BOARD QUESTION PAPER : JULY 2017 ALGEBRA

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Roderick Henry
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1 BOARD QUESTION PAPER : JULY 017 ALGEBRA Time : Hours Total Marks : 40 Note: (i) (ii) All questions are compulsory. Use of calculator is not allowed. Q.1 Attempt any five of the following sub - questions : (5) (i) Write the next two terms of the A.P., 5,8, 11,... (ii) Write the quadratic equation 7 = 4x x in the form of ax + bx + c = 0. (iii) Find the value of the determinant: (iv) A coin is tossed. For this random experiment write the sample space and find n(s). (v) Find the class mark of the class (vi) For the quadratic equation 3x + x 1 = 0, find the value of the discriminant. Q. Attempt any four of the following sub - questions : (8) (i) Write the 5 th term of an A.P. 1, 16, 0, 4,... (ii) Find k, one of the roots of the quadratic equation kx 7x + 1 = 0 is 3. (iii) For solving the following simultaneous equations by Cramer s rule, find the value of Dx and Dy : 3x y = 7; x + 4y = 11. (iv) (v) (vi) If two coins are tossed, then find the probability of the event: No head turns up. For a certain frequency distribution, the value of Mean is 101 and Median is 100. Find the value of Mode. The following pie diagram represents expenditure on different items in constructing a building. If the total construction cost of building is Rs. 5,40,000, answer the following questions: a. Find the central angle for labour expenditure. b. Find the expenditure on labour. Q.3 Attempt any three of the following sub- questions: (9) (i) Solve the following quadratic equation by completing square: z + 4z 7 = 0.

2 (ii) Two dice are thrown. Find the probability of the following events: Event A : The product of the numbers on their upper faces is 10. Event B : The sum of the numbers on their upper faces is multiple of 9. (iii) The following is the distribution of the size of certain farms from a taluka : Find the median size of farms. Size of Farm No. of Farms (in acres) (iv) (v) Area under different crops in a certain village is given below. Represent it by pie diagram: Crop Area (in hectare) Jowar 8000 Wheat 6000 Sugarcane 000 Vegetable 000 Draw the frequency polygon for the following frequency distribution: Class Frequency Q.4 Attempt any two of the following sub - questions : (8) (i) (ii) In a certain race there are three girls X, Y, Z. The winning probability of X is twice than Y and the winning probability of Y is twice than Z. If P(X) + P(Y) + P(Z) = 1, then find the winning probability of each girl. If the second term and the fourth terms of an A.P. are 1 and 0 respectively, then find the sum of first 5 terms. (iii) Solve the following simultaneous equations: = 85, + = 89. x y + 3 x y + 3 Q.5 Attempt any two of the following sub-questions: (10) (i) (ii) (iii) The product of four consecutive positive integers is 840. Find the numbers. A three digit number is equal to 17 times the sum of its digits. If 198 is added to the number, the digits are interchange. The addition of first and third digit is 1 less than middle digit. Find the number. Find the sum of all numbers from 50 to 350 which are divisible by 4. Also find 15th term.

3 BOARD ANSWER PAPER : JULY 017 ALGEBRA Q.1 Attempt any five of the following sub - questions : (5) (i) The given sequence is, 5, 8, 11, Here, t1 =, t = 5, t3 = 8, t4 = 11 The common difference between two consecutive terms is 3 t5 = = 14 t6 = = 17 The next two terms are 14 and 17. (ii) Given equation is 7 = 4x - x (iii) (iv) i.e. x - 4x + 7 = = (-3 0) - (6 8) 6 0 = 0-48 = - 48 When a coin is tossed, S = { H, T } n(s) =. + = =. (v) Class mark of 10 0 = (vi) The given equation is 3x + x - 1 = 0 Comparing it with ax + bx + c = 0, we get a = 3, b =, c = -1 = b - 4ac = () - 4(3)(-1) = = 16. Q. Attempt any four of the following sub - questions : (8) (i) The given A.P. is 1, 16, 0, 4,... Here, a = 1, d = 16-1 = 4 Since, tn = a + (n - 1)d t5 = 1 + (5-1)4 = = t5 = 108 The twenty fifth term of the given A.P. is 108.

4 (ii) x = 3 is the root of the given equation kx 7x + 1 = 0 It satisfies the given equation. k(3) 7(3) + 1 = 0 9k = 0 9k 9 = 0 9k = 9 k = 1. (iii) The given simultaneous equations are 3x y = 7 (i) and x + 4y = 11 (ii) Equations (i) and (ii) are in ax + by = c form. Here, a1 = 3, b1 = 1, c1 = 7, a = 1, b = 4, c = 11 Dx = Dy = c c a b 1 1 b c 1 1 a c 7 1 = = (7 4) (-1 11) 11 4 = = = = (3 11) (7 1) 1 11 = 33 7 = 6. (iv) When two coins are tossed, S = {HH, HT, TH, TT} n(s) = 4 Let B be the event where no head turns up. B = {TT} n(b) = 1. n( B) 1 P(B) = =. n( S) 4 (v) The inter-relation between the measures of central tendency is Mean Mode = 3(Mean Median) Mean 3(Mean Median) = Mode 101 3( ) = Mode 101 3(1) = Mode Mode = Mode = 98.

5 (vi) a. Let the central angle for labour expenditure be x o, Now, total central angle = o + 75 o + 45 o + 90 o + x o = 360 o 60 o + x o = 360 o x o = 360 o - 60 o x o = 100 o. b. Expenditure on labour = Centralanglefor labour Totalcost 360 o o = 100 5,40,000 o 360 = ` 1,50,000. Q.3 Attempt any three of the following sub- questions: (9) (i) z + 4z 7 = 0 z + 4z =7...(i) Now, third term = 1 coefficient of z 1 = 4 = 4 Adding 4 on both sides of (i), we get z + 4z + 4 = (z + ) = 11 Taking square root on both sides, we get z + = ± 11 z = - ± 11 z = or z = and are the roots of the given quadratic equation. (ii) The sample space is S = {(1, 1), (1, ), (1, 3), (1, 4), (1, 5), (1, 6), (, 1), (, ), (, 3), (, 4), (, 5), (, 6), (3, 1), (3, ), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, ), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, ), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, ), (6, 3), (6, 4), (6, 5), (6, 6)} n (S) = 36. a. Let A be the event that the product of the numbers on the upper faces is 10. A = {(, 5), (5, )} n(a) =. n( A) 1 P( A) = = = n( S) 36 18

6 b. Let B be the event that the sum of the numbers on the upper faces is a multiple of 9. B = {(3, 6), (4, 5), (5, 4), (6, 3)} n(b) = 4 n( B) 4 1 P( B) = = = n( S) 36 9 (iii) Size of Farm No. of Farms Cumulative frequency (in acres) Class Interval Frequency (less than type) fi c.f f Total Σfi = Here, Σfi = N = 100 N 100 = = 50 Cumulative frequency (less than type) which is just greater than (or equal) to 50 is 61. Median class is Now, L = 35, f = 5, c.f. = 36, h = 10. N h Median = L + c. f. f = 35 + (50 36) 10 5 = 35 + = = The median size of a farm is acres.

(iv) Crop Area (in hectare) Measure of central

Wheat 6000 6000 o 360 = 10 18, 000 Sugarcane

360 = 40 18, 000 Total 18,000 360 o o o o o

7 (iv) Crop Area (in hectare) Measure of central Angle (θ) Jowar o 360 = , 000 Wheat o 360 = 10 18, 000 Sugarcane o 360 = 40 18, 000 Vegetable o 360 = 40 18, 000 Total 18, o o o o o The pie diagram representation is as follows: (v)

8 Q.4 Attempt any two of the following sub - questions : (8) (i) Let P(Z) = x P(Y) = P(Z) (Given) P(Y) = x and P(X) = P(Y) (Given) P(X) = 4x Given, P(X) + P(Y) + P(Z) = 1 4x + x + x = 1 7x = 1 x = 1 7 P(X) = 4x = = 4 7. P(Y) = x = 1 7 = 7 P(Z) = x = 1 7 P(X) = 4 7, P(Y) = 7, P(Z) = 1 7. (ii) Given, t = 1, t4 = 0 Now, tn = a + (n - 1)d t = a + ( - 1)d 1 = a + d a + d = 1...(i) Also, t4 = a + (4-1)d 0 = a + 3d a + 3d = 0...(ii) Subtracting (i) from (ii), we get a + 3d = 0 a + d = 1 (-) (-) (-) d = 8 d = 8 = 4 Substituting d = 4 in (i), we get a + 4 = 1 a = 1-4 = 8 n Now, Sn = [a + (n - 1)d]

9 S5 = 5 [ 8 + (5-1)4] = 5 [ ] = 5 = 5 56 S5 = 1400 [ ] = 5 11 The sum of first 5 terms is (iii) The given simultaneous equations are and = 85 x y = 89 x y + 3 (i) (ii) Here, 1 x and 1 is common to both the equations. y + 3 So, let 1 x = p and 1 y + 3 = q equations (i) and (ii) become 7p + 31q = 85 and 31p + 7q = 89 7p + 31q = 85 31p + 7q = p + 58q = (p + q) = 174 p + q = Adding equations (iii) and (iv) (iii) (iv) p + q = 3 (v) Subtracting equation (iii) from (iv) 31p + 7q = 89 7p + 31q = 85 ( ) ( ) ( ) 4p 4q = 4 4(p q) = 4 p q = 4 4 p q = 1 (vi)

10 Now, adding equations (v) and (vi) p + q = 3 p q = 1 p = 4 p = 4 p = 4 p = Substituting p = in equation (vi), we get q = 1 q = 1 q = 1 q = 1 (p, q) = (, 1) Resubstituting the values of p and q, we get = 1 x and 1 = 1 y + 3 (x ) = 1 and 1(y + 3) = 1 x 4 = 1 and y + 3 = 1 x = and y = 1 3 x = 5 and y = x = 5 and y = (x, y) = 5,-. Q.5 Attempt any two of the following sub-questions: (10) (i) Let the four consecutive positive integers be x, x + 1, x + and x + 3. According to the given condition, x (x + 1) (x + ) (x + 3) = 840 [x(x + 3)] [(x + 1) (x + )] = 840 (x + 3x) (x + 3x + ) = 840 Let x + 3x = a a (a + ) = 840 a + a = 840 a + a - 840= 0 a + 30a - 8a = 0 a (a + 30) - 8 (a + 30) = 0 (a + 30)(a - 8) = 0 a + 30 = 0 or a - 8 = 0

11 a = -30 or a = 8 As x is positive integer x > 0 x > 0 x + 3x > 0 As x + 3x = a a > 0 a -30 a = 8 But, a = x + 3x x + 3x = 8 x + 3x - 8 = 0 x + 7x - 4x - 8 = 0 x(x + 7) - 4(x + 7) = 0 (x + 7)(x - 4) = 0 x + 7 = 0 or x - 4 = 0 x = -7 or x = 4 But x cannot be negative x -7 x = 4 x + 1 = = 5, x + = 4 + = 6 and x + 3 = = 7 The four consecutive positive numbers are 4, 5, 6 and 7. (ii) Let the digit in the hundredth s place be x and the digit in the unit place be y. 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 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 = (i)

99x 99y = 198 Dividing both sides, by 99 we get x y = Multiply equation (ii) by 3, we get 3x - 3y = 46 Subtracting (iii) from (i), we get 76x 3y = 7 3x 3y = 46 (-) (+) (+) 53x = 53 x = 1 Substituting

12 99x 99y = 198 Dividing both sides, by 99 we get x y = Multiply equation (ii) by 3, we get 3x - 3y = 46 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 Required Number is 153. = 110(1) + 11(3) + 10 = = 153.(ii).(iii) (iii) The numbers from 50 to 350 which are divisible by 4 are 5, 56, 60, This sequence is an A.P. with a = 5, d = 56-5 = 4, tn = 348 But tn = a + (n 1) d 348 = 5 + (n 1) = (n 1) 4 96 = (n 1) = n 1 74 = n = n n = 75 Now, Sn = n [t1 + tn] S75 = 75 [ ] = 75 (400) S75 = Also, t15 = a + (15 1) d t15 = 108 = = The sum of all numbers from 50 to 350 which are divisible by 4 is and 15th term of the A.P. is 108.

BOARD ANSWER PAPER : MARCH 2014

BOARD ANSWER PAPER : MARCH 2014 BOARD ANSWER PAPER : MARCH 014 ALGEBRA 1 Algebra 1. Attempt any five subquestions from the following: i. Given, t, t 4 1 d t 4 t 1 4 ii. (x + 5) (x ) 0 x + 5 0 or x 0 x 5 or x iii. Since, θ Valueof thecomponent