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Showing posts from March, 2022

### The Median of the Following Data is 525. Find the Values of x and y, If the Total Frequency is 100. | Bzziii

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. | Bzziii

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.

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.)

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"_3C, 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{

### All the Sides of a Parallelogram touch a Circle, show that the Parallelogram is a Rhombus. | Bzziii

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 | Bzziii

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

### Areas of two Similar traingals are Equal, Prove that they are Congruent | Bzziii.com

If the areas of two similar traingals are equal, Prove that they are congruent. SOLUTION Given, ar (\triangleABC) = ar (\trianglePQR) \triangleABC 〜 \triangleDEF 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

### 1/3x + y + 1/3x - y = 3/4 , 1/2(3x + y) - 1/2(x-y) = -1/8 - Solve the Pair of Equations by Reducing them to a Pair of Liner Equations | Bzziii.com

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) | Bzziii.com

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) | Bzziii.com

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 | Bzziii.com

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 | Bzziii.com

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\times10 + (51 - 1) 4] "S"_n = 51/2 [20 + 50\times4] "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) | Bzziii

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 | Bzziii.com

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\timesSide = 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 root 6x+2 = 0 | Bzziii.com

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

### 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. | Bzziii.com

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) 11x + 22 = 9y + 18 11x - 9y = 11 - 22 11x - 9y = -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) 6x + 18 = 5y + 15 6x - 5y = 15 - 18 6x - 5y = - 3 .......(2) Hence, our equations are 11x - 9y = -4 .......(1) 6x - 5y = - 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. | Bzziii.com

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 ROOT 7 is IRRATIONAL. - Bzziii.com

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 . (i) A King of Red Colour (ii) A Spade | Bzziiii.com

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 - cos)2 = 1 - cos/1 + cos | Bzziii.com

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) | Bzziii.com

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: 5cos2 60 + 4sec2 30-tan2 45 / sin2 30+sin2 60 - Bzziii

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.) | Bzziii.com

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. | Bzziii.com

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 ? - Bzziii.com

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