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The median of the following data is 525. Find the values of `x` and `y`, if the total frequency is 100. Class interval 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 800-900 900-1000 Frequency 2 5 x 12 17 20 y 9 7 4 SOLUTION Median = 525 Median Class = 500 – 600 Class interval Frequency (f) Cumulative frequency (cf) 0-100 2 2 100-200 5 7 200-300 x 7 + x 300-400 12 19 + x 400-500 17 36 + x 500-600 20 56 + x 600-700 y 56 + x + y 700-800 9 65 + x + y 800-900 7 72 + x + y 900-1000 4 76 + x + y N = `\sum` fi = 100   = 76 + x + y = 100 = x + y = 24 ….(i) Median = `1+\frac{\frac{n}{2}-F}{f}\times h` Since, l=500,h=100,f=20,F=36+x and N=100 Therefore, putting the value in the Median formula, we get; x = 9 y = 24 – x (from eq.i) y = 24 – 9 = 15 Therefore, the value of x = 9 and y = 15. 35 36 37.1 37.2 38 Class-10 SEBA Maths Question Paper Solution 2022

A toy in the form of cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm, find the total surface area of toy. SOLUTION Given, Radius of cone = 3.5 cm, height of cone = 15.5 – 3.5 = 12 cm Slant height of cone can be calculated as follows: `l = \sqrt{h^{2}+r^{2}}` `l = \sqrt{12^{2}+3.5^{2}}` `l = \sqrt{144^{2}+12.25^{2}}` `l = \sqrt{156.25}` = 12.5 cm Curved surface area of cone can be calculated as follows: = `\pi rl` = `22/7``\times` 3.5 `\times` 12.5 = 137.5 `"cm"^2` Curved surface area of hemispherical portion can be calculated as follows: = `2\pi r^2` = `22/7``\times` 3.5 `\times` 3.5 = 77 `"cm"^2` Hence, total surface area = 137.5 + 77 = 214.5 `"cm"^2` 35 36 37.1 37.2 38 Class-10 SEBA Maths Question Paper Solution 2022

Find the area of the shaded regionin the figure below, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral traingle OAB of side 12 cm as centre. SOLUTION Here we can see that Area of required figure = Area of circle+Area of triangle− Area of the sector ⇒A(circle)=`π"r"^2` ⇒A(circle)=`\pi` `6^2` ⇒A(circle)=36`\pi` ⇒A(triangle)=`\frac{\sqrt{3}}{4}a^2` ⇒A(triangle)=`\frac{\sqrt{3}}{4}(12^2)` ⇒A(triangle)= `36\sqrt{3}` ⇒A(sector)=`1/2``"r"^2``\theta` ⇒A(sector)=`1/2`(`"6"^2`)`\frac{\pi}{3}` ⇒A(sector)=6`\pi` Therefore, Required Area=36`\pi` + `36 \sqrt{3}` − 6`\pi` ⇒ Required Area=36`\pi` + `36 \sqrt{3}`​ 35 36 37.1 37.2 38 Class-10 SEBA Maths Question Paper Solution 2022

Construct a triangle similar to a given traingle ABC with its sides equal to `5/3` of the corresponding sides of the triangle ABC. (Write the steps of construction.) SOLUTION Steps of Construction: Step 1: Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. Step 2: From B cut off 5 arcs `"B"_1`,`"B"_2`,`"B"_3`,`"B"_4` and `"B"_5` on BX so that `"BB"_1`, = `"B"_1``"B"_2` = `"B"_3``"B"_3` = `"B"_4``"B"_5` Step 3: Join `"B"_3` to C and draw a line through `"B"_5` parallel to `"B"_3`C, interacting the extended line segment BC at C'. Step 4: Draw a line through C' parallel to CA interacting the extended line segment BA at A' (see figure). Than, A' BC' is the required triangle. Justification: Note that △ ABC 〜 △ A'BC' (Since AC || A'C') Therefore, `\frac{

If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus. SOLUTION Let ABCD be a parallelogram whose sides AB, BC, CD and DA touch a circle at the points P, Q, R and S respectively. Since the lengths of tangents drawn from an external point to a circle are equal, we have AP = AS, BP = BQ, CR = CQ and DR = DS. AB + CD = AP + BP + CR + DR = AS + BQ + CQ + DS = (AS + DS) + (BQ+CQ) = AD + BC Now, AB + CD = AD + BC ⇒ 2AB = 2BC ⇒ AB = BC AB = BC = CD = AD. Hence, ABCD is a rhombus. 33 34 35 36 37.1 Class-10 SEBA Maths Question Paper Solution 2022

The shadow of a tower standing on a level ground is found to be 40 m longer when Suns altitude is 30 than when it was 60. Find the height of the tower. (Take = `\sqrt{3}` = 1.732) SOLUTION Given tower be AB When Sun's altitude is `60^0` `\angle` ACB = `60^0` & Length of shadow = BC When Sun's altitude is `30^0` `\angle` ADB = `30^0` & Length of shadow = DB Shadow is 40 m when angle changes from 60° to 30° CD = 40 m We need to find height of tower i.e. AB Since tower is vertical to ground ∴ ∠ ABC = 90° From (1) and (2) `\frac{"AB"}{\sqrt{3}}` = `\sqrt{3}`AB - 40 AB = `\sqrt{3}` (`\sqrt{3}`AB) - 40 `\sqrt{3}` AB = 3AB - 40 `\sqrt{3}` 40 `\sqrt{3}` = 3AB - AB 40 `\sqrt{3}` = 2AB AB = `\frac{ 40 \sqrt{3}}{2}` AB = 20`\sqrt{3}` Hence, Height of the tower = AB = 20`\sqrt{3}`metre 32 33 34 35 36 Class-10 SEBA Maths Question Paper Solution 2022

If the areas of two similar traingals are equal, Prove that they are congruent. SOLUTION Given, ar (`\triangle`ABC) = ar (`\triangle`PQR) `\triangle`ABC 〜 `\triangle`DEF We Know that `\frac{\triangle "ABC"}{\triangle "PQR"}` = `"AB"^2/"PQ"^2` = `"BC"^2/"QR"^2` = `"CA"^2/"PR"^2` Now, 1 = `"AB"^2/"PQ"^2` AB = PQ Similarly, 1 = `"BC"^2/"QR"^2` BC = QR and, 1 = `"CA"^2/"PR"^2` CA = PR Thus, AB = PQ, BC = QR and CA = PR ∴ `\triangle "ABC" \cong \triangle "PQR"` 31 32 33 34 35 Class-10 SEBA Maths Question Paper Solution 2022

Solve the pair of equations by reducing them to a pair of liner equations. `\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}` `\frac{1}{2(3x+y)}+\frac{1}{2(3x-y)}=\frac{-1}{ 8}` SOLUTION `\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}` Let a= `1/3` x + y   , and b= `1/"3x"` - y a + b = 3/4       ....... (i) 4a + 4b = 3       ....... (ii) also , `\frac{1}{2(3x+y)}+\frac{1}{2(3x-y)}=\frac{-1}{ 8}` `"a"/2` - `"b"/2` = `"-1"/"8"` `"(a - b)"/"2"` = `"-1"/"8"` a - b = `"-1"/"4"` 4a - 4b = -1       ....... (ii) Adding equations (i) and (ii)  4a + 4b = 3 4a - 4b = -1 8a = 2 a = `2/8` a = `1/4` 4a - 4b = -1 4 `\times` `1/4` - 4b = -1 1 - 4b = -1 4b = 1 + 1 4b = 2 b = `2/4` b = `1/2` `"1"/"3x"` + y = a `"1"/"3x"` + y = `1/4` 3x + y = 4        ....... (iii) 1/3x - y = b 1/3x - y = 1/2 3x - y = 2        ....... (iv) Adding equations (iii) and

Find the area of the sector of a circle with radius 4 cm and angle `30^0`. Also, find the area of the corresponding major sector. (`\pi`= 3.14) SOLUTION Here, Radius (r) = 4 cm `\theta` = `30^0` Now, Area of sector = `\frac{\theta}{360} \times \pi r^2` = `30/360` ` \times` 3.14 ` \times` `(4)^2` = `1/12` ` \times` 3.14 ` \times` 4 ` \times` 4 = `1/3` ` \times` 3.14 ` \times` 4 = `12.56/3` `"cm"^2` = 4.19 `"cm"^2` Now, Area of major sector = `\frac{(360 - \theta)}{360} \times \pi r^2` = `\frac{(360 - 30)}{360} \times 3.14 \times 4^2` = `\frac{(330)}{360} \times` 3.14 `\times` 4 `\times` 4 = `11/12` `\times` 3.14 `\times` 4 `\times` 4 = 46.05 `"cm"^2` 29 30 31 32 33 Class-10 SEBA Maths Question Paper Solution 2022

Find the area of the traingle formed by joining the middle points of the sides of the traingle whose vertices are (0, -1), (2, 1) and (0, 3) SOLUTION Let the vertices of triangle be A(0,-1) B(2,1) C(0,3) Let the mid-point of AB be P BC be Q AC be R Now, joining the Points P,Q,R We get ΔPQR We need to find area of ΔABC and area of ΔPQR Now, finding area of ΔABC Area of ΔABC = `1/2` [`"x"_1("y"_2-"y"_3) + "x"_2("y"_3-"y"_1) + "x"_3("y"_1-"y"_2)`] Here, `"x"_1` = 0 , `"y"_1` = -1 `"x"_2` = 2 , `"y"_2` = 1 `"x"_3` = 0 , `"y"_3` = 3 Putting values Area of Traingle = `1/2` [`"x"_1("y"_2-"y"_3) + "x"_2("y"_3-"y"_1) + "x"_3("y"_1-"y"_2)`] Area of ΔABC = `1/2` [0 (1 - 3) + 2 (3 - (-1) + 0 ( -1-1)] = `1/2` [0 + 2(3 + 1) + 0] = `1/2` [0 + 2(4) + 0

D and E are points on the sides CA and CB respectively of a triangle of a traingal ABC right angled at C. Prove that `AE^2+BD^2`=`AB^2+DE^2` SOLUTION Given, Triangle ABC, right angled at C Two Points D and E are on the sides CA and CB To Prove that, `"AE"_2` + `"BD"_2` = `"AB"_2` + `"DE"_2` Let us join the points D with E and B and P with A Using Pythagoras theorem `"(Hypotenuse)"_2` = `"(Height)"_2` +`"(Base)"_2` Here, ACE is a right angle triangle `"AE"_2` = `"AC"_2` +`"CE"_2` .....(1) In right angle triangle DCB `"BD"_2` = `"DC"_2` +`"BC"_2` .....(2) In right angle triangle ABC `"AB"_2` = `"AC"_2` +`"BC"_2` .....(3) In right angle triangle DCE `"DE"_2` = `"DC"_2` +`"CE"_2` .....(4) L.H.S `"AE"_2` + `"BD"_2` = `"(AC"_2` + `"CE"_2)`

Find the sum of first 51 terms of an AP whose secound and third terms are 14 and 18 respectively. SOLUTION We Know that `"a"_n` = a + (n - 1) d Given 2nd term is 14 `"a"_2` = a + (2 - 1) d 14 = a + d 14 - d = a a = 14 - d .....(1) Given 3rd term is 18 `"a"_3` = a + (3 - 1) d 18 = a + d 18 - d = a a = 18 - d .....(2) From (1) & (2) 14 - d = 18 - 2d 2d - d = 18 - 14 d = 4 Putting value of d in (1) a = 14 - d a = 14 - 4 a = 10 Now, we need to find sum of first 51 terms According formula, `"S"_n` = `n/2` [2a + (n - 1) d] Now, Putting values, (n = 51, a = 10 & d = 4 ) `"S"_n` = `51/2` [2`\times`10 + (51 - 1) 4] `"S"_n` = `51/2` [20 + 50`\times`4] `"S"_n` = `51/2` [20 + 200] `"S"_n` = `51/2` `\times` 220 `"S"_n` = 51 `\times` 110 `"S"_n` = 5610 Hence, the sum of first 51 terms is 5610 26 27 28

Find the coordinates of a point A, where AB is a diameter of a circle whose centre is (2,-3) and B is (1,4). SOLUTION Let the circle be as shown with centre C (2, -3) Let AB be the diameter of the circle Since AB is the diameter , Centre C must be the mid - point of AB  Let (x,y) Since C  is the mid- point of AB x-coordinate of C = `\frac{x_{1}+x_{2}}{2}` y-coordinate of C = `\frac{y_{1}+y_{2}}{2}` Where, `x_1` = `x` `y_1` = `y`  `x_2` = 1  `y_1` = 4 `x` Coordinate of C = `\frac{x + 1}{2}` 2 = `\frac{x + 1}{2}` 2 `\times` 2 = `x` + 1 4 = `x` + 1 4 - 1= `x` 3= `x` `y` Coordinate of C = `\frac{y + 4}{2}` -3 = `\frac{y + 4}{2}` -3 `\times` 2 = `y` + 1 -6 - 4 = `y` - 10= `x` Hence, the coodinates of A (x,y) = A (3, -10) 25 26 27 28 29 Class-10 SEBA Maths Question Paper Solution 2022

Sum of the areas of two squares is `468m^2`. If the differences of their perimeters is 24 m, find the sides of two squres. Divide the polynomial p(`x`) by the polynomial g(`x`), and the find the quotient and the remainder. p(`x`)=`x^4-5x+6,` `g(x)= 2-x^2` SOLUTION Let side of square 1 be x metres Perimeter of square 1 = 4`\times`Side = 4x Now, it is given that Differences of Perimeter of square is 24 m Perimeter of square 1 - Perimeter of square 2 = 24 4x - Perimeter of square 2 4x - 24 = Perimeter of square 2 Perimeter of square 2 = 4x - 24  Now,  Perimeter of square 2 = 4x - 24  4`\times`(Side of square 2) = 4x - 24 Side of square 2 = `"4x-25"/"4"` = `"4(x-6)"/"4"` = x-6 Hence,  Side of square 1 is x & Side of square 2 is x-6 Also, given that Sum of area of square is 468`"m"^2` Area of square 1 + Area of square 2 = 468 `"(Side of square 1)"^2` + `"(Side of square 1)"^2` = 468 `x^2 + (x - 6)^`=

Solve `3x^2-2 \sqrt{6}x+2 = 0` SOLUTION  `3x^2-2 \sqrt{6}x+2 = 0` `3x^2-\sqrt{6}x-\sqrt{6}x+2=0` `3x^2-\sqrt{2\times3}x-\sqrt{2\times3}x+2=0` `3x^2-(\sqrt{2})(\sqrt{3})x-(\sqrt{2})(\sqrt{3})x+2=0` `\sqrt{3}x(\sqrt{3}x-\sqrt{2})-\sqrt{2}(\sqrt{3}x-\sqrt{2})`=0 `(\sqrt{3}x-\sqrt{2})(\sqrt{3}x-\sqrt{2})`=0 `\sqrt{2}=0` `\sqrt{3}x=\sqrt{2}` `x=\frac{\sqrt{2}}{\sqrt{3}}` `x=\sqrt{\frac{2}{3}}` `\sqrt{2}=0` `\sqrt{3}x=\sqrt{2}` `x=\frac{\sqrt{2}}{\sqrt{3}}` `x=\sqrt{\frac{2}{3}}` Hence, x=`\sqrt{\frac{2}{3}}`  and `x=\sqrt{\frac{2}{3}}` are root of the equation 23 24 25 26 27 Class-10 SEBA Maths Question Paper Solution 2022

A fraction becomes `9/11`, if 2 is added to both the numerator and denominator. If 3 is added to both the numerator and the denominator, it becomes `5/6`. Find the fraction. SOLUTION Let Numerator be x & Denominator be y So, Fraction is = `x/y` Given that, If 2 is added to both the numerators and denominator, fraction becomes `9/11` `"Numerators + 2"/"Denominator + 2"`= `9/11` `"x+ 2"/"y + 2"`= `9/11` 11 (`x + 2`) = 9 (`y + 2`) 11`x` + 22 = 9`y` + 18 11`x` - 9`y` = 11 - 22 11`x` - 9`y` = -4 .......(1) Also, Given that if 3 is added to both the numerator and the denominator, fraction become `5/6` `"Numerators + 3"/"Denominator + 3"`= `5/6` `"x+ 3"/"y + 3"`= `5/6` 6 (`x + 3`) = 5 (`y + 3`) 6`x` + 18 = 5`y` + 15 6`x` - 5`y` = 15 - 18 6`x` - 5`y` = - 3 .......(2) Hence, our equations are 11`x` - 9`y` = -4 .......(1) 6`x` - 5`y` = - 3 .......(2) From equation (1

state the Division Algorithm for polynomials. Divide the polynomial p(`x`) by the polynomial g(`x`), and the find the quotient and the remainder. SOLUTION Thus, Quotient = -`x^2` - 2 Remainder = -5x + 10 21 22 23 24 25 Class-10 SEBA Maths Question Paper Solution 2022

Prove that `\sqrt{7}` is irrational. SOLUTION Let us assume, to the contrary, that ,√7 is a rational number. Then, there exist co-prime positive integers a and b such that `\sqrt{7}` = `"a"/"b"`    b ≠ 0 So, a `\sqrt{7}`b     .....(i) Squaring both sides, we have  `"a"^2` = 7`"b"^2`   = 7 divides a So we can write a = 7c,         (where c is any integer) Putting the value of a = 7 c in (i) , we have 49 `"c"^2` = 7`"b"^2` 7 `"c"^2` = `"b"^2` It means 7 divides `"b"^2` and so  7 divides b. So, 7 is a common factor of both a and b which is a contradiction. So, our assumption that `\sqrt{7}` is a rational is wrong. Hence, we conclude that `\sqrt{7}` is a irational number. 20 21 22 23 24 Class-10 SEBA Maths Question Paper Solution 2022

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting . a   a king of red colour b   a spade SOLUTION Solution 01 Total Numbers of Cards = 52 Total Numbers of Kings of Red Colour = 2 P (`"getting a king cards of red colour"/"Total Numbers of cards"`) = `2/52` = `1/26` Solution 02 Total Numbers of Cards = 52 [4 suits - (i) spade, (ii) club, (iii) diamond and (iv) hearts] Total Numbers of Spade Cards = 13 P (`"getting a Spade Card"/"Total Numbers of cards"`) = `13/52` = `1/4` 19 20 21 22 23 Class-10 SEBA Maths Question Paper Solution 2022

Prove that `(cosec\theta-cos\theta)^2 = \frac{1 - cos\theta}{1 + cos\theta}` SOLUTION  LHS = `(cosec\theta-cos\theta)^2 = \frac{1 - cos\theta}{1 + cos\theta}` `(cosec\theta-cos\theta)^2` = `(\frac{1}{sin\theta} - \frac{cos\theta}{sin\theta})^2` `\frac{(1-cos\theta)^2}{sin^2\theta} = \frac{(1-cos\theta)^2}{1-cos\theta^2}` `\frac{(1-cos\theta)^2}{(1-cos\theta^2)(1+cos\theta^2)} = \frac{1-cos\theta^2}{1+cos\theta^2}` = RHS 18 19 20 21 22 Class-10 SEBA Maths Question Paper Solution 2022

If sec 4A=cosec (A - `20^0`), find the Value of A. (4A is an acute angle) SOLUTION Given that, sec 4A = cosec (A - `20^0`) ….(i) sec A = cosec (`90^0` - A) By using property in equation (i) we get: cosec (`90^0` - 4A) = cosec (A - `20^0`) `90^0` - 4A = A - `20^0` 5A = `110^0` A = `110^0/5` A = `22^0` 17 18 19 20 21 Class-10 SEBA Maths Question Paper Solution 2022

Evaluate: `\frac{5cos^2 60^0+4sec^2 30^0-tan^2 45^0}{sin^2 30^0+sin^2 60^0}` SOLUTION `\frac{5cos^2 60^0+4sec^2 30^0-tan^2 45^0}{sin^2 30^0+sin^2 60^0}` `\frac{5(\frac{1}{2})^2 + 4(\frac{2}{\sqrt{3}})^2-(1)^2}{(\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}` `\frac{5(\frac{1}{4})+(\frac{16}{3})-1}{\frac{1}{4}+\frac{3}{4}}` `\frac{\frac{15+54-12}{12}}{\frac{4}{4}}` `\frac{67}{12}` 16 17 18 19 20 Class-10 SEBA Maths Question Paper Solution 2022

If sec A=`13/12`, calculate sin A and cot A. (A is an acute angle.) SOLUTION  Finding cos θ   cos θ = `"1"/"sec θ "`  cos θ = `\frac{1}{\frac{13}{12}}`  cos θ = `\frac{12} {13}` `"Side adjacent to ∠θ "/"Hypotenuse"` = `\frac{12} {13}` `"AB"/"BC"` = `12/13` Let, AB = 12x & AC = 13x Using Pythagoras theorem to find BC `"(Hypotenuse)"^2` = `"(Height)"^2` +  `"(Base)"^2` `"AC"^2` = `"AB"^2` + `"BC"^2`  `"(13x)"^2` = `"(12x)"^2` +  `"(BC)"^2`  `"(BC)"^2` = `"(13x)"^2` - `"(12x)"^2`   `"(BC)"^2` = 169`"x"^2` - 144`"x"^2`  `"(BC)"^2` = 25`"x"^2` BC = `\sqrt{25"x"^{2}}` BC = `\sqrt{5^2x^{2}}` BC = 5x Now, sin θ = `"BC"/"AC"` = `"5x"/"13x"` = `"5"/"13"` cos

Find the cordinates of the point which divides the line segment joining the points (4, -3) and (8, 5) in the ratio 3:1 internally. SOLUTION Let P(x,y) be the point which divides the line segment internally. Using the section formula for the internal division, i.e. (x, y) = `(\frac{"m"_{1}"x"_{2}+"m"_{2}"x"_{1}}{"m"_{1}+"m"_{2}},\frac{"m"_{1}"y"_{2}+"m"_{2}"y"_{1}}{"m"_{1}+"m"_{2}})` Given Us, `"m"_1` = 3, `"m"_2` = 1 (`"x"_1`​,`"y"_1`​) = (4, −3) (`"x"_2`​,`"y"_2`​) = (8, 5) Putting the above values in the above formula, we get : x = `\frac{3(8)+1(4)}{3+1}`, y = `\frac{3(5)+1(-3)}{3+1}` x = `\frac{24+4}{4}`, y = `\frac{15-3}{4}` x = `\frac{28}{4}`, y = `\frac{12}{4}` x = 7, y = 3 Hence, (7,3) is the point which divides the line segment internally.

How many two-Digit numbers are divisible by 5 ? SOLUTION Two-digit numbers divisible by 5 are 10,15,20,25 …………………95 We note that here the common difference(d) is 5. Here the sequence is in arithmetic progression. We can take a = 1 0 , d = 5 , l = 9 5 ∴ `"a"_"n"` = a + (n − 1) d, 9 5 = 1 0 + (n − 1) 5, n − 1 = 85/5, n − 1 = 1 7, n = 1 8. Hence 18 two-digit numbers are divisible by 5. 13 14 15 16 17 Class-10 SEBA Maths Question Paper Solution 2022