NSMQ PAST QUESTION; MATHEMATICS pdf.

NSMQ PAST QUESTION; MATHEMATICS pdf.

Find the values of the constants a and b if the straight lines

1. ax + 5y = 3, and 4x + by = 1 intersect at the point (2, -1)

ANSWER: a = 4, b = 7

[2a – 5 = 3, 2a = 8, a = 4, 8 – b = 1, b = 7]

2. 3x + ay = 4, and bx – 5y = 6 intersect at the point (2, -2)

ANSWER: a = 1, b = -2

[6 – 2a = 4, 2a = 2, a = 1, 2b + 10 = 6, 2b = -4, b = –2]

3. ax + 2y = -3, and 2x + by = 6   meet at the point (1, 2)

ANSWER: a = -7, b = 2

[a + 4 = -3, a = -7, 2 + 2b = 6, 2b = 4, b = 2]

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Find the degree measures of the interior angles of a triangle if

1. the exterior angles are in the ratio 3: 4 : 5

ANSWER: 90°, 60°, 30°

[3x + 4x + 5x = 12x = 360, x = 30, exterior angles are 90, 120, 150, interior angles are 180 – 90 = 90, 180 – 120 = 60, 180 – 150 = 30]

2. the exterior angles are in the ratio 11: 12: 13

ANSWER: 70°, 60°, 50°

[11x + 12x + 13x = 36x = 360, x = 10, exterior angles are 110, 120, 130, interior angles are 180 – 110 = 70, 180 – 120 = 60, 180 – 130 = 50]

3. the exterior angles are in the ratio 5 : 6 : 7

ANSWER: 80°, 60°, 40°

[5x + 6x + 7x = 18x = 360, x = 20, exterior angles are 100, 120, 140, interior angles are 180 – 100 = 80, 180 – 120 = 60, 180 – 140 = 40]

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Find the values of A, B, C such that

1. 9x² + 12x + A = (3x + B)²

ANSWER: A = 4, B = 2

[ 9x² + 12x + A = 9x² + 6Bx + B², 12 = 6B, B = 2, A = B² = 2² = 4]

2. 4x² + 16x + 25 = A(x + B)² + C

ANSWER: A = 4, B = 2, C = 9

[4x² + 16x + 25 = 4(x² + 4x) + 25 = 4(x + 2)² + 25 – 16, A = 4, B = 2, C = 9]

3. 5x² – 30x – 6 = A(x + B)² + C

ANSWER: A = 5, B = -3, C = -51

[5(x² – 6x) – 6 = 5(x – 3)² – 45 – 6 = A(x + B)² + C, A = 5, B = -3, C = – 51]

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Solve the equation for x from the logarithmic equation

1. log₆x + log₆x² = 3

ANSWER: x = 6

[ log₆x + 2log₆x = 3log₆x = 3, log₆x = 1, x = 6]

2. log₃x – log₃(x – 1) = 2

ANSWER: x = 9/8

[ log₃(x/(x – 1)) = 2, x/(x – 1) = 9, x = 9(x – 1), 8x = 9, x = 9/8]

3. log₂ x = log₂ (x + 3) – 1

ANSWER: x = 3            

[log₂ (x/(x + 3)) = log₂(1/2), x/(x + 3) = ½, 2x = x + 3, x = 3]

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Find the equation of the locus of the point P (x, y) moving in the coordinate plane such that AP = BP given

1. A(-4, 2) and B(2, – 4)

ANSWER:  y = x

[(x + 4)² + (y – 2)² = (x – 2)² + (y + 4)², 8x – 4y = -4x + 8y, y = x  ]

2. A(3, -2) and B(2, – 3)

ANSWER: y = -x

[(x – 3)² + (y + 2)² = (x – 2)² + (y + 3)², -6x + 4y = -4x + 6y, -2x = 2y, y = -x]

3. A(2, 4) and B (-2, -4)

ANSWER: y = – x/2

[(x – 2)² + (y – 4)² = (x + 2)² + (y + 4)², -4x – 8y = 4x + 8y, 16y = -8x,

y = -x/2]

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Find the coordinates of the vertices of a triangle whose sides are along the lines

1. x + y = 3, x = 4, y = 5        

ANSWER: (4, 5), (4, -1), (-2, 5)

[x + y = 3 and x = 4, y = – 1, (4, -1), x + y = 3 and y = 5, x = -2, (-2, 5),

x = 4 and y = 5 gives (4, 5)]

2. x – y = 5, x = -2, y = 3

ANSWER: (8, 3), (-2, 3), (-2, -7)

[x – y = 5 and x = -2, y = -7, (-2, -7), x – y = 5 and y = 3, x = 8, (8, 3),

x = -2 and y = 3 gives (-2, 3)]

3. 2x + y = 8, x = 3, y = 4

ANSWER: (3, 2), (2, 4), (3, 4)

[2x + y = 8 and x = 3, y = 2, (3, 2), 2x + y = 8 and y = 4, gives x = 2, (2, 4),

x = 3, y = 4 gives (3, 4)]

NSMQ Past Question; Chemistry (01)

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Find the equation of the image of the given curve

1. (x – 3)² + (y + 3)² = 10, after a reflection in the x- axis,

ANSWER: (x – 3)² + (y – 3)² = 10  accept (x – 3)² + (3 – y)² = 10

[(x, y) → (x, – y), (x – 3)² + (-y + 3)² = 10, (x – 3)² + (y – 3)² = 10]

2. y² = 4x, after a reflection in the line y = x,

ANSWER: x² = 4y or y = x²/4

[(x, y) → (y, x), x² = 4y, or y = x²/4]

3. x³ – y³ = 10, after a reflection in the y-axis.

ANSWER: x³ + y³ = -10

[(x, y) → (-x, y), (-x)³ – y³ = -x³ – y³ = 10, x³ + y³ = -10]

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Find the acceleration vector of a particle of mass m acted upon by the forces

1. (3i –  4j)N, (-5i + j)N, (5i – 3j)N and  m = 0.5 kg,

ANSWER: (6i – 12j) m/s²

[(3i – 4j) + (-5i + j) + (5i – 3j) = (3i – 6j) = 0.5a, a = 2(3i – 6j) = (6i – 12j)m/s²]

2. (4i – 2j)N, (2i + 3j)N, (– 3i  + 4j)N and m = 0.2 kg

ANSWER: (15i + 25j) m/s²

[(4i – 2j) + (2i + 3j) + (-3i + 4j) = (3i + 5j) = 0.2a, a = 5(3i + 5j) = 15i + 25j m/s²]

3. (5i – 7j) N, (3i + j) N + (i – 3j) N, and m = 0.3 kg

ANSWER: (30i – 30j) m/s²      

[(5i – 7j) + (3i + j) + (i – 3j) = (9i – 9j) = 0.3a,

a = 10(9i – 9j)/3 = (30i – 30j) m/s²]

NSMQ Past Questions; Chemistry (02 and 03) pdf.

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1. Find the solution set of the equation | x |= -x

(Read as ‘absolute value of x = -x ’)

ANSWER: {x: x ≤ 0}

2. A committee of 3 is to be formed from 3 men and 3 women. In how ways can this be done if there are 2 women and 1 man on the committee

ANSWER: 9         

[3C₂ x 3C₁ = 3 x 3 = 9]

3. Describe the set of points (x, y) such that 4 < x² + y² < 9

ANSWER: Region between 2 concentric circles with center at the origin and having radii 2 and 3.

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A linear transformation is given by T: (x, y) → (2x + y, 5x + 3y). Find the coordinates of the point A(x, y) if its image under the transformation is

1. (1, 1)

ANSWER: (2, -3)    (Accept x = 2, y = -3)

[ 2x + y = 1, 5x + 3y = 1, 3(2x + y) – (5x + 3y) = 3 – 1 = 2, x = 2, y = -3]

2. (3, 2)

ANSWER: (7, -11)  (Accept x = 7, y = -11)

[2x + y = 3, 5x + 3y = 2, 3(2x + y) – (5x + 3y) = x = 9 – 2 = 7, y = -11]

3. (-2, 3)

ANSWER: (-9, 16)  (Accept x = -9, y = 16)

[2x + y = -2, 5x + 3y = 3, 3(2x + y) – (5x + 3y) = 3(-2) – 3, x = -9, y = 16]

NSMQ PAST QUESTION; MATHEMATICS

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