**Problem 1 :**

Find the area of the design as shown below.

**Solution :**

Area of rectangle = l x b

Area of semicircle = πr^{2}/2

Area of equilateral triangle = (√3/4)⋅a^{2}

= πr^{2}/2 + l x b + (√3/4)⋅a^{2}

= (1/2)⋅(22/7)⋅7^{2} + 20 ⋅ 14 + (√3/4)⋅14^{2}

= 441.368 cm^{2}

**Problem 2 :**

A cow is tied to one of the corners of a square-shaped field of sides of 10 m. The length of the rope is 7 m. How would one find the area of the field that the cow cannot graze?

**Solution :**

Area of square field = 10^{2}

Area of the filed that cow can graze = πr^{2}/4

= (1/4) ⋅ (22/7) ⋅ 7 ⋅ 7

= 77/2

Area of the filed that cow can graze = 100 - 77/2

= 61.5 m^{2}

**Problem 3 :**

Write the number 0.000001024 in scientific notation

**Solution :**

0.000001024 = 1.024 x 10^{6}

**Problem 4 :**

Write the number 1.423 × 10^{-6} in decimal form.

**Solution :**

= 1.423 × 10^{-6}

= 1.423/1000000

= 0.000001423

**Problem 5 :**

Perform the calculation and write the answer of the following in scientific notation.

(2000)^{2} ÷ (0.0001)^{4}

**Solution :**

(2000)^{2} ÷ (0.0001)^{4}

= (2x10^{3})^{2 }÷ (1x10^{-4})^{4}

= (4x10^{6}) ÷ (1x10^{-16}).

= 4x10^{6+16}

= 4x10^{22}

**Problem 6 :**

Solve for x

3 = log_{x}729

**Solution :**

x^{3} = 729

x^{3} = 3^{6}

x^{3} = (3^{2})^{3}

x = 9

So, the value of x is 9.

**Problem 7 :**

Change 2 = 64^{16} to logarithmic form.

**Solution :**

2 = 64^{16}

log_{64} 2 = 16

**Problem 8 :**

Evaluate

log_{9} (1/27)

**Solution :**

log_{9} (1/27)

= log_{9} (1/3)^{3}

= log_{9} 3^{-3}

= -1log_{9} 3

= -1/log_{3}9

= -1/log_{3}3^{2}

= -1/2log_{3}3

= -1/2

So, the answer is -1/2.

**Problem 9 :**

How many integers between 100 and 999, inclusive, have the property that some permutation of its digits is a multiple of 11 between 100 and 999 For example, both 121 and 211 have this property.

**Solution :**

110, 121, .........990

n = [(l-a)/d]+1

n = (990-110)/11 + 1

n = (880/11) + 1

n = 81

Number of three digit numbers divisible by 11 is 81.

Each number will consist of three digits, using these three digits we can make 6 possible 3 digit numbers.

81 x 6 = 486

In 486 numbers we have repeated counting.

For example, the number abc is divisible by 11, then cba is also divisible by 11. (Divisibility rule for 11)

So, 486/2 = 243

In those 243 numbers, we may have numbers consisting of one zero.

110, 220, 330, 440, 550, 660, 770, 880, 990

In this way, we can get 9 numbers. If we use the digits in the different places, that will not be divisible by 11 or it must a two digit number.

If the middle digit is 0, then

209, 307, 407, 506 by reversing the digits we will get 902, 703, 704, 605. So 8 numbers.

= 243 - (9+8)

= 226

**Problem 10 :**

Obtain the set builder representation of the set

A = {1, 1/2, 1/3, 1/4, 1/5, 1/6}

**Solution :**

A = {x: x = 1/n where n is a integer}

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