**Problem 1 :**

Find the equation of the straight line passing through the point (3, 2) and the perpendicular to the straight line joining the points (4, 5) and (1, 2).

**Solution :**

Slope of the line joining the points (4, 5) and (1, 2).

m = (y_{2}-y_{1})/(x_{2}-x_{1})

m = (2-5)/(1-4)

m = 1

Slope of the required line = -1

Equation of the required line :

y-y_{1} = (-1/m)(x-x_{1})

y-2 = -1(x-3)

x+y-2-3 = 0

x+y-5 = 0

**Problem 2 :**

Find the point of intersection of the straight lines

x - y = 3 and x + y = 5

**Solution :**

x - y = 3 ------(1)

x + y = 5 ------(2)

(1) + (2)

2x = 8

x = 4

By applying the value of x in (1), we get

4-y = 3

y = 1

So, the point of intersection of the given lines is (4, 1).

**Problem 3 :**

Find the length of the chord of a circle of radius 3 cm subtending at the center angle of 144°.

**Solution :**

In triangle OCB,

sin θ = Opposite side / Hypotenuse

sin 72 = BC/OB

0.9510 = BC/3

BC = 2.853

AB = 2BC

AB = 2(2.853)

AB = 5.706

So, the length of chord is 5.706 cm.

**Problem 4 :**

The tree stands vertically on the bank of a river. From a point on the another bank directly opposite to the tree, the angle of elevation of the top of the tree is 60°. From the point 40 m behind this point on the same bank, the angle of elevation of the top of a tree is 30°. Find the height of the tree and the width of the river.

**Solution :**

In triangle ABC,

tan θ = Opposite side / Adjacent side

tan 60 = AB/BC

√3 = AB/x

AB = x√3 -----(1)

tan 30 = AB/BD

1/√3 = AB/(x + 40)

AB = (x + 40)/√3 -----(2)

(1) = (2)

x√3 = (x + 40)/√3

3x-x = 40

x = 20

By applying the value of x in (1), we get

AB = 20√3

AB = 20(1.732)

AB = 34.64

So, height of tree and width of the river are 34.64 m and 20 m respectively.

**Problem 5 :**

Find the range of the data

27, 28, 34, 36, 39, 50

**Solution :**

To find the range of the data, we use the formula

Range = Large value - small value

Range = 50 - 27

Range = 23

**Problem 6 :**

A group of 100 candidates have their average height 163.8 cm with coefficient of variation 3.2%. What is the standard deviation of their height?

**Solution :**

Coefficient of variation (C.V) = (σ/x̄) ⋅ 100%

x̄ = 163.8, coefficient of variation = 3.2%

(σ/x̄) ⋅ 100 = 3.2

(σ/163.8) ⋅ 100 = 3.2

σ = (3.2 ⋅ 163.8)/100

σ = 5.2416

**Problem 7 :**

An integer is chosen at random from 1 to 50. Find the probability that the number is not divisible by 5.

**Solution :**

Sample space = {1, 2, 3, 4, 5, ...........50}

n(S) = 50

Let A be the event of getting a number divisible by 5.

A = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50}

n(A) = 10

p(A) = n(A)/n(S)

p(A) = 10/50 = 1/5

P(not divisible by 5) = 1 - 1/5

= 4/5

**Problem 8 :**

The height of right circular cone is 7 cm greater than its radius.The slant height is 8 cm greater than its radius. Find the curved surface area of the cone.

**Solution :**

Height = r + 7

Slant height (l) = 8 + r

Curved surface area = πrl

l^{2} = r^{2} + h^{2}

(8+r)^{2} = r^{2} + (7+r)^{2}

64+16r+r^{2} = r^{2}+49+14r+r^{2}

r^{2 }-16r + 14r + 49 - 64 = 0

r^{2 }-2r-15 = 0

(r-5)(r+3) = 0

r = 5 and r = -3 (in admissible)

l = 13

Curved surface area = π(5)(13)

= 65π cm^{3}

**Problem 9 :**

Find the L.C.M of (x^{2}-2x+1) and (x^{2}+x-2)

**Solution :**

(x^{2}-2x+1) = (x-1)(x-1)

(x^{2}+x-2) = (x+2)(x-1)

Least common multiple is (x-1)^{2}(x+2).

**Problem 10 :**

Find the sum of the series

2+4+6+8+........................+360

**Solution :**

2+4+6+8+........................+360

= 2(1+2+3+4+...........+180)

Sum of natural numbers = n(n+1)/2

= 2(180)(181/2)

= 180(181)

= 32580

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