BOARD ANSWER PAPER : MARCH 2015 ALGEBRA

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1 BOARD ANSWER PAPER : MARCH 015 ALGEBRA Algebra Q.1. Attempt any five of the following sub-questions: i. The given sequence is 1, 4, 7, 10, Here, t 1 = 1, t = 4, t 3 = 7, t 4 = 10 t t 1 = 4 1 = 3 t 3 t = 7 4 = 3 t 4 t 3 = 10 7 = 3 The difference between two consecutive terms is constant. The given sequence is an A.P. ii. S = {1,, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14, 15, 16, 17, 18, 19, 0, 1,, 3, 4, 5} iii. By adding given equations, we get 1 x + 13 y = 9 13 x + 1 y = 1 5x + 5y = 50 5 (x + y) = 50 x + y = 50 5 x + y = iv. Given, S n = For n = S 10 = 10 1 S 10 = n n1 v. The given equation is x + 3x 4 = 0 Substituting x = 1 in L.H.S of the above equation, we get L.H.S. = (1) + 3 (1) 4 = = 0 L.H.S = R.H.S 1 is the root of the given quadratic equation. vi. The given equation is x + y = 5 Substituting x = 3 in above equation, we get 3 + y = 5 y = Q.. Attempt any four of the following sub-questions: i. x 7x + 1 = 0 x 4x 3x + 1 = 0 x (x 4) 3 (x 4) = 0 (x 4) (x 3) = 0 x 4 = 0 or x 3 = 0 x = 4 or x = 3 1

2 Board Answer Paper: March 015 ii. The given A.P. is 4, 9, 14,. Here, a = 4, d = 9 4 = 5, n = 10 Since, t n = a + (n 1)d t 10 = 4 + (10 1)5 = t 10 = 49 iii. The given equation is 5x + ay = 19 The point (x, y) = (, 3) lies on the graph of the equation, hence it statifies the equation. Substituting x = and y = 3 in the given equation, we get 5 () + a (3) = a = 19 3a = 9 a = 3 iv. When a die is thrown, S = {1,, 3, 4, 5, 6} A = Event of getting an odd number A = {1, 3, 5} v. The inter-relation between the measures of central tendency is Mean Mode = 3(Mean Median) Mode = Mean 3(Mean Median) Mode = 101 3( ) Mode = 101 3(1) Mode = 98 vi. x = 1 is the root of the given equation kx 7x + 5 = 0 It satifies the given equation k(1) 7(1) + 5 = 0 k = 0 k = 0 k = Q.3. Attempt any three of the following sub-questions: i. Area (in Measures of central Crop hectares) angle () 40 Jowar = Wheat 60 Sugarcane = = Vegetables = Total: [1½]

3 The pie diagram representation is as follows: Algebra Wheat Jowar Sugarcane Vegetables [1½] ii. When two coins are tossed, S = {HH, HT, TH, TT} n(s) = 4 Let A be the event that at the most one tail turns up A = {HH, HT, TH} n(a) = 3 P(A) = n(a) n(s) P(A) = 3 4 iii. The given simultaneous equations are x + y = 7 and x y = 5.(i).(ii) From equation (i), y = 7 x x y (x, y) (0, 7) (1, 6) ( 1, 8) From equation (ii), y = x 5 x y (x, y) (0, 5) (1, 4) ( 1, 6) 3

4 Board Answer Paper: March 015 (1, 8) 8 7 Y (0, 7) Scale : (On both axes 1 cm = 1 unit) 6 (1, 6) (6, 1) [] X X 3 4 (1,4) 5 (0,5) (1,6) 6 Y The two lines intersect at point (6, 1) (6, 1) is the solution of given simultaneous equations. Solution set = {(6, 1)} 4 iv. The number of seats arranged row wise is as follows: 0,, 4,. This sequence is an A.P. with a = 0, d= 0 =, n = Now, t n = a + (n 1) d t = 0 + ( 1) = = 6 The number of seats in the twenty second row is 6.

5 v. x + 11x + 4 = 0 Algebra x + 11x = 4.(i) Now, third term = = 1 coefficient of x 1 11 = 11 4 Adding 11 4 on both sides of (i), we get x + 11x = x 11 = 5 4 Taking square root on both sides, we get x + 11 = 5 x = x = or x = 11 5 x = 6 or x = 16 x = 3 or x = 8 3 and 8 are the roots of the given quadratic equation. Q.4. Attempt any two of the following sub-questions: i. The sample space for two digit numbers without repeating the digits is S = {10, 1, 13, 14, 15, 0, 1, 3, 4, 5, 30, 31, 3, 34, 35, 40, 41, 4, 43, 45, 50, 51, 5, 53, 54} P is the event that the number so formed is even. P = {10, 1, 14, 0, 4, 30, 3, 34, 40, 4, 50, 5, 54} Q is the event that the number so formed is greater than 50. Q = {51, 5, 53, 54} R is the event that the number so formed is divisible by 3. R = {1, 15, 1, 4, 30, 4, 45, 51, 54} 5

6 Board Answer Paper: March 015 ii. Age (in years) No. of patients Cumulative frequency Class interval Frequency f i (less than type) c.f f Total: f i = 300 Here, f i = N = 300 N = 300 = 150 Cumulative frequency (less than type) which is just greater than (or equal) to 150 is 157. Median class is Now, L = 30, f = 55, c.f. = 10, h = 10. N Median = L + c.f. h f = 30 + (150 10) = = = The median age of a patient is years. iii. It is given that + = = 35 Now, ( + ) 3 = ( + ) 3 = ( + ) (5) 3 = (5) 15 = = = 90 = = 6 the required quadratic equation is x ( + )x + = 0 i.e. x 5x + 6 = 0 Q.5. Attempt any two of the following sub-questions: i. The instalments are in A.P. Here, S 10 = = 4500 Also, n = 10, d = 10 Now, S n = n [a + (n 1)d] 6

7 Algebra S 10 = 10 [a + (10 1)(10)] 4500 = 5 [a + 9 (10)] 4500 = a = a 990 = a a = 990 a = 495 Also,t n = a + (n 1)d t 10 = (10 1)(10) = (10) = t 10 = 405 The first instalment is ` 495 and the last instalment is ` 405. ii. Let there be x number of tickets each of ` 0 and y number of tickets each of ` 40. According to the first condition, x + y = 35 (i) According to the second condition, 0x + 40y = 900 0(x + y) = 900 x + y = x + y = 45 (ii) Subtracting equation (i) from (ii) x + y = 45 x + y = 35 () () () y = 10 Substituting y = 10 in equation (i), we get x + 10 = 35 x = x = 5 There were 5 tickets of ` 0 each and 10 tickets of ` 40 each sold. iii. Speed (in km/hr) No. of Students

8 Board Answer Paper: March Y Scale: On X-axis: 1 cm = 10 km/hr On Y-axis: 1 cm = 10 students [5] Histogram No. of Students Frequency Curve X 0 Y X Speed (in km/hr) 8

9 BOARD ANSWER PAPER : JULY 015 ALGEBRA Algebra Q.1. Attempt any five of the following sub-questions: i. Given, t n = n + For n = 1, t 1 = 1 + = 3 For n =, t = + = 4 ii. 3y = 10y + 7 3y 10y 7 = 0 is in the standard form. iii. iv = (4 7) ( 3) = 8 6 = When two coins are tossed, S = {HH, HT, TH, TT} v. The given sequence is 1, 3, 6, 10, Here, t 1 = 1, t = 3, t 3 = 6, t 4 = 10 t t 1 = 3 1 = t 3 t = 6 3 = 3 t t 1 t 3 t The difference between two consecutive terms is not constant. The given sequence is not an A.P. vi. Let the length of the rectangle be x cm and its breadth be y cm. According to the given condition, (x + y) = 36 x + y = 36 x + y = 18 x + y = 18 is the required equation.. Attempt any four of the following subquestions: i. x = 4 is the root of the given equation x 7x + k = 0 It satisfies the given equation (4) 7(4) + k = k = k = 0 k = 1 ii. The given A.P. is 7, 13, 19, 5, Here, a = 7, d = 13 7 = 6, n = 18 Since, t n = a + (n 1)d t 18 = 7 + (18 1)6 = t 18 = 109 iii. When a die is thrown, S = {1,, 3, 4, 5, 6} P is the event that the number is odd. P = {1, 3, 5} iv. Here, D x = 18, D y = 15 and D = 3 Using Cramer s rule, we get x = D x D 1

10 Board Answer Paper: July 015 x = 18 3 x = 6 and y = D y D y = 15 3 y = 5 x = 6 and y = 5 v. Let = 5, = 7 then, + = = 1 and = 5 (7) = 35 The required quadratic equation is x ( + ) x + = 0 x 1x + 35 = 0 vi. The inter-relation between the measures of central tendency is Mean Mode = 3 (Mean Median) Mean 180 = 3 (Mean 156) Mean 180 = 3 Mean = 3 Mean Mean 88 = Mean Mean = 88 Mean = Attempt any three of the following subquestions : i. The given equation is x + 5x + = 0 Comparing it with ax + bx + c = 0, we get a =, b = 5, c = b b 4ac x = a 5 (5) 4()() = () = = = x = = = or x = = = , are the roots of the given equation. ii. S = {1,, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14, 15, 16, 17, 18, 19, 0, 1,, 3, 4, 5, 6, 7, 8, 9, 30} n(s) = 30 A is the event that ticket drawn bears a number which is a perfect square. A = {1, 4, 9, 16, 5} n(a) = 5 iii. Given, t 18 = 5 and t 39 = 148 Now, t n = a + (n 1)d t 18 = a + (18 1)d 5 = a + 17d

11 Algebra a + 17d = 5... (i) Also, t 39 = a + (39 1)d 148 = a + 38d a + 38d = (ii) Adding (i) and (ii), we get a + 17d = 5 a + 38d = 148 a + 55d = 00 (iii) Also, S n = n [a + (n 1)d] S 56 = 56 [a + (56 1)d] = 8 [a + 55d] = 8 (00) [From (iii)] S 56 = 5600 Sum of the first 56 terms of the A.P. is iv. The given equation is 3x y = 6 3x + 6 = y y = 3x + 6 x y (x, y) (0, 6) (1, 3) (1, 9) 9 Y (1, 9) Scale : (On both axes 1 cm = 1unit) (0,6 ) 5 4 ( 1, 3) 3 1 (, 0) X x y = X [1½] The points of intersection of graph with X-axis is (, 0) and with Y-axis is (0, 6). 3

12 Board Answer Paper: July 015 v. Part of the day Percentage of electricity used Measure of central angle () Morning = Afternoon = Evening = Night = Total: The pie diagram representation is as follows: [1½] Afternoon 144 Morning Night 7 Evening [1½] 4. Attempt any two of the following subquestions: i. A card is drawn from a pack of 5 cards n(s) = 5 a. Let A be the event of getting a king card. The pack of cards consists of 4 king cards. n(a) = 4 P(A) = n(a) n(s) = 4 5 = 1 13 b. Let B be the event of getting a face card. There are 1 face cards in a pack of cards. n(b) = 1 P(B) = n(b) n(s) = 1 5 = 3 13 ii. 3x 4 13x + 10 = 0 Let x = m, The given quadratic equation becomes 3m 13m + 10 = 0 3m 3m 10m + 10 = 0 3m(m 1) 10(m 1) = 0 (m 1)(3m 10) = 0 m 1 = 0 or 3m 10 = 0 m = 1 or m = 10 3 But x = m x = 1 or x =

13 Algebra x = 1 or x = 10 3 iii The solution set is 1, 1,, 3 3 Bowling speed (km/hr) Class interval No. of players Frequency f i f f m f Here, the maximum frequency is 11. Modal class is Now, L = 100, f m = 11, f 1 = 9, f = 8, h = 15 fm f1 Mode = L + h fm f1 f 11 9 = = = = The modal bowling speed of a player is 106 km/hr. 5. Attempt any two of the following subquestions: i. Let there be x number of rows and y number of students in each row total number of students = xy According to the first condition, (y 3)(x + 10) = xy xy 3x + 10y 30 = xy 3x + 10y 30 = xy xy 3x + 10y = x + 10y = 30 3x 10y = 30 (i) According to the second condition, (y + 5)(x 10) = xy xy + 5x 10y 50 = xy 5x 10y 50 = xy xy 5x 10y = x 10y = 50 (ii) Subtracting equation (i) from (ii), we get 5x 10y = 50 3x 10y = 30 () (+) (+) x = 80 x = 80 Substituting the value of x in equation (i), we get 3(40) 10y = y = 30 10y = y = 150 x = 40 5

14 Board Answer Paper: July y = 150 y = y = 15 The total number of students = xy = = 600. The total number of students participating in the drill is 600. ii. Let the temperatures from Monday to Friday in A.P. be a d, a d, a, a + d, a + d. According to the first condition, a d + a d + a = 0 3a 3d = 0 a d = 0 a = d According to the second condition, a + d + a + d = 15 a + 3d = 15 a + 3a = 15 [ d = a] 5a = 15 a = 3 d = 3 [ d = a] Thus, a d = 3 3 = 3 a d = 3 3 = 0 a = 3 a + d = = 6 a + d = = 9 The temperatures of each of the five days are 3 C, 0 C, 3 C, 6 C and 9 C respectively. iii. House Rent (in ` per month) Number of families Y Scale: On X-axis: cm = ` 00 On Y-axis: 1 cm = 50 Families Number of families X Y House rent per month in (`) X [5] 6

15 BOARD ANSWER PAPER : MARCH 016 ALGEBRA Q.1. Attempt any five of the following sub-questions: i. Given, t n = 3n 4 For n = 1, t 1 = 3(1) 4 = 1 For n =, t = 3() 4 = 6 4 = Algebra ii. Given equation is x x 3 = 0 Comparing it with ax + bx + c = 0, we get a =, b = 1, c = 3 iii. Let = and = 3 the required quadratic equation is x ( + ) x + = 0 i.e., x ( 3) x + ( ) ( 3) = 0 i.e., x + 5x + 6 = 0 iv = (4 1) ( 3) = = 10 v. S = {Sunday, Monday, Tuesday, Wednesday, Thrusday, Friday, Saturday} vi. Class mark of 0 30 = Class mark of = = = 35. Attempt any four of the following subquestions: i. Given, a = 11, d = t 1 = a = 11 t = t 1 + d = 11 + ( ) = 9 t 3 = t + d = 9 + ( ) = 7 The first three terms of the A.P. are 11, 9 and 7. ii. x + 11x + 4 = 0 x + 8x + 3x + 4 = 0 x (x + 8) + 3 (x + 8) = 0 (x + 8) (x + 3) = 0 x + 8 = 0 or x + 3 = 0 x = 8 or x = 3 iii. x 5 = (x) 3 ( 5) = 31 4x + 15 = 31 4x = x = 16 x = 4 iv. When a die a thrown S = {1,, 3, 4, 5, 6} n (S) = 6 1

16 Board Answer Paper : March 016 a. A is the event of getting number divisible by 3 A = {3,6} n(a) = P(A) = n(a) 1 n(s) 6 3 b. B is the event of getting number less than 5. B = {1,, 3, 4} n(b) = 4 P(B) = n(b) 4 n(s) 6 3 v. Number of words Class mark (x i ) No. of candidates frequency (f i ) Total: f i = 100 f i x i = f i x i fx f i i Mean x i = 1100 The mean number of words written are vi. Subject Marks Measure of central angle() Marathi = Hindi = English = Mathematics = Total The pie diagram representation is as follows: Hindi English Marathi Mathematics

17 Algebra 3. Attempt any three of the following subquestions : i. The given equation is 6x 7x 1 = 0 Comparing it with ax + bx + c = 0, we get a = 6, b = 7, c = 1 x = x = = b b 4ac a ( 7) ( 7) 4(6)( 1) (6) , = are the roots of the given equation. ii. Let the three boys be B 1, B, B 3 and the two girls be G 1 and G. A committee of two is to be formed. The sample space is S = {B 1 B, B 1 B 3, B 1 G 1, B 1 G, B B 3, B G 1, B G, B 3 G 1, B 3 G, G 1 G } n(s) = 10 a. A is the event that the committee contains at least one boy. A={ B 1 B, B 1 B 3, B 1 G 1, B 1 G, B B 3, B G 1, B G, B 3 G 1, B 3 G } n(a) = 9 P(A) = n(a) n(s) = 9 10 b. B is the event that the committee contains one boy and one girl. B = {B 1 G 1, B 1 G, B G 1, B G, B 3 G 1, B 3 G } n(b) = 6 iii. P(B) = n(b) n(s) = 6 10 = 3 5 Diameter (in mm) Class interval Class mark (x i ) d i = x i A d i = x i 40 No. of screws Frequency (f i ) A Total: f i = 98 f i d i = 51 f i d i [1½] d = fid f Mean x = A + d i i = 51 = = x = 40.5 The mean diameter of the head of a screw is 40.5 mm. 3

18 Board Answer Paper : March 016 iv. Marks Scored Number of students Y Scale: On X-axis: cm = 0 marks On Y-axis: 1 cm = Students 18 No. of students X X Y Marks [3] v. Rainfall (in mm) Class mark (x i ) No. of Years Frequency (f i )

19 Algebra Y Scale: On X-axis: 1 cm = 5 mm On Y-axis: 1 cm = years 1 10 No. of Years X Y Rainfall (in mm) X [3] 4. Attempt any two of the following subquestions: i. Given, t 11 = 16, t 1 = 9 a. Since, t n = a + (n 1)d t 11 = a + (11 1)d 16 = a + 10d a + 10d = (i) Also, t 1 = a + (1 1)d 9 = a + 0d a + 0d = 9... (ii) Subtracting (i) from (ii), we get a + 0d = 9 a + 10d = 16 () () () 10d = 13 d = Substituting d = 13 in (i), we get 10 a = 16 a + 13 = 16 a = a = 3 a = 3 and d = = 1.3 The 1 st term is 3 and the common difference is 1.3 b. Now, t n = a + (n 1)d t 34 = 3 + (34 1)1.3 = = t 34 = 45.9 The 34 th term is

20 Board Answer Paper : March 016 c. Given, t n = 55 Since, t n = a + (n 1)d 55 = 3 + (n 1) = (n 1)1.3 5 = (n 1) = n = n 1 40 = n 1 n = = 41 t n = 55 for n = 41 ii. The given simultaneous equations are 7 13 x1 y = 7... (i) 13 7 x1 y = (ii) 1 Let x1 1 y be q equations (i) and (ii) becomes 7p + 13q = 7 13p + 7q = 33 Adding equations (iii) and (iv), 7p + 13q = 7 13p + 7q = (iii)... (iv) 0p + 0q = 60 0 (p + q) = 60 p + q = 60 0 p + q = 3... (v) Subtracting equation (iii) from (iv), 13p + 7q = 33 7p + 13q = 7 () () () 6p 6q = 6 6(p q) = 6 p q = 6 6 p q = 1... (vi) Now, adding equations (v) and (vi), p + q = 3 p q = 1 p = 4 6 p = 4 p = Substituting p = in equation (v), we get + q = 3 q = 3 q = 1 (p, q) = (,1)

21 Resubstituting the values of p and q, we get 1 = x1 and 1 = 1 y (x + 1) = 1 and 1(y + ) = 1 4x + = 1 and y + = 1 4x = 1 and y = 1 4x = 1 and y = 1 1 x = 4 and y = 1 Algebra (x, y) = 1, 1 4 iii. Let P(C) be x P(B) = P(C) (Given) P(B) = x and P(A) = P(B) (Given) P(A) = 4x Given, P(A) + P(B) + P(C) = 1 4x + x + x = 1 7x = 1 x = 1 7 P(A) = 4x = = 4 7, P(B) = x = 1 7 = 7 P(C) = x = 1 7 P(A) = 4 7, P(B) = 7, P(C) = Attempt any two of the following subquestions: i. Let the divisor be x quotient = x and remainder = 1 x = x Now, Divisor Quotient + Remainder = Dividend. x x x + = 613 x + x = 613 x + x = 613.[Multiplying both sides by ] x + x = 146 x + x 146 = 0 Comparing with ax + bx + c = 0, we get a =, b = 1, c = 146 b b 4ac x = a = = 4 7

22 Board Answer Paper : March = = 4 = = 31 4 or or x = 78 or x = 157 But x 157 as a divisor cannot be negative as well as a fraction. x = 78 The divisor is 78. ii. The numbers from 50 to 350 which are divisible by 6 are 54, 60, 66, This sequence is an A.P. with a = 54, d = = 6, t n = 348 But t n = a + (n 1) d 348 = 54 + (n 1) = (n 1) 6 94 = (n 1) = n 1 49 = n = n n = 50 Now, S n = n [t 1 + t n ] S 50 = 50 [ ] = 5 (40) S 50 = Also, t 15 = a + (15 1) d = = t 15 = 138 The sum of all numbers from 50 to 350 which are divisible by 6 is and 15 th term of the A.P. is 138. iii. Let the digit in the hundreath s place be x and the digit in the unit place be y. Digit H T U Original Number x x + y + 1 y Reversed Number y y + x + 1 x 8 The original number = 100x + 10 (x + y + 1) + y = 100x + 10x + 10y y = 110x + 11y + 10 According to the first condition, 110x + 11y + 10 = 17 [(x) + (x + y + 1) + y] 110x + 11y + 10 = 17 (x + y + 1) 110x + 11y + 10 = 34x + 34y x 34x + 11y 34y = x 3y = 7. (i)

23 Algebra The reversed number = 100y + 10 (y + x + 1) + x = 100y + 10y + 10x x = 11x + 110y + 10 According to the second condition. (110x + 11y + 10) = 11x + 110y x 11x + 11y 110y = x 99y = 198 Dividing both sides, by 99 we get x y =. (ii) Multiply equation (ii) by 3, we get 3x 3y = 46. (iii) Subtracting (iii) from (i), we get 76x 3y = 7 3x 3y = 46 () (+) (+) 53x = 53 x = 1 Substituting value of x in equation (ii), we get 1 y = 1 + = y y = 3 Required Number = 110x + 11y + 10 = 110(1) + 11(3) + 10 = = 153 Required Number is 153 9

24 BOARD ANSWER PAPER : JULY 016 ALGEBRA Q.1. Attempt any five of the following sub-questions: i. By adding given equations, we get 3x + y = 10 x + 3y = 15 5x + 5y = 5 5(x + y) = 5 x + y = 5 5 x + y = 5 Algebra ii. The given A.P. is 3, 5, 7, t 1 = 3, t = 5 d = t t 1 = 5 3 d = iii. When two coins are tossed, S = {HH, HT, TH, TT} iv. x (x + 3) = 7 x + 3x = 7 x + 3x 7 = 0 is standard form. v. mean fixi x = f = mean x = 4 i vi. Given equation is x + 18 = 6x i.e., x 6x + 18 = 0 Comparing it with ax + bx + c = 0, we get a =, b = 6, c = 18. Attempt any four of the following subquestions: i. The given A.P. is 1, 7, 13, 19,... Here, a = 1, d = 7 1 = 6 Since, t n = a + (n 1)d t 18 = 1 + (18 1)6 = = t 18 = 103 The eighteenth term of the given A.P. is 103. ii. Let = 3 and = 8 then, + = 11 and = 4 the required quadratic equation is x ( + )x + = 0 i.e., x 11x + 4 = 0 1

25 Board Answer Paper: July 016 iii. The given simultaneous equations are 4x + 3y = 4 6x + 5y = 8 (i) (ii) Equations (i) and (ii) are in ax + by = c form. D = 4 3 = (4 5) (3 6) 6 5 = 0 18 = 0 D x = 4 3 = (4 5) (3 8) 8 5 = 0 4 = 4 D y = 4 4 = (4 8) (4 6) 6 8 = 3 4 = 8 By Cramer s rule, we get x = D x and y = D y D D 4 x = and y = 8 x = and y = 4 x = and y = 4 is the solution of the given simultaneous equations. iv. The sample space for two digit numbers without repetition of digit is S = {10, 1, 13, 14, 0, 1, 3, 4, 30, 31, 3, 34, 40, 41, 4, 43} n(s) = 16 P is the event that the number so formed is an even. P = {10, 1, 14, 0, 4, 30, 3, 34, 40, 4} n(p) = 10 v. The inter-relation between the measures of central tendency is Mean Mode = 3(Mean Median) = 3(54.6 Median) 0.6 = 3 (54.6 Median) = 54.6 Median 0. = 54.6 Median Median = Median = 54.4 vi. Subject Marks Measure of central angle() Marathi = English = Science = Mathematics = Total:

26 The pie diagram representation is as follows: Algebra English Science Marathi Mathematics 3. Attempt any three of the following subquestions : i. Price of sugar per kg (in `) Class interval Number of weeks Frequency (f i ) Y Scale: On X- axis: 1 cm = ` On Y- axis: 1 cm = weeks 0 Number of weeks X 0 Y Price of sugar per kg. (in `) X [3] 3

27 Board Answer Paper: July 016 ii. The sample space is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} n(s) = 8 a. A is the event of getting head on the middle coin. A = {HHH, HHT, THH, THT} n(a) = 4 P(A) = n(a) n(s) = 4 8 = 1 b. B is the event of getting exactly one tail. B = {HHT, HTH, THH} n(b) = 3 P(B) = n(b) n(s) = 3 8 iii. Age (in years) Class interval No. of patients Frequency f i Cumulative frequency (less than type) c.f f Total: f i = 300 Here, f i = N = 300 N = 300 = 150 Cumulative frequency (less than type) which is just greater than (or equal) to 150 is 157. Median class is Now, L = 30, f = 55, c.f. = 10, h = 10. N h Median = L + c.f. f = 30 + (150 10) 10 = = = The median age of a patient is years. iv. The given equation is 7y 5y = 0 Comparing it with ay + by + c = 0, we get a = 7, b = 5, c = b b 4ac y = a ( 5) ( 5) 4(7)( ) = (7) = = 5 81 = y = 5 9 = = 1 or y = = = , are the roots of given equation. 7 4

28 v. Algebra Rainfall (in cm) No. of Years Frequency (f i ) Y Scale: On X-axis: 1 cm = 5 cm On Y-axis: 1 cm = years 1 10 No. of Years X Y Rainfall (in cm) X 4. Attempt any two of the following subquestions: i. Given, t 11 = 16, t 1 = 9 a. Since, t n = a + (n 1)d t 11 = a + (11 1)d 16 = a + 10d a + 10d = (i) Also, t 1 = a + (1 1)d 9 = a + 0d a + 0d = 9... (ii) Subtracting (i) from (ii), we get a + 0d = 9 a + 10d = 16 () () () d = d = 13 5

29 Board Answer Paper: July Substituting d = 13 in (i), we get 10 a = 16 a + 13 = 16 a = a = 3 a = 3 and d = = 1.3 The 1 st term is 3 and the common difference is 1.3 b. Now, t n = a + (n 1)d t 34 = 3 + (34 1)1.3 = = t 34 = 45.9 The 34 th term is 45.9 ii. Since, (3, 1) is the point of intersection of the lines ax + by = 7 and bx + ay = 5 the point (x, y) = (3, 1) satisfies the two equations. ax + by = 7 (i) and bx + ay = 5 (ii) Putting x = 3 and y = 1 in the above equations, we get 3a + b = 7 (iii) and 3b + a = 5 (iv) Multiplying equation (iv) by 3, we get 3a + 9b = 15 (v) Subtracting equation (v) from (iii), we get 3a + b = 7 3a + 9b = 15 () () () 8b = 8 8b = 8 b = 1 Putting b = 1 in equation (iv), we get 3(1) + a = a = 5 a = 5 3 a = a = and b = 1. iii. Let P(C) = x P(B) = P(C) (Given) P(B) = x and P(A) = P(B) (Given) P(A) = 4x Given, P(A) + P(B) + P(C) = 1 4x + x + x = 1 7x = 1 x = 1 7

30 Algebra P(A) = 4x = = 4 7, P(B) = x = 1 7 = 7 P(C) = x = 1 7 P(A) = 4 7, P(B) = 7, P(C) = Attempt any two of the following subquestions: i. The numbers from 50 to 50 which are divisible by 6 are 54, 60, 66, This sequence is an A.P. with a = 54, d = = 6, t n = 46 But t n = a + (n 1) d 46 = 54 + (n 1) = (n 1) 6 19 = (n 1) = n 1 3 = n = n n = 33 Now, S n = n [t 1 + t n ] S 33 = 33 [ ] = 33 (300) = 33 (150) S 33 = 4950 Also, t 13 = a + (13 1) d = (6) = t 13 = 16 The sum of all numbers from 50 to 50 which are divisible by 6 is 4950 and 13 th term of the A.P. is ii. 9 x x 3 1 x x 0 = 0 1 x x = 1 x x Now, + The equation becomes 1 1 9x 3 x x x 0 = 0 Put 1 x = m x (i) 9(m + ) 3m 0 = 0 9m m 0 = 0 9m 3m = 0 9m + 3m 6m = 0 3m(3m + 1) (3m + 1) = 0 (3m + 1)(3m ) = 0 3m + 1 = 0 or 3m = 0 m = 1 3 or m = 3 7

31 Board Answer Paper: July When m =, from equation (i), we get x = x 3 Multiplying both sides by 3x, we get 3x 3 = x 3x + x 3 = 0 Here, a = 3, b = 1, c = 3 b b 4ac 1 (1) 4(3)( 3) x = = a (3) = = 6 6 When m =, from equation (i), we get 3 1 x = x 3 Multiplying both sides by 3x, we get 3x 3 = x 3x x 3 = 0 Here, a = 3, b =, c = 3 x = b b 4ac a ( ) ( ) 4(3)( 3) x = (3) = = 40 6 = 10 6 = (1 10) 6 = The solution set is,,, iii. Let the digit in the hundreath s place be x and the digit in the unit place be y. Digit H T U Original number x x + y + 1 y Reversed Number y y + x + 1 x 8 The original number = 100x + 10 (x + y + 1) + y = 100x + 10x + 10y y = 110x + 11y + 10 According to the first condition, 110x + 11y + 10 = 17 [(x) + (x + y + 1) + y] 110x + 11y + 10 = 17 (x + y + 1) 110x + 11y + 10 = 34x + 34y x 34x + 11y 34y = x 3y = 7. (i) The reversed number = 100y + 10 (y + x + 1) + x = 100y + 10y + 10x x = 11x + 110y + 10

32 Algebra According to the second condition. (110x + 11y + 10) = 11x + 110y x 11x + 11y 110y = x 99y = 198 Dividing both sides, by 99 we get x y =.(ii) Multiply equation (ii) by 3, we get 3x 3y = 46.(iii) Subtracting (iii) from (i), we get 76x 3y = 7 3x 3y = 46 () (+) (+) 53x = 53 x = 1 Substituting value of x in equation (ii), we get 1 y = 1 + = y y = 3 Required Number = 110x + 11y + 10 = 110(1) + 11(3) + 10 = = 153 Required Number is 153 9

33 BOARD ANSWER PAPER : MARCH 015 GEOMETRY Geometry 1. Solve any five sub-questions: A( ABC) i. = AB A( DCB) DC A(ΔABC) A(ΔDCB) = [Ratio of the areas of two triangles having equal base is equal to the ratio of their corresponding heights.] ii. Equation of the line is: y = x + 3 comparing it with y = mx + c, we get m = and c = 3 slope = and y-intercept = 3 iii. sin = sin = BC AC opposite side hypotenuse sin = 1 iv. Diagonal of a square = (side) = (10) Diagonal of a square = 10 cm v. Volume of a cube = l 3 But, volume of cube = 1000 cm 3 l 3 = 1000 l = 10 cm ----[Taking cube root on both sides] vi. If two circles touch internally, then the distance between their centres is difference of their radii. the distance between their centres = 5 3 = cm. Solve any four sub-questions: i. sin = 5 13 We know that, sin + cos = [Given] cos = 1 cos = cos = cos = [Taking square root on both sides] 1

34 Board Answer Paper: March 015 ii. A D 115 B Draw angle of 115 Draw arcs Obtain point D Join BD (angle bisector of ABC) C iii. Let C (3, 5) (x 1, y 1 ), D (, 3) (x, y ) slope of line CD = y x y x 1 1 = = slope of line CD = 8 5 iv. Radius (r) = 5 cm Length of arc (l) = 10 cm Area of sector = radius length of arc = 5 10 = 5 5 = 5 Area of sector is 5 cm. v. In PQR, Seg RS is the angle bisector of PRQ ---- [Given] PS PR SQ QR ---- [By property of angle bisector of a triangle] QR QR = QR = 0

35 Geometry vi. D A B Y C E X maeb = 1 m(arc AYC) ---- [Inscribed angle theorem] maeb = 1 40 = (i) mead = 1 m(arc DXE) ---- [Inscribed angle theorem] mead = = (ii) With respect to BAE, EAD is the exterior angle. EAD = ABE + AEB ---- [Remote interior angle theorem] 50 = ABE [From (i) and (ii)] ABE = 30 i.e., DBE = [D A B] 3. Solve any three sub-questions: i. MPN = [Given] QMP = 90 QNP = 90 In MQNP, MPN + QMP + QNP + MQN = MQN = MQN = 360 MQN = MQN = 140 ii [Tangent is perpendicular to radius] ---- [Sum of the measures of the angles of a quadrilateral is 360] P M.8 cm C 7 cm L Q Draw the circle of radius.8 cm Plot point L such that ML = 7 cm Draw perpendicular bisector of seg ML Draw circle with centre C Draw tangents LP and LQ 3

36 Board Answer Paper: March 015 iii. Let A 1 and A be the areas of larger triangle and smaller triangle respectively and h 1 and h be their corresponding heights. A1 A = (i) [Given] 5 h 1 = (ii) [Given] A1 A = h [Ratio of the areas of two triangles having equal base h is equal to the ratio of their corresponding heights.] 6 9 = 5 h ---- [From (i) and (ii)] h = h = 15 h = 7.5 cm The corresponding height of the smaller triangle is 7.5 cm. iv. C A 45 M 30 m B 10 m D Let AB and CD represent the height of the two buildings. AB = 30 m Let BD represent the width of the road. BD = 10 m Draw seg AM seg CD. CAM is the angle of elevation. CAM = 45 In ABDM, B = D = 90 M = [Construction] A = [Remaining angle of ABDM] ABDM is a rectangle ---- [Each angle is 90] AM = BD = 10 m ---- (i) DM = AB = 30 m ---- (ii) In right angled AMC, tan CAM = tan 45 tan 45 = CM AM 1 = CM [From (i)] CM = 10 m ---- (iii) CD = CM + DM ---- [CMD] CD = = 40 m ---- [From (ii) and (iii)] The height of the second building is 40 m. 4 [Opposite sides of a rectangle]

37 v. Given: For the sphere, radius (r) = 4. cm To find: Volume, total surface area Solution: Volume of the sphere = 4 3 r3 Geometry = = = cm 3 Total surface area of sphere = 4r = = = 1.76 cm The volume of the sphere is cm 3 and the surface area of the sphere is 1.76 cm. 4. Solve any two sub-questions: i. Given: ABCD is a cyclic quadrilateral. A D O B C To prove: BAD + BCD = 180, ABC + ADC = 180 Proof: mbad = 1 m(arc BCD) ---- (i) [Inscribed angle theorem] m BCD = 1 m(arc BAD) ---- (ii) [Inscribed angle theorem] Adding equations (i) and (ii), we get mbad + mbcd = 1 m(arc BCD) + 1 m(arc BAD) mbad + mbcd = 1 [m(arc BCD) + m(arc BAD)] BAD + BCD = [Measure of a circle is 360] BAD + BCD = (iii) In ABCD, BCD +BAD +ABC +ADC = [Sum of measures of angles of a quadrilateral] ABC + ADC = [From (iii)] ABC + ADC = ABC + ADC = 180 Hence, the opposite angles of a cyclic quadrilateral are supplementary. ii. L.H.S. = sin 6 + cos 6 = (sin ) 3 + (cos ) 3 = (sin + cos ) [(sin ) sin cos + (cos ) ] ---- [ a 3 + b 3 = (a + b) (a ab + b )] = 1[sin 4 sin cos + cos 4 ] ---- [ sin + cos = 1] = sin 4 + sin cos 3sin cos + cos 4 = sin 4 + sin cos + cos 4 3 sin cos 5

38 Board Answer Paper: March 015 = (sin + cos ) 3sin cos ----[ (a + b) = a + ab + b ] = (1) 3sin cos = 1 3sin cos = R.H.S. sin 6 + cos 6 = 1 3sin cos iii. Given: For the cylindrical part: Diameter = 0 mm = 0 10 = cm radius (r 1 ) = diameter = = 1 cm height (h 1 ) = height of test tube height of hemispherical portion h 1 = 15 1 = 14 cm For the hemispherical portion: Radius (r ) = 1 cm To find: Capacity (volume) of the test tube Solution: Capacity of test tube = volume of cylindrical part + volume of hemispherical portion + 3 = rh r = ( ) + ( 1 1 1) 3 = 14 = 3.14( ) 3 = = cm 3 Capacity of the test tube is cm Solve any two sub-questions: i. E A B D C 6 Given: In ABC, ray AD bisects BAC To Prove that: BD DC = AB AC Construction: Draw a line parallel to ray AD, passing through point C. Extend BA to intersect the line at E. Proof: In BEC, seg AD side EC ---- [By construction] BD AB DC AE ---- (i) [By B.P.T.] line AD line EC on transversal BE BAD AEC ---- (ii) [Corresponding angles]

39 line AD line EC on transversal AC. Geometry CAD ACE ---- (iii) [Alternate angles] Also, BAD CAD ---- (iv) [ Ray AD bisects BAC] AEC ACE ---- (v) [From (ii), (iii) and (iv)] In AEC, AEC ACE ---- [From (v)] AE = AC ---- (vi) [Sides opposite to congruent angles] BD AB = DC AC ---- [From (i) and (vi)] ii. l m = n = 3 A (, 6) P (x, y) B (3, 4) Let A (, 6) (x 1, y 1 ), B (3, 4) (x, y ) m : n = : 3 Let P divide AB internally in the ratio : 3 By internal division section formula, mx nx1 my ny1 P, mn mn (3) 3( ) ( 4) 3(6) P, 3 3 P 6 6, P 010, 5 5 P (0, ) Equation of line with slope (m) = 3 and passing through the point P (0, ) (x 1, y 1 ) in slope point form y y 1 = m(x x 1 ) is y = 3 (x 0) (y ) = 3x y 4 = 3x 3x y + 4 = 0 The equation of the line is 3x y + 4 = 0. iii. RST ~ UAY S = A = [c. a. s. t] RS ST RT ---- (i) [c. s. s. t] UA AY UY Since, corresponding sides are in the ratio 5 : 4 RS ST RT [From (i)] UA AY UY (ii) UA AY 4 7

40 Board Answer Paper: March Consider, ---- [From (ii)] UA 4 5 UA = 6 4 UA = UA = 4.8 cm Now, Consider AY [From (ii)] AY 5 = AY = AY = 6 cm In UAY, UA = 4.8 cm, AY = 6 cm and A = 50 Y U 4.8 cm 50 A 8

41 BOARD ANSWER PAPER : JULY 015 GEOMETRY Geometry 1. Solve any five sub-questions: AABE i. = BE A( BAD) AD A ABE = 6 A( BAD) 9 A ΔABE A(ΔBAD) = 3...[Ratio of areas of two triangles having equal base is equal to the ratio of their corresponding heights.] ii. Diagonal of a square = (side) = (16) Diagonal of a square = 16 cm iii. If two circles touch internally, then the distance between their centres is difference of their radii. the distance between their centres = 8 3 = 5 cm iv. Given, cos = 3 But cos 30 = cos = cos 30 = 30 3 v. Equation of line in slope-intercept form is y = mx + c. Here, m = and c = 5 Equation of line is y = x + 5. vi. Total surface area of cube = 6 l = 6 (9) Total surface area of cube = 486 cm. Solve any four sub-questions: i. In ABC, line l side BC AP PB = AY YC 4 8 = 5 x x = x = 10 units... [Given]... [By B.P.T.] 1

42 Board Answer Paper : July 015 ii. MPN = [Given] QMP = 90 QNP = 90 In MQNP, MPN + QMP + QNP + MQN = MQN = MQN = 360 MQN = MQN = [Tangent is perpendicular to radius]... [Sum of the measures of the angles of a quadrilateral is 360] iii. P 3.5 cm R Draw a circle of radius 3.5 cm Extend the line passing through R Draw the perpendicular at point R iv. Y P X O A X Y The initial arm rotates by 0 in clockwise direction. The angle is more than 180 and less than 70 AOP lies between 70and 180. The terminal arm lies in quadrant II.

43 Geometry v. Given, radius (r) = 3 cm height (h) = 7 cm Curved surface area of cylinder = rh = 3 7 = 13 cm 7 The curved surface area of the cylinder is 13 cm. vi. Given, radius (r) = 10 cm, central angle () = 7 Area of sector = r 360 = = Area of sector = 6.8 cm 3. Solve any three sub-questions: i. seg AQ is the median on side BC. BQ = QC = 1 BC... [Q is the midpoint on side BC] = 1 10 = 5 units In ABC, seg AQ is the median AB + AC = AQ + BQ... [By Apollonius theorem] 1 = AQ + (5) 1 = AQ = AQ 7 = AQ AQ = 7 AQ = 36 AQ = 6 units... [Taking square root on both sides] The length of the median on side BC is 6 units. ii. Line CM is a tangent at M and line CA is a secant. CM = CA CB... (i) [Tangent secant property] Line CN is a tangent at N and line CA is a secant. CN = CA CB... (ii) [Tangent secant property] CM = CN... [From (i) and (ii)] CM = CN... [Taking square root on both sides] 3

44 Board Answer Paper : July 015 iii. T O P cm 70 M Draw PMT of given measure Draw the perpendicular bisectors of side PM and side TM Draw circumcircle by taking O as centre iv. L.H.S. = sec + cosec = 1 1 cos sin sin cos = cos sin 1 = cos sin 1 = cos 1 sin cos,sin sec cosec... [ sin +cos = 1] = sec cosec = R.H.S. sec + cosec = sec cosec 4 v. Let A (3, 11), B (6, ), C (k, 4) y y Slope of a line = 1 x x1 For line AB: Let A (3, 11) (x 1, y 1 ), B (6, ) (x, y ) Slope of line AB = 11 = 9 6 ( 3) = (i) For line AC: Let A (3, 11) (x 1, y 1 ), C (k, 4) (x, y ) Slope of line AC = 4 11 k ( 3) = 7 k 3... (ii)

45 Since, points A, B and C are collinear, Slope of AB = Slope of AC 7 1 = k 3 k 3 = 7 k = k = 4 k = 4 The value of k is Solve any two sub-questions: i. Given: ABCD is a cyclic quadrilateral. A... [From (i) and (ii)] Geometry B O D C To prove: BAD + BCD = 180, ABC + ADC = 180 Proof: mbad = 1 m(arc BCD)...(i) [Inscribed angle theorem] m BCD = 1 m(arc BAD)...(ii) [Inscribed angle theorem] Adding equations (i) and (ii), we get mbad + mbcd = 1 m(arc BCD) + 1 m(arc BAD) mbad + mbcd = 1 [m(arc BCD) + m(arc BAD)] BAD + BCD = [Measure of a circle is 360] BAD + BCD = (iii) In ABCD, BCD +BAD +ABC +ADC = [Sum of measures of angles of a quadrilateral] ABC + ADC = [From (iii)] ABC + ADC = ABC + ADC = 180 Hence, the opposite angles of a cyclic quadrilateral are supplementary. ii. A P E 30 C B 45 D 4 m 5

46 Board Answer Paper : July 015 Let AB represent the lighthouse and CD be the ship. CD = 4 m The distance of the ship from lighthouse is BD Draw ray AP seg BD and CE AB PAC and PAD are the angles of depression PAC = 30 and PAD = 45 Also, ACE = PAC = 30 ADB = PAD = 45...[Alternate angles] In ECDB B = D = 90 E = 90...[construction] C = 90...[Remaining angle of ECDB] ECDB is a rectangle BE = CD = 4 cm...(i) EC = BD...(ii) In ABD, tan ADB = tan 45 tan 45 = AB BD 1 = AB BD AB = BD...(iii) In AEC, tan ACE = tan 30 tan 30 = AE EC 13 = AE EC AB BE = 13 EC BD 4 = 13 BD BD = 3 (BD 4) BD = 3 BD BD BD = 4 3 BD 3 1 = [From (i), (ii) and (iii)] 6 BD = = 4 3 ( 3 1) = = 13 3 = = (1.73) = BD = The distance of the ship from lighthouse is m.

47 iii. A(4,7) Geometry F E B(, 3) D C(0, 1) Let A (4, 7) (x 1, y 1 ); B (, 3) (x, y ) and C (0, 1) (x 3, y 3 ) By midpoint formula, x x3 y y D (x, y) =, =, 4 =, = (1, ) x3 x1 y3 y E (x, y) =, =, 4 8 =, = (, 4) x1x y1 y F (x, y) =, =, 4 10 =, = 10, = (1, 5) For Median AD: A (4, 7) (x 1, y 1 ) and D (1, ) (x 4, y 4 ) Equation of median AD in two-point form is: x x1 y y1 x1x4 y1 y4 x = y 7 7 x 4 y 7 = 41 5 x 4 y 7 = 5 5 x 4 = y 7...[Multiplying both sides by 5] x y = 0 x y + 3 = 0 For median BE: B (, 3) (x, y ) and E (, 4) (x 5, y 5 ) Equation of median BE in two-point form is: x x y y x x5 y y5 x y 3 x y 3 = = x + = 4 (y 3) x + = 4y 1 x 4y = 0 x 4y + 14 = 0 For median CF: C (0, 1) (x 3, y 3 ) and F (1, 5) (x 6, y 6 ) Equation of median CF in two-point form is: x x3 y y3 x x y y

48 Board Answer Paper : July 015 x 0 y 1 = x y 1 = 1 4 4x = y 1 4x y + 1= 0 The equations of medians are x y + 3 = 0; x 4y + 14 = 0; 4x y + 1 = Solve any two sub-questions: i. Consider, ab = A(XYZ) YZ = ( 1 XZ YP) YZ ab = XZ YP YZ... (i) Consider, b 4 + 4a = YZ 4 + 4[A(XYZ)] = YZ XY YZ YZ XY YZ = YZ 4 + XY YZ b 4 + 4a = YZ [YZ + XY ] 4 b 4a = YZ + XY YZ... (ii) In XYZ, XYZ = 90 YZ + XY = XZ... (iii) [By Pythagoras theorem] b 4 + 4a = YZ XZ... [From (ii) and (iii)] 4 b 4a = 4 b 4a YP = YZ XZ... [Taking square root on both sides] = YZ XZ... (iv) ab XZYP YZ =... [Dividing (i) by (iv)] 4 b 4a YZ XZ ab b +4a 4 8 ii. In ABC, A + B + C = [Sum of the measures of all angles of a triangle is 180] A = 180 A + 10 = 180 A = A = 60 ABC LMN... [Given] A = L = [c.a.s.t] B = M = AB LM = BC MN = AC LN... (i) [c.s.s.t] Since, AC LN = (ii) [Given] AB LM = BC MN = [From (i) and (ii)] AB LM = LM = LM = LM =

49 Geometry LM = 8.5 cm In LMN, LM = 8.5 cm, M = 55, L = 60 N L cm 55 M iii. Volume of cylinder = r h Now, volume of ink filled in the cylindrical container = 71% of volume of cylindrical ink container = 71% of rh 1 1 = = cm 3... (i) 100 Also, the volume of ink filled in the refill = 84 % of volume of the cylindrical refill = 84% of rh = = cm (ii) Number of refills that can be filled with ink volumeof inkfilledin cylindricalcontainer = volumeof ink filledin therefill = [From (i) and (ii)] = = = 3550 The number of refills that can be filled with ink is

50 BOARD ANSWER PAPER : MARCH 016 GEOMETRY 1. Solve any five sub-questions: i. DEF ~ MNK...[Given] A( DEF) DE A( MNK) MN...[By theorem on areas of similar triangles] Geometry A( DEF) () A( MNK) (5) A(ΔDEF) 4 A(ΔMNK) 5 ii. In ABC, A = 30, C = 60, B = 90 ABC is a traingle...[given] BC = 1 AC...[side opposite to 30] BC = 1 16 BC = 8 iii. PQS = 1 m (arc PMQ)...[Tangent secant theorem] PQS = [Given] PQS = 55 iv. = 30...[Given] cos ( 30) = cos 30...[ cos ( ) = cos ] cos ( 30) = 3 v. Slope = tan = tan 60...[ = 60] Slope = 3 vi. By using Euler s formula, F + V = E V = 10 + V = 1 6 V = 6 1

51 Board Answer Paper : March 016. Solve any four sub-questions: i. In PRQ Seg RS is the angle bisector of PRQ...[Given] PS SQ = PR...[By property of angle bisector of a triangle] QR QR QR = QR = 1 units ii. C A 1 P Line AP is a tangent to the circle at A and line PC is a secant. AP = CP BP... [Tangent secant property] (1) = CP = CP 9 CP = CP = 16 units BP + BC = CP... [PBC] 9 + BC = 16 BC = 16 9 BC = 7 units iii. B 9 A O E B D C Draw ABC of given measure Draw the perpendicular bisectors of side BC and side AC Draw circumcircle by taking O as centre

52 iv. Y Geometry P X O A X Y The initial arm rotates by 130 in anti-clockwise direction. The angle is more than 90 and less than 180 AOP lies between 90and 180. The terminal arm lies in quadrant II. v. Radius (r) = 9 cm length of arc (l) = 16 cm Area of sector = radius length of arc = 9 16 = 9 8 = 7 Area of sector is 7 cm. vi. Radius (r) = 1.4 cm Surface area of the sphere = 4r = = Surface area of the sphere = 4.64 cm 3. Solve any three sub-questions: i. 17 cm A D 11 cm O B C Let ABCD be the parallelogram and its diagonals AC and BD intersect each other at O. AB = 11 cm, AD = 17 cm, BD = 6 cm OB = 1 BD... [Diagonals of a parallelogram bisect each other] OB = 1 6 OB = 13 cm... (i) In ABD, O is the midpoint of diagonal BD... [Diagonals of a parallelogram bisect each other] seg AO is the median of ABD AB + AD = OA + OB... [By Apollonius theorem] (11) + (17) = OA + (13) = OA = (OA) = OA 7 = OA 7 = OA 3

53 Board Answer Paper : March 016 OA = 36 OA = 6 cm... (ii) [Taking square root on both sides] OA = 1 AC... [Diagonals of a parallelogram bisect each other] 6 = 1 AC... [From (ii)] AC = 1 cm The length of the other diagonal is 1 cm. ii. a. mpqr = 1 m(arc PCR)... [Inscribed angle theorem] mpqr = [Given] mpqr = 13...(i) b. mspq = 1 m(arc QDS)... [Inscribed angle theorem] mspq = 1 48 mspq = 4 SRQ SPQ... [Angles subtended by the same arc are congruent] SRQ = 4... (ii) c. With respect to ARQ, SRQ is the exterior angle. SRQ = RAQ + AQR... [Remote interior angle theorem] 4 = RAQ [From (i) and (ii)] RAQ = 4 13 mraq = 11 iii. A l O B K 4 P i. For drawing a circle of radius 3.5 cm ii. For drawing chord BK iii. For drawing BAK iv. For drawing BKP

54 Geometry iv. sec = cos = 3 1 sec = 1 = [Given] 1... cos sec 3 cos = sin + cos = 1 sin + 3 = 1 sin = 1 sin = sin = sin = 1 4 sin = 1... [Taking square root on both sides] But, lies in the quadrant IV. sin is negative. sin = 1 cosec = 1 sin cosec = 1 = 1 cosec = Consider, 1cosec 1cosec ( ) = 1 1 ( ) 1 1 1cosec 1cosec = 3 v. Let P (, 3) (x 1, y 1 ), Q (4, 7) (x, y ) The line passes through points P and Q. Equation of line in two-point form is: x x1 x x = y y1 y y 1 x y 3 = x y 3 = 4 4 (x ) = (y 3) 4x + 8 = y + 6 y = 4x y = 4x y = x 1... [Dividing both sides by ] y = x 1 is the required equation of line which is of the form y = mx + c. 5

55 Board Answer Paper : March Solve any two sub-questions: i. Given: A circle with centre O, an external point P of the circle. The two tangents through the point P touch the circle at the points A and B. To prove: PA = PB Construction: Draw seg OA, seg OB and seg OP. A O P Proof: PAO = PBO = 90...[Tangent is perpendicular to radius.] In the right angled PAO and the right angled PBO, seg OA seg OB...[Radii of the same circle] hypotenuse PO hypotenuse PO...[Common side] PAO PBO...[Hypotenuse side test] seg PA seg PB...[c.s.c.t.] ii. PA = PB Hence, the lengths of the two tangent segments drawn to a circle from an external point are equal. A B B D C 40 m Let AB represent the height of the tree and BD represent the width of river. D and C are the initial and final positions of the observer. DC = 40 m ADB and ACD are the angles of elevation. ADB = 60 and ACD = 30 In right angled ABD, tan 60 = AB BD 3 = AB BD AB = 3 BD... (i) In right angled ABC, tan 30 = AB BC 13 = AB BD DC 13 = AB BD [BDC]

56 Geometry BD 40 =AB 3... (ii) BD 40 3 BD = 3... [From (i) and (ii)] 3 BD 3 = BD BD = BD BD BD = 40 BD = 40 BD = 40 BD = 0 m Now, AB = 3 BD = [From (i)] AB = 0 3 = AB = 34.6 m The height of the tree is 34.6 m and width of the river is 0 m. iii. A(5,4) P B(3,) D(x, y) C(1,8) Let A (5, 4) (x 1, y 1 ); B (3, ) (x, y ) and C (1, 8) (x 3, y 3 ). D (x, y) is the midpoint of BC. By midpoint formula, 3 3 D (x, y) = x x, y y 31 8 =, = 10, = (1, 5) Let D (1, 5) (x 4, y 4 ) For Median AD: The equation of the median AD in two-point form is: x x1 y y1 x x y y x x 5 = 51 x 5 6 x 5 = y y 4 = 9 = y 4 3 y (x 5) = (y 4) 3x 15 = y 8... [Multiplying both the sides by 3] 7

57 Board Answer Paper : March x y = 0 3x y 7 = 0 y1 y Slope of line AC = = x x = 4 8 = = 3 The slope of parallel lines are equal. Slope of line (l) = Slope of line AC = 3 By slope-point form, equation of line l passing through B (3, ) (x 1, y 1 ) and having slope (m) = 3 is: y y 1 = m (x x 1 ) y () = 3 [x (3)] y + = 3 (x + 3) y + = 3x + 9 3x + 9 y = 0 3x y + 7 = 0. The equation of median AD is 3x y 7 = 0 and the equation of the line parallel to AC and passing through point B is 3x y + 7 = Solve any two sub-questions: i. In ABF, EB = EF = a E is midpoint of seg BF.... [Given] seg AE is the median AB + AF = AE + EB... [By Apollonius theorem] AB + a = a + a AB + a = 4a AB = 4a a AB = 3a AB = 3 a... [Taking square root on both sides] In ACE FC = FE = a... [Given] F is the midpoint of seg EC seg AF is the median AE + AC = AF + FC... [By Apollonius theorem] a + AC = a + a AC + a = 4a AC = 4a a AC = 3a AC = 3 a... [Taking square root on both sides] AB = AC = 3 a

58 ii. U Geometry U R S 4.5 cm H V R Analytical figure S 4.5 cm H V S 1 S o S 3 S 4 o S5 X Draw SHR of given measure Draw a ray making an acute angle at S with side SV, mark points S 1, S,., S 5 such that SS 1 = S 1 S = S S 3 = S 3 S 4 = S 4 S 5 Join S 3 and H and draw seg S 5 V parallel to S 3 H, where V is the point on extended SH Draw VU side HR SVU is the required triangle similar to SHR. iii. For the cylindrical pipe: diameter = 0 mm radius = diameter = 0 = 10 mm = 1cm Rate of flow of water through the pipe = 15 m/minute = cm/minute = 1500 cm/minute Water flow s through a distance (h) of 1500 cm in a minute 9

59 Board Answer Paper : March 016 Volume of water flowing through the pipe in 1 minute = r h = = 1500 cm 3 For the conical vessel: diameter = 40 cm radius = diameter = 40 = 0 cm depth (h) = 45 cm Volume of conical vessel = 1 3 r h = cm3 volumeof conical vessel Time taken to fill the conical vessel = volumeof water flowing through pipe in1minute = = = = 4 min Time taken to fill the conical vessel is 4 min. 10

60 BOARD ANSWER PAPER : JULY 016 GEOMETRY Geometry Time: Hours Max. Marks: 40 Note: i. Solve All questions. Draw diagrams wherever necessary. ii. Use of calculator is not allowed. iii. Figures to the right indicate full marks. iv. Marks of constructions should be distinct. They should not be rubbed off. v. Diagram is essential for writing the proof of the theorem. Q.P. SET CODE A 1. Solve any five sub-questions: i. RP : PK = 3 : ----[Given] A( TRP) = RP ---- [Ratio of the areas of two triangles having equal heights A( TPK) PK is equal to the ratio of their corresponding bases.] A(ΔTRP) A(ΔTPK) = 3 ii. If two circles touch externally, then the distance between their centres is sum of their radii. the distance between their centres = = 7 cm iii. slope = tan = tan 45 slope of the line is 1 iv. By using Euler s formula, F + V = E V = 30 + V = 3 1 V = 0 v. Diagonal of a square = (side) = (8) Diagonal of a square = 8 cm vi. Y X O A X P Y The initial arm rotates by 305in anticlockwise direction. The angle is more than 70 and less than 360. AOP lies between 70 and 360. The terminal arm lies in quadrant IV. 1

61 Board Answer Paper : July 016. Solve any four sub-questions: i. M A P B N Draw seg AB of length 9.7 cm Take point P at a distance 3.5 cm from A Draw MN AB through point P ii. The terminal arm passes through (3, 4) x = 3 and y = 4 r = x + y r = (3) + (4) r = r = 5 r = [Taking square root on both sides] Let the angle formed be. By definition of trigonometric ratios in standard position, we get y 4 sin = = r 5 iii. DMN and AQR are similar. Reason: In DMN and AQR, DMN AQR ---- [Each is 55] DNM ARQ ---- [Each is of same measure] DMN AQR ---- [By AA test of similarity] iv. m (arc AKC) + m (arc BMC) = m (arc BMC) = 180 m (arc BMC) = m (arc BMC) = 140 v. Slope of line (m) = 6 7 and the line passes through the point P(0, 6) (x 1, y 1 ) Equation of line in slope point form is: y y 1 = m(x x 1 ) y 6 = 6 (x 0) 7 y 6 = 6 7 x 7(y 6) = 6x 7y 4 = 6x 6x 7y + 4 = 0 The equation of line is 6x 7y + 4 = 0.

62 vi. radius (r) = 1 cm, central angle () = 60 Area of sector = r 360 Geometry o 60 = = = Area of sector = 31 cm 3. Solve any three sub-questions: i. Line CB is a tangent to the circle at B and line AC is a secant CB = CA CD.(i) [Tangent secant property] Since, line CB is tangent at B and AB is a diameter ABC = 90 In right angled ABC AB + BC = AC.[By Pythagoras Theorem] AB + CA CD = AC.[From (i)] (r) + AC CD = AC (r) = AC AC CD 4r = AC (AC CD) 4r = AC AD.[A D C] i.e. 4(radius) = AC AD ii. In PQR, Ray PT is the angle bisector of QPR. PQ PR = QT ---- [By property of angle bisector of a triangle] TR 3.6 x = = 4 x = x x = 4.5 cm PR = 4.5 cm ---- [ PR = x] Now, QR = QT + TR ---- [QTR] QR = QR = 9 cm Perimeter of PQR = PQ + QR + PR = = 17.1 cm The value of x is 4.5 cm and the perimeter of PQR is 17.1 cm. iii. l : b : h = 5 : 4 : ---- [Given] Let the common multiple be x length (l) = 5x, breadth (b) = 4x, height (h) = x Total surface area of cuboid = (lb + bh + hl) But, total surface area of cuboid = 116 cm ---- [Given] (lb + bh + hl) = 116 (lb+ bh + hl) = 116 3

63 Board Answer Paper : July 016 5x 4x + 4x x + x 5x = 608 0x + 8x + 10x = x = 608 x = x = 16 x = [Taking square root on both sides] length = 5x = 5 4 = 0 cm breadth = 4x = 4 4 = 16 cm height = x = 4 = 8 cm The dimensions of the cuboid are (0 16 8) cm. iv. R I S M 7 cm T Draw RST of given measure Draw the angle bisectors of S and T Draw incircle by taking I as centre v. L.H.S. = 1 cosa 1 cosa = 1cosA 1cosA 1cosA 1 cosa ---- [On rationalising the denominator] = (1 cos A) 1 cos A = (1 cos A) sin A ----[ sin = 1 cos ] 4 4 = 1 cosa sin A = 1 sina cosa sin A = cosec A cot A = R.H.S.[ 1 sin 1 cosa = cosec A cot A 1 cosa = cosec ]

64 4. Solve any two sub-questions: i. Given: ABCD is a cyclic quadrilateral. B A O C D To prove: BAD + BCD = 180, ABC + ADC = 180 Proof: Geometry mbad = 1 m(arc BCD) ---- (i) [Inscribed angle theorem] m BCD = 1 m(arc BAD) ---- (ii) [Inscribed angle theorem] Adding equations (i) and (ii), we get mbad + mbcd = 1 m(arc BCD) + 1 m(arc BAD) mbad + mbcd = 1 [m(arc BCD) + m(arc BAD)] BAD + BCD = [Measure of a circle is 360] BAD + BCD = (iii) In ABCD, BCD +BAD +ABC +ADC = [Sum of measures of angles of a quadrilateral] ABC + ADC = [From (iii)] ABC + ADC = ABC + ADC = 180 Hence, the opposite angles of a cyclic quadrilateral are supplementary. ii. A B D 40 m C Let AB represent the height of the tree and BD represent the width of river. D and C are the initial and final positions of the observer. DC = 40 m ADB and ACD are the angles of elevation. ADB = 60 and ACD = 30 In right angled ABD, tan 60 = AB BD 3 = AB BD AB = 3 BD ---- (i) 5

65 Board Answer Paper : July 016 In right angled ABC, tan 30 = AB BC 13 = AB BD DC ---- [BDC] 13 = AB BD 40 BD 40 =AB (ii) BD 40 3 BD = [From (i) and (ii)] 3 BD 3 = BD BD = BD BD BD = 40 BD = 40 BD = 40 BD = 0 m Now, AB = 3 BD = [From (i)] AB = 0 3 = AB = 34.6 m The height of the tree is 34.6 m and width of the river is 0 m. iii. Given, diameter = 0.9 m radius (r) = diameter = 0.9 = 0.45 m height (h) = length of roller = 1.8 m Curved surface area of roller = rh = = = m Area of ground pressed by roller in 1 revolution = curved surface area of roller = m Area of ground pressed by roller in 500 revolutions = = = = m. 0 Area of the ground pressed by the roller in 500 revolutions is m. 5. Solve any two sub-questions: i. Given: In PQR, line l side QR. Line l intersects side PQ and side PR in points M and N respectively, such that PMQ and PNR.

66 Geometry P M N l Q R To Prove that: PM MQ = PN NR Construction: Draw seg QN and seg RM. Proof: In PMN and QMN, where PMQ, A( PMN) = PM ---- (i) A( QMN) MQ In PMN and RMN, where PNR, A( PMN) = PN ---- (ii) A( RMN) NR [Ratio of the areas of two triangles having equal heights is equal to the ratio of their corresponding bases.] [Ratio of the areas of two triangles having equal heights is equal to the ratio of their corresponding bases.] A(QMN) = A(RMN) ---- (iii) [Areas of two triangles having equal bases and equal heights are equal.] A( PMN) A( QMN) = A( PMN) A( RMN) ---- (iv) [From (i), (ii) and (iii)] PM MQ = PN NR ---- [From (i), (ii) and (iv)] ii. Let A (4, 8) (x 1, y 1 ); B (5, 5) (x, y ); C (, 4) (x 3, y 3 ) and D (1, 7) (x 4, y 4 ) y y Slope of a line AB = 1 = 5 8 x x1 5 4 = 3 = (i) y3 y Slope of line BC = = 4 5 x3 x (ii) y4 y3 Slope of line DC = = 7 4 = 3 = (iii) x4 x3 1 1 y y Slope of line AD = 4 1 x4 x = 7 8 = (iv) Slope of line AB = Slope of line DC ---- [From (i) and (iii)] line AB line DC ---- (v) Slope of line BC = Slope of line AD ---- [From (ii) and (iv)] line BC line AD ---- (vi) ABCD is a parallelogram ---- [From (v) and (vi)] Points (4, 8), (5, 5), (, 4) and (1, 7) are the vertices of a parallelogram. 7

67 Board Answer Paper : July 016 iii. T E T E 10 A H M 6.3 cm Analytical figure A cm H M A 1 A A 3 A 4 A 5 A 6 A 7 X Draw AMT of given measure Draw a ray making an acute angle at A with side AM, mark points A 1, A,., A 7 such that AA 1 = A 1 A = A A 3 = A 3 A 4 = A 4 A 5 = A 5 A 6 = A 6 A 7 Join A 7 and M and draw seg A 5 H parallel to A 7 M, where H is the point on AM Draw HE side MT AHE is the required triangle similar to AMT. 8 8

68 BOARD ANSWER PAPER : MARCH 015 SCIENCE AND TECHNOLOGY SECTION A Science and Technology 1. (A) (a) Rewrite the following statements with suitable words in the blanks: i. The device used for producing electric current is called a generator. ii. Stratosphere, the second layer of the atmosphere reaches 48 km above the earth s surface. (b) Rewrite the following table so as to match the second column with the first column: Column A Column B i. eosin synthetic indicator ii. oxidation 1 losing hydrogen (B) (c) CaOCl Rewrite the following statements by selecting the correct options: i. (C) When phenolphthalein is added to NaOH, the colour of the solution will become pink. ii. (D) If the potential difference across the ends of a conductor is 0 V and the resistance of the conductor is 44 (ohm), then the current flowing through is 5 A. iii. (B) 1 A = 10 3 ma iv. (B) The distance between principal focus and optical centre of the lens is focal length. v. (B) When rays of light are incident on a glass slab, then the incident ray and emergent ray are parallel to each other.. Answer any five of the following: i. a. Scattering of light depends upon the particle size and wavelength of the incident light. b. In the visible range of light, red light has greater wavelength. c. Red light can travel larger distance without getting scattered. d. This enables viewer to spot red colour distinctly even from a distance. Hence, danger signals are red in colour. ii. CuO + HCl CuCl + H O [] Copper Water chloride iii. Newlands Octaves law states that When the elements are arranged in an increasing order of their atomic masses, the properties of the eighth element are similar to the first. [] iv. Given: 1 = m/s = m/s To find: 1 1 Formula: 1 = From Formula, = = The refractive index of the medium with respect to air is. 1

69 Board Answer Paper : March 015 v. Resistances in series a. The voltage across each resistor is different. b. The current through each resistor is the same. c. This combination is used to increase the effective resistance of the circuit. d. This combination decreases the current in the circuit. Resistances in parallel The voltage across each resistor is same. The current in various resistors are inversely proportional to the resistances. This combination is used to decrease the effective resistance of the circuit. This combination increases the current in the circuit. [] vi. A B 1 C B F P A 1 Object (AB) between centre of curvature and focus Image (A 1 B 1 ) beyond centre of curvature, real, inverted and magnified. 3. Answer any five of the following: i. The role of citizen in pollution control: a. Citizens should plant trees and develop gardens, parks and open grounds in the locality. b. Citizens should save fossil fuels and reduce pollution. c. They should minimize electricity consumption. d. They should use public transport instead of private vehicles. e. Citizens should use non-conventional source of energy like solar, wind energy, tidal energy. f. They should maintain vehicles in well tuned condition. g. Citizens should keep home and public places clean and should keep their own locality free from pollution. [Any six points: ½ mark each] [3] ii. a. The band of coloured components of a light beam is called as spectrum. b. White light is composed of seven colours. When white light is incident a prism, it splits up into its constituent colours. c. Each colour bends through different angles with respect to incident ray. So the rays of each colour emerge along different path and become distinct. Hence, we get a spectrum of seven colours when white light is dispersed by a prism. iii. a. The four most common electrical appliances based on heating effect of electric current are electric iron, electric heater, electric toaster and electric oven. [] b. Finely heated platinum wire is used in surgery for cutting tissues much more efficiently than a knife. [] iv. a. Gypsum is obtained on mixing Plaster of Paris with water. CaSO 4.H O + 3H O CaSO 4.H O + Heat. Plaster of Paris Water Gypsum b. Gypsum is used as a raw material in manufacturing cement. c. POP is used in making statues and decorating roofs. It is also used in surgical bandages.

70 v. Metals Non-metals Metalloids Mg, Hg C, S Si, As Science and Technology vi. a. = b. 4 = 4 c. = 3 4. Attempt any one of the following: (A) a. Short circuiting takes place if a live wire and a neutral wire come in direct contact or touch each other. b. During a short circuit, the resistance of the circuit becomes very small. c. A huge amount of current flows through the circuit during a short circuit. d. The flow of a large amount of current in the circuit beyond the permissible value (of current) is called overloading. It may cause fire. e. The effects of overloading can be avoided by not connecting many appliances (especially of high power rating) at a time in the circuit. (B) a. Negative power indicates that 8 students use spectacles having concave lens. b. Positive power indicates that students use spectacles having convex lens. c. Generally, most of the students use the spectacles having concave lenses of suitable focal length. d. Most of the students suffer from myopia or near-sightedness. e. Two possible reasons for myopia: 1. The ciliary muscles are unable to relax sufficiently.. Increase in the distance between eye lens and retina due to lengthening of eyeball or curved lens. SECTION B 5. (A) (a) Find the correlation in the given pair and rewrite the answer: i. Tinning : Tin :: Galvanizing : Zinc. ii. Mammals : Reptiles :: Amphibia : Fishes. (b) State True or False: i. True ii. False iii. True (B) Rewrite the following statements by selecting the correct options: i. (A) The molecular formula of acetic acid is CH 3 COOH. ii. (C) Carbon dioxide enters into the leaves through tiny pores present on the surface of the leaf called stomata. iii. (A) CuSO 4 solution is blue in colour. iv. (D) Yeast reproduces by budding. v. (B) Raisins put in water absorb water by the process of osmosis. 6. Solve any five of the following: i. a. Common salt is an ionic compound having strong force of attraction between the oppositely charged Na + and Cl ions. b. So, a large amount of heat energy is required to break these forces of attraction and to melt or boil the common salt. Thus, common salt has high melting point and boiling point. [3] 3

71 Board Answer Paper : March 015 ii. The pancreas with their associated structures: Bile duct Common duct Pancreatic duct Gall bladder Stomach Pancreas [] iii. a. Connecting link between Peripatus and Annelida: Segmental nephridia, thin cuticle and parapodia like appendages. b. Connecting link between Peripatus and Arthropoda: Trachea and open circulation. iv. v. Two plant hormones and their functions: a. Gibberellins: Help in growth of stem b. Cytokinins: Promote cell division Toilet soap Laundry soap i. High quality of fats and oils are used as raw materials. Cheaper quality of fats and oils are used as raw materials. ii. Expensive perfumes are added. Cheaper perfumes are added. iii. Toilet soaps do not contain free alkalies which are harmful to skin. Laundry soaps contain free alkalies which contribute towards its cleaning action. [Any two points :1 mark each] [] vi. Objectives of sustainable development are : a. Reduce pollution by using eco-friendly technology. b. Restrain the use of natural resources to ensure availability for the future generation. c. Protection of environment. d. Social equality in accessing resources. e. Continuous economic growth. [Any four points :½ mark each] [] 7. Answer any five of the following: i. Alloy: a. An alloy is a homogenous mixture of two or more metals or a metal and a non-metal in definite proportion. b. Alloys do not corrode easily. Example: a. Brass (copper and zinc) b. Bronze (copper and tin) 4 ii. a. Cells that assist the neuron in their function Neuroglia b. The small gap between the consecutive neurons Synapse c. Part of the brain that co-ordinates the voluntary functions Forebrain (Cerebrum) iii. Fertilization: a. During the process of fertilization, sperms enter through the vaginal passage, travel upwards and reach the oviduct. b. In oviduct, one of the sperm fuses with egg and the fertilization is completed. Development: a. After fertilization, the egg cell (zygote) begins to divide and redivide to form a ball of cells called a blastocyst. b. This blastocyst implants itself in the wall of the uterus.

72 Science and Technology c. The embryo develops inside the uterus. It obtains all the nutrients and oxygen from its mother s blood through the umbilical cord. d. Development of the foetus takes place for nine months. Birth: a. Nine months onwards, the baby is ready to take birth. It begins to move down towards the vaginal passage. b. During birth, the cervix gradually opens and the baby is released through the vagina. iv. Vestigial organs: Vestigial organs are the non-functional organs in some organisms but have essential functions in other organisms. Examples of vestigial organs in human beings: Vermiform appendix, wisdom teeth. Examples of vestigial organs in plants: Scale like leaves on Indian pipe plant, Stamens which lack anthers in some flowers. v. a. Recycling is a type of green technology that uses old material to make new products. Many waste products from the industries such as paper, glass, plastics and metals can be recycled. b. Example: Old used tyres are recycled to create play ground flooring in parks to provide soft surfaces that increase the safety of the children playing there. c. Advantages of recycling: 1. It conserves energy and raw materials.. It saves space used in landfills. 3. It protects environment by effective handling of waste materials. 4. It reduces the cost of production. [Any two points:½ mark each] vi. a. Sexual reproduction gives rise to variation. b. Importance of variation in survival of species: Variations give rise to variety and diversity. They enable organisms to adapt and survive in the changing environmental conditions. Variations also help to prevent the complete extinction of plant and animal species. [] 8. (A) i. Anaerobic reaction = CO + Ethanol ii. Reaction in human muscles = Lactic acid iii. Aerobic respiration = CO + H O iv. Reaction in plant cells = Starch v. Reaction in liver = Glycogen (B) i. The other two names of ethanol: Ethyl alcohol, Spirit. ii. The structural formula of ethanol: H H iii. H C C OH H H Properties of ethanol: i. It is a colourless liquid. ii. It is combustible and burns with a blue flame. iv. Action of phosphorus trichloride on ethanol: When ethanol reacts with phosphorus trichloride (PCl 3 ), it forms ethyl chloride and phosphorus acid. 3C H 5 OH + PCl 3 3C H 5 Cl + H 3 PO 3 Ethanol Phosphorus Ethyl Phosphorus trichloride chloride acid [] 5

73 BOARD ANSWER PAPER : JULY 015 SCIENCE AND TECHNOLOGY SECTION A Science and Technology 1. (A) (a) Fill in the blanks: i. Very fine particles mainly scatter blue light. ii 1 ma = 10 3 A. (b) Match the column A with column B : Column A Column B i. Eka-boron. Scandium ii. Eka-Aluminium 3. Gallium iii. Eka-Silicon 1. Germanium (B) Choose the correct alternative and rewrite the following: i. Mirror used by a dental surgeon is concave. ii. (D) If the potential difference across it is doubled and the resistance is halved, current passing through a resistance becomes four times. iii. (B) If three equal resistances are given, they can be arranged in four combinations. iv. (B) Inside water, an air bubble behaves always like a concave lens. v. (D) A glass slab is placed in the path of convergent light. The point of convergence of light undergoes lateral shift.. Answer the following questions (any five): i. Dobereiner s triads: i. In 189, Dobereiner classified existing elements in a tabular form by placing three elements having similar properties in a group called triad. ii. In each triad, the elements were placed according to increasing order of their atomic masses. iii. The atomic mass of the middle element in each triad was approximately the mean of the atomic masses of other two elements. Eg. In the triad of Lithium, Sodium and Potassium, the atomic mass of Sodium (3) is the mean of atomic masses of Lithium (6.9) and Potassium (39). ii. a. When edible oil is stored in an iron or tin container for a long time, it undergoes oxidation reaction. b. Due to oxidation reaction, the taste and smell of edible oil changes and it becomes rancid, if food is cooked in this oil, its taste changes. Hence, edible oil is not allowed to stand for a long time in an iron or tin container. iii. Given: Power (P) = 750 W Time (t) = hrs / day Number of days (n) = 30 To find: Energy consumed =? Formula: Energy consumed = P t n Solution: From formula, Energy consumed = = W h = 45 kwh Energy consumed = 45 kwh 1

74 Board Answer Paper : July 015 iv. Conductors Insulators i. Substances having very low resistance Substances having infinite are called conductors. resistance are called insulators. ii. They allow charges to flow freely. They do not allow the charges to flow. iii. They contain large number of free electrons. They contain practically no free electrons. iv. Generally, conductors are metals. Generally, insulators are non metals. Eg. Copper, silver, gold, iron etc. Rubber, plastic, glass, ebonite etc. [Any two points: 1 mark each] [] v. Uses of sodium carbonate (washing soda): a. Washing soda is used in washing clothes as a cleansing agent. b. It is very useful in manufacturing detergent powder, paper and glass. c. It is also used to refine petroleum. d. It plays an important role in making the water soft and potable. vi. Screen Dispersion of light Diagram: Labels: 3. Answer the following questions (any five): i. a. Silver chloride (AgCl) b. Double displacement reaction. c. AgNO 3(aq) + NaCl (aq) AgCl (s) + NaNO 3(aq) Silver nitrate Sodium chloride Silver chloride Sodium nitrate ii. a. When two or more substances combine to form a single product, then the chemical reaction is known as combination reaction. b. The chemical reaction which is accompanied by absorption of heat is called endothermic reaction. c. The chemical reaction in which reactants gain oxygen atom or lose hydrogen atom to form products is known as oxidation reaction. iii. iv. Concave mirrors are used: a. in torches and headlights. b. in flood lights. c. as reflecting mirrors for projector. d. to collect heat radiations in solar devices. e. in shaving mirror, dentist s mirror. f. in solar furnaces. The domestic appliances based on heating effect of electric current are: a. Electric iron b. Electric heater c. Electric toaster d. Electric oven e. Electric kettle f. Electric bulb

75 Science and Technology v. Formation of rainbow: a. The beautiful phenomenon of the rainbow is a combination of different phenomena like dispersion, refraction and reflection of light. b. The rainbow appears in the sky after a rain shower. When sunlight enters the water droplets present in the atmosphere, water droplets act as small prism. c. These drops refract and disperse the incident sunlight. d. This dispersed light gets reflected inside the droplet and again is refracted. As a collective effect of all these phenomena, the seven coloured rainbow is observed as shown in figure. Sunlight Rain drop Red Violet Formation of rainbow vi. a. A convex lens forms real, diminished and inverted image when an object is placed beyond F 1. b. The image is formed between F and F. c. The formation of image is as shown in the figure. L A B F 1 F 1 O F B 1 F A 1 Object (AB) beyond F 1 Image (A 1 B 1 ) between F and F, Diminished, real and inverted. 4. Answer the following question (any one): i. a. B C N A D S Iron core Armature coil B 1 R 1 Slip rings R B Axle G Electric AC generator [] b. Principle of electric generator: Electric generator works on the principle of electromagnetic induction. When the coil of electric generator rotates in a magnetic field, magnetic field induces a current in this coil. This induced current then flows into circuit connected to the coil. c. Function of slip rings: The ends of armature coil are connected to two slip rings. Slip rings rotate along with the armature coil. 3

76 Board Answer Paper : July 015 ii. d. Any two uses of a generator: 1. Generators are used in cinema halls, office buildings, hospitals, etc to provide electricity in case of power failure in the city.. Generators provide backup electrical power to household appliances such as refrigerators, water heaters, etc. Sources of noise pollution: a. Industrial activity b. construction activity c. generator sets d. loud speakers e. public address systems f. music systems g. vehicular horns h. other mechanical devices, etc. [Any six sources : ½ mark each] [3] Impact of noise pollution on human body: i. Impact of noise pollution on human beings depends on the noise intensity, frequency and exposure duration. ii. Noise pollution can cause auditory fatigue and deafness. iii. It can cause communication interference, sleep interference, concentration interference, ill temper, annoyance, violent behaviour, mental disorientation, bickering and loss of working efficiency. iv. Noise pollution can cause physiological effects such as nausea, fatigue, anxiety, visual disturbances, insomnia, hypertension, cardio vascular disease. 4

77 BOARD ANSWER PAPER : JULY 015 SCIENCE AND TECHNOLOGY SECTION B Science and Technology 1. (A) (a) Fill in the blanks: i. Chlorine is a greenish coloured poisonous gas. ii. Compounds with identical molecular formula but different structures, are called isomers. (b) (c) Name the following: Neuroglia Match the following: Column I Column II i. Inhibits plant growth 5. Abscisic acid ii. Cytokinins 4. Promote cell division iii. Cellular respiration 3. Mitochondria iv. Bile 1. Breaks large fat globules into smaller ones (B) Choose the correct alternative and rewrite the following: i. Oxygen is released in plants during the process of photosynthesis. ii. To observe the hydra bud clearly, Raju should see it first under the low power lens and then under the high power lens in order to see all of the above. iii. Reaction of iron nails with copper sulphate solution is an example of displacement reaction. iv. CuSO 4 is blue in colour. v. Acetic acid turns blue litmus red.. Answer the following questions (any five): i. a. When a piece of calcium is placed in water, initially it sinks in water as its density is greater than that of water. b. Calcium reacts with water less vigorously to form calcium hydroxide and hydrogen gas. Ca (s) + H O (l) Ca(OH) (aq) + H (g) Calcium Water Calcium Hydrogen hydroxide c. Since sufficient heat is not evolved during the reaction, hydrogen does not catch fire. d. Instead, calcium starts floating because the bubbles of hydrogen gas formed stick to the surface of calcium metal. Hence, calcium floats over water during its reaction with water. ii. Toilet soap i. High quality of fats and oils are used as raw materials. Laundry soap Cheaper quality of fats and oils are used as raw materials. ii. Expensive perfumes are added. Cheaper perfumes are added. iii. Toilet soaps do not contain free alkalies which are harmful to skin. Laundry soaps contain free alkalies which contribute towards its cleaning action. [Any two points: 1 mark each] [] iii. a. There is no definite excretory system or organs present in plants for the removal of excretory products. Gaseous excretory materials are eliminated by diffusion. b. Many waste products are stored in the vacuoles of leaves, flowers, fruits and bark which fall off later on. Whereas other waste products are stored as resins and gum in old xylem. 1

78 Board Answer Paper : July 015 iv. c. Plants also excrete some waste substances into the soil around them. In some plants, waste is in the form of calcium oxalate crystals called raphides. d. Rubber, latex, gum, resins and essential oils like eucalyptus or sandalwood oil are plant wastes useful to human beings. Dendrites Cyton (cell body) Nucleus Axon Neuron Diagram: Any two labels: v. a. Regeneration is the process of asexual division in some multicellular organisms, through which they can reconstruct the entire body from the isolated body cells. b. Regeneration is carried out by specialised regenerative cells. These cells proliferate and make large number of cells. c. These later develop into various cell types and tissues and help in the production of new organisms. d. The capacity to regenerate is very high among some animals. In Planaria, when the body is cut into many pieces, each piece develops into a whole new organism. vi. a. Organisms which are structurally intermediate between two different groups are called Connecting Links. b. Example: Peripatus is considered as a connecting link between Annelida and Arthropoda. 3. Answer the following questions (any five): i. Mechanism of Mendel s monohybrid cross: a. Monohybrid cross involves crossing of two plants with one pair of contrasting characters. Mendel selected dominant red flowered (RR) and recessive white flowered (rr) pea plants as parents for crossing. b. After crossing of the parents, Mendel obtained the F 1 generation, which possessed red coloured flowers, but they were different from red-flowered plants of parental generation, because these plants outwardly showed the same characteristics (i.e. red flowers) but were genetically different. This happened because the F 1 generation also contained white recessive coloured factors. c. From this, Mendel concluded that red was dominant over white and the F 1 generation was phenotypically red and genotypically a hybrid of red and white. Pure red Pure white flowers P 1 generation RR R rr r Gametes Rr Plants with red flowers F 1 generation

79 Science and Technology d. The F 1 generation red flowered plant possess Rr genotype and thus, produces two types of gametes R and r. e. The F 1 plants were allowed to self-pollinate to produce second filial generation. (F generation). The F generation can be represented in Punnett square as: R r R RR Homozygous red Rr Heterozygous red r Rr Heterozygous red rr Homozygous white f. From F generation, Mendel observed that the phenotypic ratio of the offsprings was approximately 3 red (dominant):1 white (recessive), whereas the genotypic ratio was 1 dominant (RR): hybrid (Rr):1 recessive (rr), which is based on genetic constitution. ii. Parents: XY XX Gametes: X Y X X Children: iii. a. Seismonastic movement: The movement of plant part in response to the stimulus of touch is called as seismonastic movement. b. Type of movements in plants: 1. Growth dependent movement: e.g. movement of root system of plants in response to the stimulus of water and gravity i.e. hydrotropic and gravitropic movements respectively.. Growth independent movement: e.g. Opening of lotus petals in the morning and tuberose at night. iv. a. Metal A is more reactive than B as it has to lose only one electron from the outermost orbit while metal B has to lose two electrons to get stable electronic configuration. b. Metal A with electronic configuration of, 8, 1 is sodium (Na). Metal B with electronic configuration of, 8, is Magnesium (Mg). c. Magnesium reacts with dil. HCl to form magnesium chloride and hydrogen gas. Mg (s) + HCl (aq) MgCl (aq) + H (g) Magnesium XX XX XY XY Girl Boy Sex determination in human beings Hydrochloric acid Magnesium chloride Hydrogen v. a. Allotropes are two or more forms of same element that differ from each other in their physical properties but have same or similar chemical properties. b. Two allotropic forms of carbon: Diamond and graphite Uses of diamond: a. Diamonds are used as precious stones in jewellery. b. Black diamonds are used for cutting glass. [Any one use: ½ mark] 3

80 Board Answer Paper : July 015 Uses of graphite: a. Graphite is used in making carbon electrodes. b. It is used as lubricants and in lead pencils. [Any one use: ½ mark] vi. i. Green plants synthesize their food by the process of photosynthesis. ii. The factors which take part in this process are carbon dioxide, water, chlorophyll and sunlight. iii. Chemical equation of photosynthesis: 6CO + 6H O C 6 H 1 O 6 + 6O Chlorophyll Sunlight 4. Answer the following question (any one): i. a. The organs of the human male reproductive system: 1. Testes. Epididymis 3. Vas deferens 4. Seminal vesicle 5. Prostate gland 6. Penis. b. Functions: 1. Testes: Testes produce sperms. They secrete hormone testosterone, which brings about changes in boys during puberty.. Epididymis: Sperms are stored and matured in the epididymis. 3. Vas deferens: It is a passage through which the sperms travel towards the urethra. 4. Seminal vesicles and prostate gland: They produce ejaculatory fluid which helps the sperm in transport and provides nutrition. 5. Penis: It delivers the sperms to the site of fertilization. [Write function of any two organs: 1 mark each] [] ii. The functions of the Maharashtra Pollution Control Board: i. MPCB plans comprehensive programmes for the prevention, control or abatement of pollution. ii. MPCB inspects sewage or trade effluent treatment and disposal facilities. iii. MPCB supports and encourages the developments in the fields of pollution control, iv. waste recycle reuse, eco-friendly practices, etc. It educates and guides the entrepreneurs in improving environment by suggesting appropriate pollution control technologies and techniques. v. It creates public awareness about the clean and healthy environment and considers the public complaints regarding pollution. 4

81 BOARD ANSWER PAPER : MARCH 016 SCIENCE AND TECHNOLOGY SECTION A Science and Technology 1. (A) (a) Fill in the blanks: i. The modern periodic table consists of seven periods. ii. The formulae of chloride of metal M is MCl. The metal M belongs to nd or II A iii. Corrosion can be prevented by using anti rust solution. (b) Find the odd one out: i. Thermometer: Others work on magnetic effect of electric current. ii. Bar magnet: Others are appliances which work on magnetic effect of electric current. (B) Choose the correct alternative and rewrite the following: i. (C) Reaction of iron nails with copper sulphate solution is an example of Displacement reaction. ii. (A) Dilute NaOH can be tested with red litmus paper. iii. (B) In series combination current remains constant. iv. (D) An object of 10 cm is placed in front of a plane mirror. The height of image will be 10 cm. v. (C) When a ray of light travels from air to glass slab and strikes the surface of separation at 90, then it passes unbent.. Answer the following questions (any five): i. a. The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to another. 1joule 1 volt = 1coulomb b. Electric power is defined as the rate at which electrical energy is consumed. ii. The major harm done to human beings by air pollution are: a. Short term effects of air pollution on human beings: 1. Irritation of eyes, nose, mouth and throat.. Respiratory infections such as bronchitis, pneumonia. 3. Headache, nausea and allergy 4. Asthma attacks 5. Reduced lung functioning. b. Long term effects of air pollution on human beings: 1. Chronic pulmonary disease. Cardio vascular disease 3. Lung cancer 4. Premature death 1

82 Board Answer Paper : March 016 iii. Power of lens = + D P = 1 f f = 1 P f = 1 = 0.5 m = 50 cm Focal length of the lens = 50 cm. iv. Applications of Sodium bicarbonate (baking soda): a. Sodium bicarbonate is used to prepare light and spongy breads, cakes and dhokalas. b. It also helps to reduce acidity in stomach. c. It is very useful in preparing CO gas and is one of the contents of fire extinguishers. [Any two: 1 mark each] [] v. a. When an object is placed within the focal length of convex lens, one gets a magnified and erect image of the object. b. Thus, the watch repairer can see the minute parts of the watch more clearly with the help of simple microscope than the naked eye, without any strain on the eye. A magnification of about 0 times is obtained by simple microscope. Hence, the watch repairers make use of simple microscope while repairing watches. vi. NEUTRAL 0 ACIDIC 7 BASIC ph scale 14 [] 3. Answer the following questions (any five): i. Features of Mendeleev s periodic table: a. The horizontal rows in the periodic table are called periods. There are seven periods numbered from 1 to 7. b. Properties of elements in a particular period show regular gradation from left to right. c. Vertical columns in the periodic table are called groups. There are eight groups numbered from I to VIII. Groups Ito VII are further divided into Aand B subgroups. ii. Redox reactions: Redox reaction is a chemical reaction in which both oxidation and reduction take place simultaneously. Eg. Reduction (loss of oxygen) BaSO 4 + 4C BaS + 4CO Oxidation (gain of oxygen) In the above example, BaSO 4 is reduced to BaS by removing oxygen atom and carbon is oxidized to carbon monoxide. iii. a. Defect of human eye shown in the figure is hypermetropia. In this defect person can see distant objects clearly but cannot see nearby object clearly. b. Reason for hypermetropia: 1. Weak action of ciliary muscles cause low converging power of eye lens.. The distance between the eye lens and the retina decreases on account of either shortening of eye ball or flattening of lens. In this case focal length of eye lens is too long. Correction: i. Hypermetropia can be corrected by using spectacles having convex lenses of suitable focal length.

83 ii. Science and Technology The convex lens produces convergence of the light rays passing through it and then light rays are converged by eye lens. As a result the image of the nearby object is formed on the retina. O O L 1 L Image at retina iv. Remedy of hypermetropia Refraction of light: The phenomenon of change in the direction of light when it passes from one transparent medium to another is called refraction of light. Relationship between refraction of light and refractive index: a. The extent of change in the direction of light ray is different for different media and depends upon the relative speed of propagation of light in different media. The relative speed of propagation of light is basically measured with the help of refractive index of medium. b. The refractive index ( 1 ) of second medium with respect to the first is given by the ratio of the magnitude of velocity of light in first medium to that in second medium, velocityof lightin mediumfirst i.e., 1 = velocityof lightin mediumsecond. Thus, the refraction of light depends upon refractive index of the material. v. a. The band of coloured components of a light beam is called as spectrum. b. White light is composed of seven colours. When white light is incident a prism, it splits up into its constituent colours. c. Each colour bends through different angles with respect to incident ray. So the rays of each colour emerge along different path and become distinct. Hence, we get a spectrum of seven colours when white light is dispersed by a prism. vi. a. 1. Use sound proof tiles on walls of classroom or curtains to avoid spread of noise pollution.. Teacher should engage students in interesting activities during free time due to which students will not make noise. b. 1. Purchase energy efficient products and operate them efficiently.. Use daylighting in your home by using energy efficient windows. 3. Use solar water heater rather than electric water heater. 4. Turn off lights, fans, air conditioners, T.V., Computers etc. when not in use. [Any two points: ½ mark each] c. 1. Bursting of fire crackers near silence zones such as near hospitals and schools must be avoided.. Fire crackers should be avoided. 3. Fire crackers which do not produce loud sound and cause less pollution must be preferred. [Any two points: ½ mark each] 4. Answer the following question (any one): [5] i. Expression for equivalent resistance in series: + V I C D R 1 R R 3 A + + K E Resistors in series combination 3

84 Board Answer Paper : March 016 a. Let R 1, R and R 3 be the three resistors connected in series between points C and D as shown in the circuit diagram. b. An electric circuit is completed by connecting an ammeter (A), a voltmeter (V) a plug key (K) and a battery (E). c. Suppose I is the current and V is the P.D. across points C and D. d. Suppose V 1, V and V 3 are the P.D.s across resistors R 1, R and R 3 respectively, such that V = V 1 + V + V 3.(1) e. If R s is the equivalent resistance, then using Ohm s law, V = IR s and V 1 = IR 1 ; V = IR ; V 3 = IR 3 Substituting the values in equation (1) we get, IR s = IR 1 + IR + IR 3 IR s = I(R 1 + R + R 3 ) R s = R 1 + R + R 3.() Equation () represents equivalent resistance in series combination of resistors. For n number of resistors connected in series, R s = R 1 + R + R R n Characteristics of series combination of resistors: a. Resistance of the combination of resistors is equal to the sum of the resistances of individual resistors. R s = R 1 + R + R R n b. The effective resistance in a series combination is greater than the individual resistances. c. Hence, this combination is used to increase resistance of the circuit. d. Current is the same in every part of the circuit. e. The total voltage across the combination is equal to the sum of the voltage across the separate resistors. [Any two characteristics: 1 mark each] [] ii. a. B C N A D S Iron core Armature coil R 1 R B 1 B Split ring Axle Electric motor [] b. Electric motor works on the principle that a current carrying conductor placed on a magnetic field experiences a force. c. Appliances in which electric motor is used : 1. Electric fans. Hair dryers 3. Electric cranes 4. Rolling mills 4

85 BOARD ANSWER PAPER : MARCH 016 SCIENCE AND TECHNOLOGY SECTION B Science and Technology 1. (A) Answer the following sub-questions : i. Liver is the largest gland in the body. ii. iii. Vas deferens is a part of male reproductive system, rest all are parts of female reproductive system. False: Aquatic animals breathe at a faster rate than the terrestrial animals. iv. rr : Homozygous : : Rr : Heterozygous v. Main ore of aluminium-bauxite. (B) Choose the correct alternative and rewrite the following: i. Hydrogen is liberated when acetic acid reacts with sodium metal. ii. For binary fission, Amoeba requires one parent cells. iii. A solution of CuSO 4 in water is blue in colour. iv. Raisins are formed by drying grapes. The process that takes place during formation of raisins from grapes is dehydration. v. Ethanoic acid has a pungent odour.. Answer the following questions (any five): i. Voluntary movements i. Voluntary movements are the movements which can be controlled by our will. ii. Voluntary movements are controlled by motor areas in the fore brain and cerebellum in the hind brain. Eg. Riding a bicycle, picking up an object from the floor, etc. Involuntary movements Involuntary movements are the movements which cannot be controlled by our will. Involuntary movements are controlled by the mid-brain and the medulla oblongata of the hind brain. Breathing, sneezing, etc. ii. a. Cellular DNA is the source of information for synthesizing proteins in the cell. b. DNA possesses a double helical structure, which consists of two strands coiled around each other. c. A fragment of DNA that provides complete information about one protein is referred to as Gene for that protein. d. During sexual reproduction, each of the parent contributes equally to the DNA of the progeny. 1

86 Board Answer Paper : March 016 iii. Style Stigma Petal Carpel Ovary Anther Filament Stamen Sepal [] Longitudinal section of a typical flower iv. i. The root system of the plants are specialized for absorbing water and minerals from the soil. ii. The roots respond to the stimuli of gravity and water and grow towards the soil. Thus, the root system of plants show gravitropic movement. Hence, the roots of plants grow away from light. v. a. Drinking water sources such as rivers, lakes should be protected from water pollution. b. Rain water harvesting plant should be constructed to conserve rain water. c. Waste water should be taken to sewage treatment sites where it undergoes various methods of treatment. Such water can be used for agriculture and other purposes. [Any two: 1 mark each] [] vi. a. Fossils are the dead remains of plants and animals which existed in the past. b. They are formed by various methods. c. Sometimes, impressions of plants and animals are formed on mud which get converted into fossils at later stage. d. At other times, plants and animals get burried in the soil and the soft part of the body gets decayed, while the hard part (bones) remains in the soil in the form of fossils. 3. Answer the following questions (any five): i. Alloy: a. An alloy is a homogenous mixture of two or more metals or a metal and a non-metal in definite proportion. Eg. a. Brass (copper and zinc) b. Bronze (copper and tin) ii. a. In human males, two dissimilar chromosomes are present in the 3 rd pair, the longer X and the shorter Y. b. In females, the 3 rd pair contains two similar X chromosomes. c. All children inherit one X chromosome from their mother. The other chromosome is inherited from the father. d. If the chromosome inherited from the father is X, then the offspring is a daughter and if it is Y, then the offspring is a son. Thus, the sex of the child is determined by the male sex chromosome. Parents: XY XX Gametes: X Y X X Children: XX XX XY XY Girl Boy Sex determination in human beings