**Problem 1 : **

Find the 5^{th} term of the Geometric progression

64, 16, 4………?

**Solution :**

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

a = 64, d = 16 - 64 ==> -48 and n = 5

a_{5} = 64 + (5 - 1)(-48)

a_{5} = 64 + 4(-48)

a_{5} = 64 - 192

a_{5} = -128

So, the 5^{th} term of the given sequence is -128.

**Problem 2 :**

Find the sum of first 11 terms of the following A.P

3, 8, 13,……………...

**Solution :**

a = 3, d = 8 - 3 ==> 5 and n = 11

S_{n} = (n/2) [2a + (n - 1)d]

S_{11} = (11/2) [2(3) + (11-1)5]

S_{11} = (11/2) [6 + 50]

S_{11} = (11/2) (56)

S_{11} = 11 (28)

S_{11} = 308

**Problem 3 :**

Find the sum of

11+12+13+………..+31

**Solution :**

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

11+12+13+………..+31 = (1+2+3+ ..... +31) - (1+2+3+.....+10)

= 31(32)/2 - 10(11)/2

= 496 - 55

= 441

So the sum of the given series is 441.

**Problem 4 :**

Find the total area of the squares whose sides are

20 cm, 21 cm ….........27 cm

respectively

**Solution :**

Total area of squares using side length are

20^{2} + 21^{2} + ........ + 27^{2}

= (1^{2}+2^{2}+3^{2}+.......+27^{2}) - (1^{2}+2^{2}+3^{2}+.......+19^{2})

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

= (27⋅28⋅55)/6 - (19⋅20⋅39)/6

= 6930 - 2470

= 4460

So the sum of total surface area of given squares is 4460 cm^{2}.

**Problem 5 :**

The radius of the top of a bucket is 18 cm and that of the bottom is 6 cm.Its depth is 24 cm.Find the capacity of the bucket.

**Solution :**

Volume of frustum cone = (1/3) π h (R^{2} + r ^{2} + R r)

R = 18 cm, r = 6 cm and height (h) = 24 cm

= (1/3) (22/8) 24 [18^{2} + 6 ^{2} + 18(6)]

= 22[324 + 36 + 108]

= 10296 cm^{3}

So, the capacity of the frustum cone is 10296 cm^{3}.

**Problem 6 :**

A hemispherical bowl of radius 30 cm is filled with soap paste.If that paste is made into cylindrical soap cakes each of radius 5 cm and height 2 cm, how many cakes do we get?

**Solution :**

Radius of cylinder = 5 cm and height of cylinder = 2 cm

Volume of soap past filled in the hemispherical bowl

= n (Volume of one cylindrical soap cakes)

Radius of hemisphere = 30 cm

Volume of soap paste in hemispherical bowl = (2/3)π r^{3}

= (2/3)π (30)^{3 }------(1)

Volume of one cylindrical soap = π r^{2}h

= π 5^{2 }(2)^{ }------(2)

(1) / (2)

n = (2/3)π (30)^{3 }/ π 5^{2 }(2)^{ }

n = 360

So, the number of soaps made is 360.

**Problem 7 :**

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {2, 4, 6},B = {1, 2, 3, 5} find (AUB)’

**Solution :**

A u B = {1, 2, 3, 4, 5, 6}

(A u B)' = {7, 8, 9, 10}

**Problem 8 :**

In a class of 35 students 18 speak French,12 speak Hindi and 15 speak English.2 students speak French and Hindi.4 Hindi and English and 5 speak English and French. Calculate the number of students who speak all three languages.Also find the number of students Hindi and English but not French.

**Solution :**

Total number of students = 35

Number of students who speaks atleast one one language

11+x+2-x+6+x+x+5-x+4-x+6+x = 35

34 + x = 35

x = 1

Number of students who speaks Hindi and English not French = 4 - x

= 4 - 1

= 3

**Problem 9 :**

A bag contains ten, five and two dollar currencies. The total number of currencies is 20 and the total value of money is $125.If the second and third sorts of currencies are interchanged the value will be decreased by $6.Find the number of currency in each sort.

**Solution :**

Let x, y and z be the number of number of currencies in 10, 5 and 2 dollars respectively.

x+y+z = 20 -----(1)

10x+5y+2z = 125 -----(2)

10x+2y+5z = 119 -----(3)

(1) ⋅ 10 ==> 10x+10y+10z = 200

(1) - (2) ==> -10x-5y-2z = -125

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

5y+8z = 75 ----(4)

(2)-(3)

10x+5y+2z = 125

-10x-2y-5z = -119

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

3y-3z = 6

y-z = 2 ----(5)

(4)+8(5) ==> 13y = 75+16

13y = 91

y = 7

By applying the value of y in (5), we get

7-z = 2

z = 5

By applying the values of y and z in (1), we get

x + 7 + 5 = 20

x = 20-12

x = 8

So, the number of currencies in 5, 10 and 2 dollars are 8, 2 and 5 respectively.

**Problem 10 :**

Factories the cubic polynomial

2x^{3} + x^{2} - 5x + 2

**Solution :**

(x-1) is a factor. So, we can get the other two factors by factoring the quadratic polynomial.

2x^{2} + 3x - 2

(2x-1) (x+2)

So, the three factors are (2x-1) (x+2) and (x-1).

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