2 \[1em] \Rightarrow 2 - 2 \[1em] \Rightarrow 0.⇒2×(21​)3+7×(21​)2+2×21​−3⇒2×81​+7×41​+1−3⇒41​+47​−2⇒48​−2⇒2−2⇒0.Since, remainder is equal to zero.∴ x - 12\dfrac{1}{2}21​ is factor of polynomial,⇒ x - 12\dfrac{1}{2}21​ = 0⇒ x = 12\dfrac{1}{2}21​⇒ 2x = 1⇒ 2x - 1 is factor of polynomial.Dividing 2x3 + 7x2 + 2x - 3 by 2x - 1, we get :2x−1)x2+4x+32x−1)2x3+7x2+2x−3‾2x−1))+−2x3−+x2‾2x−1+2x3+18x2+2x2x−1)+2x31+−8x2−+4x‾2x−1)+2x31+8x2126x−32x−1)+2x31+8x21+−6x−+3‾2x−1)+2x31+8x21+6×\begin{array}{l} \phantom{2x - 1)}{\quad x^2 + 4x + 3} \ 2x - 1\overline{\smash{\big)}\quad 2x^3 + 7x^2 + 2x - 3} \ \phantom{2x - 1)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \ \phantom{{2x - 1}+2x^3 + 1}8x^2 + 2x \ \phantom{{2x - 1)}+2x^31}\underline{\underset{-}{+}8x^2 \underset{+}{-} 4x} \ \phantom{{2x - 1)}+2x^31+8x^212}6x - 3 \ \phantom{{2x - 1)}+2x^31+8x^21}\underline{\underset{-}{+}6x \underset{+}{-} 3} \ \phantom{{2x - 1)}+2x^31+8x^21+6}\times \end{array}2x−1)x2+4x+32x−1)2x3+7x2+2x−3​2x−1))−+​2x3+−​x2​2x−1+2x3+18x2+2x2x−1)+2x31−+​8x2+−​4x​2x−1)+2x31+8x2126x−32x−1)+2x31+8x21−+​6x+−​3​2x−1)+2x31+8x21+6×​2x3 + 7x2 + 2x - 3 by 2x - 1 = (2x - 1)(x2 + 4x + 3)= (2x - 1)[x2 + 3x + x + 3]= (2x - 1)[x(x + 3) + 1(x + 3)]= (2x - 1)(x + 1)(x + 3).Hence, 2x3 + 7x2 + 2x - 3 = (2x - 1)(x + 1)(x + 3).

Question 7

Question 2(ii)A line segment joining P (2, -3) and Q (0, -1) is cut by the x-axis at the point R. A line AB cuts the y-axis at T(0, 6) and is perpendicular to PQ at S. Find the :(a) equation of line PQ(b) equation of line AB(c) coordinates of points R and S.

Accepted Answer

(a) By formula,Slope of line = (y2−y1x2−x1)\Big(\dfrac{y_2 - y_1}{x_2 - x_1}\Big)(x2​−x1​y2​−y1​​)Substituting values we get :Slope of line PQ=(−1−(−3)0−2)=−1+3−2=2−2=−1.\text{Slope of line PQ} = \Big(\dfrac{-1 - (-3)}{0 - 2}\Big) \[1em] = \dfrac{-1 + 3}{-2} \[1em] = \dfrac{2}{-2} \[1em] = -1.Slope of line PQ=(0−2−1−(−3)​)=−2−1+3​=−22​=−1.Equation of line :⇒ y - y1 = m(x - x1)⇒ y - (-3) = -1(x - 2)⇒ y + 3 = -x + 2⇒ x + y + 3 - 2 = 0⇒ x + y + 1 = 0.Hence, equation of line PQ is x + y + 1 = 0.(b) We know that,Product of slope of perpendicular lines = -1.∴ Slope of PQ × Slope of AB = -1⇒ -1 × Slope of AB = -1⇒ Slope of AB = −1−1\dfrac{-1}{-1}−1−1​ = 1.Equation of line :⇒ y - y1 = m(x - x1)Equation of line AB :⇒ y - 6 = 1(x - 0)⇒ y - 6 = x⇒ x - y + 6 = 0Hence, equation of line AB is x - y + 6 = 0.(c) Given,Line PQ cuts x-axis at point R.Let R = (a, 0)Equation of line PQ = x + y + 1 = 0Since, point R lies on line PQ,⇒ a + 0 + 1 = 0⇒ a + 1 = 0⇒ a = -1.R = (a, 0) = (-1, 0)Given,AB is perpendicular to PQ at point S.∴ Point S is the intersection point of AB and PQ.PQ : x + y + 1 = 0AB : y - x = 6 or y = x + 6Substituting value of y from equation AB in equation PQ, we get :⇒ x + (x + 6) + 1 = 0⇒ 2x + 7 = 0⇒ 2x = -7⇒ x = −72-\dfrac{7}{2}−27​Substituting value of x in equation AB, we get :y = −72+6=−7+122=52-\dfrac{7}{2} + 6 = \dfrac{-7 + 12}{2} = \dfrac{5}{2}−27​+6=2−7+12​=25​.S = (−72,52)\Big(-\dfrac{7}{2}, \dfrac{5}{2}\Big)(−27​,25​).Hence, coordinates of R = (-1, 0) and S = (−72,52)\Big(-\dfrac{7}{2}, \dfrac{5}{2}\Big)(−27​,25​).

Question 8

Question 3(ii)In the adjoining diagram, a tilted right circular cylindrical vessel with base diameter 7 cm contains a liquid. When placed vertically, the height of the liquid in the vessel is the mean of two heights shown in the diagram. Find the area of wet surface, when the cylinder is placed vertically on a horizontal surface. (Use π=227\pi = \dfrac{22}{7}π=722​ )

Accepted Answer

When vertically placed,Height of liquid (h) = 1+62=72\dfrac{1 + 6}{2} = \dfrac{7}{2}21+6​=27​ cmDiameter of base = 7 cmRadius (r) = 72\dfrac{7}{2}27​ cmArea of wet surface=πr2+2πrh=πr(r+2h)=227×72(72+2×72)=11×(3.5+7)=11×10.5=115.5 cm2.\text{Area of wet surface} = πr^2 + 2πrh \[1em] = πr(r + 2h) \[1em] = \dfrac{22}{7} \times \dfrac{7}{2}\Big(\dfrac{7}{2} + 2\times \dfrac{7}{2}\Big) \[1em] = 11 \times (3.5 + 7) \[1em] = 11 \times 10.5 \[1em] = 115.5 \text{ cm}^2.Area of wet surface=πr2+2πrh=πr(r+2h)=722​×27​(27​+2×27​)=11×(3.5+7)=11×10.5=115.5 cm2.Hence, area of wet surface = 115.5 cm2.

Question 9

Question 3(iii)Study the graph and answer each of the following :(a) Write the coordinates of points A, B, C and D.(b) Given that, point C is the image of point A. Name and write the equation of the line of reflection.(c) Write the coordinates of the image of the point D under reflection in y-axis.(d) What is the name given to a point whose image is the point itself ?(e) On joining the points A, B, C, D and A in order, a figure is formed. Name the closed figure.

Accepted Answer

(a) From graph,Coordinates of A = (3, 3), B = (-2, 1), C = (3, -1) and D = (0, 1).(b) From graph,C is the image of point A in the line y = 1.Hence, the line is BD and its equation is y = 1.(c) Since, point D lies on y-axis, thus it is invariant under reflection in y-axis.Hence, coordinates of point D on reflection in y-axis is (0, 1).(d) A point whose image is the point itself is called invariant point.(e) On joining the points A, B, C, D and A in order the figure formed is a concave quadrilateral or arrowhead.

Question 10

Question 4(i)A man buys 250, ten-rupee shares each at ₹12.50. If the rate of dividend is 7%, find the :(a) dividend he receives annually.(b) percentage return on his investment.

Accepted Answer

(a) Nominal Value of 1 share = ₹10Market Value of 1 share = ₹12.50Number of shares purchased = 250Nominal Value of 250 shares = 250 x 10 = ₹2500Rate of dividend = 7%∴ Dividend received = 7% of 2500= 7100×2500\dfrac{7}{100} \times 25001007​×2500= ₹175.Hence, annual dividend = ₹175.(b) Amount Invested = No. of shares x Market Value= 250 x 12.50= ₹3125Return percentage =Dividend receivedInvestment×100=1753125×100=175003125=5.6\text{Return percentage } = \dfrac{\text{Dividend received}}{\text{Investment}} \times 100 \[1em] = \dfrac{175}{3125} \times 100 \[1em] = \dfrac{17500}{3125} \[1em] = 5.6%.Return percentage =InvestmentDividend received​×100=3125175​×100=312517500​=5.6Hence, return percentage = 5.6%.

Question 11

Question 4(ii)Solve the following inequation, write the solution set and represent it on the real number line.5x - 21 < 5x7−6≤−337+x\dfrac{5x}{7} - 6 \le -3\dfrac{3}{7} + x75x​−6≤−373​+x, x ∈ R.

Accepted Answer

Given, inequation : 5x - 21 < 5x7−6≤−337+x\dfrac{5x}{7} - 6 \le -3\dfrac{3}{7} + x75x​−6≤−373​+xSolving L.H.S. of the inequation :⇒5x−21<5x7−6⇒5x−5x7<21−6⇒35x−5x7<15⇒30x7<15⇒x<7×1530⇒x<72 ..........(1)\Rightarrow 5x - 21 \lt \dfrac{5x}{7} - 6\[1em] \Rightarrow 5x - \dfrac{5x}{7} \lt 21 - 6 \[1em] \Rightarrow \dfrac{35x - 5x}{7} \lt 15 \[1em] \Rightarrow \dfrac{30x}{7} \lt 15 \[1em] \Rightarrow x \lt \dfrac{7 \times 15}{30} \[1em] \Rightarrow x \lt \dfrac{7}{2} \text{ ..........(1)}⇒5x−21<75x​−6⇒5x−75x​<21−6⇒735x−5x​<15⇒730x​<15⇒x<307×15​⇒x<27​ ..........(1)Solving R.H.S. of the inequation :⇒5x7−6≤−337+x⇒5x7−6≤−247+x⇒x−5x7≥−6+247⇒7x−5x7≥−42+247⇒2x7≥−187⇒2x≥−18⇒x≥−182⇒x≥−9 ..........(2)\Rightarrow \dfrac{5x}{7} - 6 \le -3\dfrac{3}{7} + x \[1em] \Rightarrow \dfrac{5x}{7} - 6 \le -\dfrac{24}{7} + x \[1em] \Rightarrow x - \dfrac{5x}{7} \ge -6 + \dfrac{24}{7} \[1em] \Rightarrow \dfrac{7x - 5x}{7} \ge \dfrac{-42 + 24}{7} \[1em] \Rightarrow \dfrac{2x}{7} \ge \dfrac{-18}{7} \[1em] \Rightarrow 2x \ge -18 \[1em] \Rightarrow x \ge \dfrac{-18}{2} \[1em] \Rightarrow x \ge -9 \text{ ..........(2)}⇒75x​−6≤−373​+x⇒75x​−6≤−724​+x⇒x−75x​≥−6+724​⇒77x−5x​≥7−42+24​⇒72x​≥7−18​⇒2x≥−18⇒x≥2−18​⇒x≥−9 ..........(2)From equation (1) and (2),Solution set = {x : -9 ≤ x < 72\dfrac{7}{2}27​, x ∈ R}Hence, solution set = {x : -9 ≤ x < 72\dfrac{7}{2}27​, x ∈ R}.

Question 12

Question 5(i)In the given figure (drawn not to scale) chords AD and BC intersect at P, where AB = 9 cm, PB = 3 cm and PD = 2 cm.(a) Prove that △ APB ~ △ CPD(b) Find the length of CD(c) Find area △ APB : area △ CPD.

Accepted Answer

(a) In △ APB and △ CPD,⇒ ∠APB = ∠CPD (Vertically opposite angles are equal)⇒ ∠BAP = ∠DCP (Angles in same segment are equal)∴ △ APB ~ △ CPD (By A.A. axiom)Hence, proved that △ APB ~ △ CPD.(b) We know that,Corresponding sides of similar triangles are proportional.∴CDAB=PDPB⇒CD9=23⇒CD=9×23⇒CD=6 cm.\therefore \dfrac{CD}{AB} = \dfrac{PD}{PB} \[1em] \Rightarrow \dfrac{CD}{9} = \dfrac{2}{3} \[1em] \Rightarrow CD = 9 \times \dfrac{2}{3} \[1em] \Rightarrow CD = 6\text{ cm}.∴ABCD​=PBPD​⇒9CD​=32​⇒CD=9×32​⇒CD=6 cm.Hence, CD = 6 cm.(c) We know that,Ratio of area of similar triangles is equal to the ratio of square of the corresponding sides.∴Area of △APBArea of △CPD=PB2PD2=3222=94=9:4.\therefore \dfrac{\text{Area of △APB}}{\text{Area of △CPD}} = \dfrac{PB^2}{PD^2} \[1em] = \dfrac{3^2}{2^2} \[1em] = \dfrac{9}{4} \[1em] = 9 : 4.∴Area of △CPDArea of △APB​=PD2PB2​=2232​=49​=9:4.Hence, area △ APB : area △ CPD = 9 : 4.

Question 13

Question 5(ii)Mr. Sameer has a recurring deposit account and deposits ₹ 600 per month for 2 years. If he gets ₹ 15600 at the time of maturity, find the rate of interest earned by him.

Accepted Answer

Let rate of interest be r%.Given,P = ₹ 600/monthn = 2 years or 24 monthsM.V. = ₹ 15600By formula,M.V. = P×n+P×n(n+1)2×12×r100P \times n + P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}P×n+P×2×12n(n+1)​×100r​Substituting values we get :⇒15600=600×24+600×24×(24+1)2×12×r100⇒15600=14400+6×25×r⇒15600−14400=6×25×r⇒150r=1200⇒r=1200150=8\Rightarrow 15600 = 600 \times 24 + 600 \times \dfrac{24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \[1em] \Rightarrow 15600 = 14400 + 6 \times 25 \times r \[1em] \Rightarrow 15600 - 14400 = 6 \times 25 \times r \[1em] \Rightarrow 150r = 1200 \[1em] \Rightarrow r = \dfrac{1200}{150} = 8%.⇒15600=600×24+600×2×1224×(24+1)​×100r​⇒15600=14400+6×25×r⇒15600−14400=6×25×r⇒150r=1200⇒r=1501200​=8Hence, rate of interest = 8%.

Question 14

Question 6(i)Find the coordinates of the centroid P of the △ ABC, whose vertices are A(-1, 3), B(3, -1) and C(0, 0). Hence, find the equation of a line passing through P and parallel to AB.

Accepted Answer

By formula,Centroid of triangle = (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​​)Substituting values we get :⇒Centroid of △ ABC=(−1+3+03,3+(−1)+03)⇒P=(23,23).\Rightarrow \text{Centroid of △ ABC} = \Big(\dfrac{-1 + 3 + 0}{3}, \dfrac{3 + (-1) + 0}{3}\Big) \[1em] \Rightarrow P = \Big(\dfrac{2}{3}, \dfrac{2}{3}\Big).⇒Centroid of △ ABC=(3−1+3+0​,33+(−1)+0​)⇒P=(32​,32​).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=−1−33−(−1)=−44=−1.\text{Slope of AB} = \dfrac{-1 - 3}{3 - (-1)} \[1em] = \dfrac{-4}{4} \[1em] = -1.Slope of AB=3−(−1)−1−3​=4−4​=−1.We know that,Slope of parallel lines are equal.By point-slope form,Equation of line : y - y1 = m(x - x1)Substituting values we get :Equation of line passing through P and parallel to AB :⇒y−23=−1(x−23)⇒3y−23=−1×3x−23⇒3y−2=−1(3x−2)⇒3y−2=−3x+2⇒3y+3x=2+2⇒3y+3x=4.\Rightarrow y - \dfrac{2}{3} = -1\Big(x - \dfrac{2}{3}\Big) \[1em] \Rightarrow \dfrac{3y - 2}{3} = -1 \times \dfrac{3x - 2}{3} \[1em] \Rightarrow 3y - 2 = -1(3x - 2) \[1em] \Rightarrow 3y - 2 = -3x + 2 \[1em] \Rightarrow 3y + 3x = 2 + 2 \[1em] \Rightarrow 3y + 3x = 4.⇒y−32​=−1(x−32​)⇒33y−2​=−1×33x−2​⇒3y−2=−1(3x−2)⇒3y−2=−3x+2⇒3y+3x=2+2⇒3y+3x=4.Hence, required equation is 3x + 3y = 4.

Question 15

Question 6(ii)In the given figure PT is a tangent to the circle. Chord BA produced meets the tangent PT at P. Given PT = 20 cm and PA = 16 cm.(a) Prove △ PTB ~ △ PAT(b) Find the length of AB.

Accepted Answer

(a) In △ PTB and △ PAT,⇒ ∠PTA = ∠PBT (Alternate segment theorem)⇒ ∠TPA = ∠BPT (Common angle)∴ △ PTB ~ △ PAT (By A.A. axiom)Hence, proved that △ PTB ~ △ PAT.(b) We know that,If a chord and a tangent intersect externally, then the product of the lengths of 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.⇒ PA × PB = PT2⇒ PA × (PA + AB) = PT2⇒ 16 × (16 + AB) = 202⇒ 16 × (16 + AB) = 400⇒ 16 + AB = 25⇒ AB = 25 - 16 = 9 cm.Hence, AB = 9 cm.

Question 16

Question 6(iii)The following bill shows the GST rate and the marked price of articles :Rajdhani Departmental StoreS.No.ItemMarked priceDiscountRate of GST(a)Dry fruits (1 kg)₹ 1200₹ 10012%(b)Packed Wheat Flour (5 kg)₹ 286Nil5%(c)Bakery Products₹ 50010%12%Find the total amount to be paid (including GST) for the above bill.

Accepted Answer

For Dry fruits :M.P. = ₹ 1200Discount = ₹ 100Discounted price = ₹ 1200 - ₹ 100 = ₹ 1100GST = 12%Tax = 12100×1100\dfrac{12}{100} 	imes 110010012​×1100 = ₹ 132Price including tax = ₹ 1100 + ₹ 132 = ₹ 1232.For Wheat flour :M.P. = ₹ 286Discount = NilGST = 5%Tax = 5100×286\dfrac{5}{100} 	imes 2861005​×286 = ₹ 14.30Price including tax = ₹ 286 + ₹ 14.30 = 300.30For bakery products :M.P. = ₹ 500Discount = 10% = 10100×500\dfrac{10}{100} 	imes 50010010​×500 = ₹ 50.Discounted price = ₹ 500 - ₹ 50 = ₹ 450GST = 12%Tax = 12100×450\dfrac{12}{100} 	imes 45010012​×450 = ₹ 54Price including tax = ₹ 450 + ₹ 54 = ₹ 504.Total price = ₹ 1232 + ₹ 300.30 + ₹ 504 = ₹ 2036.30Hence, total amount to be paid = ₹ 2036.30

Question 17

Question 8(i)In a T.V. show, a contestant opt for video call a friend life line to get an answer from three of his friends, named Amar, Akbar and Anthony. The question which he asks from one of his friends has four options. Find the probability that :(a) Akbar is chosen for the call.(b) Akbar couldn't give the correct answer.

Accepted Answer

(a) Since, there are three people which can be called.Hence, probability that Akbar is chosen for the call = 13\dfrac{1}{3}31​.(b) Since, there are four options out of which one is correct.Hence, probability that Akbar couldn't give the correct answer = 34\dfrac{3}{4}43​.

Question 18

Question 8(ii)If x, y and z are in continued proportion, prove that :xy2.z2+yz2.x2+zx2.y2=1x3+1y3+1z3\dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}y2.z2x​+z2.x2y​+x2.y2z​=x31​+y31​+z31​

Accepted Answer

Given,x, y and z are in continued proportion.∴xy=yz⇒y2=xz\therefore \dfrac{x}{y} = \dfrac{y}{z} \[1em] \Rightarrow y^2 = xz∴yx​=zy​⇒y2=xzTo prove :xy2.z2+yz2.x2+zx2.y2=1x3+1y3+1z3\dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}y2.z2x​+z2.x2y​+x2.y2z​=x31​+y31​+z31​Solving L.H.S.,⇒xy2.z2+yz2.x2+zx2.y2⇒x3+y3+z3x2.y2.z2⇒x3+y3+z3x2.xz.z2⇒x3+y3+z3x3.z3⇒x3x3.z3+y3x3z3+z3x3.z3⇒1z3+y3(xz)3+1x3⇒1z3+y3(y2)3+1x3⇒1z3+y3y6+1x3⇒1z3+1y3+1x3.\Rightarrow \dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} \[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^2.y^2.z^2} \[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^2.xz.z^2} \[1em] \Rightarrow \dfrac{x^3 + y^3 + z^3}{x^3.z^3} \[1em] \Rightarrow \dfrac{x^3}{x^3.z^3} + \dfrac{y^3}{x^3z^3} + \dfrac{z^3}{x^3.z^3} \[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{(xz)^3} + \dfrac{1}{x^3} \[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{(y^2)^3} + \dfrac{1}{x^3} \[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{y^3}{y^6} + \dfrac{1}{x^3} \[1em] \Rightarrow \dfrac{1}{z^3} + \dfrac{1}{y^3} + \dfrac{1}{x^3}.⇒y2.z2x​+z2.x2y​+x2.y2z​⇒x2.y2.z2x3+y3+z3​⇒x2.xz.z2x3+y3+z3​⇒x3.z3x3+y3+z3​⇒x3.z3x3​+x3z3y3​+x3.z3z3​⇒z31​+(xz)3y3​+x31​⇒z31​+(y2)3y3​+x31​⇒z31​+y6y3​+x31​⇒z31​+y31​+x31​.Since, L.H.S. = R.H.S.Hence, proved that xy2.z2+yz2.x2+zx2.y2=1x3+1y3+1z3\dfrac{x}{y^2.z^2} + \dfrac{y}{z^2.x^2} + \dfrac{z}{x^2.y^2} = \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}y2.z2x​+z2.x2y​+x2.y2z​=x31​+y31​+z31​.

Question 19

Question 8(iii)A manufacturing company prepares spherical ball bearings, each of radius 7 mm and mass 4 gm. These ball bearings are packed into boxes. Each box can have maximum of 2156 cm3 of ball bearings. Find the :(a) maximum number of ball bearings that each box can have.(b) mass of each box of ball bearings in kg.(use π=227)\Big(	ext{use } \pi = \dfrac{22}{7}\Big)(use π=722​)

Accepted Answer

(a) Given,Radius of ball bearings = 7 mmVolume of box = 2156 cm3 = 2156 × 103 mm3Number of ball bearings that each box can have (N)= Volume of boxVolume of each ball\dfrac{\text{Volume of box}}{\text{Volume of each ball}}Volume of each ballVolume of box​Substituting values we get :N=2156×10343×227×73=2156×103883×72=2156×3×100088×49=64680004312=1500.N = \dfrac{2156 \times 10^3}{\dfrac{4}{3} \times \dfrac{22}{7} \times 7^3} \[1em] = \dfrac{2156 \times 10^3}{\dfrac{88}{3} \times 7^2} \[1em] = \dfrac{2156 \times 3 \times 1000}{88 \times 49} \[1em] = \dfrac{6468000}{4312} \[1em] = 1500.N=34​×722​×732156×103​=388​×722156×103​=88×492156×3×1000​=43126468000​=1500.Hence, maximum no. of ball bearings in a box = 1500.(b) Mass of each box = No. of balls × Mass of each ball= 1500 × 4 gm= 6000 gm= 60001000\dfrac{6000}{1000}10006000​ = 6 kg.Hence, mass of each box = 6 kg.

Question 20

Question 10(i)A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed. If it takes 3 hours to complete the total journey then form a quadratic equation and solve it to find its original average speed.

Accepted Answer

Given,A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed.Total time taken to complete the journey = 3 hours∴72x+81x+6=3⇒72(x+6)+81xx(x+6)=3⇒72(x+6)+81x=3x(x+6)⇒72x+432+81x=3x2+18x⇒3x2+18x−72x−81x−432=0⇒3x2−135x−432=0⇒3(x2−45x−144)=0⇒x2−45x−144=0⇒x2−48x+3x−144=0⇒x(x−48)+3(x−48)=0⇒(x+3)(x−48)=0⇒x+3=0 or x−48=0⇒x=−3 or x=48.\therefore \dfrac{72}{x} + \dfrac{81}{x + 6} = 3 \[1em] \Rightarrow \dfrac{72(x + 6) + 81x}{x(x + 6)} = 3 \[1em] \Rightarrow 72(x + 6) + 81x = 3x(x + 6) \[1em] \Rightarrow 72x + 432 + 81x = 3x^2 + 18x \[1em] \Rightarrow 3x^2 + 18x - 72x - 81x - 432 = 0 \[1em] \Rightarrow 3x^2 - 135x - 432 = 0 \[1em] \Rightarrow 3(x^2 - 45x - 144) = 0 \[1em] \Rightarrow x^2 - 45x - 144 = 0 \[1em] \Rightarrow x^2 - 48x + 3x - 144 = 0 \[1em] \Rightarrow x(x - 48) + 3(x - 48) = 0 \[1em] \Rightarrow (x + 3)(x - 48) = 0 \[1em] \Rightarrow x + 3 = 0 \text{ or } x - 48 = 0 \[1em] \Rightarrow x = -3 \text{ or } x = 48.∴x72​+x+681​=3⇒x(x+6)72(x+6)+81x​=3⇒72(x+6)+81x=3x(x+6)⇒72x+432+81x=3x2+18x⇒3x2+18x−72x−81x−432=0⇒3x2−135x−432=0⇒3(x2−45x−144)=0⇒x2−45x−144=0⇒x2−48x+3x−144=0⇒x(x−48)+3(x−48)=0⇒(x+3)(x−48)=0⇒x+3=0 or x−48=0⇒x=−3 or x=48.As speed cannot be negative in this case,Hence, the original speed = 48 km/hr.

Solved 2025 Specimen Paper ICSE Class 10 Mathematics

Section A

Section B