**Problem 1 :**

Find the 12^{th} term of an arithmetic progression

6, 1, -4, ...........

**Solution :**

a = 6, d = 1 - 6 ==> -5

a_{n} = a + (n-1)d

a_{12} = 6 + (12-1)(-5)

a_{12} = 6 + 11(-5)

a_{12} = 6 - 55

a_{12} = - 49

**Problem 2 :**

If a clock strikes appropriate number of times at each hour how many times will it strike in a day ?

**Solution :**

Clock will strike 1 time at 1'o clock, 2 times at 2'o clock and so on.

1+2+3+ ......+12

Number of times that will strike

= 2[1+2+3 +....... +12]

S_{n} = (n/2) [a+l]

= 2 ⋅ (12/2) [1+12]

= 12(13)

= 156

So, total number of strikes is 156.

**Problem 3 :**

Find the sum of the series

400 + 441 + .......... +16000

**Solution :**

400 + 441 + .......... +16000

= 20^{2} + 21^{2} + ......+40^{2}

(1^{2} + 2^{2} + .......+40^{2}) - (1^{2} + 2^{2} + .......+19^{2})

Sum of squares = n(n+1)(2n+1)/6

= 40(41)(81)/6 - 15(16)(31)/6

= 22140 - 1240

= 22900

**Problem 4 :**

A circus tent is cylindrical to a height of 3m and conical above it. If the base radius is 52.5 m and slant height of the cone is 53 m, find the area of canvas required to make the tent?

**Solution :**

Area of canvas = πrl

r = 52.5 m, l = 53 m

= π(52.5) (53)

= 2782.5π m^{2}

**Problem 5 :**

In a cultural program 24 students took part in dance 11 in drama, 25 in a group songs, 7 students in dance and drama, 4 students in drama and group songs, 12 students in dance and group songs and 3 participated in all three. If total of 50 students were there in the class,find how many did not participated in the program?

**Solution :**

Total number of students = 50

Number of students participated in any one of the competition = 8+4+3+3+9+1+12

= 43

Number of students who did not participate any one of the competition

= 50 - 43

= 7

**Problem 6 :**

In a triangle ABC m∠C is 20° greater than m∠A. The sum of m∠A and m∠C is twice the m∠B.Find three angles A, B and C.

**Solution :**

<C = 20+<A ---(1)

<A+<C = 2<B ----(2)

Applying the value of <C in (2), we get

<A+20+<A = 2<B

2<A - 2<B = -20

<B = 10 + <A

In a triangle, the sum of interior angles of triangle is 180.

<A+<B+<C = 180

<A+10+<A+20+<A = 180

3<A + 30 = 180

3<A = 150

<A = 50

<B = 10+50

<B = 60

<C = 20+50

<C = 70

**Problem 7 :**

A motor boat whose speed is 15 km/hr in still water goes 30 km downstream and comes back in 4 hours 30 minutes.Determine the speed of water.

**Solution :**

If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then:

Speed of boat (u) = 15 km/hr

Speed downstream = (u+v) km/hr = 15 + v

Speed upstream = (u-v) km/hr = 15 - v

Distance to be covered = 30 km

Time = Distance / Speed

30/(15 + v) + 30/(15 - v) = 4 1/2

30[(15-v+15+v)/225-v^{2}] = 9/2

30(30)/225-v^{2} = 9/2

1800 = 9(225-v^{2})

200 = 225-v^{2}

v^{2 } = 225-200

v = 5

Speed of water is 5 km/hr.

**Problem 8 :**

If the points A(2, -2), B(8, 4), C(5, 7) are the vertices of the parallelogram ABCD taken in order, find the fourth vertex D.

**Solution :**

Midpoint of diagonal AC = Midpoint of diagonal BD

Midpoint = (x_{1} + x_{2})/2, (y_{1} + y_{2})/2

(4 + 8)/2, (6 + 2)/2 = (10+b)/2, (4 + 4)/2

12/2, 8/2 = (10+b)/2, 8/2

(6, 4) = ((10+b)/2, 4)

Equating x coordinates,

6 = (10 + b)/2

12 = 10 + b

b = 12 - 10

b = 2

So, the value of b is 2.

**Problem 9 :**

Find the value of a for which the straight lines 2x+y-1 = 0, 2x+ay-3 = 0 and 3x+2y-2 = 0 are concurrent.

**Solution :**

2x+y-1 = 0 -----(1)

2x+ay-3 = 0 -----(2)

3x+2y-2 = 0 -----(3)

(1) x 2 ==> 4x + 2y - 2 = 0

(1) - (2) -3x - 2y + 2 = 0

---------------------

x = 0

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

0 + y - 1 = 0

y = 1

The point of intersection of lines is (0, 1).

Since the given lines are concurrent, the other line will pass through the point (0, 1).

2x+ay-3 = 0

2(0)+a(1)-3 = 0

a = 3

So, the value of a is 3.

**Problem 10 :**

A person stands at a distance of 40 m from a building and observes the top and bottom of a flag pole on the building at angles of elevation 60° and 45°. Find the height of the building and the height of the flag pole.

**Solution :**

In triangle ABC,

tan ϑ = AB/BC

tan 60 = AB/(40 - x)

√3(40 - x) = AB

AB = √3(40 - x)-----(1)

In triangle ABD,

tan 45 = AB/BD

1 = AB/40

AB = 40 -----(2)

(1) = (2)

40 = √3(40 - x)

40 = 40√3 - x√3

x = 40(√3 - 1)/√3

x = 16.90

So, the height of the building is 16.9 m.

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