ALGEBRA BOARD PAPER MARCH 2014 WITH COMPLETE SOLUTION.

Algebra Board Question Paper: March 2014 - Solutions

Original Question Paper

Q.1. Attempt any five of the following sub-questions:

1. Given: For an A.P., t₃ = 8 and t₄ = 12.

   Solution: The common difference (d) of an A.P. is the difference between consecutive terms.

   d = t₄ - t₃

   d = 12 - 8 = 4

   Answer: The common difference d is 4.

2. Given: (x + 5)(x - 2) = 0

   Solution:

   x + 5 = 0  or  x - 2 = 0

   x = -5  or  x = 2

   Answer: The roots of the quadratic equation are -5 and 2.

3. Given: Number of students using Bus = 100.

   Solution:

   Total number of students = 140 + 100 + 70 + 40 + 10 = 360

   Central angle (θ) for 'Bus' = (Number of students using Bus / Total students) × 360°

   θ = (100 / 360) × 360° = 100°

   Answer: The central angle for the Bus is 100°.

4. Given: A coin is tossed.

   Answer: The sample space S = {H, T}.

5. Given: Σf_ix_i = 75 and Σf_i = 15.

   Solution:

   Mean (X̄) = Σf_ix_i / Σf_i

   Mean (X̄) = 75 / 15 = 5

   Answer: The mean is 5.

6. Given equation: 3x² = 10x + 7

   Solution: The standard form of a quadratic equation is ax² + bx + c = 0.

   Bringing all terms to one side: 3x² - 10x - 7 = 0

   Answer: The standard form is 3x² - 10x - 7 = 0.

Q.2. Attempt any four of the following sub-questions:

1. Given sequence: 1, 4, 7, 10, ...

   Solution: Let t₁ = 1, t₂ = 4, t₃ = 7, t₄ = 10.

   t₂ - t₁ = 4 - 1 = 3

   t₃ - t₂ = 7 - 4 = 3

   t₄ - t₃ = 10 - 7 = 3

   Since the difference between consecutive terms is constant (d = 3), the given sequence is an A.P.

2. Given equation: 4x² - 9 = 0

   Solution:

   (2x)² - (3)² = 0

   (2x - 3)(2x + 3) = 0    [Using a² - b² = (a - b)(a + b)]

   2x - 3 = 0  or  2x + 3 = 0

   2x = 3  or  2x = -3

   x = 3/2  or  x = -3/2

3. Given: The point (a, 3) lies on the equation 5x + 2y = -4.

   Solution: Substitute x = a and y = 3 into the equation:

   5(a) + 2(3) = -4

   5a + 6 = -4

   5a = -4 - 6 = -10

   a = -10 / 5 = -2

   Answer: a = -2

4. Given equations:

   7x + 5y = 11    ...(1)

   5x + 7y = 13    ...(2)

   Solution: Adding equation (1) and (2):

   12x + 12y = 24

   Dividing throughout by 12, we get:

   x + y = 2

5. Given: A die is thrown.

   Solution:

   Sample space (S) = {1, 2, 3, 4, 5, 6}

   Number of sample points n(S) = 6

   Event A: Getting a number multiple of 3 on the upper face.

   A = {3, 6}

   n(A) = 2

6. Given: Mean = 15, Mode = 9.

   Solution: Using the empirical relationship:

   Mode = 3(Median) - 2(Mean)

   9 = 3(Median) - 2(15)

   9 = 3(Median) - 30

   3(Median) = 9 + 30 = 39

   Median = 39 / 3 = 13

   Answer: The median is 13.

Q.3. Attempt any three of the following sub-questions:

1. Given equation: 4x² + 7x + 2 = 0

   Solution: Comparing with standard form ax² + bx + c = 0, we have:

   a = 4, b = 7, c = 2

   Δ = b² - 4ac = (7)² - 4(4)(2) = 49 - 32 = 17

   By formula method: x = [-b ± √(Δ)] / 2a

   x = [-7 ± √17] / [2(4)]

   x = (-7 ± √17) / 8

   Answer: The roots are (-7 + √17)/8 and (-7 - √17)/8.

2. Given equations: 3x + 2y = -11 and 7x - 4y = 9

   Solution:

   D = (3 × -4) - (2 × 7) = -12 - 14 = -26

   D_x = (-11 × -4) - (2 × 9) = 44 - 18 = 26

   D_y = (3 × 9) - (-11 × 7) = 27 - (-77) = 27 + 77 = 104

   x = D_x / D = 26 / -26 = -1

   y = D_y / D = 104 / -26 = -4

   Answer: x = -1, y = -4.

3. Given: Two coins are tossed simultaneously.

   Solution:

   Sample space S = {HH, HT, TH, TT}

   n(S) = 4

   a. Event A: getting at most one tail.

   A = {HH, HT, TH}  ;  n(A) = 3

   b. Event D: getting no head.

   D = {TT}  ;  n(D) = 1

4. Solution (Assumed Mean Method):

   Let Assumed Mean (A) = 25

   • Class 0-10: x_i = 5, d_i = -20, f_i = 5, f_id_i = -100
   • Class 10-20: x_i = 15, d_i = -10, f_i = 7, f_id_i = -70
   • Class 20-30: x_i = 25 (A), d_i = 0, f_i = 5, f_id_i = 0
   • Class 30-40: x_i = 35, d_i = 10, f_i = 2, f_id_i = 20
   • Class 40-50: x_i = 45, d_i = 20, f_i = 6, f_id_i = 120

   Σf_i = 25 and Σf_id_i = -30

   d̄ = Σf_id_i / Σf_i = -30 / 25 = -1.2

   Mean (X̄) = A + d̄ = 25 + (-1.2) = 23.8

   Answer: The mean money collected is ∞ 23.8.

5. Solution: To draw the histogram, plot "Price of Sugar (in ∞)" on the X-axis (classes 18-20, 20-22, etc.) and "Number of Weeks" on the Y-axis. Construct adjacent rectangles for each class with heights proportional to their respective frequencies (4, 8, 22, 12, 8, 6).

Q.4. Attempt any two of the following sub-questions:

1. Given: Principal = 1000, Interest = 140. Total Amount (S₁₂) = 1000 + 140 = 1140.

   Number of instalments (n) = 12. Common difference (d) = -10.

   Solution: Using sum of A.P. formula:

   S_n = (n/2) [2a + (n - 1)d]

   1140 = (12/2) [2a + (12 - 1)(-10)]

   1140 = 6 [2a - 110]

   1140 / 6 = 2a - 110

   190 = 2a - 110

   2a = 190 + 110 = 300

   a = 150

   Answer: The first instalment should be ∞ 150.

2. Given: 3 boys (B1, B2, B3) and 2 girls (G1, G2). A committee of two is formed.

   Solution: Sample space S = {B1B2, B1B3, B2B3, B1G1, B1G2, B2G1, B2G2, B3G1, B3G2, G1G2}

   n(S) = 10

   a. At least one girl: Event A = {B1G1, B1G2, B2G1, B2G2, B3G1, B3G2, G1G2}

   n(A) = 7. Probability P(A) = 7/10

   b. One boy and one girl: Event B = {B1G1, B1G2, B2G1, B2G2, B3G1, B3G2}

   n(B) = 6. Probability P(B) = 6/10 = 3/5

   c. Only boys: Event C = {B1B2, B1B3, B2B3}

   n(C) = 3. Probability P(C) = 3/10

3. Given: Salesman A's central angle = 90°. Total sale by A = 18,000.

   Solution:

   a. Total sale: Let total sale be x.

   (90° / 360°) × x = 18,000 &implies; (1/4) × x = 18,000 &implies; x = 72,000

   b. Sale of each salesman:

   • Salesman B (120°): (120/360) × 72,000 = 24,000
   • Salesman C (80°): (80/360) × 72,000 = 16,000
   • Salesman D (70°): (70/360) × 72,000 = 14,000
   • Salesman A: 18,000 (Given)

   c. Highest sale: Salesman B with 24,000.

   d. Difference between highest and lowest sale:

   Highest (B) = 24,000. Lowest (D) = 14,000.

   Difference = 24,000 - 14,000 = 10,000.

Q.5. Attempt any two of the following sub-questions:

1. Given: m(t_m) = n(t_n)

   Solution: The general term is t_k = a + (k - 1)d.

   m[a + (m - 1)d] = n[a + (n - 1)d]

   ma + m(m - 1)d = na + n(n - 1)d

   ma - na + (m² - m)d - (n² - n)d = 0

   a(m - n) + [m² - n² - (m - n)]d = 0

   a(m - n) + [(m - n)(m + n) - (m - n)]d = 0

   a(m - n) + (m - n)[m + n - 1]d = 0

   Since m ≠ n, divide the entire equation by (m - n):

   a + (m + n - 1)d = 0

   By formula, t_(m+n) = a + (m + n - 1)d

   Therefore, t_(m+n) = 0. (Hence Proved)

2. Given: Product of four consecutive natural numbers which are multiples of five is 15,000.

   Solution: Let the consecutive natural numbers which are multiples of five be 5x, 5(x+1), 5(x+2), and 5(x+3).

   [5x] [5(x+1)] [5(x+2)] [5(x+3)] = 15000

   625[x(x+1)(x+2)(x+3)] = 15000

   x(x+1)(x+2)(x+3) = 15000 / 625 = 24

   [x(x+3)] [(x+1)(x+2)] = 24

   (x² + 3x) (x² + 3x + 2) = 24

   Let y = x² + 3x. The equation becomes:

   y(y + 2) = 24

   y² + 2y - 24 = 0

   (y + 6)(y - 4) = 0 &implies; y = -6 or y = 4

   Case 1: x² + 3x = -6 &implies; x² + 3x + 6 = 0 (Roots are not real natural numbers).

   Case 2: x² + 3x = 4 &implies; x² + 3x - 4 = 0

   (x + 4)(x - 1) = 0 &implies; x = -4 or x = 1

   Since the numbers are natural, x = 1.

   The numbers are: 5(1)=5, 5(2)=10, 5(3)=15, 5(4)=20.

   Answer: The four natural numbers are 5, 10, 15, and 20.

3. Given equations: 4x + 3y = 24 and 3y = 4x + 24

   Solution:

   Equation 1: 4x + 3y = 24. Intercepts: If x = 0, y = 8 → Point (0, 8). If y = 0, x = 6 → Point (6, 0).

   Equation 2: -4x + 3y = 24. Intercepts: If x = 0, y = 8 → Point (0, 8). If y = 0, x = -6 → Point (-6, 0).

   Intersection point: Both lines intersect at (0, 8) on the Y-axis.

   Triangle formed with X-axis: The vertices of the triangle on the graph are (6, 0), (-6, 0), and (0, 8).

   Base of the triangle = Distance between (6,0) and (-6,0) = 12 units.

   Height of the triangle = Y-coordinate of intersection = 8 units.

   Area of triangle = (1/2) × base × height = (1/2) × 12 × 8 = 48 sq. units.

   Answer: The intersection point is (0, 8) and the area of the triangle is 48 sq. units.