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Roll No:8 , Page :13
[38. i)] Coordinates of $$P = (4, 6)$$ Coordinates of $$Q = (3, 2)$$ Coordinates of $$R = (6, 5)$$ [38. ii)] let coordinates of which divides line segment $$PR$$ into $$2:1$$ ratio be $$A(x, y)$$ Section formula $$A(x, y) = \left[ \frac{12+4}{2+1}, \frac{10+6}{2+1} \right]$$ $$A(x, y) = \left[ \fra... | |
Roll No:8 , Page :12
[37.] Loan taken by Rishi = $$₹ 11,80,000$$ Amount paid by $$1^{st}$$ instalment = $$₹ 10,000$$ Amount increased for every next instalment = $$₹ 1000$$. $$a = ₹ 10,000$$ $$d = ₹ 1,000$$ [i] $$a_n = a + (n - 1)d$$ $$a_{30} = a + (30 - 1)d = 10000 + 29(1000) = 10000 + 29000 = ₹ 39000$$. $$ herefore$$... | |
Roll No:8 , Page :11
[ii)] A red face card\nNo. of red faced cards $$= 6$$\n$$P(E) = \frac{6}{52} = \frac{3}{26}$$. \n\n[iv)] Jack of hearts\nNo. of jacks of hearts $$= 1$$\n$$P(E) = \frac{1}{52}$$. \n\n[v)] A spade\nNo. of spades $$= 13$$\n$$P(E) = \frac{13}{52} = \frac{1}{4}$$. \n\nSection-f\n[36 i] The shape of path... | |
Roll No:8 , Page :10
[34.] Given, $$C.I \quad f_i \quad x_i \quad d_i \quad f_id_i$$ $$100-150 \quad 4 \quad 125 \quad -100 \quad -400$$ $$150-200 \quad 5 \quad 175 \quad -50 \quad -250$$ $$200-250 \quad 12 \quad 225 \text{ } a \quad 0 \quad 0$$ $$250-300 \quad 2 \quad 275 \quad 50 \quad 100$$ $$300-350 \quad 2 \quad 3... | |
Roll No:8 , Page :9
[33] let , the line segment joining the points $$(4,1)$$ $$(-2,-3)$$ be $$A, B$$ respectively. 1 : 2 2 : 1 A P Q B (4,1) (x,y) (x_1, y_1) (-2,-3) Considering point P, Section Formula $$P(x,y) = \left[ \frac{-2+8}{1+2}, \frac{-3+2}{1+2} \right]$$ $$P(x,y) = \left[ \frac{6}{3}, \frac{-1}{3} \right]$$ ... | |
Roll No:8 , Page :8
[ii)] Number blw 2 and 6 No's blw 2 and 6 = 3, 4, 5 $$P(E) = \frac{3}{6} = \frac{1}{2}$$ [iii)] Odd Number Odd No's = 1, 3, 5 $$P(E) = \frac{3}{6} = \frac{1}{2}$$ Section-D [32.] let us assume $$\sqrt{5}$$ is rational. i.e, $$\sqrt{5} = \frac{a}{b}$$ S.O.B.S $$5 = \frac{a^2}{b^2}$$ $$a^2 = 5b^2$$ $$... | |
Roll No:8 , Page :7
[30] Given,
$$\begin{array}{|c|c|c|} \hline C.I & f & c.f \\ \hline 1-3 & 7 & 7 \quad c.f \\ l \quad 3-5 & 8 \quad f & 15 \quad M.C \\ 5-7 & 2 & 17 \\ 7-9 & 2 & 19 \\ 9-11 & 1 & 20 \\ \hline \end{array}$$
$$l = 3$$
$$c.f = 7$$
$$f = 8$$
$$\frac{n}{2} = 10$$
$$h = 2$$
$$\text{Mode} = l + \left(\frac{... | |
Roll No:8 , Page :6
$$a_{31} = a + (31-1) d$$ $$= 3 + 30(5)$$ $$\therefore a_{31} = 3 + 150 = 153.$$ [29.] Given, vertices of parallelogram are $$(1, 2), (4, y), (x, 6), (3, 5)$$ taken in order. let , they be $$A, B, C, D$$ respectively $$A = (1, 2) \quad B = (4, y) \quad C = (x, 6) \quad D = (3, 5)$$ WKT , Midpoint of... | |
Roll No:8 , Page :5
$$LHS \neq RHS$$ \therefore Our assumption is wrong.. $$5-\sqrt{3}$$ is 'irrational'. [27.] let, $$x^2 + 7x + 10 = 0$$ $$x^2 + 2x + 5x + 10 = 0$$ $$x(x+2) + 5(x+2) = 0$$ $$(x+2)(x+5) = 0$$ $$x+2=0$$ (or) $$x+5=0$$ $$x=-2$$ (or) $$x=-5.$$ $$x = -2, -5.$$ let, $$-2$$ be $$\alpha$$ and $$-5$$ be $$\bet... | |
Roll No:8 , Page :4
[25] Given, No. of red balls = 3 No. of black balls = 5 Total no. of balls = 8. $$P(E) = \frac{\text{no. of favourable outcomes}}{\text{total no. of outcomes}}$$ i) Red $$P(E) = \frac{3}{8}$$ ii) Not red $$P(E) + P(\bar{E}) = 1$$ $$\frac{3}{8} + P(\bar{E}) = 1$$ $$P(\bar{E}) = 1 - \frac{3}{8}$$ $$P(... | |
Roll No:8 , Page :3
$$(-1, 6) = \left[ \frac{6m_1 - 3m_2}{m_1 + m_2}, \frac{-8m_1 + 10m_2}{m_1 + m_2} \right]$$ equating $$x$$-coordinate, $$\frac{6m_1 - 3m_2}{m_1 + m_2} = -1.$$ $$6m_1 - 3m_2 = -m_1 - m_2$$ $$7m_1 = 2m_2$$ $$\frac{m_1}{m_2} = \frac{2}{7}$$ $$\therefore$$ The ratio is $$2:7$$. [24.] Given, C.I fi xi fi... | |
Roll No:8 , Page :2
[22] Given, $$11^{th}$$ term $$(a_{11}) = 38 \rightarrow (1)$$ $$16^{th}$$ term $$(a_{16}) = 73 \rightarrow (2)$$ Subtracting eq (1) from (2), we get $$a_{16} = a + 15d = 73$$ $$(-) a_{11} = a + 10d = 38$$ $$(-)$$ $$(-)$$ $$(-)$$ $$5d = 35$$ $$d = \frac{35}{5} = 7.$$ Substitute $$d$$ value in eq (1)... | |
Roll No:8 , Page :1
Section-A [1] A [2] B [3] D [4] D [5] C [6] B [7] C [8] C [9] A. [10] B [11] A [12] C [13] C [14] B [15] B [16] B [17] C [18] B [19] B [20] B Section-B [21] Given, Sum of zero's and product of zero's are $$-3, 2$$ respectively. Quadratic Polynomial, $$k (x^2 - (\text{sum of zero's}) x + (\text{produ... | |
Roll No:4 , Page :15
[38) iii)] $$PQ = \sqrt{17}$$ units ; $$QR = \sqrt{18}$$ units $$PR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$= \sqrt{(6 - 4)^2 + (5 - 6)^2}$$ $$= \sqrt{4 + 1} = \sqrt{5}$$ units $$\therefore \Delta PQR$$ is not an isosceles triangle. | |
Roll No:32 , Page :15
left by mistake. | |
Page :16
Section - E
[36]
i) degree of $$p(x) = 3$$
degree of $$q(x) = 3$$
ii) $$x+2=0$$ $$x+2 \begin{array}{r} 2x^2 - 11x + 19 \\ \hline ) 2x^3 - 7x^2 - 3x + 18 \\ \underline{- (2x^3 + 4x^2)} \\ -11x^2 - 3x \\ \underline{- (-11x^2 - 22x)} \\ +19x + 18 \\ \underline{- (19x + 38)} \\ -20 \end{array}$$
remainder $$= -20.... | |
Roll No:32 , Page :0
32, 9-F Tanvi Shetty scale on: x axis - $$2 \text{ unit} = 10 \text{ marks}$$ y axis - $$2 \text{ unit} = 2 \text{ students}$$ y axis 14 12 10 8 6 4 2 no. of students 0 10 20 30 40 50 60 70 x axis marks FOR EDUCATIONAL USE ONLY bigcolors | |
Roll No:32 , Page :17
[iii)] $$p(x) = x^3 + 13x^2 + 32x + 20 ; x=1 , x=-1$$ $$\therefore x=1$$ $$p(1) = (1)^3 + 13(1)^2 + 32(1) + 20$$ $$= 1 + 13 + 32 + 20$$ $$= 14+52$$ $$p(1) = 66$$ $$\therefore x=-1$$ $$p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20$$ $$= -1 + 13(1) + -32 + 20$$ $$= -1 + 13 - 32 + 20$$ $$= 12 - 32 + 20$$ $... | |
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[37 i)] In $$\Delta ABC$$ & $$\Delta AFE$$, $$AF = AB$$ [$$ABDF$$ is a square] $$\angle AFE = \angle ABC$$ [$$90^{\circ}$$; $$ABDF$$ is square] $$FE = BC$$ [given] Thus, $$\Delta ABC \cong \Delta AFE$$ by SAS Rule. [ii)] Yes, As $$DF = DB$$ [sides of square] $$DF - EF = DB - BC$$ [$$EF = BC$$] $$DE = DC$$ [iii... | |
Roll No:32 , Page :19
38. [i)] $$\angle PAR + \angle PQR = 180^{\circ}$$ [sum of opp. angles of cyclic quad. is $$180^{\circ}$$]
$$\angle PAR = 180^{\circ} - 120^{\circ}$$
$$\angle PAR = 60^{\circ}$$
[ii)] If $$\angle PAR$$ is $$60^{\circ}$$, then $$\angle POR = 120^{\circ}$$ [angle subtended by an arc at centre of ci... | |
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Section-C [28] $$x - 2 = 0$$ $$x = 2$$ $$\therefore 2x^3 + 3x^2 - a = ax^3 - 5x + 2$$ [equating both the $$p(x)$$] $$\therefore 2(2)^3 + 3(2)^2 - a = a(2)^3 - 5(2) + 2$$ $$\therefore 2(8) + 3(4) - a = a(8) - 10 + 2$$ $$\therefore 16 + 12 - a = 8a - 8$$ $$\therefore 28 - a = 8a - 8$$ $$\therefore 28 + 8 = 8a + ... | |
Roll No:32 , Page :21
[30] $$\angle incidence = \angle reflection$$ $$\therefore \angle DAO = \angle CAO$$ $$\therefore \angle DBA = \angle OAB$$ [AIA.] $$AC \parallel BD$$ [converse of AIA] | |
Roll No:6 , Page :11
[33] $$(4, -1) \text{ --- } (-2, -3)$$\nA = $$(4, -1)$$\nB = $$(-2, -3)$$\nlets take $$m:n$$ as $$1:2$$\nSection formula = $$(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$$\n= $$(\frac{1(-2)+2(4)}{1+2}, \frac{1(-3)+2(-1)}{1+2})$$\n= $$(\frac{-2+8}{3}, \frac{-3-2}{3})$$\n= $$(\frac{6}{3}, -\frac{5}... | |
Roll No:5 , Page :2
Name: T. Sunitha Subject: Mathematics Block: oxford. [4] $$a^{\frac{3}{2}} = 2a^{\frac{5}{4}}$$ $$\frac{a^{\frac{3}{2}}}{a^{\frac{5}{4}}} = 2$$ $$a^{\frac{3}{2} - \frac{5}{4}} = 2$$ $$a^{\frac{6}{4} - \frac{5}{4}} = 2$$ $$a^{\frac{1}{4}} = 2$$ $$a = 2^{4}$$ $$a = 16$$ [5] By observing $$1^{2} = 1$$ ... | |
Roll No:5 , Page :3
$$= \frac{b+a}{ab} \frac{b+a}{ab}$$ $$= \frac{b+a}{ab}$$ $$= \frac{ab}{b+a}$$ [7.] False $$\pi$$ is not rational number It is irrational number [8.] $$(2x-4) (5x-3)$$ $$= 10x^2 - 6x - 20x - 12$$ $$= 10x^2 - 26x - 12$$ $$= -26x$$ [9)] ITALY [10)] $$30 + 0.3 \neq 0.003$$ $$30 + 0.3 - 0.003$$ $$= 30.29... | |
Roll No:5 , Page :4
Name: J. sunitha
Subject: Mathematics
Block: oxford.
$$= \frac{2}{4}$$
percent Increase $$= \frac{2}{4} \times 100 \%$$
$$= 2 \times 25$$
$$= 50\%$$
[2] $$(a+b)^2 = a^2 + 2ab + b^2$$
$$(a-b)^2 = a^2 - 2ab + b^2$$
$$(a^2 - b^2) = (a+b)(a-b)$$
$$(x+a)(x+b) = x^2 + bx + ax + ab$$
$$= x^2 + x(a+b) + a... | |
Roll No:5 , Page :5
By arranging the given data in ascending order
$$6, 8, 10, 10, 15, 15, (15), 50, 80, 100, 200$$
$$median = 15$$
[5] Flight number | Departure time from mumbai | Arrival time at delhi | Time duration
[5] G8-142 | $$17-55=(5:55P.M)$$ | $$19-45 \ 7:45P.M$$ |
[5] A1-605 | $$21-30 \ 9:30P.m$$ | $$23-20 ... | |
Roll No:5 , Page :6
Name: T. sunitha Subject: Mathematics Block: oxford. Time duration of flight number Ai - 605 9:30p.m 30min 10p.m 1h 11pm 20min 11:20p.m $$= 30min + 1h + 20min$$ $$= 1h + 50min$$ $$= 1h 50min$$ $$= 1hr 50min$$ [III] [1] $$a = b \rightarrow (1)$$ $$a^2 = ab$$ $$a^2 - b^2 = ab - b^2$$ we know that $$(a... | |
Roll No:5 , Page :7
Name: J. Sunitha Subject: Mathematics Block: oxford. $$= \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 14 cm$$ $$= \frac{1}{3} \times 22 \times 7 \times 14 cm$$ $$= \frac{2156}{3}$$ $$= 718.66 m^3$$ III [4] 5005 branches to 5 and 1001, 1001 branches to 7 and 143, 143 branches to 11 and 13... | |
Roll No:5 , Page :8
Name: J. Sunitha Subject: Mathematics Block: oxford. [2] I have one chocolate. I want to share with two of my friends what I did here. I share the chocolate equally for two my friend. each part of a chocolate is called a fraction. Fractions means A part from whole is called fraction Aids: Pizza, cho... | |
Roll No:5 , Page :9
Name: J. Sunitha Subject: Mathematics Block: Oxford [ii)] Distance of car from sea level $$= 233 ext{ m} = 56 ext{ m}$$ [iii)] Distance of submarine from the sea level $$= 233 ext{ m}$$ [III 8)] $$1^{st} ext{ person} = 40 ext{kg weight}$$ $$2^{nd} ext{ person} = 40 + 40 ext{kg} = 80 ext{kg}$$ $$= ... | |
Roll No:5 , Page :10
Name: J. Sunitha Subject: Mathematics Block: oxford. - 2 small triangles - 1 Square - 1 Parallelogram Rule: All these shapes should be joined together to get the required shape without any overlapping. * It helps in understanding sides, angles, shapes. * It helps in problem solving fractions, creat... | |
Roll No:5 , Page :11
[III] [5] He is a great Greek mathematician and Engineer He worked in Alexandria. His works on Geometery and Alexandria formula. Formula to find the area of triangle only by using the sides of triangle $$\sqrt{s(s-a)(s-b)(s-c)}$$ where $$s$$ is a perimeter [II] [3] class V VI VII VIII IX No of chil... | |
Roll No:5 , Page :13
Hence, the point $$c$$ is $$(2, -\frac{5}{3})$$ and wkt, point $$D$$ is the midpoint of $$CB$$. $$D(x_4, y_4) = D \left[ \frac{x_3 + x_2}{2}, \frac{y_3 + y_2}{2} \right]$$ $$D(x_4, y_4) = D \left[ \frac{2 - 2}{2}, \frac{\frac{-5}{3} - 3}{2} \right]$$ $$D(x_4, y_4) = D \left[ 0, -\frac{7}{3} \right]... | |
Roll No:5 , Page :14
[34] C.I | $$f_i$$ | $$x_i$$ | $$d_i = x_i - a$$ | $$f_i d_i$$ 100-150 | 4 | $$a(125)$$ | 0 | 0 150-200 | 5 | 175 | 50 | 250 200-250 | 12 | 225 | 100 | 1200 250-300 | 2 | 275 | 150 | 300 300-350 | 2 | 325 | 200 | 400 $$25 \Rightarrow \sum f_i$$ $$2150 = \sum f_i d_i$$ $$(\overline{x}) = a + \frac{\... | |
Roll No:5 , Page :15
[35] i. Total no. of cards = $$52$$
King of red colour = $$2$$
$$P(E) = \frac{\text{no. of favourable outcome}}{\text{Total no. of possible outcome}}$$
$$P(\text{king of red colour}) = \frac{2}{52} = \frac{1}{26}$$
[35] ii) Total no. of cards = $$52$$
no. of face cards = $$12$$
$$P(E) = \fra... | |
Roll No:5 , Page :16
[iv] Total no. of Cards = $$52$$ no of jack of hearts = $$1$$ $$P(E) = \frac{\text{no. of favourable outcomes}}{\text{Total no. of possible outcomes}}$$ $$P(\text{jack of hearts}) = \frac{1}{52}$$ [v] Total no. of Cards = $$52$$ no. of spade Cards = $$13$$ $$P(E) = \frac{\text{no. of favourable out... | |
Roll No:5 , Page :17
Section - E [36.i.] The shape of poses shown is parabola. [36.ii.] $$P(x) = ax^2 + bx + c$$ parabola open downwards if, $$a < 0$$ parabola open upwards if, $$a > 0$$ (or) ii. zero's $$\alpha = -2$$ $$\beta = 3$$ $$\alpha + \beta = -2 + 3 = 1$$ $$\alpha\beta = (-2)(3) = -6$$ $$P(x) = k[x^2 - (\alpha... | |
Roll No:5 , Page :18
[37 i.] $$a_{30} = a + 29d$$ $$a = 10,000$$ $$d = 1000$$ $$a_{30} = 10,000 + (29 \times 1000)$$ $$a_{30} = 10,000 + 29000$$ $$a_{30} = 39,000$$ The $$30^{th}$$ installment is $$39,000/-$$ [37 ii.] $$S_{30} = 15 [2(10,000) + 29(1000)]$$ $$S_n = \frac{n}{2} [2a + (n-1)d]$$ $$S_{30} = 15 [20,000 + 290... | |
Roll No:5 , Page :20
[iii] $$P(4,6)$$ $$Q(3,2)$$ $$R(6,5)$$
WKT,
$$PQ = \sqrt{17}m$$
$$QR = \sqrt{18}m$$
PR distance is
$$P(4,6)$$ $$R(6,5)$$
$$x_1 y_1$$ $$x_2 y_2$$
$$PR = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$
$$= \sqrt{(6-4)^2 + (5-6)^2}$$
$$= \sqrt{(2)^2 + (-1)^2}$$
$$= \sqrt{4+1}$$
$$= \sqrt{5}m$$
NO, $$\Delta PQR$$ i... | |
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$$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$
$$\Delta = \sqrt{15(15-12)(15-12)(15-6)}$$
$$\Delta = \sqrt{15 \times 3 \times 3 \times 9}$$
$$\Delta = \sqrt{5 \times 3 \times 3 \times 3 \times 3 \times 3}$$
$$\Delta = 3 \times 3 \sqrt{5 \times 3}$$
$$\Delta = 9\sqrt{15} \text{ cm}^2$$
[Q-22] In $$\Delta ABC$$
$$\angle ... | |
Page :22
[Q-21] Area of rectangle = $$l \times b$$ \n = $$(2\sqrt{3} + \sqrt{10})(10 + \sqrt{3})$$ \n = $$20\sqrt{3} + (2 \times 3) + 10\sqrt{10} + \sqrt{30}$$ \n = $$20\sqrt{3} + 6 + 10\sqrt{10} + \sqrt{30}$$ \n = $$30\sqrt{13} + 6\sqrt{30}$$ \n = $$36\sqrt{43}$$ \n 43 \n 360 \n 220 \n - 140 \n [Q-20] D - A is false b... | |
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[Q-12] $$C - AC = \frac{1}{2} AB$$ [Q-11] $$A - 1$$ [Q-10] $$D - x = -1.5y$$ [Q-9] $$A - 6$$ [Q-14] $$A - 10\sqrt{3}$$ [Q-13] $$C - 12, 12, 8$$ [Q-8] $$A - 1$$ [Q-7] C - Fourth quadrant [Q-6] B - The perpendicular distance of point $$(5, 7)$$ on the $$x$$-axis is 5 [Q-5] $$B - (-2b, 2b)$$ $$9 - 2p - 7 = 0$$ $$... | |
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[Q-4] $$C - \frac{27}{8}$$ [Q-3] $$C - 2\sqrt{2}$$ [Q-2] $$A - 1$$ [Q-1] $$C - 2$$ | |
Roll No:18 , Page :12
[36] i) The shape of poses shown is parabolic. [36] ii) parabola opens upward = $$a > 0$$ parabola opens downward = $$a < 0$$. [36] iii) The graph shown in fig cuts the x-axis at $$(-2, 0)$$ and $$(3, 0)$$, so the zeroes are $$(-2, 3)$$. [37] i) $$a = 10,000$$ $$d = 10,000$$ Amount paid in 30th In... | |
Roll No:6 , Page :2
Name: k. Shirisha Subject : mathematics Block : Oxford Dealing classes: IV & V [I(4)] Given $$a^{\frac{3}{2}} = 2 a^{\frac{5}{4}}$$ $$\frac{a^{\frac{3}{2}}}{a^{\frac{5}{4}}} = 2 \quad \left[ \frac{a^m}{a^n} = a^{m-n} \right]$$ $$a^{\frac{3}{2} - \frac{5}{4}} = 2 \quad \left[ Lcm \text{ of } 2, 4 = 4... | |
Roll No:6 , Page :3
Name : k. Shirisha Subject : maths Block : Oxford Dealing classes : IV & V [I 7] false. $$\pi$$ is not a rational number. It is an irrational number because it is non terminating $$\pi = 3.14$$ [I 8] $$(2x - 4) (5x - 3)$$ $$10x - 6x - 20x + 12$$ $$10x - 26x + 12$$ Middle Term is $$-26x$$ [I 9] 9 | 2... | |
Roll No:6 , Page :4
Name : k. Shirisha
Subject : maths
Block : Oxford
Dealing classes: IV & V
$$ = \frac{6-4}{4} $$
$$ = \frac{2}{4} $$
Percent Increase $$ = \frac{2}{4} \times 100\% $$
$$ = 50\% $$
II [2] Four Identities use in the royal subject algebra are
(i) $$(a+b)^2 = a^2 + 2ab + b^2$$
(ii) $$(a-b)^2 = a^2 - 2a... | |
Roll No:6 , Page :5
Name: k-shirisha Subject: maths Block: oxford Dealing classes: $$\mathrm{IV} \ \& \ \mathrm{V}$$ $$Mean = 39$$ By arranging the given data in Ascending Order we get $$6, 8, 10, 10, 15, (15), 15, 50, 80, 100, 120$$ $$Median = 15$$ [5] Flight Number | Departure time from mumbai | Arrival time at Delhi... | |
Roll No:6 , Page :6
Name : k. Shirisha Subject : maths Block : Oxford Dealing classes : IV & V [III (1)] $$a = b \rightarrow (1)$$ $$a^2 = ab$$ [multiply both sides with '$$a$$'] $$a^2 - b^2 = ab - b^2$$ [subtract $$b^2$$ on both sides] we know that $$(a + b)(a - b) = a^2 - b^2$$ $$(a + b)(a - b) = b(a - b)$$ $$a + b =... | |
Roll No:6 , Page :8
Name: K. Shirisha Subject: maths Block: Oxford Dealing classes: IV & V Now $$a+b+c = \sqrt{7}-\sqrt{5} + \sqrt{5}-\sqrt{3} + \sqrt{3}-\sqrt{7}$$ $$= 0$$ $$\therefore a+b+c = 0$$ Now $$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ $$= 0(a^2+b^2+c^2-ab-bc-ca)$$ $$= 0$$ [10] (i) Distance of Helico... | |
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Name ! k. Shirisha Subject : maths Block : Oxford Dealing classes: IV & V $$= (12 - 1) \times 40 \text{ kg}$$ $$= 11 \times 40 \text{ kg}$$ $$= 440 \text{ kg}$$ $$\therefore \text{The last person experiences a total of 440 kg force}$$ [III-7] Amount paid for a workman for each day $$= P$$ No. of day... | |
Roll No:6 , Page :10
Name: k. Shirisha
Subject: maths
Block: oxford
Dealing classes: IV & V
* 2 large triangles
* 1 medium triangle
* 2 small triangles
* 1 square
* 1 parallelogram
Rule:
All these seven pieces should be joined to get required shape without any overlapping.
It helps in understanding sides, angles, sha... | |
Roll No:6 , Page :11
Name: K. Shirisha Subject: maths Block: Oxford Dealing classes: IV & V [III (2)] To Introduce fractions I would like to draw a pizza on board and I ask them what happens If this is divided into 2 pieces. So that the child will come to know about halves, Numerator Denominator ($$\frac{1}{2}$$) Then ... | |
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Name : k.shrisha Subject : maths Block : Oxford Dealing classes: IV & V [III-5] Heron of Alexandria He is a famous mathematician and engineer. He lived and worked in Alexandria. He is widely prominent for his work in Geometry and in the discovery of Heron's Formula. Formula to find the Area of tria... | |
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[35] (i) $$\frac{2}{52} = \frac{1}{26} = \frac{1}{26}$$ (ii) $$\frac{16}{52} = \frac{8}{26} = \frac{4}{13} = \frac{4}{13}$$ (iii) $$\frac{8}{52} = \frac{4}{26} = \frac{2}{13} = \frac{2}{13}$$ (iv) $$\frac{1}{52}$$ (v) $$\frac{13^1}{52_4} = \frac{1}{4}$$ [36] (i) The shape is parabola (ii) if $$\alp... | |
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[38.(i)] $$P(\overset{x_1}{4}, \overset{y_1}{6})$$ $$Q(\overset{x_2}{3}, \overset{y_2}{2})$$ $$R(\overset{x_3}{6}, \overset{y_3}{5})$$ [ii)] PQ distance, $$P(\overset{x_1}{4}, \overset{y_1}{6})$$ $$Q(\overset{x_2}{3}, \overset{y_2}{2})$$ $$\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$ $$= \sqrt{(3-4)^2 + (2-6... | |
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Section D [Q32. (B)] $$x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$$ Conjugate: $$\sqrt{3} + \sqrt{2}$$ $$x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}$$ $$x = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2}$$ $$x = \frac{3 + 2 + ... | |
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$$= 49 + 20\sqrt{6} + \frac{1}{49 + 20\sqrt{6}}$$ Conjugate: $$49 - 20\sqrt{6}$$ $$= 49 + 20\sqrt{6} + \frac{1}{49 + 20\sqrt{6}} \times \frac{49 - 20\sqrt{6}}{49 - 20\sqrt{6}}$$ $$= 49 + 20\sqrt{6} + \frac{49 - 20\sqrt{6}}{(49)^2 - (20\sqrt{6})^2}$$ $$= 49 + 20\sqrt{6} + 49 - 20\sqrt{6}$$ $$= 98$$ $$\therefore... | |
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$$ = (98)^2 - 2 $$ \n $$ = 9604 - 2 $$ \n $$ = 9602 $$ \n\n [Q33.] \n (i) $$ 5x + 8y = 320 $$ \n\n (ii) $$ 5x + 8y = 320 $$ \n $$ x = 40 $$ \n $$ 5(40) + 8y = 320 $$ \n $$ 8y = 120 $$ \n $$ y = 15 $$ \n $$ \therefore $$ 15 orange trees \n\n (iii) $$ 2x - 5y = 12 $$ \n\n $$\begin{array}{|c|c|c|} \h... | |
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[(a)] $$2x - 5y = 12$$ $$x = 0$$ $$2(0) - 5y = 12$$ $$y = \frac{-12}{5}$$ $$\therefore (0, \frac{-12}{5})$$ [(b)] $$2x - 5y = 12$$ $$x = 6$$ $$2(6) - 5y = 12$$ $$-5y = 0$$ $$y = 0$$ $$\therefore (6, 0)$$ [Q34. (A)] Marks | No. of students | Class width | Proportional Frequency 0-10 | 8 | 10 | $$\frac{8}{10} \ti... | |
Roll No:17 , Page :1
Labhesh Patil
Roll no. 17
Marks obtained by a students in a class Test
Scale:
On x-axis:
$$1 \text{ unit} = 10 \text{ marks}$$
On y-axis:
$$1 \text{ unit} = 1 \text{ student}$$
No. of students (y-axis)
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Marks (x-axis)
0, 10, 20, 30, 40, 45, 50, 60
#15/12
FOR EDU... | |
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I am Harsh Patil [Q35.] Given: $$AC = 240m$$ $$AB = 360m$$ $$CB = 200m$$ $$CD = 320m$$ $$AD = 400m$$ $$E$$ is mid pt of $$AD$$ To find: Area of $$\Delta AEC$$, $$\Delta ABC$$ & $$\Delta DEC$$ Sol^n: As $$E$$ is mid pt of $$AD$$ $$AE = ED = \frac{1}{2} AD = 200m$$ $$\Delta ACB$$, $$s = \frac{a+b+c}{2} = \frac{2... | |
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ELPRO INTERNATIONAL SCHOOL A Powerhouse of Learning = $$ \sqrt{800 \times 200 \times 560 \times 440} $$ = $$ \sqrt{8 \times 6 \times 100^2 \times 56 \times 44 \times 10^2} $$ = $$ 100 \times 10 \sqrt{8 \times 6 \times 8 \times 7 \times 4 \times 11} $$ = $$ 100 \times 10 \times 8 \sqrt{6 \times 7 \times 4 \time... | |
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As $$CE$$ is a median, it divides $$\Delta ACD$$ in 2 equal halves of same area Area of $$\Delta ACE = Area of \Delta ECD = x$$ $$2x = 38.4 \text{ hectares } 38400 m^2$$ $$x = 19.4 \text{ hectares } 19200 m^2$$ Total area $$= \Delta ACB + \Delta ACD$$ $$= 38.4 + 16. \quad 38400 + 16000 \sqrt{2}$$ ... | |
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Section E
[Q36. (i)]
Degree of $$p(x) = 3$$
Degree of $$q(x) = 3$$
[Q36. (ii)]
$$q(x) = 2x^3 - 7x^2 - 3x + 18$$
Let $$g(x) = x + 2$$
$$0 = x + 2$$
$$x = -2$$
$$q(-2) = 2(-2)^3 - 7(-2)^2 - 3(-2) + 18$$
$$= 2(-8) - 7(4) - 3(-2) + 18$$
$$= -16 - 28 + 6 + 18$$
$$= -44 + 24$$
$$= -20$$
$$\therefore \text{Remainde... | |
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[iii(A)] $$p(x) = x^3 + 13x^2 + 32x + 20$$ (1) $$x = 1$$ $$p(1) = (1)^3 + 13(1)^2 + 32(1) + 20$$ $$= 1 + 13 + 32 + 20$$ $$= 66$$ (2) $$x = -1$$ $$p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20$$ $$= -1 + 13 - 32 + 20$$ $$= 33 - 33$$ $$= 0$$ $$\therefore x = 1$$ is not a zero of $$p(x)$$ $$\therefore x = -1$$ is a zer... | |
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[Q37.] Given: $$BC = FE$$ $$AB \perp BC$$ To find : (i) Is $$\triangle ABC \cong \triangle AFE$$? (ii) Is $$DE = DC$$? (iii) If $$AB = 15 cm, AC = 17 cm, BC = ?$$ Sol: (i) In $$\triangle AFE$$ & $$\triangle ABC$$, $$AB = AF$$ (Sides of a square) $$\angle F = \angle B = 90^{\circ}$$ $$FE = BC$$ $$\therefore \tr... | |
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[(iii)(A)] In $$ \Delta AGC $$ $$ AG = 15\text{cm} $$ $$ AC = 17\text{cm} $$ By Pythagoras Theorem, $$ AG^2 + GC^2 = AC^2 $$ $$ 15^2 + (GC)^2 = (17)^2 $$ $$ GC^2 = 289 - 225 $$ $$ GC^2 = 64 $$ $$ GC = 8\text{cm} $$ As $$ AC = AE $$ (CPCT) because $$ \Delta ABE \cong \Delta ADC $$ $$ \therefore \De... | |
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$$\therefore \Delta AEG \cong \Delta AGC$$ (AAS) $$\therefore EG = GC$$ (CPCT) $$EG + GC = EC$$ $$2GC = EC$$ $$2(8) = EC$$ $$\therefore EC = 16\text{cm}$$ [Q38.] Given: $$AP = PR = AR = 30\text{m}$$ $$\angle PQR = 120^\circ$$ To find: (i) $$\angle PAR$$ (ii) reflex $$\angle POR$$ (iii) $$\Delta PAR, PA = 30\te... | |
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$$120^\circ = (\frac{1}{2} \angle POR) \text{ reflex}$$ $$240^\circ = \text{reflex } \angle POR$$ $$\angle POR = 360^\circ - \text{reflex } \angle POR$$ $$= 360^\circ - 240^\circ$$ $$= 120^\circ$$ $$\angle PAR = \frac{1}{2} \angle POR$$ (Angle subtended at the centre is double than subtended in the opposite se... | |
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[ii] $$\Delta PAR$$ is an equilateral $$\Delta$$ $$a = 30m$$ Area of eq $$\Delta = \frac{\sqrt{3}}{4} a^2$$ $$= \frac{\sqrt{3}}{4} (30)^2$$ $$= \frac{\sqrt{3}}{4} \times 900 \quad 225$$ $$= 225 \sqrt{3} m^2$$ Rough $$\sqrt{400(200)(160)(40)}$$ $$(40 \times 10 \times 40 \times 5 \times 40 \times 40 \times 1)$$ ... | |
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S. Alekhya IX Hubble => $$16x^2 - 42x + 5$$ $$\therefore$$ The Quadratic polynomial whose sum and product of zeroes are $$\frac{21}{8}$$ and $$\frac{5}{16}$$ is $$16x^2 - 42x + 5$$ [10.] Given, Quadratic equation: $$15x^2 + 10\sqrt{6}x + 10 = 0$$ $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$D = b^2 ... | |
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[11.] Given, $$\sqrt{3}$$ is irrational
Required to prove:- $$2-5\sqrt{3}$$ is irrational
let $$P: 2-5\sqrt{3}$$ is irrational
and $$not \ P: 2-5\sqrt{5}$$ is rational
let $$2-5\sqrt{3} = \frac{P}{q} (P, q \in Z, q \neq 0, (P, q) = 1)$$
$$-5\sqrt{3} = \frac{P}{q} - 2$$
$$-5\sqrt{3} = \frac{P-2q}{q}$... | |
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S. Alekhya IX Hubble Section-C III [12] Given, cartons of coke $$= 144$$ cartons of pepsi $$= 90$$ $$144 = 2 \times 2 \times 2 \times 13 = 2^3 \times 13$$ $$90 = 3 \times 3 \times 5 \times 2 = 3^2 \times 5 \times 2$$ To find greatest no-of cartons in each stack: $$HCF = 2^3 \times 13 \tim... | |
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[13] Given, polynomial: $$4x^2 + 4x - 3$$ let $$P(x) = 4x^2 + 4x - 3$$ i) let $$x = \frac{1}{2}$$ $$P\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) - 3$$ $$= \frac{4}{4} + \frac{4}{2} - 3$$ $$= 1 + 2 - 3$$ $$= 3 - 3$$ $$= 0$$ $$\therefore P(x) = 0$$ $$\therefore \... | |
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$$= 9 + (-6) - 3$$
$$= 9 - 6 - 3$$
$$= 9 - 9$$
$$= 0$$
$$\therefore P(x) = 0$$
$$\therefore -\frac{3}{2}$$ is also a zero of $$P(x)$$
$$\alpha + \beta = -\frac{b}{a}$$ and $$\alpha\beta = \frac{c}{a}$$
from $$4x^2 + 4x - 3$$
$$a = 4, b = 4, c = -3$$
$$\alpha + \beta = -\frac{4}{4} = -1$$ | $$\frac{1... | |
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[14] Given, eq(1) $$= 4x - y = 4$$ eq(2) $$: 3x + 2y = 14$$ i) $$4x - y = 4$$ $$4x = 4 + y$$ $$x = \frac{4+y}{4}$$ y | 0 | 4 | 8 x | 1 | 2 | 3 i.e, Points: $$(1, 0), (2, 4)$$ and $$(3, 8)$$ ii) $$3x + 2y = 14$$ $$2y = 14 - 3x$$ $$y = \frac{14-3x}{2}$$ x | 2 | 4 | 5 y | 4 | 1 | -1 i.e, Points: $$(2, ... | |
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After plotting the points on the graph, we can see that the lines intersect at one point i.e Point $$(2,4)$$ $$\therefore x=2, y=4$$ Section-D [IV] [16] Required to prove : $$\sqrt{5}$$ is irrational let $$P : \sqrt{5}$$ is irrational let us assume not $$P : \sqrt{5}$$ is rational let $$\sqrt{5} = \... | |
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[37. i)] $$a_{30} = a + 29d$$ $$a = ₹ 10,000$$ $$d = ₹ 1000$$ $$a_{30} = 10000 + 29 \times 1000$$ $$a_{30} = 29000 + 10000$$ $$a_{30} = 39,000$$ The 30th installements is ₹ 39,000 [37. ii)] $$S_{30} = \frac{n}{2} [2a + (n-1)d]$$ $$= 15 [20000 + 29000]$$ $$= 15 \times 49000$$ $$= ₹ 735000$$ The amou... | |
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Name:- Joseph Class:- 9th Section:- Agni subject:- Maths. [IV] [i.] No. of favourable outcomes = 37 Total No. of possible outcomes = 74 $$\frac{\text{No. of favourable outcomes}}{\text{Total No. of possible outcomes}} = \frac{37}{74}$$ $$= \frac{37}{74} = \frac{37}{74}$$ [ii.] No. of favourable outc... | |
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Name: Joseph. Class: 9th Section: Agni. Subject: Maths. III [5] Median = 50 Marks Frequency Cumulativity frequency 20-30 $$p$$ $$p$$ 30-40 $$15 (cf)$$ $$p+15$$ 40-50 $$25 (f)$$ $$p+15+25 = 40+p$$ Median Class 50-60 $$20$$ $$p+15+25+20 = 60+p$$ 60-70 $$q$$ $$p+15+25+20+q = 60+p+q$$ 70-80 $$8$$ $$p+15... | |
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Name;- Joseph class?- 9th Section!- Agni Subject;- Maths [4] proof:- To prove that $$2-3\sqrt{5}$$ is irrational. Let us Assume that $$2-3\sqrt{5}$$ is rational. $$2-3\sqrt{5} = \frac{a}{b}$$ where $$b \neq 0$$, $$a$$ and $$b$$ are Integers a and b do not have a factor expect $$1$$. $$2-3\sqrt{5} = ... | |
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Name: R. Joseph peter
Class: 9th
Section: Agni
Subject: Maths
II
[3] class Frequency $$x_i$$ $$f_i x_i$$
0-10 5 $$\frac{0+10}{2} = \frac{10}{2} = 5$$ 25
10-20 18 $$\frac{10+20}{2} = \frac{30}{2} = 15$$ 270
20-30 15 $$\frac{20+30}{2} = \frac{50}{2} = 25$$ 375
30-40 f $$\frac{30+40}{2} = \frac{70}{2}... | |
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Name:- R. Joseph peter. class:- $$9^{th}$$ Section:- Agni. Subject:- Maths. I [1] If Three unbiased coins are tossed- The outcomes are:- Head and tail. i. No. of favourable outcomes = 1 Total No. of possible outcomes = 2 $$\frac{\text{No. of favourable outcomes}}{\text{Total No. of possible outcomes... | |
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[13] [Given, $$\alpha = \frac{-3}{2}, \beta = \frac{1}{2}$$] Let $$p(x) = 4x^2 + 4x - 3$$ Let us assume $$p(x) = 0$$ to get zeroes, $$p(x) = 4x^2 + 6x - 2x - 3$$ [Splitting the middle term] $$p(x) = 2x(2x+3) - 1(2x+3)$$ $$p(x) = (2x-1)(2x+3)$$ $$0 = (2x-1)(2x+3)$$ By using zero product rule, $$2x - ... | |
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we know, Sum of zeroes $$( ext{α}+ ext{β}) = rac{-( ext{coefficient of } u)}{( ext{coefficient of } u^2)}$$ $$⋅ rac{-3}{2} + rac{1}{2} = rac{-(4)}{4}$$ $$ rac{-3+1}{2} = -1$$ $$ rac{-2}{2} = -1$$ $$-1 = -1$$ we also know, product of zeroes $$( ext{α} ext{β}) = rac{c}{a}$$ $$ rac{-3}{2} imes r... | |
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[14] Let us redraw the diagram with the given values. $$AB = 5cm, BE = 7cm, BC = u-y, CD = u+y$$. Given that perimeter = $$27 cm$$ we know, Perimeter = $$AB + BC + CD + DE + AE$$ $$27 = 5 + (u-y) + (u+y) + (u-y) [BC || DE] + 5$$ $$27 = 10 + 3u - y$$ $$17 = 3u - y \rightarrow ①$$ AS $$CD || BE$$, $$C... | |
Roll No:4 , Page :10
$$17 + 7 = (3u + u) + (-y + y)$$
$$24 = 4u$$
$$\leftarrow$$
$$u = \frac{24}{4}$$
$$u = 6$$
Substituting '$$u$$' in equation ②,
$$u + y = 7$$
$$6 + y = 7 \Rightarrow y = 7 - 6$$
$$y = 1$$
Therefore, $$u = 6$$ and $$y = 1$$
[15] $$\frac{1}{u} - \frac{1}{u-2} = 3$$
Multi
$$\frac{(u-2) - (u)}{u(u-2)} ... | |
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$$au^2+bu+c$$. we get, $$a=3, b=-6, c=2$$ we know, $$D = b^2-4ac$$ $$= 36-24$$ $$D=12$$ As $$D>0$$, It has real roots. we know, $$u = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$u = \frac{-b + \sqrt{b^2-4ac}}{2a}$$ $$u = \frac{-b - \sqrt{b^2-4ac}}{2a}$$ $$u = \frac{-(-6) + \sqrt{(6)^2-4(3)(2)}}{2(3)}$$ $$... | |
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Therefore, $$x = \frac{3 \pm \sqrt{3}}{3}$$ Section-D IV) [16] Let us assume $$\sqrt{5}$$ is not an irrational number. That is, $$\sqrt{5}$$ is rational. Let, $$\sqrt{5} = \frac{p}{q}$$ (where $$p, q \in \mathbb{Z}, q \neq 0, (p, q) = 1$$) -- (1) $$5 = \frac{p^2}{q^2}$$ [Squaring on both sides] $$5... | |
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$$q^2 = 5k^2 \rightarrow (3)$$ So, 5 is a factor of $$q^2$$ then 5 is also a factor of $$q$$. From (2) and (3), 5 is a factor of $$p$$ and $$q$$. $$\therefore$$ It is a contradiction from (1). $$\therefore$$ Our assumption is wrong. $$\therefore \sqrt{5}$$ is an irrational number. [17] Given, $$cu ... | |
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First, let us take $$\frac{a_1}{a_2} = \frac{b_1}{b_2}$$ $$\Rightarrow \frac{c}{12} = \frac{3}{c}$$ $$c^2 = 36$$ $$c = \sqrt{36}$$ $$c = 6$$ Therefore, the value of $$c$$ for which the system of equations has infinitely many solutions is '6'. Section-E [V)] [18)i)] Yes, kavita's claim is correct an... | |
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Given,
$$\alpha = \frac{-2}{\sqrt{3}}$$ , $$\beta = \frac{\sqrt{3}}{4}$$
Sum of zeroes $$(\alpha + \beta) = \frac{-2}{\sqrt{3}} + \frac{\sqrt{3}}{4}$$
$$= \frac{(-2)(4) + (\sqrt{3})(\sqrt{3})}{4\sqrt{3}} [L.C.M]$$
$$= \frac{-8+3}{4\sqrt{3}}$$
Sum of zeroes $$(\alpha + \beta) = \frac{-5}{4\sqrt{3}}$... | |
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Multiplying with $$4\sqrt{3}$$ to remove denominator, required polynomial $$= 4\sqrt{3}(u^2) + (4\sqrt{3})\frac{5u}{4\sqrt{3}} - \frac{(\sqrt{3})^2}{4\sqrt{3}} x$$ required polynomial $$= 4\sqrt{3}u^2 + 5u - 2\sqrt{3}$$ [iii] Given, $$\alpha = \frac{-2}{\sqrt{3}}, \beta = \frac{\sqrt{3}}{4}$$ $$\Ri... | |
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$$= \frac{(\frac{-5}{4\sqrt{3}})^2 + 1}{\frac{1}{4}}$$ $$= \frac{\frac{25}{16 \times 3} + 1}{\frac{1}{4}} = \frac{\frac{25}{48} + 1}{\frac{1}{4}}$$ $$= (\frac{25+48}{48}) \times 4$$ $$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{73}{12}$$ Section-C [Extra] [III] [12] Given that three bells toll a... | |
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Here, common prime factors are is only 3. So, $$HCF = 3$$ Section-E [v) 18) iii)] $$(\frac{\alpha^{\frac{1}{3}}}{\beta^{\frac{2}{3}}} + \frac{\beta^{\frac{1}{3}}}{\alpha^{\frac{2}{3}}})^6$$ $$= \frac{\alpha^{\frac{6}{3}}}{\beta}$$ Section-B [II) 10)] Let $$\alpha = \frac{2}{a}, \beta = a$$ $$\Right... | |
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$$\Rightarrow (\frac{2}{a})^2 + 4(\frac{2}{a}) + 2a = 0$$ $$\frac{4}{a^2} + \frac{8}{a} + 2a = 0$$ $$\Rightarrow (a)^2 + 4(a) + 2a = 0$$ $$a^2 + 4a + 2a = 0$$ $$a^2 + 6a = 0$$ $$a(a+6) = 0$$ By zero product rule $$a = 0$$ | $$a+6=0$$ $$a = -6$$ $$a = 0, -6$$ | |
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Section - A [1] (a) [2] (b) [3] (d) [4] (d) [5] (c) [6] (b) [7] (c) [8] (b) [9] (a) [10] (c) [11] (a) [12] (b) [13] (c) [14] (b) [15] (b) [16] (b) [17] (c) [18] (b) [19] (a) [20] (b) Section - B [21] Given, Sum of zeroes of the polynomial $$= -3$$ Product of zeroes of the polynomial $$= 2$$ | |
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$$\alpha + \beta = -3$$\n$$\alpha \beta = 2$$\n$$p(x) = k[x^2 - (\alpha + \beta)x + \alpha \beta]$$\n$$= K[x^2 - (-3)x + (2)]$$\n$$= K[x^2 + 3x + 2]$$\nlet $$K = 1$$ then\n$$\therefore p(x) = x^2 + 3x + 2$$\n\n[22] Given\n$$a_{11} = 38$$\n$$a_{16} = 73$$\n$$a_{16} - a_{11} = 73 - 38$$\n$$a + 15d - (... |
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