Question 1

Question 50A shopkeeper marked a pressure cooker at ₹ 1800. The rate of GST on pressure cooker is 12%. The customer has only ₹ 1792 with him and he requests the shopkeeper to reduce the price so that he can buy the cooker in ₹ 1792. What percent discount must the shopkeeper give ?

Accepted Answer

M.P. of pressure cooker = ₹ 1800.Let after discount the price of cooker be ₹ x.GST on pressure cooker = 12%According to question,∴x+12100×x=1792⇒x+0.12x=1792⇒1.12x=1792⇒x=17921.12=₹ 1600.\therefore x + \dfrac{12}{100} \times x = 1792 \[1em] \Rightarrow x + 0.12x = 1792 \[1em] \Rightarrow 1.12x = 1792 \[1em] \Rightarrow x = \dfrac{1792}{1.12} = \text{₹ 1600}.∴x+10012​×x=1792⇒x+0.12x=1792⇒1.12x=1792⇒x=1.121792​=₹ 1600.Discount given by shopkeeper = ₹ 1800 - ₹ 1600 = ₹ 200.Discount % = DiscountM.P.×100\dfrac{\text{Discount}}{\text{M.P.}} \times 100M.P.Discount​×100=2001800×100=19×100=11.11= \dfrac{200}{1800} \times 100 \[1em] = \dfrac{1}{9} \times 100 \[1em] = 11.11 %.=1800200​×100=91​×100=11.11Hence, discount offered = 11.11 %.

Question 2

Question 51A man opened a recurring deposit account in a branch of PNB. The man deposits certain amount of money per month such that after 2 years, the interest accumulated is equal to his monthly deposits. Find the rate of interest per annum that the bank was paying for the recurring deposit account.

Accepted Answer

Given,Time (n) = 2 years or 24 monthsRate = r% (let)P = ₹ x/monthI = ₹ xBy formula,I = 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 :⇒x=x×24×(24+1)2×12×r100⇒x=x×24×2524×r100⇒1=r4⇒r=4\Rightarrow x = \dfrac{x \times 24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \[1em] \Rightarrow x = \dfrac{x \times 24 \times 25}{24} \times \dfrac{r}{100} \[1em] \Rightarrow 1 = \dfrac{r}{4} \[1em] \Rightarrow r = 4%.⇒x=2×12x×24×(24+1)​×100r​⇒x=24x×24×25​×100r​⇒1=4r​⇒r=4Hence, rate of interest = 4%.

Question 3

Question 52Akshay buys 350 shares of ₹ 50 par value of a company. The dividend declared by the company is 14%. If his return percent from the shares is 10%, find the market value of each share.

Accepted Answer

Given,Rate of dividend = 14%Dividend on ₹ 50 = 14100×50\dfrac{14}{100} 	imes 5010014​×50 = ₹ 7.Given,Return percent from the shares is 10%.∴ Interest on ₹ 100 = 10100×100\dfrac{10}{100} 	imes 10010010​×100 = ₹ 10.∴ ₹ 7 will be interest on 10010×7\dfrac{100}{10} 	imes 710100​×7 = ₹ 70.Hence, market value of each share = ₹ 70.

Question 4

Question 53Solve the following inequation and answer the questions given below.12(2x−1)≤2x+12≤512+x\dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x21​(2x−1)≤2x+21​≤521​+x(a) Write the maximum and minimum values of x for x ∈ R.(b) What will be the change in maximum and minimum values of x if x ∈ W.

Accepted Answer

(a) Given,12(2x−1)≤2x+12≤512+x\dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x21​(2x−1)≤2x+21​≤521​+xSolving L.H.S. of the inequation, we get :⇒12(2x−1)≤2x+12⇒x−12≤2x+12⇒2x−x≥−12−12⇒x≥−1 ...........(1)\Rightarrow \dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \[1em] \Rightarrow x - \dfrac{1}{2} \le 2x + \dfrac{1}{2} \[1em] \Rightarrow 2x - x \ge -\dfrac{1}{2} - \dfrac{1}{2} \[1em] \Rightarrow x \ge -1 \text{ ...........(1)}⇒21​(2x−1)≤2x+21​⇒x−21​≤2x+21​⇒2x−x≥−21​−21​⇒x≥−1 ...........(1)Solving R.H.S. of the inequation, we get :⇒2x+12≤512+x⇒2x+12≤112+x⇒2x−x≤112−12⇒x≤102⇒x≤5 ...........(2)\Rightarrow 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x \[1em] \Rightarrow 2x + \dfrac{1}{2} \le \dfrac{11}{2} + x \[1em] \Rightarrow 2x - x \le \dfrac{11}{2} - \dfrac{1}{2} \[1em] \Rightarrow x \le \dfrac{10}{2} \[1em] \Rightarrow x \le 5 \text{ ...........(2)}⇒2x+21​≤521​+x⇒2x+21​≤211​+x⇒2x−x≤211​−21​⇒x≤210​⇒x≤5 ...........(2)From equation (1) and (2), we get :-1 ≤ x ≤ 5 and x ∈ R.Hence, minimum and maximum value of x is -1 and 5 respectively.(b) If x ∈ W.Then, minimum value = 0 and maximum value = 5.Hence, minimum and maximum value of x is 0 and 5 respectively, when x is a whole number.

Question 5

Question 54Solve for x, if 5x+43=23x2\dfrac{5}{x} + 4\sqrt{3} = \dfrac{2\sqrt{3}}{x^2}x5​+43​=x223​​, x ≠ 0

Accepted Answer

Given,5x+43=23x2\dfrac{5}{x} + 4\sqrt{3} = \dfrac{2\sqrt{3}}{x^2}x5​+43​=x223​​Substituting 1x\dfrac{1}{x}x1​ = t in the given equation, we get :⇒5t+43=23t2⇒23t2−5t−43=0⇒23t2−8t+3t−43=0⇒2t(3t−4)+3(3t−4)=0⇒(2t+3)(3t−4)=0⇒2t+3=0 or 3t−4=0⇒2t=−3 or 3t=4⇒t=−32 or t=43⇒1x=−32 or 1x=43⇒x=−23 or x=34.\Rightarrow 5t + 4\sqrt{3} = 2\sqrt{3}t^2 \[1em] \Rightarrow 2\sqrt{3}t^2 - 5t - 4\sqrt{3} = 0 \[1em] \Rightarrow 2\sqrt{3}t^2 - 8t + 3t - 4\sqrt{3}= 0 \[1em] \Rightarrow 2t(\sqrt{3}t - 4) + \sqrt{3}(\sqrt{3}t - 4) = 0 \[1em] \Rightarrow (2t + \sqrt{3})(\sqrt{3}t - 4) = 0 \[1em] \Rightarrow 2t + \sqrt{3} = 0 \text{ or } \sqrt{3}t - 4 = 0 \[1em] \Rightarrow 2t = -\sqrt{3} \text{ or } \sqrt{3}t = 4 \[1em] \Rightarrow t = -\dfrac{\sqrt{3}}{2} \text{ or } t = \dfrac{4}{\sqrt{3}} \[1em] \Rightarrow \dfrac{1}{x} = -\dfrac{\sqrt{3}}{2} \text{ or } \dfrac{1}{x} = \dfrac{4}{\sqrt{3}} \[1em] \Rightarrow x = -\dfrac{2}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4}.⇒5t+43​=23​t2⇒23​t2−5t−43​=0⇒23​t2−8t+3t−43​=0⇒2t(3​t−4)+3​(3​t−4)=0⇒(2t+3​)(3​t−4)=0⇒2t+3​=0 or 3​t−4=0⇒2t=−3​ or 3​t=4⇒t=−23​​ or t=3​4​⇒x1​=−23​​ or x1​=3​4​⇒x=−3​2​ or x=43​​.Hence, x = −23 or x=34.-\dfrac{2}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4}.−3​2​ or x=43​​.

Question 6

Question 55The marked price of a toy is same as the percentage of GST that is charged. The price of the toy is ₹ 24 including GST. Taking the marked price as x, form an equation and solve it to find x.

Accepted Answer

Given,M.P. of toy = ₹ xG.S.T. charged = x%Price including G.S.T. = ₹ 24∴24=x+x100×x⇒24=x+x2100⇒24=100x+x2100⇒2400=x2+100x⇒x2+100x−2400=0⇒x2+120x−20x−2400=0⇒x(x+120)−20(x+120)=0⇒(x−20)(x+120)=0⇒x−20=0 or x+120=0⇒x=20 or x=−120\therefore 24 = x + \dfrac{x}{100} \times x \[1em] \Rightarrow 24 = x + \dfrac{x^2}{100} \[1em] \Rightarrow 24 = \dfrac{100x + x^2}{100} \[1em] \Rightarrow 2400 = x^2 + 100x \[1em] \Rightarrow x^2 + 100x - 2400 = 0 \[1em] \Rightarrow x^2 + 120x - 20x - 2400 = 0 \[1em] \Rightarrow x(x + 120) - 20(x + 120) = 0 \[1em] \Rightarrow (x - 20)(x + 120) = 0 \[1em] \Rightarrow x - 20 = 0 \text{ or } x + 120 = 0 \[1em] \Rightarrow x = 20 \text{ or } x = -120∴24=x+100x​×x⇒24=x+100x2​⇒24=100100x+x2​⇒2400=x2+100x⇒x2+100x−2400=0⇒x2+120x−20x−2400=0⇒x(x+120)−20(x+120)=0⇒(x−20)(x+120)=0⇒x−20=0 or x+120=0⇒x=20 or x=−120Since, M.P. cannot be negative.∴ M.P. = ₹ 20 and G.S.T. = 20%.Hence, the marked price of toy = ₹ 20 and G.S.T. = 20%.

Question 7

Question 56The mean proportion between two numbers is 6 and their third proportion is 48. Find two numbers.

Accepted Answer

Let the numbers be a and b.Given,Mean proportion between the numbers is 6.∴a6=6b⇒ab=36⇒a=36b .........(1)\therefore \dfrac{a}{6} = \dfrac{6}{b} \[1em] \Rightarrow ab = 36 \[1em] \Rightarrow a = \dfrac{36}{b}\text{ .........(1)}∴6a​=b6​⇒ab=36⇒a=b36​ .........(1)Third proportion of the numbers is 48.∴ab=b48⇒b2=48a⇒b2=48×36b⇒b3=1728⇒b=17283⇒b=12.\therefore \dfrac{a}{b} = \dfrac{b}{48} \[1em] \Rightarrow b^2 = 48a \[1em] \Rightarrow b^2 = 48 \times \dfrac{36}{b} \[1em] \Rightarrow b^3 = 1728 \[1em] \Rightarrow b = \sqrt[3]{1728} \[1em] \Rightarrow b = 12.∴ba​=48b​⇒b2=48a⇒b2=48×b36​⇒b3=1728⇒b=31728​⇒b=12.Substituting value of b in equation (1), we get :⇒a=3612\Rightarrow a = \dfrac{36}{12}⇒a=1236​ = 3.Hence, the numbers are 3 and 12.

Question 8

Question 57Pamela factorized the following polynomial :2x3 + 3x2 - 3x - 2She found the result as (x + 2)(x - 1)(x - 2). Using remainder and factor theorem, verify whether her result is correct. If incorrect, give the correct result.

Accepted Answer

Dividing the polynomial f(x) = 2x3 + 3x2 - 3x - 2 by (x + 2), we get :⇒ x + 2 = 0⇒ x = -2f(-2) = 2(-2)3 + 3(-2)2 - 3(-2) - 2= 2(-8) + 3(4) + 6 - 2= -16 + 12 + 6 - 2= -18 + 18= 0.Dividing the polynomial f(x) = 2x3 + 3x2 - 3x - 2 by (x - 1), we get :⇒ x - 1 = 0⇒ x = 1f(1) = 2(1)3 + 3(1)2 - 3(1) - 2= 2(1) + 3(1) - 3 - 2= 2 + 3 - 3 - 2= 0.Dividing the polynomial f(x) = 2x3 + 3x2 - 3x - 2 by (x - 2), we get :⇒ x - 2 = 0⇒ x = 2f(2) = 2(2)3 + 3(2)2 - 3(2) - 2= 2(8) + 3(4) - 6 - 2= 16 + 12 - 6 - 2= 20.Since, f(2) ≠ 0∴ x - 2 does not divide the polynomial 2x3 + 3x2 - 3x - 2.Dividing the polynomial f(x) by (x + 2)(x - 1) or by (x2 + x - 2), we get :x2+x−2)2x+1x2+x−2)2x3+3x2−3x−2‾x2+x−2))+−2x3+−2x2−+4x‾x2+x−22x3+20x3x2+x−2x2+x−2)2x3+x3+−x2+−x−+2‾x2+x−2)x3−2x2(31)x×\begin{array}{l} \phantom{x^2 + x - 2)}{\quad 2x + 1} \ x^2 + x - 2\overline{\smash{\big)}\quad 2x^3 + 3x^2 - 3x - 2} \ \phantom{x^2 + x - 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}2x^2 \underset{+}{-}4x} \ \phantom{{x^2 + x - 2}2x^3 + 20x^3}x^2 + x - 2 \ \phantom{{x^2 + x - 2)}2x^3 + x^3}\underline{\underset{-}{+}x^2 \underset{-}{+} x \underset{+}{-} 2} \ \phantom{{x^2 + x - 2)}{x^3-2x^{2}(31)}{x}}	imes \end{array}x2+x−2)2x+1x2+x−2)2x3+3x2−3x−2​x2+x−2))−+​2x3−+​2x2+−​4x​x2+x−22x3+20x3x2+x−2x2+x−2)2x3+x3−+​x2−+​x+−​2​x2+x−2)x3−2x2(31)x×​∴ 2x3 + 3x2 - 3x - 2 = (x2 + x - 2)(2x + 1)= (x + 2)(x - 1)(2x + 1).Hence, 2x3 + 3x2 - 3x - 2 = (x + 2)(x - 1)(2x + 1).

Question 9

Question 58A=[−6042] and B=[1013]A = \begin{bmatrix*}[r] -6 & 0 \ 4 & 2 \end{bmatrix*} 	ext{ and } B = \begin{bmatrix*}[r] 1 & 0 \ 1 & 3 \end{bmatrix*}A=[−64​02​] and B=[11​03​].Find matrix M, if M = 12A−2B+5l\dfrac{1}{2}A - 2B + 5l21​A−2B+5l, where l is the identity matrix.

Accepted Answer

Given,⇒M=12A−2B+5l=12[−6042]−2[1013]+5[1001]=[−3021]−[2026]+[5005]=[−3−2+50−0+02−2+01−6+5]=[0000].\Rightarrow M = \dfrac{1}{2}A - 2B + 5l \[1em] = \dfrac{1}{2}\begin{bmatrix*}[r] -6 & 0 \ 4 & 2 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 0 \ 1 & 3 \end{bmatrix*} + 5\begin{bmatrix*}[r] 1 & 0 \ 0 & 1 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] -3 & 0 \ 2 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 0 \ 2 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] 5 & 0 \ 0 & 5 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] -3 - 2 + 5 & 0 - 0 + 0 \ 2 - 2 + 0 & 1 - 6 + 5 \end{bmatrix*} \[1em] = \begin{bmatrix*}[r] 0 & 0 \ 0 & 0 \end{bmatrix*}.⇒M=21​A−2B+5l=21​[−64​02​]−2[11​03​]+5[10​01​]=[−32​01​]−[22​06​]+[50​05​]=[−3−2+52−2+0​0−0+01−6+5​]=[00​00​].Hence, M = [0000].\begin{bmatrix*}[r] 0 & 0 \ 0 & 0 \end{bmatrix*}.[00​00​].

Question 10

Question 59(a) Write the nth term (Tn) of an Arithmetic Progression (A.P.) consisting of all whole numbers which are divisible by 3 and 7.(b) How many of these are two-digit numbers? Write them.(c) Find the sum of first 10 terms of this A.P.

Accepted Answer

(a) A.P. = 21, 42, 63, .........The above sequence is an A.P. with first term (a) = 21 and common difference (d) = 21.By formula,Tn = a + (n - 1)d= 21 + (n - 1)21= 21 + 21n - 21= 21n.Hence, the nth term = 21n.(b) A.P. = 21, 42, 63, 84, 105, ........Hence, there are four two digit numbers i.e. 21, 42, 63, 84 in the A.P.(c) By formula,Sum of A.P. = n2(a+an)\dfrac{n}{2}(a + a_n)2n​(a+an​)Sum of first 10 terms of A.P.=102(a+a10)=5[a+a+(n−1)d]=5[2a+(n−1)d]=5[2×21+(10−1)×21]=5[42+9×21]=5[42+189]=5×231=1155.\text{Sum of first 10 terms of A.P.} = \dfrac{10}{2}(a + a_{10}) \[1em] = 5[a + a + (n - 1)d] \[1em] = 5[2a + (n - 1)d] \[1em] = 5[2 \times 21 + (10 - 1) \times 21] \[1em] = 5[42 + 9 \times 21] \[1em] = 5[42 + 189] \[1em] = 5 \times 231 \[1em] = 1155.Sum of first 10 terms of A.P.=210​(a+a10​)=5[a+a+(n−1)d]=5[2a+(n−1)d]=5[2×21+(10−1)×21]=5[42+9×21]=5[42+189]=5×231=1155.Hence, sum of first 10 terms of the A.P. = 1155.

Question 11

Question 60Write the first five terms of the sequence given by (3)n(\sqrt{3})^n(3​)n, n ∈ N.(a) Is the sequence an A.P. or G.P.?(b) If the sum of its first ten terms is p(3+3)p(3 + \sqrt{3})p(3+3​), find the value of p.

Accepted Answer

Terms of the sequence given by (3)n(\sqrt{3})^n(3​)n are :⇒ (3)1,(3)2,(3)3,(3)4,(3)5(\sqrt{3})^1, (\sqrt{3})^2, (\sqrt{3})^3, (\sqrt{3})^4, (\sqrt{3})^5(3​)1,(3​)2,(3​)3,(3​)4,(3​)5, .......⇒ 3,3,33,9,93\sqrt{3}, 3, 3\sqrt{3}, 9, 9\sqrt{3}3​,3,33​,9,93​(a) Ratio between terms = 33=3\dfrac{3}{\sqrt{3}} = \sqrt{3}3​3​=3​.Hence, the sequence is a G.P. with common ratio = 3\sqrt{3}3​.(b) By formula,Sum of G.P. (S) = a(rn−1)(r−1)\dfrac{a(r^n - 1)}{(r - 1)}(r−1)a(rn−1)​Substituting values we get :⇒S10=3[(3)10−1]3−1⇒p(3+3)=3(243−1)3−1⇒p=3(243−1)(3−1)(3+3)⇒p=3×242(3−1)3(3+1)⇒p=2423−1⇒p=2422=121.\Rightarrow S_{10} = \dfrac{\sqrt{3}[(\sqrt{3})^{10} - 1]}{\sqrt{3} - 1} \[1em] \Rightarrow p(3 + \sqrt{3})= \dfrac{\sqrt{3}(243 - 1)}{\sqrt{3} - 1} \[1em] \Rightarrow p = \dfrac{\sqrt{3}(243 - 1)}{(\sqrt{3} - 1)(3 + \sqrt{3})} \[1em] \Rightarrow p = \dfrac{\sqrt{3} \times 242}{(\sqrt{3} - 1)\sqrt{3}(\sqrt{3} + 1)} \[1em] \Rightarrow p = \dfrac{242}{3 - 1} \[1em] \Rightarrow p = \dfrac{242}{2} = 121.⇒S10​=3​−13​[(3​)10−1]​⇒p(3+3​)=3​−13​(243−1)​⇒p=(3​−1)(3+3​)3​(243−1)​⇒p=(3​−1)3​(3​+1)3​×242​⇒p=3−1242​⇒p=2242​=121.Hence, p = 121.

Question 12

Question 61ABC is a triangle as shown in the figure below.(a) Write down the coordinates of A, B and C on reflecting through the origin.(b) Write down the coordinates of the point/s which remain invariant on reflecting the triangle ABC on the x-axis and y-axis respectively.

Accepted Answer

(a) From figure,On reflecting in origin A, B and C becomes A', B' and C'.Coordinates of A' = (-4, -5), B' = (0, -3) and C' = (-3, 0).Hence, on reflecting A, B and C in origin the coordinates are (-4, -5), (0, -3) and (-3, 0) respectively.(b) Point C(3, 0) lies on x-axis.∴ It remains invariant on reflecting in x-axis.Point B(0, 3) lies on y-axis.∴ It remains invariant on reflecting in y-axis.Hence, the invariant point for x-axis is point C and for y-axis is point B.

Question 13

Question 62Determine the ratio in which the line y = 2 + 3x divides the line segment AB joining the points A(-3, 9) and B(4, 2).

Accepted Answer

By two point form,Equation of line :⇒ y - y1 = y2−y1x2−x1\dfrac{y_2 - y_1}{x_2 - x_1}x2​−x1​y2​−y1​​(x - x1)Equation of AB :⇒y−9=2−94−(−3)[x−(−3)]⇒y−9=−74+3[x+3]⇒y−9=−77[x+3]⇒y−9=−1[x+3]⇒y−9=−x−3⇒y+x=−3+9⇒x+y=6.\Rightarrow y - 9 = \dfrac{2 - 9}{4 - (-3)}[x - (-3)] \[1em] \Rightarrow y - 9 = \dfrac{-7}{4 + 3}[x + 3] \[1em] \Rightarrow y - 9 = \dfrac{-7}{7}[x + 3] \[1em] \Rightarrow y - 9 = -1[x + 3] \[1em] \Rightarrow y - 9 = -x - 3 \[1em] \Rightarrow y + x = -3 + 9 \[1em] \Rightarrow x + y = 6.⇒y−9=4−(−3)2−9​[x−(−3)]⇒y−9=4+3−7​[x+3]⇒y−9=7−7​[x+3]⇒y−9=−1[x+3]⇒y−9=−x−3⇒y+x=−3+9⇒x+y=6.Solving equation y = 2 + 3x and x + y = 6 simultaneously,⇒ x + y = 6 .......(1)⇒ y = 2 + 3x .......(2)Substituting value of y from equation (2) in (1), we get :⇒ x + (2 + 3x) = 6⇒ 4x + 2 = 6⇒ 4x = 6 - 2⇒ 4x = 4⇒ x = 44\dfrac{4}{4}44​⇒ x = 1.Substituting value of x in equation (2), we get :⇒ y = 2 + 3(1) = 2 + 3 = 5.Let (1, 5) divide the line AB in the ratio k : 1.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 :⇒(1,5)=(k×4+1×−3k+1,k×2+1×9k+1)⇒(1,5)=(4k−3k+1,2k+9k+1)⇒1=4k−3k+1 or 5=2k+9k+1⇒1(k+1)=4k−3 or 5(k+1)=2k+9⇒k+1=4k−3 or 5k+5=2k+9⇒4k−k=1+3 or 5k−2k=9−5⇒3k=4 or 3k=4⇒k=43⇒k:1=43:1=4:3.\Rightarrow (1, 5) = \Big(\dfrac{k \times 4 + 1 \times -3}{k + 1}, \dfrac{k \times 2 + 1 \times 9}{k + 1}\Big) \[1em] \Rightarrow (1, 5) = \Big(\dfrac{4k - 3}{k + 1}, \dfrac{2k + 9}{k + 1}\Big) \[1em] \Rightarrow 1 = \dfrac{4k - 3}{k + 1} \text{ or } 5 = \dfrac{2k + 9}{k + 1} \[1em] \Rightarrow 1(k + 1) = 4k - 3 \text{ or } 5(k + 1) = 2k + 9 \[1em] \Rightarrow k + 1 = 4k - 3 \text{ or } 5k + 5 = 2k + 9 \[1em] \Rightarrow 4k - k = 1 + 3 \text{ or } 5k - 2k = 9 - 5 \[1em] \Rightarrow 3k = 4 \text{ or } 3k = 4 \[1em] \Rightarrow k = \dfrac{4}{3} \[1em] \Rightarrow k : 1 = \dfrac{4}{3} : 1 = 4 : 3.⇒(1,5)=(k+1k×4+1×−3​,k+1k×2+1×9​)⇒(1,5)=(k+14k−3​,k+12k+9​)⇒1=k+14k−3​ or 5=k+12k+9​⇒1(k+1)=4k−3 or 5(k+1)=2k+9⇒k+1=4k−3 or 5k+5=2k+9⇒4k−k=1+3 or 5k−2k=9−5⇒3k=4 or 3k=4⇒k=34​⇒k:1=34​:1=4:3.Hence, the line y = 2 + 3x divides the line segment AB in the ratio 4 : 3.

Question 14

Question 63Square ABCD lies in the third quadrant of a XY plane such that its vertex A is at (-3, -1) and the diagonal DB produced is equally inclined to both the axes. The diagonals AC and BD meets at P(-2, -2). Find the :(a) Slope of BD(b) equation of AC

Accepted Answer

(a) Given,BD is equally inclined to both axes.∴ Slope of BD = tan 45° = 1.Hence, slope of BD = 1.(b) Since, diagonals of a square are perpendicular and product of slope of perpendicular lines equals to -1.∴ Slope of BD × Slope of AC = -1⇒ 1 × Slope of AC = -1⇒ Slope of AC = -1.Since, diagonals meet at point (-2, -2).By point-slope form,⇒ y - y1 = m(x - x1)⇒ y - (-2) = -1[x - (-2)]⇒ y + 2 = -1[x + 2]⇒ y + 2 = -x - 2⇒ x + y + 2 + 2 = 0⇒ x + y + 4 = 0.Hence, equation of AC is x + y + 4 = 0.

Question 15

Question 64ABCD is a rectangle where side BC is twice side AB. If △ACQ ~ △BAP, find area of △BAP : area of △ACQ.

Accepted Answer

Given,ABCD is a rectangle where side BC is twice side AB.⇒ BC = 2ABIn right angled triangle ABC,By pythagoras theorem,⇒ AC2 = AB2 + BC2⇒ AC2 = AB2 + (2AB)2⇒ AC2 = AB2 + 4AB2⇒ AC2 = 5AB2⇒ AC = 5\sqrt{5}5​ AB.We know that,The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.∴area of △BAParea of △ACQ=BA2AC2⇒area of △BAParea of △ACQ=BA2(5BA)2⇒area of △BAParea of △ACQ=BA25BA2⇒area of △BAParea of △ACQ=15\therefore \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{AC^2} \[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{(\sqrt{5}BA)^2} \[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{5BA^2} \[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{1}{5} \[1em]∴area of △ACQarea of △BAP​=AC2BA2​⇒area of △ACQarea of △BAP​=(5​BA)2BA2​⇒area of △ACQarea of △BAP​=5BA2BA2​⇒area of △ACQarea of △BAP​=51​Area of △BAP : Area of △ACQ = 1 : 5.Hence, Area of △BAP : Area of △ACQ = 1 : 5.

Question 16

Question 65Given a triangle ABC, and D is a point on BC such that BD = 4 cm and DC = x cm. If ∠BAD = ∠C and AB = 8 cm, then,(a) prove that triangle ABD is similar to triangle CBA.(b) find the value of 'x'.

Accepted Answer

(a) In △ ABD and △ CBA,⇒ ∠ABD = ∠CBA (Common angle)⇒ ∠BAD = ∠ACB (Given)∴ △ ABD ~ △ CBA (By A.A. axiom)Hence, proved that △ ABD ~ △ CBA.(b) We know that,Corresponding sides of similar triangle are proportional.∴ABBC=BDBA⇒8x+4=48⇒4(x+4)=8×8⇒4x+16=64⇒4x=64−16⇒4x=48⇒x=484= 12 cm.\therefore \dfrac{AB}{BC} = \dfrac{BD}{BA} \[1em] \Rightarrow \dfrac{8}{x + 4} = \dfrac{4}{8} \[1em] \Rightarrow 4(x + 4) = 8 \times 8 \[1em] \Rightarrow 4x + 16 = 64 \[1em] \Rightarrow 4x = 64 - 16 \[1em] \Rightarrow 4x = 48 \[1em] \Rightarrow x = \dfrac{48}{4} = \text{ 12 cm}.∴BCAB​=BABD​⇒x+48​=84​⇒4(x+4)=8×8⇒4x+16=64⇒4x=64−16⇒4x=48⇒x=448​= 12 cm.Hence, x = 12 cm.

Question 17

Question 66In the extract of Survey of India map G43S7, prepared on a scale of 2 cm to 1 km, a child finds the length of the cart track between two settlements is 7.6 cm. Find :(a) the actual length of the cart track on the ground.(b) actual area of a grid square, if each has an area of 4 cm2.

Accepted Answer

Scale factor (k) = 2 cm to 1 km or 2 cm to 100000 cm = 2100000=150000\dfrac{2}{100000} = \dfrac{1}{50000}1000002​=500001​(a) Length of cart track on map = k × Length of cart track on ground7.6=1500007.6 = \dfrac{1}{50000}7.6=500001​ × Length of cart track on groundLength of cart track on ground = 7.6 × 50000 = 380000 cm = 380000100000\dfrac{380000}{100000}100000380000​ = 3.8 kmHence, length of actual cart track on ground = 3.8 km(b) Area of model grid square = k2 × Area of actual grid square⇒ 4 = (150000)2\Big(\dfrac{1}{50000}\Big)^2(500001​)2 × Area of actual grid square⇒ 4 = (125×108)\Big(\dfrac{1}{25 	imes 10^8}\Big)(25×1081​) × Area of actual grid square⇒ Area of actual grid square = 4×25×1084 	imes 25 	imes 10^84×25×108⇒ Area of actual grid square = 100 × 108 = 102 × 108 = 1010 cm2= 10101010\dfrac{10^{10}}{10^{10}}10101010​ km2 = 1 km2.Hence, actual area of a grid square = 1 km2.

Question 18

Question 67Construct a triangle ABC such that AB = 7 cm, BC = 6 cm and CA = 5 cm.(a) Draw the locus of the points such that(i) it is equidistant from BC and BA.(ii) it is equidistant from points A and B.(b) Mark P where the loci (i) and (ii) meet, measure and write length of PA.

Accepted Answer

We know that,Locus of the points equidistant from two sides is the angle bisector between two sides.Locus of the points equidistant from two points is the perpendicular bisector of the line joining two points.Steps of construction :Draw a line segment AB = 7 cm.With A and B as center and radii 5 cm and 6 cm respectively draw arcs intersecting at C.Join AC and BC.Draw BD, angle bisector of B.Draw XY perpendicular bisector of AB.Mark point P, the intersection point of XY and BD.Measure AP.Hence, AP = 3.8 cm.

Question 19

Question 68In the given figure O is the centre of the circle. ABCD is a quadrilateral where sides AB, BC, CD and DA touch the circle at E, F, G and H respectively. If AB = 15 cm, BC = 18 cm and AD = 24 cm, find the length of CD.

Accepted Answer

We know that,The length of tangents drawn from an external point to the circle are equal.⇒ AE = AH = x (let)⇒ BE = BF = y (let)⇒ CF = CG = z (let)⇒ DG = DH = k (let)Given,⇒ AB = 15⇒ AE + EB = 15⇒ x + y = 15 .......(1)⇒ BC = 18⇒ BF + CF = 18⇒ y + z = 18 ........(2)⇒ AD = 24⇒ AH + DH = 24⇒ x + k = 24 ..........(3)Adding equations (2) and (3), we get :⇒ y + z + x + k = 18 + 24⇒ x + y + z + k = 42⇒ 15 + z + k = 42 [From equation (1)]⇒ z + k = 42 - 15⇒ z + k = 27.⇒ CD = CG + DG = k + z = 27 cm.Hence, length of CD = 27 cm.

Question 20

Question 69In the given diagram, ABCDEF is a regular hexagon inscribed in a circle with centre O. PQ is a tangent to the circle at D. Find the value of :(a) ∠FAG(b) ∠BCD(c) ∠PDE

Accepted Answer

(a) By formula,Each interior angle of a n sided polygon = (n−2)×180°n\dfrac{(n - 2) \times 180°}{n}n(n−2)×180°​=(6−2)×180°6=4×180°6=23×180°=120°.= \dfrac{(6 - 2) \times 180°}{6} \[1em] = \dfrac{4 \times 180°}{6} \[1em] = \dfrac{2}{3} \times 180° \[1em] = 120°.=6(6−2)×180°​=64×180°​=32​×180°=120°.∴ ∠FAB = 120°From figure,∠FAG and ∠FAB form a linear pair.∴ ∠FAB + ∠FAG = 180°⇒ 120° + ∠FAG = 180°⇒ ∠FAG = 180° - 120° = 60°.Hence, ∠FAG = 60°.(ii) Each interior angle of regular hexagon = 120°.Hence, ∠BCD = 120°.(iii) Join FC.We know that,Angle in a semicircle is a right angle.∴ ∠FBC = 90°In triangle BCD,⇒ BC = CD (Sides of regular hexagon)⇒ ∠CBD = ∠CDB = x (Angles opposite to equal sides are equal)By angle sum property of triangle,⇒ ∠BCD + ∠CDB + ∠CBD = 180°⇒ 120° + x + x = 180°⇒ 2x = 180° - 120°⇒ 2x = 60°⇒ x = 60°2\dfrac{60°}{2}260°​ = 30°.From figure,⇒ ∠FBD = ∠FBC - ∠CBD = 90° - x = 90° - 30° = 60°.We know that,The angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.⇒ ∠PDE = ∠FBD = 60°.Hence, ∠PDE = 60°.

Short Answer Questions 1