# NSMQ PAST QUESTION; MATHEMATICS pdf.

## NSMQ PAST QUESTION; MATHEMATICS pdf.

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

- 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]

- 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]

- 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]

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

- 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]

- 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]

- 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]

#### Find the values of A, B, C such that

- 9x
^{2}+ 12x + A = (3x + B)^{2}

**ANSWER: A = 4, B = 2**

[ 9x^{2} + 12x + A = 9x^{2} + 6Bx + B^{2}, 12 = 6B, B = 2, A = B^{2} = 2^{2} = 4]

- 4x
^{2}+ 16x + 25 = A(x + B)^{2}+ C

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

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

- 5x
^{2}– 30x – 6 = A(x + B)^{2}+ C

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

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

#### Solve the equation for x from the logarithmic equation

- log
_{6}x + log_{6}x^{2}= 3

**ANSWER: x = 6**

[ log_{6}x + 2log_{6}x = 3log_{6}x = 3, log_{6}x = 1, x = 6]

- log
_{3}x – log_{3}(x – 1) = 2

**ANSWER: x = 9/8**

[ log_{3}(x/(x – 1)) = 2, x/(x – 1) = 9, x = 9(x – 1), 8x = 9, x = 9/8]

- log
_{2 }x = log_{2}(x + 3) – 1

**ANSWER: x = 3 **

[log_{2} (x/(x + 3)) = log_{2}(1/2), x/(x + 3) = ½, 2x = x + 3, x = 3]

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

- 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 ]

- 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]

- 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]

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

- 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)]

- 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)]

- 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)

#### Find the equation of the image of the given curve

- (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]

- y
^{2}= 4x, after a reflection in the line y = x,

**ANSWER: x ^{2} = 4y or y = x^{2}/4**

[(x, y) → (y, x), x^{2} = 4y, or y = x^{2}/4]

- x
^{3}– y^{3}= 10, after a reflection in the y-axis.

**ANSWER: x ^{3} + y^{3} = -10**

[(x, y) → (-x, y), (-x)^{3} – y^{3} = -x^{3} – y^{3} = 10, x^{3} + y^{3} = -10]

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

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

**ANSWER: (6i – 12j) m/s ^{2}**

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

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

**ANSWER: (15i + 25j) m/s ^{2}**

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

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

**ANSWER: (30i – 30j) m/s ^{2} **

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

a = 10(9i – 9j)/3 = (30i – 30j) m/s^{2}]

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

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

(Read as ‘absolute value of x = -x ’)

**ANSWER: {x: x ****≤**** 0}**

- 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 **

[3C_{2} x 3C_{1} = 3 x 3 = 9]

- Describe the set of points (x, y) such that 4 < x
^{2}+ y^{2}< 9

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

##### 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)

**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]

- (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]

- (-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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