Q: Question 82Determine whether the following quadratic equation has real roots.5𝑥2 − 9𝑥 + 4 = 0(a) Give reasons for your answer.(b) If the equation has real roots, identify them.

(a) Given, equation : 5𝑥2 − 9𝑥 + 4 = 0Comparing above equation with ax2 + bx + c = 0, we get :a = 5, b = -9 and c = 4.Discriminant (D) = b2 - 4ac = (-9)2 - 4 × 5 × 4 = 81 - 80 = 1.Since, D > 0 and a perfect square.Hence, equation 5𝑥2 − 9𝑥 + 4 = 0 has real roots.(b) Solving,⇒ 5𝑥2 − 9𝑥 + 4 = 0⇒ 5x2 - 5x - 4x + 4 = 0⇒ 5x(x - 1) - 4(x - 1) = 0⇒ (5x - 4)(x - 1) = 0⇒ 5x - 4 = 0 or x - 1 = 0⇒ 5x = 4 or x = 1⇒ x = 45\dfrac{4}{5}54 or x = 1.Hence, roots of the equation are 1 and 45\dfrac{4}{5}54.

Question 1

Question 78The following bill shows the GST rate and the marked price of items :S.No.ItemMarked price (₹)QuantityRate of GST1Wheat Flour (unpacked)35.005 kgx%2Basmati Rice (Branded & packed)180.005 kg5%3Surf Excel Quick Wash Detergent220.00y kg18%Find :(a) the value of x if the total GST on wheat flour and basmati rice is ₹ 45.(b) the value of y, if CGST paid for detergent powder is ₹ 39.60(c) total amount to be paid (including GST) for the above bill.

Accepted Answer

(a) Given, total GST on wheat flour and basmati rice is ₹ 45.Total price of wheat flour = 35 × 5 = ₹ 175.Total price of basmati rice = 180 × 5 = ₹ 900.Given,Total GST on wheat flour and basmati rice is ₹ 45.∴x100×175+5100×900=45⇒1.75x+45=45⇒1.75x=45−45⇒1.75x=0⇒x=0.\therefore \dfrac{x}{100} \times 175 + \dfrac{5}{100} \times 900 = 45 \[1em] \Rightarrow 1.75x + 45 = 45 \[1em] \Rightarrow 1.75x = 45 - 45 \[1em] \Rightarrow 1.75x = 0 \[1em] \Rightarrow x = 0.∴100x​×175+1005​×900=45⇒1.75x+45=45⇒1.75x=45−45⇒1.75x=0⇒x=0.Hence, x = 0%.(b) From table,Total price of surf excel detergent = 220 × y = ₹ 220y.GST on detergent powder = 18%CGST = GST2=182\dfrac{\text{GST}}{2} = \dfrac{18}{2}2GST​=218​ = 9%.Given,CGST paid for detergent powder is ₹ 39.60∴9100×220y=39.60⇒9×220y=39.60×100⇒y=39.60×1009×220⇒y=39601980=2.\therefore \dfrac{9}{100} \times 220y = 39.60 \[1em] \Rightarrow 9 \times 220y = 39.60 \times 100 \[1em] \Rightarrow y = \dfrac{39.60 \times 100}{9 \times 220} \[1em] \Rightarrow y = \dfrac{3960}{1980} = 2.∴1009​×220y=39.60⇒9×220y=39.60×100⇒y=9×22039.60×100​⇒y=19803960​=2.Hence, y = 2 kg.(c) For wheat flour,Total price = 35 × 5 = ₹ 175GST = 0%So, total price = ₹ 175.For basmati rice,Total price = 180 × 5 = ₹ 900GST = 5%So, total price = ₹ 900 + 5100×900\dfrac{5}{100} \times 9001005​×900 = ₹ 900 + ₹ 45 = ₹ 945.For detergent powder,Total price = 220 × 2 = ₹ 440GST = 18%So, total price = ₹ 440 + 18100×440\dfrac{18}{100} \times 44010018​×440 = ₹ 440 + ₹ 79.20 = ₹ 519.20Total bill = ₹ 175 + ₹ 945 + ₹ 519.20 = ₹ 1639.20Hence, total bill = ₹ 1639.20

Question 2

Question 79Amit deposited ₹ 600 per month in a recurring deposit account. The bank pays a simple interest of 12% p.a. Calculate the:(a) number of monthly installments Amit deposits to get a maturity amount of ₹ 11826?(b) total interest paid by the bank.(c) total amount deposited by him.

Accepted Answer

(a) Let money be deposited for n months.By formula,M.V. = P × n + P×n(n+1)2×12×r100\dfrac{P \times n(n + 1)}{2 \times 12} \times \dfrac{r}{100}2×12P×n(n+1)​×100r​Substituting values we get :⇒11826=600×n+600×n(n+1)2×12×12100⇒11826=600n+600(n2+n)200⇒11826=600n+3(n2+n)⇒11826=600n+3n2+3n⇒3n2+603n=11826⇒3(n2+201n)=11826⇒n2+201n=118263⇒n2+201n=3942⇒n2+201n−3942=0⇒n2+219n−18n−3942=0⇒n(n+219)−18(n+219)=0⇒(n−18)(n+219)=0⇒n−18=0 or n+219=0⇒n=18 or n=−219.\Rightarrow 11826 = 600 \times n + \dfrac{600 \times n(n + 1)}{2 \times 12} \times \dfrac{12}{100} \[1em] \Rightarrow 11826 = 600n + \dfrac{600(n^2 + n)}{200} \[1em] \Rightarrow 11826 = 600n + 3(n^2 + n) \[1em] \Rightarrow 11826 = 600n + 3n^2 + 3n \[1em] \Rightarrow 3n^2 + 603n = 11826 \[1em] \Rightarrow 3(n^2 + 201n) = 11826 \[1em] \Rightarrow n^2 + 201n = \dfrac{11826}{3} \[1em] \Rightarrow n^2 + 201n = 3942 \[1em] \Rightarrow n^2 + 201n - 3942 = 0 \[1em] \Rightarrow n^2 + 219n - 18n - 3942 = 0 \[1em] \Rightarrow n(n + 219) - 18(n + 219) = 0 \[1em] \Rightarrow (n - 18)(n + 219) = 0 \[1em] \Rightarrow n - 18 = 0 \text{ or } n + 219 = 0 \[1em] \Rightarrow n = 18 \text{ or } n = -219.⇒11826=600×n+2×12600×n(n+1)​×10012​⇒11826=600n+200600(n2+n)​⇒11826=600n+3(n2+n)⇒11826=600n+3n2+3n⇒3n2+603n=11826⇒3(n2+201n)=11826⇒n2+201n=311826​⇒n2+201n=3942⇒n2+201n−3942=0⇒n2+219n−18n−3942=0⇒n(n+219)−18(n+219)=0⇒(n−18)(n+219)=0⇒n−18=0 or n+219=0⇒n=18 or n=−219.Since, no. of months cannot be negative.∴ n = 18.Hence, number of monthly installments = 18.(b) By formula,Interest = P×n(n+1)2×12×r100\dfrac{P \times n(n + 1)}{2 \times 12} \times \dfrac{r}{100}2×12P×n(n+1)​×100r​Substituting values we get :Interest =600×18(18+1)2×12×12100=600×18×19200=3×18×19=₹ 1026.\text{Interest } = \dfrac{600 \times 18(18 + 1)}{2 \times 12} \times \dfrac{12}{100} \[1em] = \dfrac{600 \times 18 \times 19}{200} \[1em] = 3 \times 18 \times 19 \[1em] = \text{₹ 1026}.Interest =2×12600×18(18+1)​×10012​=200600×18×19​=3×18×19=₹ 1026.Hence, total interest paid = ₹ 1026.(c) Total amount deposited by Amit = P × n= ₹ 600 × 18= ₹ 10800.Hence, total amount deposited by Amit = ₹ 10800.

Question 3

Question 80Aman has 500, ₹ 100 shares of a company quoted at ₹ 120, paying a 10% dividend. When the share price rises to ₹ 200 each, he sells all his shares. He invests half of the sale proceeds in ₹ 10, 12% shares at ₹ 25, and the remaining sale proceeds in ₹ 400, 9% shares at ₹ 500.Find his:(a) sales proceeds.(b) investment in ₹ 10, 12% shares at ₹ 25.(c) original income.(d) change in income.

Accepted Answer

(a) No. of shares Aman sells = 500Aman sells the share when they rise to ₹ 200.Sale proceeds = 500 × ₹ 200 = ₹ 1,00,000.Hence, sale proceeds = ₹ 1,00,000.(b) Given,Aman invests half of the sale proceeds in ₹ 10, 12% shares at ₹ 25.∴ Investment = Sale proceeds2=1000002\dfrac{	ext{Sale proceeds}}{2} = \dfrac{100000}{2}2Sale proceeds​=2100000​ = ₹ 50,000.Hence, investment in ₹ 10, 12% shares at ₹ 25 = ₹ 50,000.(c) By formula,Income = No. of shares × Rate of dividend100\dfrac{	ext{Rate of dividend}}{100}100Rate of dividend​ × Nominal value of share= 500 × 10100\dfrac{10}{100}10010​ × 100= ₹ 5,000.Hence, original income = ₹ 5,000.(d) Aman invests ₹ 50,000 in each of the new shares.For 1st share :N.V. = ₹ 10Dividend = 12%M.V. = ₹ 25No. of shares bought = 5000025\dfrac{50000}{25}2550000​ = 2000.Income = No. of shares × Rate of dividend100\dfrac{	ext{Rate of dividend}}{100}100Rate of dividend​ × Nominal value of share= 2000 × 12100×10\dfrac{12}{100} 	imes 1010012​×10= ₹ 2400.For 2nd share :N.V. = ₹ 400Dividend = 9%M.V. = ₹ 500No. of shares bought = 50000500\dfrac{50000}{500}50050000​ = 100.Income = No. of shares × Rate of dividend100\dfrac{	ext{Rate of dividend}}{100}100Rate of dividend​ × Nominal value of share= 100 × 9100×400\dfrac{9}{100} 	imes 4001009​×400= ₹ 3,600.New income = ₹ 3,600 + ₹ 2,400 = ₹ 6000Change in income = New income - Original income = ₹ 6,000 - ₹ 5,000 = ₹ 1,000.Hence, change in income = ₹ 1,000 (increase).

Question 4

Question 81Solve the following inequation.11+3x5≥3−x>−32\dfrac{11 + 3x}{5} \ge 3 - x \gt -\dfrac{3}{2}511+3x​≥3−x>−23​, x ∈ R(a) Write the solution set.(b) Represent the solution on the number line.

Accepted Answer

Given,11+3x5≥3−x>−32\dfrac{11 + 3x}{5} \ge 3 - x \gt -\dfrac{3}{2}511+3x≥3−x>−23Solving L.H.S. of the equation,⇒11+3x5≥3−x⇒11+3x≥5(3−x)⇒11+3x≥15−5x⇒3x+5x≥15−11⇒8x≥4⇒x≥48⇒x≥12 ...........(1)\Rightarrow \dfrac{11 + 3x}{5} \ge 3 - x \[1em] \Rightarrow 11 + 3x \ge 5(3 - x) \[1em] \Rightarrow 11 + 3x \ge 15 - 5x \[1em] \Rightarrow 3x + 5x \ge 15 - 11 \[1em] \Rightarrow 8x \ge 4 \[1em] \Rightarrow x \ge \dfrac{4}{8} \[1em] \Rightarrow x \ge \dfrac{1}{2} \text{ ...........(1)}⇒511+3x≥3−x⇒11+3x≥5(3−x)⇒11+3x≥15−5x⇒3x+5x≥15−11⇒8x≥4⇒x≥84⇒x≥21 ...........(1)Solving R.H.S. of the equation,⇒3−x>−32⇒x<3+32⇒x<6+32⇒x<92 .......(2)\Rightarrow 3 - x \gt -\dfrac{3}{2} \[1em] \Rightarrow x \lt 3 + \dfrac{3}{2} \[1em] \Rightarrow x \lt \dfrac{6 + 3}{2} \[1em] \Rightarrow x \lt \dfrac{9}{2} \text{ .......(2)}⇒3−x>−23⇒x<3+23⇒x<26+3⇒x<29 .......(2)From equation (1) and (2), we get :⇒12≤x<92\Rightarrow \dfrac{1}{2} \le x \lt \dfrac{9}{2}⇒21≤x<29Hence, solution set = {x:12≤x<92,x∈Rx : \dfrac{1}{2} \le x \lt \dfrac{9}{2}, x ∈ Rx:21≤x<29,x∈R}

Question 5

Question 82Determine whether the following quadratic equation has real roots.5𝑥2 − 9𝑥 + 4 = 0(a) Give reasons for your answer.(b) If the equation has real roots, identify them.

Accepted Answer

(a) Given, equation : 5𝑥2 − 9𝑥 + 4 = 0Comparing above equation with ax2 + bx + c = 0, we get :a = 5, b = -9 and c = 4.Discriminant (D) = b2 - 4ac = (-9)2 - 4 × 5 × 4 = 81 - 80 = 1.Since, D > 0 and a perfect square.Hence, equation 5𝑥2 − 9𝑥 + 4 = 0 has real roots.(b) Solving,⇒ 5𝑥2 − 9𝑥 + 4 = 0⇒ 5x2 - 5x - 4x + 4 = 0⇒ 5x(x - 1) - 4(x - 1) = 0⇒ (5x - 4)(x - 1) = 0⇒ 5x - 4 = 0 or x - 1 = 0⇒ 5x = 4 or x = 1⇒ x = 45\dfrac{4}{5}54​ or x = 1.Hence, roots of the equation are 1 and 45\dfrac{4}{5}54​.

Question 6

Question 83The profit in rupees in a local restaurant and the number of customers who visited the restaurant are tabulated below for each week for one month.Week numberWeek 1Week 2Week 3Week 4Number of customers14005600x3212Profit in ₹2800011200032140yFind :(a) if the number of customers and profit per week in continued proportion or not? Justify your answer.(b) the value of x and y.

Accepted Answer

(a) Since,140028000=5600112000=120\dfrac{1400}{28000} = \dfrac{5600}{112000} = \dfrac{1}{20}280001400​=1120005600​=201​.Hence, number of customers and profit per week are in proportion but not in continued proportion.(b) As,Number of customers and profit per week are in proportion.∴x32140=120⇒x=32140×120=1607\therefore \dfrac{x}{32140} = \dfrac{1}{20} \[1em] \Rightarrow x = 32140 \times \dfrac{1}{20} = 1607 \[1em]∴32140x​=201​⇒x=32140×201​=1607and∴3212y=120⇒y=3212×20=64240.\therefore \dfrac{3212}{y} = \dfrac{1}{20} \[1em] \Rightarrow y = 3212 \times 20 = 64240.∴y3212​=201​⇒y=3212×20=64240.Hence, x = 1607 and y = 64240.

Question 7

Question 84Given, 9𝑥2 - 4 is a factor of 9𝑥3 - m𝑥2 - n𝑥 + 8 :(a) find the value of m and n using the remainder and factor theorem.(b) factorise the given polynomial completely.

Accepted Answer

(a) Simplifying 9𝑥2 - 4, we get :⇒ 9𝑥2 - 4⇒ (3𝑥)2 - 22⇒ (3𝑥 + 2)(3𝑥 - 2).We know that,If (x - a) is a factor of f(x), then f(a) = 0.Given,9𝑥2 - 4 is a factor of 9𝑥3 - m𝑥2 - n𝑥 + 8.∴ (3𝑥 + 2) and (3𝑥 - 2) are the factors of 9𝑥3 - m𝑥2 - n𝑥 + 8.⇒ 3𝑥 + 2 = 0⇒ 3𝑥 = -2⇒ 𝑥 = −23-\dfrac{2}{3}−32​Substituting x = −23-\dfrac{2}{3}−32​ in 9𝑥3 - mx2 - nx + 8, we get remainder = 0.⇒9×(−23)3−m×(−23)2−n×(−23)+8=0⇒9×−827−m×49+2n3+8=0⇒−83−4m9+2n3+8=0⇒−24−4m+6n+729=0⇒6n−4m+489=0⇒6n−4m+48=0⇒2(3n−2m+24)=0⇒3n−2m+24=0⇒3n−2m=−24 ..........(1)\Rightarrow 9 \times \Big(-\dfrac{2}{3}\Big)^3 - m \times \Big(-\dfrac{2}{3}\Big)^2 - n \times \Big(-\dfrac{2}{3}\Big) + 8 = 0 \[1em] \Rightarrow 9 \times -\dfrac{8}{27} - m \times \dfrac{4}{9} + \dfrac{2n}{3} + 8 = 0 \[1em] \Rightarrow -\dfrac{8}{3} - \dfrac{4m}{9} + \dfrac{2n}{3} + 8 = 0 \[1em] \Rightarrow \dfrac{-24 - 4m + 6n + 72}{9} = 0 \[1em] \Rightarrow \dfrac{6n - 4m + 48}{9} = 0 \[1em] \Rightarrow 6n - 4m + 48 = 0 \[1em] \Rightarrow 2(3n - 2m + 24) = 0 \[1em] \Rightarrow 3n - 2m + 24 = 0 \[1em] \Rightarrow 3n - 2m = -24 \text{ ..........(1)}⇒9×(−32​)3−m×(−32​)2−n×(−32​)+8=0⇒9×−278​−m×94​+32n​+8=0⇒−38​−94m​+32n​+8=0⇒9−24−4m+6n+72​=0⇒96n−4m+48​=0⇒6n−4m+48=0⇒2(3n−2m+24)=0⇒3n−2m+24=0⇒3n−2m=−24 ..........(1)⇒ 3𝑥 - 2 = 0⇒ 3𝑥 = 2⇒ 𝑥 = 23\dfrac{2}{3}32​Substituting 𝑥 = 23\dfrac{2}{3}32​ in 9𝑥3 - m𝑥2 - n𝑥 + 8, we get remainder = 0.⇒9×(23)3−m×(23)2−n×(23)+8=0⇒9×827−m×49−2n3+8=0⇒83−4m9−2n3+8=0⇒24−4m−6n+729=0⇒96−6n−4m9=0⇒96−6n−4m=0⇒4m+6n=96⇒2(2m+3n)=96⇒2m+3n=962⇒2m+3n=48 ..........(2)\Rightarrow 9 \times \Big(\dfrac{2}{3}\Big)^3 - m \times \Big(\dfrac{2}{3}\Big)^2 - n \times \Big(\dfrac{2}{3}\Big) + 8 = 0 \[1em] \Rightarrow 9 \times \dfrac{8}{27} - m \times \dfrac{4}{9} - \dfrac{2n}{3} + 8 = 0 \[1em] \Rightarrow \dfrac{8}{3} - \dfrac{4m}{9} - \dfrac{2n}{3} + 8 = 0 \[1em] \Rightarrow \dfrac{24 - 4m - 6n + 72}{9} = 0 \[1em] \Rightarrow \dfrac{96 - 6n - 4m}{9} = 0 \[1em] \Rightarrow 96 - 6n - 4m = 0 \[1em] \Rightarrow 4m + 6n = 96 \[1em] \Rightarrow 2(2m + 3n) = 96 \[1em] \Rightarrow 2m + 3n = \dfrac{96}{2} \[1em] \Rightarrow 2m + 3n = 48 \text{ ..........(2)}⇒9×(32​)3−m×(32​)2−n×(32​)+8=0⇒9×278​−m×94​−32n​+8=0⇒38​−94m​−32n​+8=0⇒924−4m−6n+72​=0⇒996−6n−4m​=0⇒96−6n−4m=0⇒4m+6n=96⇒2(2m+3n)=96⇒2m+3n=296​⇒2m+3n=48 ..........(2)Adding equation (1) and (2), we get :⇒ 3n - 2m + 2m + 3n = -24 + 48⇒ 6n = 24⇒ n = 246\dfrac{24}{6}624​⇒ n = 4.Substituting value of n in equation (1), we get :⇒ 3(4) - 2m = -24⇒ 12 - 2m = -24⇒ -2m = -24 - 12⇒ -2m = -36⇒ m = −36−2\dfrac{-36}{-2}−2−36​ = 18.Hence, m = 18 and n = 4.(b) Substituting value of m and n in 9𝑥3 - m𝑥2 - n𝑥 + 8, we get :9𝑥3 - 18𝑥2 - 4𝑥 + 8Dividing 9𝑥3 - 18𝑥2 - 4𝑥 + 8 by 9𝑥2 - 4, we get :9x2−4)x−29x2−4)9x3−18x2−4x+8‾9x2−4))+−9x3−18x2−+4x‾9x2−4+9x3−−18x2−4x+89x2−4)+9x3−−+18x2−4x+−8‾9x2−4)x3−2x2(31)x×\begin{array}{l} \phantom{9x^2 - 4)}{\quad x - 2} \ 9x^2 - 4\overline{\smash{\big)}\quad 9x^3 - 18x^2 - 4x + 8} \ \phantom{9x^2 - 4)}\phantom{)}\underline{\underset{-}{+}9x^3 \phantom{- 18x^2} \underset{+}{-}4x} \ \phantom{{9x^2 - 4}{+9x^3 - }}-18x^2 \phantom{- 4x} + 8 \ \phantom{{9x^2 - 4)}{+9x^3 - }}\underline{\underset{+}{-}18x^2 \phantom{- 4x} \underset{-}{+} 8} \ \phantom{{9x^2 - 4)}{x^3-2x^{2}(31)}{x}}\times \end{array}9x2−4)x−29x2−4)9x3−18x2−4x+8​9x2−4))−+​9x3−18x2+−​4x​9x2−4+9x3−−18x2−4x+89x2−4)+9x3−+−​18x2−4x−+​8​9x2−4)x3−2x2(31)x×​∴ 9𝑥3 - 18𝑥2 - 4𝑥 + 8 = (9𝑥2 - 4)(𝑥 - 2)= (3𝑥 + 2)(3𝑥 - 2)(𝑥 - 2).Hence, factors are (3𝑥 + 2), (3𝑥 - 2) and (𝑥 - 2).

Question 8

Question 85The marks scored by 100 students are given below:Marks scoredNo. of students0-10410-20520-30930-40740-501350-601260-701570-801180-901490-10010A student in the class is selected at random. Find the probability that the student has scored:(a) less than 20(b) below 60 but 30 or more(c) more than or equal to 70(d) above 89.

Accepted Answer

Marks scoredNo. of studentsCumulative frequency0-104410-205920-3091830-4072540-50133850-60125060-70156570-80117680-90149090-10010100(a) By formula,Probability that student has scored less than 20= No. of students who scored less than 20Total no. of students=9100\dfrac{	ext{No. of students who scored less than 20}}{	ext{Total no. of students}} = \dfrac{9}{100}Total no. of studentsNo. of students who scored less than 20​=1009​.Hence, probability that the student has scored less than 20 = 9100\dfrac{9}{100}1009​.(b) From table,No. of students who scored less than 60 = 50No. of students who scored less than 30 = 18∴ No. of students who score below 60 but 30 or more = 50 - 18 = 32.Probability that student has scored below 60 but 30 or more = No. of students scoring between 60 and 30Total no. of students=32100=825\dfrac{	ext{No. of students scoring between 60 and 30}}{	ext{Total no. of students}} = \dfrac{32}{100} = \dfrac{8}{25}Total no. of studentsNo. of students scoring between 60 and 30​=10032​=258​.Hence, probability that the student has scored below 60 but 30 or more = 825\dfrac{8}{25}258​.(c) From table,No. of students who scored less than 70 = 65Total no. of students = 100∴ No. of students who score more than or equal to 70 = 100 - 65 = 35.Probability that student has scored more than or equal to 70 = No. of students scoring 70 or moreTotal no. of students=35100=720\dfrac{	ext{No. of students scoring 70 or more}}{	ext{Total no. of students}} = \dfrac{35}{100} = \dfrac{7}{20}Total no. of studentsNo. of students scoring 70 or more​=10035​=207​.Hence, probability that the student has scored more than or equal to 70 = 720\dfrac{7}{20}207​.(d) No. of students those who have scored more than 89 = No. of students who has scored between 90-100 = 10.Probability that student has scored more than 89 = No. of students scoring >89Total no. of students=10100=110\dfrac{	ext{No. of students scoring 	extgreater 89}}{	ext{Total no. of students}} = \dfrac{10}{100} = \dfrac{1}{10}Total no. of studentsNo. of students scoring >89​=10010​=101​.Hence, probability that the student has scored more than 89 = 110\dfrac{1}{10}101​.

Question 9

Question 86Given, matrix A = [x1y2] and B=[xx−2]\begin{bmatrix*}[r] x & 1 \ y & 2 \end{bmatrix*} 	ext{ and } B = \begin{bmatrix*}[r] x \ x - 2 \end{bmatrix*}[xy​12​] and B=[xx−2​] such that AB is a null matrix. Find :(a) order of the null matrix(b) possible values of x and y.

Accepted Answer

Given,AB is a null matrix.Let AB = X, where X is a null matrix of order a × b.⇒ AB = X⇒ A2 × 2 × B2 × 1 = Xa × bWe know that,The resultant matrix has no. of rows equal to the rows in the first matrix and no. of columns equal to the no. of columns in the second matrix.∴ a = 2 and b = 1.∴[x1y2]2×2[xx−2]2×1=[00]⇒[x2+1(x−2)xy+2(x−2)]=[00]⇒[x2+x−2xy+2x−4]=[00] .............(1)\therefore \begin{bmatrix*}[r] x & 1 \ y & 2 \end{bmatrix*}_{2 \times 2} \begin{bmatrix*}[r] x \ x - 2 \end{bmatrix*}_{2 \times 1} = \begin{bmatrix*}[r] 0 \ 0 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] x^2 + 1(x - 2) \ xy + 2(x - 2) \end{bmatrix*} = \begin{bmatrix*}[r] 0 \ 0 \end{bmatrix*} \[1em] \Rightarrow \begin{bmatrix*}[r] x^2 + x - 2 \ xy + 2x - 4 \end{bmatrix*} = \begin{bmatrix*}[r] 0 \ 0 \end{bmatrix*} \text{ .............(1)}∴[xy​12​]2×2​[xx−2​]2×1​=[00​]⇒[x2+1(x−2)xy+2(x−2)​]=[00​]⇒[x2+x−2xy+2x−4​]=[00​] .............(1)From equation (1) :⇒ x2 + x - 2 = 0⇒ x2 + 2x - x - 2 = 0⇒ x(x + 2) - 1(x + 2) = 0⇒ (x - 1)(x + 2) = 0⇒ x - 1 = 0 or x + 2 = 0⇒ x = 1 or x = -2.From equation (1) :⇒ xy + 2x - 4 = 0Substituting x = 1, we get :⇒ 1.y + 2.1 - 4 = 0⇒ y + 2 - 4 = 0⇒ y - 2 = 0⇒ y = 2.Substituting x = -2, we get :⇒ (-2).y + 2.(-2) - 4 = 0⇒ -2y - 4 - 4 = 0⇒ -2y - 8 = 0⇒ -2y = 8⇒ y = 8−2\dfrac{8}{-2}−28​ = -4.Hence, x = 1, y = 2 or x = -2, y = -4.

Question 10

Question 87The sum of a certain number of terms of the Arithmetic Progression (A.P.) 20, 17, 14, ....... is 65. Find the:(a) number of terms.(b) last term.

Accepted Answer

(a) Let no. of terms be n.By formula,Sum of A.P. = n2[2a+(n−1)d]\dfrac{n}{2}[2a + (n - 1)d]2n​[2a+(n−1)d]Substituting values we get :⇒65=n2[2×20+(n−1)×(−3)]⇒65=n2[40−3n+3]⇒65=n2[43−3n]⇒65×2=n(43−3n)⇒130=43n−3n2⇒3n2−43n+130=0⇒3n2−30n−13n+130=0⇒3n(n−10)−13(n−10)=0⇒(3n−13)(n−10)=0⇒3n−13=0 or n−10=0⇒3n=13 or n=10⇒n=133 or n=10.\Rightarrow 65 = \dfrac{n}{2}[2 \times 20 + (n - 1) \times (-3)] \[1em] \Rightarrow 65 = \dfrac{n}{2}[40 - 3n + 3] \[1em] \Rightarrow 65 = \dfrac{n}{2}[43 - 3n] \[1em] \Rightarrow 65 \times 2 = n(43 - 3n) \[1em] \Rightarrow 130 = 43n - 3n^2 \[1em] \Rightarrow 3n^2 - 43n + 130 = 0 \[1em] \Rightarrow 3n^2 - 30n - 13n + 130 = 0 \[1em] \Rightarrow 3n(n - 10) - 13(n - 10) = 0 \[1em] \Rightarrow (3n - 13)(n - 10) = 0 \[1em] \Rightarrow 3n - 13 = 0 \text{ or } n - 10 = 0 \[1em] \Rightarrow 3n = 13 \text{ or } n = 10 \[1em] \Rightarrow n = \dfrac{13}{3} \text{ or } n = 10.⇒65=2n​[2×20+(n−1)×(−3)]⇒65=2n​[40−3n+3]⇒65=2n​[43−3n]⇒65×2=n(43−3n)⇒130=43n−3n2⇒3n2−43n+130=0⇒3n2−30n−13n+130=0⇒3n(n−10)−13(n−10)=0⇒(3n−13)(n−10)=0⇒3n−13=0 or n−10=0⇒3n=13 or n=10⇒n=313​ or n=10.Since, no. of terms cannot be in fraction.∴ n = 10.Hence, no. of terms = 10.(b) By formula,Last term (l) = a + (n - 1)d= 20 + (10 - 1) × (-3)= 20 + 9 × (-3)= 20 - 27= -7.Hence, last term = -7.

Question 11

Question 88(a) Point P(2, -3) on reflection becomes P'(2, 3). Name the line of reflection (say L1).(b) Point P' is reflected to P'' along the line (𝐿2), which is perpendicular to the line 𝐿1 and passes through the point, which is invariant along both axes. Write the coordinates of P''.(c) Name and write the coordinates of the point of intersection of the lines 𝐿1 and 𝐿2.(d) Point P is reflected to P''' on reflection through the point named in the answer of part I of this question. Write the coordinates of P'''. Comment on the location of the points P'' and P'''.

Accepted Answer

(a) P(2, -3) ⇒ P'(2, 3)Since, sign of y-coordinate changes.∴ L1 = x-axisHence, point P becomes P' on reflection in x-axis.(b) Origin remains invariant on reflection along both the axes.L2 is perpendicular to L1.It means L2 is perpendicular to x-axis and passes through (0, 0).∴ L2 is y-axis.P'(2, 3) on reflection in y-axis becomes P''(-2, 3).Hence, coordinates of P'' = (-2, 3).(c) x-axis and y-axis intersect at origin.Hence, coordinates of intersection of lines L1 and L2 is (0, 0).(d) P(2, -3) on reflection in origin becomes P'''(-2, 3).Since, P'' and P''' have similar co-ordinates.Hence, P'' and P''' are coincident points.

Question 12

Question 89In the given figure, if the line segment AB is intercepted by the y-axis and x-axis at C and D, respectively, such that AC : AD = 1 : 4 and D is the midpoint of CB. Find the coordinates of D, C and B.

Accepted Answer

Let coordinates of C be (0, b) and D be (a, 0).Given,AC : AD = 1 : 4Let AC = x and AD = 4x.From figure,⇒ AD = AC + CD⇒ 4x = x + CD⇒ CD = 4x - x = 3x.AC : CD = 1 : 3.By section formula,⇒(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)⇒(0,b)=(1×a+3×−21+3,1×0+3×61+3)⇒(0,b)=(a−64,0+184)⇒(0,b)=(a−64,184)⇒a−64=0 and b=184⇒a−6=0 and b=92⇒a=6 and b=92.\Rightarrow (x, y) = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \[1em] \Rightarrow (0, b) = \Big(\dfrac{1 \times a + 3 \times -2}{1 + 3}, \dfrac{1 \times 0 + 3 \times 6}{1 + 3}\Big) \[1em] \Rightarrow (0, b) = \Big(\dfrac{a - 6}{4}, \dfrac{0 + 18}{4}\Big) \[1em] \Rightarrow (0, b) = \Big(\dfrac{a - 6}{4}, \dfrac{18}{4}\Big) \[1em] \Rightarrow \dfrac{a - 6}{4} = 0 \text{ and } b = \dfrac{18}{4} \[1em] \Rightarrow a - 6 = 0 \text{ and } b = \dfrac{9}{2} \[1em] \Rightarrow a = 6 \text{ and } b = \dfrac{9}{2}.⇒(x,y)=(m1​+m2​m1​x2​+m2​x1​​,m1​+m2​m1​y2​+m2​y1​​)⇒(0,b)=(1+31×a+3×−2​,1+31×0+3×6​)⇒(0,b)=(4a−6​,40+18​)⇒(0,b)=(4a−6​,418​)⇒4a−6​=0 and b=418​⇒a−6=0 and b=29​⇒a=6 and b=29​.C = (0 , b) = (0,92)\Big(0, \dfrac{9}{2}\Big)(0,29​) and D = (6, 0).Given, D is the mid-point of CB.∴(6,0)=(0+p2,92+q2)⇒(6,0)=(p2,9+2q2×2)⇒(6,0)=(p2,9+2q4)⇒p2=6 and 9+2q4=0⇒p=12 and 9+2q=0⇒p=12 and 2q=−9⇒p=12 and q=−92⇒B=(p,q)=(12,−92).\therefore (6, 0) = \Big(\dfrac{0 + p}{2}, \dfrac{\dfrac{9}{2} + q}{2}\Big) \[1em] \Rightarrow (6, 0) = \Big(\dfrac{p}{2}, \dfrac{9 + 2q}{2 \times 2}\Big) \[1em] \Rightarrow (6, 0) = \Big(\dfrac{p}{2}, \dfrac{9 + 2q}{4}\Big) \[1em] \Rightarrow \dfrac{p}{2} = 6 \text{ and } \dfrac{9 + 2q}{4} = 0 \[1em] \Rightarrow p = 12 \text{ and } 9 + 2q = 0 \[1em] \Rightarrow p = 12 \text{ and } 2q = -9 \[1em] \Rightarrow p = 12 \text{ and } q = -\dfrac{9}{2} \[1em] \Rightarrow B = (p, q) = \Big(12, -\dfrac{9}{2}\Big).∴(6,0)=(20+p​,229​+q​)⇒(6,0)=(2p​,2×29+2q​)⇒(6,0)=(2p​,49+2q​)⇒2p​=6 and 49+2q​=0⇒p=12 and 9+2q=0⇒p=12 and 2q=−9⇒p=12 and q=−29​⇒B=(p,q)=(12,−29​).Hence, coordinates of B = (12,−92),C=(0,92)\Big(12, -\dfrac{9}{2}\Big), C = \Big(0, \dfrac{9}{2}\Big)(12,−29​),C=(0,29​) and D = (6, 0).

Question 13

Question 90Find the equation of the straight line perpendicular to the line x + 2y = 4, which cuts an intercept of 2 units from the positive y-axis. Hence, find the intersection point of the two lines.

Accepted Answer

Given,Equation : x + 2y = 4⇒ 2y = -x + 4⇒ y = −12x+4-\dfrac{1}{2}x + 4−21​x+4Comparing above equation with y = mx + c we get :⇒ m = −12-\dfrac{1}{2}−21​Let slope of line perpendicular to line x + 2y = 4 be m1.We know that,Product of slope of perpendicular lines = -1.⇒m1×−12=−1⇒m1=−1×2−1=2.\Rightarrow m_1 \times -\dfrac{1}{2} = -1 \[1em] \Rightarrow m_1 = \dfrac{-1 \times 2}{-1} = 2.⇒m1​×−21​=−1⇒m1​=−1−1×2​=2.Substituting values we get :⇒ y = mx + c⇒ y = 2x + 2.Simultaneously solving equation :⇒ x + 2y = 4 .........(1)⇒ y = 2x + 2 .......(2)Substituting value of y from equation (2) in (1), we get :⇒ x + 2(2x + 2) = 4⇒ x + 4x + 4 = 4⇒ 5x = 4 - 4⇒ 5x = 0⇒ x = 05\dfrac{0}{5}50​ = 0.Substituting value of x in equation (2), we get :⇒ y = 2(0) + 2 = 2.Hence, equation of required line is y = 2x + 2 and point of intersection = (0, 2).

Question 14

Question 91While preparing a PowerPoint presentation, ∆ ABC is enlarged along the side BC to ∆ AB'C', as shown in the diagram, such that BC ∶ B'C' is 3 ∶ 5. Find :(a) AB ∶ BB'(b) length AB, if BB' = 4 cm.(c) Is ∆ ABC ~ ∆ AB'C' ? Justify your answer.(d) ar (∆ ABC) : ar (quad. BB'C'C).

Accepted Answer

Since, ∆ ABC is enlarged along the side BC to ∆ AB'C'.∴ ∆ ABC and ∆ AB'C' are similar triangles.(a) We know that,Ratio of corresponding sides of similar triangles are proportional.∴ABAB′=BCB′C′⇒ABAB′=35\therefore \dfrac{AB}{AB'} = \dfrac{BC}{B'C'} \[1em] \Rightarrow \dfrac{AB}{AB'} = \dfrac{3}{5}∴AB′AB​=B′C′BC​⇒AB′AB​=53​Let AB = 3x and AB' = 5xFrom figure,⇒ AB' = AB + BB'⇒ 5x = 3x + BB'⇒ BB' = 5x - 3x = 2x.⇒ AB : BB' = 3x : 2x = 3 : 2.Hence, AB : BB' = 3 : 2.(b) As,⇒ABBB′=32⇒AB4=32⇒AB=32×4⇒AB=6 cm.\Rightarrow \dfrac{AB}{BB'} = \dfrac{3}{2} \[1em] \Rightarrow \dfrac{AB}{4} = \dfrac{3}{2} \[1em] \Rightarrow AB = \dfrac{3}{2} \times 4 \[1em] \Rightarrow AB = 6 \text{ cm}.⇒BB′AB​=23​⇒4AB​=23​⇒AB=23​×4⇒AB=6 cm.Hence, AB = 6 cm.(c) Since, ∆ ABC is enlarged along the side BC to ∆ AB'C'.∴ BC || B'C'In ∆ ABC and ∆ AB'C',⇒ ∠BAC = ∠B'AC' (Common angle)⇒ ∠ABC = ∠AB'C' (Corresponding angle are equal)∴ ∆ ABC ~ ∆ AB'C' (By A.A. axiom).Hence, proved that ∆ ABC ~ ∆ AB'C'.(d) We know that,The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.∴Area of ∆ ABCArea of ∆ AB’C’=(ABAB′)2⇒Area of ∆ ABCArea of ∆ AB’C’=(35)2⇒Area of ∆ ABCArea of ∆ AB’C’=925.\therefore \dfrac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\dfrac{AB}{AB'}\Big)^2 \[1em] \Rightarrow \dfrac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\dfrac{3}{5}\Big)^2 \[1em] \Rightarrow \dfrac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \dfrac{9}{25}.∴Area of ∆ AB’C’Area of ∆ ABC​=(AB′AB​)2⇒Area of ∆ AB’C’Area of ∆ ABC​=(53​)2⇒Area of ∆ AB’C’Area of ∆ ABC​=259​.Let area of ∆ ABC = 9a and area of ∆ AB'C' = 25a.Area of quadrilateral BB'C'C = Area of ∆ AB'C' - Area of ∆ ABC = 25a - 9a = 16a.∴ Area of ∆ ABC : Area of quadrilateral BB'C'C = 9a : 16a = 9 : 16.Hence, area of ∆ ABC : area of quadrilateral BB'C'C = 9 : 16.

Question 15

Question 92The approximate volume of a human eye is 6.5 cm3. The volume of a laboratory model (excluding base and stand) of the human eye is 1404 cm3.(a) State whether the scale factor k is less than, equals to or greater than 1.(b) Calculate the:(i) value of k(ii) diameter of the human eye if the radius of the model is 7.2 cm.(iii) the external surface area of the human eye if the surface area of the model is 651.6 cm2.

Accepted Answer

(a) Scale factor is greater than 1 as volume of model is greater than the original eye.(b)(i) By formula,⇒k3=Volume of modelVolume of original human eye⇒k3=14046.5⇒k3=2161⇒k3=63⇒k=6.\Rightarrow k^3 = \dfrac{\text{Volume of model}}{\text{Volume of original human eye}} \[1em] \Rightarrow k^3 = \dfrac{1404}{6.5} \[1em] \Rightarrow k^3 = \dfrac{216}{1} \[1em] \Rightarrow k^3 = 6^3 \[1em] \Rightarrow k = 6.⇒k3=Volume of original human eyeVolume of model​⇒k3=6.51404​⇒k3=1216​⇒k3=63⇒k=6.Hence, scale factor (k) = 6.(ii) By formula,⇒k=Radius of modelRadius of human eye⇒6=7.2Radius of human eye⇒Radius of human eye=7.26=1.2 cm\Rightarrow k = \dfrac{\text{Radius of model}}{\text{Radius of human eye}} \[1em] \Rightarrow 6 = \dfrac{7.2}{\text{Radius of human eye}} \[1em] \Rightarrow \text{Radius of human eye} = \dfrac{7.2}{6} = 1.2 \text{ cm}⇒k=Radius of human eyeRadius of model​⇒6=Radius of human eye7.2​⇒Radius of human eye=67.2​=1.2 cmDiameter of human eye = 1.2 × 2 = 2.4 cmHence, diameter of human eye = 2.4 cm.(iii) By formula,⇒k2=Surface area of modelSurface area of human eye⇒62=651.6Surface area of human eye⇒Surface area of human eye=651.662⇒Surface area of human eye=651.636=18.1 cm2.\Rightarrow k^2 = \dfrac{\text{Surface area of model}}{\text{Surface area of human eye}} \[1em] \Rightarrow 6^2 = \dfrac{651.6}{\text{Surface area of human eye}} \[1em] \Rightarrow \text{Surface area of human eye} = \dfrac{651.6}{6^2} \[1em] \Rightarrow \text{Surface area of human eye} = \dfrac{651.6}{36} = 18.1 \text{ cm}^2.⇒k2=Surface area of human eyeSurface area of model​⇒62=Surface area of human eye651.6​⇒Surface area of human eye=62651.6​⇒Surface area of human eye=36651.6​=18.1 cm2.Hence, surface area of human eye = 18.1 cm2.

Question 16

Question 93In the adjoining diagram PQ, PR and ST are the tangents to the circle with centre O and radius 7 cm. Given OP = 25 cm. Find :(a) length of ST(b) value of ∠OPQ, i.e. θ(c) ∠QUR, in nearest degree

Accepted Answer

(a) We know that,The radius of a circle and tangent are perpendicular at the point of contact.∴ ∠PQO = 90°In right-angled triangle PQO,⇒ PO2 = PQ2 + OQ2⇒ 252 = PQ2 + 72⇒ 625 = PQ2 + 49⇒ PQ2 = 625 - 49⇒ PQ2 = 576⇒ PQ = 576\sqrt{576}576​ = 24 cm.In △ POQ,⇒ tan θ = OQPQ\dfrac{OQ}{PQ}PQOQ​⇒ tan θ = 724\dfrac{7}{24}247​ ...............(1)From figure,⇒ ∠PAS = 90°⇒ PA = OP - OA = 25 - 7 = 18 cm.⇒ tan θ = ASPA\dfrac{AS}{PA}PAAS​⇒ tan θ = AS18\dfrac{AS}{18}18AS​ ...........(2)From equation (1) and (2), we get :⇒724=AS18⇒AS=724×18⇒AS=7×34=214 cm.\Rightarrow \dfrac{7}{24} = \dfrac{AS}{18} \[1em] \Rightarrow AS = \dfrac{7}{24} \times 18 \[1em] \Rightarrow AS = \dfrac{7 \times 3}{4} = \dfrac{21}{4} \text{ cm}.⇒247​=18AS​⇒AS=247​×18⇒AS=47×3​=421​ cm.From figure,⇒ ST = 2 × AS = 2×214=2122 \times \dfrac{21}{4} = \dfrac{21}{2}2×421​=221​ = 10.5 cmHence, ST = 10.5 cm.(b) From equation (1),⇒ tan θ = 724\dfrac{7}{24}247​⇒ tan θ = 0.292⇒ tan θ = tan 16° 16'⇒ θ = 16° 16'.Hence, θ = 16° 16'.(c) In △ POQ,⇒ ∠POQ + ∠PQO + ∠QPO = 180°⇒ ∠POQ + 90° + 16° 16' = 180°⇒ ∠POQ + 106° 16' = 180°⇒ ∠POQ = 180° - 106° 16' = 73° 44' = 74°.We know that,Tangent from an external point to the circle are equal in length.∴ PQ = PR.Also,OR = OQ (Both equal to radius of circle)In △ POQ and △ POR,⇒ PQ = PR (Proved above)⇒ OQ = OR (Radius of same circle)⇒ PO = PO (Common side)∴ △ POQ ≅ △ POR (By S.S.S. axiom)We know that,Corresponding parts of congruent triangles are equal.∴ ∠POR = ∠POQ = 74°From figure,⇒ ∠QOR = ∠POR + ∠POQ = 74° + 74° = 148°.We know that,The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.⇒ ∠QOR = 2∠QUR⇒ ∠QUR = ∠QOR2=148°2\dfrac{∠QOR}{2} = \dfrac{148°}{2}2∠QOR​=2148°​ = 74°.Hence, ∠QUR = 74°.

Question 17

Question 94Use ruler and compass to answer this question. Construct a triangle ABC where AB = 5.5 cm, BC = 4.5 cm and angle ABC = 135°. Construct the circumcircle to the triangle ABC. Measure and write down the length of AC.

Accepted Answer

Steps of construction :Draw a line segment BC = 4.5 cmConstruct XB such that ∠XBC = 135°.Cut AB = 5.5 cm from XB.Join and measure AC.Draw PQ and RS the perpendicular bisectors of BC and AB.Mark point O the intersection of PQ and RS.With O as center and radius as OA, draw a circle touching the vertices A, B and C.On measuring AC = 9.1 cm and radius = 6.5 cm.

Question 18

Question 95The curved surface area of a right circular cone is half of another right circular cone. If the ratio of their slant heights is 2 : 1 and that of their volumes is 3 : 1, find ratio of their:(a) radii(b) heights

Accepted Answer

(a) Let radius of smaller and larger cone be r and R, respectively.Let height of smaller and larger cone be h and H, respectively.Let slant height of smaller and larger cone be l and L respectively.Given,Ratio of their slant heights is 2 : 1.∴ l : L = 2 : 1.Given,Curved surface area of a right circular cone is half of the other right circular cone.∴ πrl = 12πRL\dfrac{1}{2}πRL21​πRL⇒ rl = 12RL\dfrac{1}{2}RL21​RL⇒ rR=L2l=12×12\dfrac{r}{R} = \dfrac{L}{2l} = \dfrac{1}{2} \times \dfrac{1}{2}Rr​=2lL​=21​×21​⇒ rR=14\dfrac{r}{R} = \dfrac{1}{4}Rr​=41​⇒ r : R = 1 : 4Hence, ratio of the radii = 1 : 4.(b) Let volume of cone with smaller curved surface area be v and that with larger curved surface area be V.Given,Volumes are in the ratio 3 : 1.∴ v : V = 3 : 1⇒Volume of larger CSA coneVolume of smaller CSA cone=Vv⇒13πR2H13πr2h=13⇒4212×Hh=13⇒161×Hh=13⇒Hh=13×16⇒Hh=148⇒hH=481.\Rightarrow \dfrac{\text{Volume of larger CSA cone}}{\text{Volume of smaller CSA cone}} = \dfrac{V}{v} \[1em] \Rightarrow \dfrac{\dfrac{1}{3}πR^2H}{\dfrac{1}{3}πr^2h} = \dfrac{1}{3} \[1em] \Rightarrow \dfrac{4^2}{1^2} \times \dfrac{H}{h} = \dfrac{1}{3} \[1em] \Rightarrow \dfrac{16}{1} \times \dfrac{H}{h} = \dfrac{1}{3} \[1em] \Rightarrow \dfrac{H}{h} = \dfrac{1}{3 \times 16} \[1em] \Rightarrow \dfrac{H}{h} = \dfrac{1}{48} \[1em] \Rightarrow \dfrac{h}{H} = \dfrac{48}{1}.⇒Volume of smaller CSA coneVolume of larger CSA cone​=vV​⇒31​πr2h31​πR2H​=31​⇒1242​×hH​=31​⇒116​×hH​=31​⇒hH​=3×161​⇒hH​=481​⇒Hh​=148​.Hence, ratio of heights = 48 : 1.

Question 19

Question 96A cylindrical drum is unloaded from a truck by rolling it down along a wooden plank. The length of the plank is 10 m and it is making an angle of 10° with the horizontal ground. Find the height from which the cylindrical drum was rolled down. Give your answer correct to 3 significant figures.

Accepted Answer

Let AC be the plank and BC be the height from where the drum is rolled down.⇒ sin θ = PerpendicularHypotenuse\dfrac{	ext{Perpendicular}}{	ext{Hypotenuse}}HypotenusePerpendicular​⇒ sin 10° = BCAC\dfrac{BC}{AC}ACBC​⇒ 0.1736 = BC10\dfrac{BC}{10}10BC​⇒ BC = 1.736 ~ 1.74 mHence, height from which the cylindrical drum was rolled down = 1.74 m.

Question 20

Question 97The data given below shows the marks of 12 students in a test, arranged in ascending order :2, 3, 3, 3, 4, x, x + 2, 8, p, q, 8, 9.If the given value of the median and mode is 6 and 8 respectively, then find the values of x, p, q.

Accepted Answer

By formula,Median = n2 th term+(n2+1) th term2\dfrac{\dfrac{n}{2}\text{ th term} + \Big(\dfrac{n}{2} + 1\Big)\text{ th term}}{2}22n​ th term+(2n​+1) th term​Substituting values we get :⇒6=122 th term+(122+1) th term2⇒6×2=6th term + 7th term⇒12=x+x+2⇒12=2x+2⇒12−2=2x⇒2x=10⇒x=102=5.\Rightarrow 6 = \dfrac{\dfrac{12}{2}\text{ th term} + \Big(\dfrac{12}{2} + 1\Big)\text{ th term}}{2} \[1em] \Rightarrow 6 \times 2 = \text{6th term + 7th term} \[1em] \Rightarrow 12 = x + x + 2 \[1em] \Rightarrow 12 = 2x + 2 \[1em] \Rightarrow 12 - 2 = 2x \[1em] \Rightarrow 2x = 10 \[1em] \Rightarrow x = \dfrac{10}{2} = 5.⇒6=2212​ th term+(212​+1) th term​⇒6×2=6th term + 7th term⇒12=x+x+2⇒12=2x+2⇒12−2=2x⇒2x=10⇒x=210​=5.Numbers :2, 3, 3, 3, 4, 5, 7, 8, p, q, 8, 9.Since, mode = 8, it means 8 occurs for the most times in the series.Since, 3 occurs 3 times in the series,∴ 8 must occur for atleast 4 times in order to be the mode.∴ p = q = 8.Hence, x = 5, p = 8 and q = 8.

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