SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available 10 th STD. MARCH - 017 Public Exam Question Paper with Answers MATHEMATICS [Time Allowed : ½ Hrs.] [Maximum Marks : 100] SECTION - I 7. Slope of the straight line which is perpendicular to the straight line joining the points (, 6) and Note: (i) Answer all the 15 questions. (4, ) is equal to : (ii) Choose the correct answer from the 1 (a) (b) (c) (d) 1 given four alternatives and write the option code and the corresponding answer. [15 1 15]. If the points (, 5), (4, 6) and (a, a) are collinear, 1. If f(x) x then the value of a is equal to : + 5, than f( 4) (a) 6 (b) 1 (a) (b) 4 (c) 0 (d) 0 (c) 4 (d). If k +, 4k 6, k are the three consecutive terms of an A.P., then the value of k is : 9. The perimeters of two similar triangles are 4 cm and 1 cm respectively. If one side of (a) (b) (c) 4 (d) 5 the first triangle is cm, then corresponding. If the product of the first four consecutive terms side of the other triangle is : of a G.P. is 56 and if the common ratio is 4 (a) 4 cm (b) cm and the first term is positive, then its rd term (c) 9 cm (d) 6 cm is : 1 10. ABC is a right angled triangle where 1 (a) (b) (c) (d) 16 B 90º and BD AC. If BD cm, 16 4. The remainder when x x + 7 is divided by x + 4 is : (a) (b) 9 (c) 0 (d) 1 5. The common root of the equations x bx + c 0 and x + bx a 0 is : (a) c+ a (b) c a b b (c) c+ b a (d) a + b c 7 1 0 6. If A 1 and A + B 4 then the matrix B 1 0 (a) 0 1 (c) 1 7 (b) 6 1 (d) 1 7 AD 4 cm, then CD is : (a) 4 cm (b) 16 cm (c) cm (d) cm 11. In the adjoining figure ABC (a) 45º C (b) 0º (c) 60º (d) 50º 1. 9 tan q 9 sec q (a) 1 (b) 0 (c) 9 (d) 9 1. If the surface area of a sphere is 100 π cm, then its radius is equal to : (a) 5 cm (b) 100 cm (c) 5 cm 100 m A (d) 10 cm 100m B [1] http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available Sura s Mathematics - X Std. March 017 Question Paper with Answers 14. Standard deviation of a collection of a data is. If each value is multiplied by, then the standard deviation of the new data is : (a) 1 (b) 4 (c) 6 (d) 9 15. A card is drawn from a pack of 5 cards at random. The probability of getting neither an ace nor a king card is: (a) (c) 1 4 1 (b) 11 1 (d) 1 SECTION - II Note: (i) Answer 10 questions (ii) Question number 0 is compulsory. 7. The total surface area of a solid right circular cylinder is 1540 cm. If the height is four times the radius of the base, then find the height of the cylinder. Select any 9 questions from the first. The smallest value of a collection of data is 1 14 questions. [10 0] and the range is 59. Find the largest value of 16. Given, A {1,,, 4, 5}, B {, 4, 5, 6} and the collection of data. C {5, 6, 7, }, show that 9. In tossing a fair coin twice, find the probability AÈ(BÈC) (AÈ B)ÈC of getting : 17. The following table represents a function from (i) Two heads (ii) Exactly one tail A {5, 6,, 10} to B {19, 15, 9, 11} where f (x) x 1. Find the values of a and b. 0. (a) If the volume of a solid sphere is x 5 6 10 f(x) a 11 b 19 1. If 7, m, 7 (m + ) are in G.P. find the values of m. 19. Solve by elimination method : 1x + 11y 70, 11x + 1y 74. 0. Simplify : 6 x + 9 x. x 1x 1. Construct a matrix A [a ij ] whose elements are given by a ij i j.. Let A 5 1 and B matrix C, if C A + B. 1 4. Find the. Find the coordinates of the point which divides the line segment joining (, 5) and (4, 9) in the ratio 1 : 6 internally. 4. The points (0, a), a > 0 lie on x - axis for all a. Justify the truthness of the statement. 5. In PQR, AB QR. If AB is cm, PB is cm and PR is 6 cm, then find the length of QR. 6. The angle of elevation of the top of a tower as seen by an observer is 0º. The observer is at a distance of 0 m from the tower. If the eye level of the observer is 1.5 m above the ground level, then find the height of the tower. 741 1 7 cu.cm, then find its radius. Take π 7 [OR] (b) If x a secθ + b tanθ and y a tanθ + b sec θ, then prove that x y a b. SECTION - III Note: (i) Answer 9 questions. (ii) Question No. 45 is Compulsory. Select any questions from the 14 questions. [9 5 45] 1. Let A {a, b, c, d, e, f. g, x, y, z}, B {1,, c, d, e} and C {d, e, f, g,, y}. Verify A\ (BÈC) (A \ B) (A\ C). http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available Sura s Mathematics - X Std. March 017 Question Paper with Answers. Let A {6, 9, 15, 1, 1}; B {1,, 4, 5, 6} and f : A B be defined by f(x) x. Represent f by : (i) an arrow diagram (ii) a set of ordered pairs (iii) a table (iv) a graph. Find the sum of the first n terms of the series 1 + 4 +... 4. Find the sum of first n terms of the series 7 + 77 + 777 +... 5. The speed of a boat in still water is 15 km/hr. (b) A straight line cuts the coordinate axes It goes 0 km upstream and return downstream at A and B. If the mid point of AB is (, ), then find the equation of AB. to the original point in 4 hrs. 0 minutes. Find the speed of the stream. SECTION - IV 6. Find the values of a and b if Note: Answer both the questions choosing either 16x 4 4x + (a 1) x + (b +1) x + 49 is a of the alternatives. [ 10 0] perfect square. 46. (a) Draw the two tangents from a point which is 10 cm away from the centre of a circle 5 1 7. If A 7 and B 1 1 verify that of radius 6 cm. Also, measure the lengths of the tangents. (AB) T B T A T.. Find the area of the quadrilateral formed by the points ( 4, ), (, 5), (, ) and (, ). 9. State and prove Pythagoras theorem. 40. A flag post stands on the top of a building. From a point on the ground, the angles of elevation of the top and bottom of the flag post are 60º and 45º respectively. If the height of the flag post is 10m, find the height of the building. 17. ( ) 41. The perimeter of the ends of a frustum of a cone are 44 cm and.4 π cm. If the depth is 14 cm, then find its volume. 4. The length, breadth and height of a solid metallic cuboid are 44 cm, 1 cm and 1 cm respectively. It is melted and a solid cone is made out of it. If the height of the cone is 4 cm, then find the diameter of its base. 4. Find the coefficient of variation of the following data. 1, 0, 15, 1, 5 44. If a die is rolled twice, find the probability of getting an even number in the first time or a total of. 45. (a) Find the GCD of the following polynomials x 4 + 6x 1x 4x and 4x 4 + 14x + x x. [OR] [OR] (b) Construct a cyclic quadrilateral ABCD, given AB 6 cm, ÐABC 70, BC 5 cm and ÐACD 0. 47. (a) Solve graphically x +x 60. [OR] (b) Draw the graph of xy 0, x, y > 0. Use the graph to find y when x 5, and to find x when y 10. http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available 4 Sura s Mathematics - X Std. March 017 Question Paper with Answers ANSWERS SECTION I 1. (b). (b). (a) 4. (d) 5. (a) 6. (c) 7. (c). (d) 9. (d) 10. (b) 11. (c) 1. (d) 1. (c) 14. (c) 15. (b) SECTION II Solving () & (4) x y...(4) 16. Solution: x y 6...() Now, B C {, 4, 5, 6} {5, 6, 7, } x 4 x {, 4, 5, 6, 7, } Substituting in () + y 6 y 4 A (B C) {1,,, 4, 5} The required the solution is (, 4) {, 4, 5, 6, 7, } 0. Solution : {1,,, 4, 5, 6, 7, }...(1) Now, A B { 1,,, 4, 5} {, 4, 5, 6} Hint : 6x + 9x x( x+ ) {1,,, 4, 5, 6} x 1x x( x 4) Only products (A B) C {1,,, 4, 5, 6} {5, 6, 7, } can be cancelled. x + {1,,, 4, 5, 6, 7, }...() x 4 From (1) and (), we have A (B C) (A B) C. 1. Solution : aij i j 17. Solution: a y f (x) x 1 11 () 1 1 1 1 a f (5) 5 1 9 a1 () 1 0 b f () 1 15 a1 ( ) 1 4 1 1. Solution: a ( ) 4 If 7, m, 7 (m + ) are in G.P., find the values of m. Since the given terms are in G.P. their Common ratio is the same. m 7 ( m + ) 7 m 19. Solution : 1x + 11y 70, 11x + 1y 74 1x + 11y 70...(1) 11x + 1y 74...() Adding (1) & () 4x + 4y 144 4 x + y 6...() Subtracting (1) & () x y 4 x y....(4) 1 0 Þ A. Solution: C A + B 1 5 1 + 4 m 7 ( m + ) 7 m m + m m 0 (m ) + (m + 1) 0 m or 1. 6 4 1 10 + 14 4 14 5 http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available Sura s Mathematics - X Std. March 017 Question Paper with Answers 5. Solution: (x 1, y 1 ) is (, 5) (x, y ) is (4, 9); m 1, m 6 (, 5) A 1 P ( x, y) 6 (4, 9) B D P mx + nx m+ n my + ny m+ n 1 1 ( xy, ), 4 1 9+ 0 (, ),, + + 14 1 1 6 1 6 7 7 P xy 4. Solution: 14 1, (, ) 7 7 For all a>0 (0, a) kind of points will be on the positive side of y-axis. Hence, the given statement is false. and P is common PAB PQR (AA similarity criterion) Since corresponding sides are proportional, AB PB P QR PR A 6 B QR AB PR PB 6 Thus, QR 9 cm. 6. Solution: Let BD be the height of the tower and AE be the distance of the eye level of the observer from the ground level. Q R E 0 C 0 m A 0 m B Draw EC parallel to AB such that AB EC. Given AB EC 0 m and AE BC 1.5 m In right angled DEC CD tan 0º EC CD EC tan 0º 0 CD 0 m 5. Solution: Thus, the height of the tower, Given AB is cm, PB is cm PR is 6 cm and BD BC + CD AB QR In PAB and PQR 1.5 + 0 1.5 m 7. Solution: PAB PQR (corresponding angles) 1. 5m Height 4(radius) h 4r T.S.A. of a circular cylinder 1540 cm (i.e.) πr (h + r) 1540 cm 5 70 140 r[ 4r + r] 1540 7. Solution: r ( r) 7 5 5 7 Given, S 1; R 59 r 7 7 r 7 cm. h 4r 4 7 cm R L S Þ 59 L 1 Þ L 59 + 1 71 9. Solution: In tossing a coin twice, the sample space S {HH, HT, TH, TT} n(s) 4. http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available 6 Sura s Mathematics - X Std. March 017 Question Paper with Answers (i) Let A be the event of getting two heads. Then A {HH}. Thus, n(a) 1. P(A) n ( A) n( S ) 1 4 (ii) Let C be the event of getting exactly one tail. Then C {HT, TH} Thus, n(c). P(C) n ( C) n( S ) 4 1 0. Solution: (a) Let r and V be the radius and volume of the solid sphere respectively. r 741 1 x 7 cm f(x) Given that V 741 1 f(6) 6 cu. cm 1 7 4 506 f(9) 9 6 πr 7 4 f(15) 15 1 7 506 4 r 7 f(1) 1 15 r 506 7 5 7 4 f(1) 1 1 17 4 6 Thus, the radius of the sphere, r 1 cm. (i) A f B (or) (b) L.H.S x y Hint : (a + b) a + ab + b (a secθ + b tanθ) (a tanθ + secθ) a sec θ + ab secθ tanθ + b tan θ (a tan θ + ab secθ tanθ + b sec θ) a (sec θ tan θ ) b (sec θ tan θ) a (1) b (1) a b R.H.S. 1. Solution: SECTION III First, we find B C {1,, c,d,e} {d, e, f, g,, y} {1,, c, d, e, f, g, y}. Then A \ (B C) {a, b, c, d, e, f, g, x, y, z} \ {1,, c, d, e, f, g, y} {a, b, x, z}....(1) Next, we have A \ B {a, b, f, g, x, y, z} and A \ C a, b, c, x, z} and so (A \ B) (A \ C) {a, b, x, z}....() Hence, from (1) and () it follows that A \ (B C) (A \ B) (A \ C).. Solution: 6 1 9 15 4 1 5 1 6 (ii) f {(6, 1), (9, ), (15, 4), (1, 5), (1, 6)} (iii) f 6 9 15 1 1 f (x) 1 4 5 6 http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available Sura s Mathematics - X Std. March 017 Question Paper with Answers 7 (iv) 6 5 4 1 (9,) (6,1) (1,6) (1,5) (15,4) 6 9 1 15 1 1. Solution: We want to find 1 + 4 +... to n terms 1 4 + 9 16 + 5... to n terms (1 4) + (9 16) + (5 6) +... to n terms. (after grouping) 4. Solution: n [ 6 4n + 4] n [ 4n ] n (n + 1) n (n + 1) ( ) 7 n 10 10 1 9 10 1 ( ) 7 n 10 10 1 9 9 5. Solution: n n 70 1 10 n 1 n ( ) 79 Speed in still water 15km/hr. Total Time taken 4 hrs 0 m 4 1 9 hrs. Let speed of the stream Speed of the boat upstream x km/hr 15 x + ( 7) + ( 11) +... n terms Speed of the boat down stream 15 + x Now, the above series is in an A.P. with first 0 term a and common difference d 4 Time taken for upstream T 1 15 x Therefore, the required sum n 0 [a + (n 1) d] Time taken for downstream T 15 + x n [( ) + (n 1) ( 4)] T 1 + T 9 0 15 x + 0 15 + x 9 + + 0 10 ( 15 x) ( 15 x) + 9 ( 15 x)( 15 x) Sn 7 + 77 + 777 +... + (n terms) Sn 7(1 + 11 + 111 +... + n terms) 7 (9 + 99 + 999 +... + n terms) 9 7 9 [(10 1) + (100 1) + (1000 1) +... + (10 n 1)] 7 9 [(10 + 10 + 10 +... + 10 n ) (1 + 1 + 1... n terms)] (5 x ) 0 0 675 x 600 x 75 x 75 5 x 5 km/hr. (i.e.) Speed of the stream 5 km/hr. Aliter : 5 x 00 x 5 x 5 x 5 http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available Sura s Mathematics - X Std. March 017 Question Paper with Answers 6. Solution: Since the given expression is a perfect square, the remainder is zero. 1 B b + 1 + a 10 1 1 1 BT 1 1 6 0...(1) 5 1 AB 7 1 1 10 5+ 14 7+ a 10 49 0...() 11 4 a 10 From () 49 7 (AB) T 11 a 10 4... (1) 7 a 10 56 a 66 1 5 7 B T A T Substituting a 66 in (1) we get, 1 1 b + 1 + 56 6 0 b + 1 + 4 0 b 4 4x x x x a 10 6x + 4x a 10 x + 16x 4 4x + (a 1) x + (b + 1) x + 49 16x 4 4x + (a 1) x (+) ( ) 4x + 9 x (a 10) x + (b + 1) x + 49 (+) ( ) ( ) (a 10) x a 10 + 6x + a 10 a 10 a 10 a 10 b + 1 + 6 x + 49 10 14 11 5+ 7+ 4... () (1) () (AB) T B T A T 7. Solution: 5 A 7 AT 5 7 http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available Sura s Mathematics - X Std. March 017 Question Paper with Answers 9. Solution: A ( 4, ) B (, 5) y O D (, ) x C (, ) Let us plot the points roughly and take the vertices in counter clock-wise direction. Let the vertices be A( 4, ), B(, 5), C(, ) and D(, ) Area of the quadrilateral ABCD { In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Given : In a right angled ABC, A 90º. To prove : BC AB + AC Construction : Draw AD BC Proof : In triangles ABC and DBA, B is the common angle. Also, we have BAC ADB 90º. ABC DBA (AA similarity criterion) Thus, their corresponding sides are proportional. Hence, AB BC DB BA AB DB BC...(1) { Similarly, we have ABC DAC. Thus, BC AC AC DC AC BC DC...() Adding (1) and () we get, AB + AC BD BC + BC DC BC (BD + DC) BC BC BC Thus, BC AB + AC Hence the Pythagoras theorem. 40. Solution: Let A be the point of observation and B be the 1 foot of the building. {(x 1 y + x y + x y 4 + x 4 y 1 ) (x y 1 + Let BC denote the height of the building and CD denote height of the flag post. x y + x 4 y + x 1 y 4 ) 1 Given that CAB 45º, DAB 60º and {0 + 6 + 9 4) (6 15 4 1)} CD 10 m Let BC h metres and AB x metres. 1 4 4 Now, in the right angled CAB, 5 1 tan 45º BC {1 + 5} sq. units. AB Thus, AB BC i.e., x h...(1) 9. Pythagoras theorem: Also, in the right angled DAB, tan 60º BD AB h + 10 AB tan 60º x h + 10 From (1) and (), we get h h + 10 h h 10 h ( ) 10 + 1 1 10 1 5 (.7) 1.66 m. + 1 + 1...() Hence, the height of the building is 1.66 m. A B D C A 60 45 x D 10 m B C h http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available 10 Sura s Mathematics - X Std. March 017 Question Paper with Answers 41. Solution: Depth, h 14 cm. Given, πr 44 cm R 44 R 7 R 7 cm 4. Also given πr. 4 π 44 7 Þ r 4. cm Þ r 7 7 Þ r 1 Þ r 1 cm Diameter of its base 4 cm. 4. Solution: Let us calculate the A.M of the given data. A.M. x 1 + 15 + 1 + 0 + 5 5 Volume of the frustum 1 πh[ R + r + Rr] 140.6 cm 4. Solution: Cuboid Cone l 44 cm radius r? b 1 cm height h 4 cm h 1 cm Given, Volume of the Cuboid Volume of the Cone 90 5 1. 1 x d x 1 d 14 [ 7 + ( 4.) + 7 4.] 7 1 6 6 15 9 44 49 17 64 9 4 [ +. +.] 1 0 0 0 4 44 (96.04) 140.5 cm 5 7 49 d 0 d 9 σ d n 9 5 19. 6 4.4 The coefficient of variation σ x 100 l b h 1 πr h 44 1 1 1 7 r 4 4. 4 1 100 44. 1 The coefficient of variation is 4.6 44 1 1 7 4 r http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available Sura s Mathematics - X Std. March 017 Question Paper with Answers 11 44. Solution: Sample space { (1, 1), (1, ), (1, ), (1, 4), (1, 5), (1, 6) (, 1), (, ), (, ), (, 4), (, 5), (, 6) (, 1), (, ), (, ), (, 4), (, 5), (, 6) (4, 1), (4, ), (4, ), (4, 4), (4, 5), (4, 6) (5, 1), (5, ), (5, ), (5, 4), (5, 5), (5, 6) (6, 1), (6, ), (6, ), (6, 4), (6, 5), (6, 6)} n(s) 6 Let A be the event of getting an even number in the first die and B be the event of getting a total of. (A) { (, 1), (, ), (, ), (, 4), (, 5), (, 6) (4, 1), (4, ), (4, ), (4, 4), (4, 5), (4, 6) (6, 1), (6, ), (6, ), (6, 4), (6, 5), (6, 6)} remainder ( 0) n (A) 1 P (A) n ( A) n( S ) 1 6 B { (, 6), (6, ), (5, ), (, 5), (4, 4)} n (B) 5 P (B) n ( B) n( S ) 5 6 A B { (, 6), (4, 4), (6, )} P (A B) n ( A B) n( S) 6 n (A B) By using the addition theorem on probabilities: P (A B) P (A) + P (B) (A B) 45.(a) Solution: Let f(x) x 4 + 6x 1x 4x x (x + x 4x ) Let g(x) 4x 4 + 14x + x x x (x + 7x + 4x 4) Let us find the GCD for the polynomials x + x 4x and x + 7x + 4x 4 We choose the divisor to be x + x 4x remainder Common factor of x + x 4x and x + 7x + 4x 4 is x + 4x + 4 Also common factor of x and x is x. Thus, GCD ( f(x), g(x) x (x + 4x + 4). 1 6 + 5 0 6 6 6 5 9 (OR) http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available 1 Sura s Mathematics - X Std. March 017 Question Paper with Answers (b) Let the x and y intercepts be a and b A(a, 0) and B (0, b) are the points on x and y axes B(0,b) Equating x and y co-ordinates on both sides a a 6 b b 4 Hint: Any point on x-axis will have its y co-ordinate 0, and any point on y-axis, the x coordinate 0. (, ) B(a,0) M is the mid-pt. of AB, (, ) (, ) a + 0 0 + b, a b, SECTION IV 46. (a) Solution: Given: Radius of the circle 6 cm. OP 10 cm Rough Diagram T The required equation is x a x y x + 1 + y 1 6 4 1 y + 1 b Multiplying by 1 x + y 1 The required equation of AB is x + y 1 0. 6 cm O 6 cm 10 cm P T http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available Sura s Mathematics - X Std. March 017 Question Paper with Answers 1 Fair Diagram T 6 cm cm O 10 cm M T T Steps of Construction: (i) With O as the centre draw a circle of radius 6 cm. (ii) Mark a point P at a distance of 10 cm from O and join OP. (iii) Draw the perpendicular bisector of OP. Let it meet OP at M. (iv) With M as centre and MO as radius, draw another circle. (v) (iv) Let the two circles intersect at T and T. 6 cm cm Join PT and P T. They are the required tangents. Length of the tangent, PT cm. P Verification: In the right angled triangle OPT, PT OP OT 10 6 100 6 64 cm PT PT cm (OR) http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available 14 Sura s Mathematics - X Std. March 017 Question Paper with Answers (b) Given: In the cyclic quadrilateral ABCD, AB 6 cm, ABC 70º, BC 5 cm and ACD 0º Rough Diagram C D 0 Fair Diagram Y Steps of Construction: A X C D 0 6 cm O (i) Draw a rough diagram and mark the measurements. Draw a line segment AB 6 cm. (ii) From B draw BX such that ABX 70º (iii) With B as centre and radius 5 cm, draw an arc intersecting BX at C. (iv) Join BC. (v) Draw the perpendicular bisectors of AB and BC intersecting each other at O. (vi) With O as centre and OA OB OC as radius, draw the circumcircle of ABC. (vii) From C, draw CY such that ACD 0º which intersects the circle at D. (viii) Join AD and CD. Now, ABCD is the required cyclic quadrilateral. 5 cm 70 B 70 A 6 cm B 5 cm http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available Sura s Mathematics - X Std. March 017 Question Paper with Answers 15 47. (a) Solution: x +x 6 0 Þ y x +x 6 First, let us form the following table by assigning integer values for x from to and finding the corresponding values of y x +x 6 Table x 1 0 1 x 9 4 1 0 1 4 9 x 1 0 1 x 1 0 1 6 6 6 6 6 6 6 6 y 9 0 5 6 4 15 Plot the points (, 9), (, 0), ( 1, 5), (0, 6), (1, ), (, 4) and (, 15) on the graph. Draw the graph by joining the points by a smooth curve. msî Â l«scale: xm R x axis 1br.Û cm 1 myf 1 unit ym R y axis 1br.Û 1 cm myffÿ units x y O y y x + x - 6 x Join the points by a smooth curve. The curve, thus obtained, is the graph of y x + x 6. The curve cuts the x-axis at the points (, 0) and (1.5, 0). The x-coordinates of the above points are and 1.5. Hence, the solution set is {, 1.5 }. http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available 16 Sura s Mathematics - X Std. March 017 Question Paper with Answers 47. (b) Solution: x 1 4 5 y 0 10 5 4 From the table we observe that as x increases y decreases. This type of variation is called indirect variation. y α 1/x or xy k where k is a constant of proportionality. Also from the table we find that, 1 0 10 4 5 0 k We get k 0 Plot the points (1, 0), (, 10), (4, 5) and (5, 4) and join them. y Y 0 (1,0) 1 16 14 1 10 (,10) 6 4 (4,5) (5,4) x xy 0 0 1 4 5 6 10 1 Scale: x axis cm 1 unit y axis 1 cm units x y \ The relation xy 0 is a rectangular hyperbola as exhibited in the graph. From the graph we find, (i) When x 5, y 4 (ii) When y 10, x. http://www.trbtnpsc.com/016/0/10th-1th-public-exam-answer-keys.html