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If the prime factorization of the denominator of a fraction has only factors as 2 or 5 or a combination of 2 and 5 then the decimal form of that fractional will be a terminating decimal form. 

Consider the fractions 17/20 and 19/6

Now, 20 = 2 x 2 x 5, and 6 = 2 x 3 

∴ 17/20 is terminating decimal form while 19/6 is recurring decimal form.

{3 + (6 x 7) + (9 ÷ 3)} + {- 8 + 8 x 9} 

Note: The above problem has many solutions. Students may write solution other than the one given.

C) L.C.M × H.C.F

If the prime factorization of the denominator of a fraction has only factors as 2 or 5 or a combination of 2 and 5 then the decimal form of that fractional will be a terminating decimal form.

Consider the fractions 17/20 and 19/6

Now, 20 = 2 x 2 x 5, and 6 = 2 x 3

∴ 17/20 is terminating decimal form while 19/6 is recurring decimal form.

i. 2 is the smallest prime number. 

ii. There are 15 prime numbers from 1 to 50. 

They are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

iii. [17], 15, 4, [3], 1, [2], 12, [23], 27, 35, [41], [43], 58, 51, 72, [79], 91, [97].

Given number is 8430. 

The given number has zero in the ones place. 

So, 8430 is divisible by 2. 

And the unit sum is8 + 4 + 3 + 0 = 15 is a multiple of 3. 

So, 8430 is divisible by 3.

If a number is divisible by both 2 and 3, then only it is divisible by 6. 

8430 is divisible by both 2 and 3. 

Therefore 8430 is divisible by 6.

2 is the even prime number.

Correct option is  D) 11

17 and 71; 37 and 73; 79 and 97.

Correct option is  B) 40

The greatest prime number between 50 and 100 is 97.

The greatest four digit number is 9999. 

Sum of the digits of 9999 = 9 + 9 + 9 + 9 = 36 is divisible by 9 

So, 9999 is divisible by 9. 

Sum of the digits of 9999 is also multiple of 3. 

Therefore, 9999 is also divisible by 3. 

We notice that the numbers which are divisible by 9 are always divisible by 3.

Correct option is  A) 81

i. Factors of 28 = 1, 2, 4, 7, 14, 28 

Factors of 42 = 1,2, 3, 6, 7, 14, 21, 42 

∴ HCF of 28 and 42 = 14

ii. Factors of 51 = 1, 3, 17, 51 

Factors of 27 = 1, 3, 9, 27 

∴ HCF of 51 and 27 = 3

iii. Factors of 25 = 1, 5, 25 

Factors of 15 = 1, 3, 5, 15 

Factors of 35 = 1, 5, 7, 35 

∴ HCF of 25, 15 and 35 = 5

D) 1, 2, 3, 4, 6, 12

i. Factors of 8: 1, 2, 4, 8 

Factors of 14: 1, 2, 7, 14 

∴ Common factors of 8 and 14: 1,2 

∴ 8 and 14 are not a pair of co-prime numbers.

ii. Factors of 4: 1, 4 

Factors of 5: 1, 5 

∴ Common factors of 4 and 5: 1 

∴ 4 and 5 are a pair of co-prime numbers.

iii. Factors of 17: 1, 17 

Factors of 19: 1, 19 

∴ Common factors of 17 and 19: 1 

∴ 17 and 19 are a pair of co-prime numbers

iv. Factors of 27: 1, 3, 9, 27 

Factors of 15: 1, 3, 5, 15

∴ Common factors of 27 and 15 : 1,3 

∴ 27 and 15 are not a pair of co-prime numbers.

Correct option is  C) (3, 5)

Correct answer is

(C) 40, 20

Hint: 40 = 2 x 2 x 2 x 5 

20 = 2 x 2 x 5 

∴ HCF of 40 and 20 = 2 x 5 = 10

Correct option is  C) 1

(a) 504210.

Since the HCF of each pair = 7, let the four numbers be 7a, 7b, 7c, 7d. 

Also, LCM = abcd × HCF 

⇒ abcd \(\frac{1470}{7} = 210\)

∴ Product of the numbers 

= 7a × 7b × 7c × 7d 

= 7⁴ × abcd = 7⁴ × 210 = 504210.

Answer is (d) 420

LCM = 28 HCF 

Also, LCM + HCF = 1740

⇒ 28 HCF + HCF = 1740 

⇒ 29 HCF = 1740 ⇒ HCF = \(\frac{1740}{29}=60\)

⇒ LCM = 28 × 60 = 1680 

Since, one number = 240 

∴ Other number = \(\frac{HCF \times LCM}{One\,number}\) 

= \(\frac{60 \times 1680}{240}\) = 420

i. 14 = 2 x 7 

28 = 2 x 14 

= 2 x 2 x 7 

∴ HCF of 14 and 28 = 2 x 7 

= 14

LCM of 14 and 28 = 2 x 2 x 7 

= 28

ii. 32 = 2 x 16 

= 2 x 2 x 8 

= 2 x 2 x 2 x 4 

= 2 x 2 x 2 x 2 x 2 

16 = 2 x 8 

= 2 x 2 x 4 

= 2 x 2 x 2 x 2 

∴ HCF of 32 and 16 = 2 x 2 x 2 x 2 

= 16 

∴ LCM of 32 and 16 = 2 x 2 x 2 x 2 x 2 

= 32

iii. 17 = 17 x 1 

102 = 2 x 51 

= 2 x 3 x 17 

170 = 2 x 85 

= 2 x 5 x 17 

∴ HCF of 17, 102 and 170 = 17

∴ LCM of 17, 102 and 170 = 17 x 2 x 3 x 5

= 510

Correct option is  C) 90

(c) 225

Let the numbers be a and b. Then, 

a – b = 20           .... (i) 

ab = 56.25 × 20 = 1125

⇒ b = \(\frac{1125}{a}\) 

∴ Putting the value of b in (i) we get 

a – \(\frac{1125}{a}\) = 20 ⇒ a² – 1125 = 20a 

⇒ a² – 20a – 1125 = 0 

⇒ a² – 45a + 25a – 1125 = 0 

⇒ a (a – 45) + 25 (a – 45) = 0 

⇒ (a – 45) (a + 25) = 0 

⇒ a = 45 or –25

Neglecting the negative value of a, b = 45 – 20 = 25 

Now, a = 45 = 3² × 5, b = 25 = 5² 

∴ LCM of (45, 25) = 3² × 5² = 225

Correct option is  D) 8

Correct Answer is  Factor

Prime factorization

i) 24 

24 = 1 × 24 

24 = 3 × 8 

24 = 2 × 12 

24 = 4 × 6 

Factors of 24 are 1,2, 3, 4, 6, 8, 12 and 24.

ii) 56

56 = 1 × 56 

56 = 7 × 8 

56 = 2 × 28 

56 = 4 × 14 

Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56.

iii) 80 

80 = 5 × 16 

80 = 2 × 40 

80 = 8 × 10 

80 = 4 × 20 

Factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80.

iv) 98 

98 = 7 × 14 

98 = 2 × 49

Factors of 98 are 1, 2, 7, 14, 49, 98.

Correct Answer is  Composite

Correct Answer is  2

Perfect number

Difference = 24 – 15 = 9 is the multiple of 3. 

Yes, 3 is a factor of difference of 15 and 24.

Correct Answer is  3

Co-prime (or) relatively prime

Correct Answer is  9

Correct option is (A) Taluka

y = e^{x^x}

take log both sides, log y = log_e e^{x^x}

log y = x^x log^e_e

log y = x^x (∵ log e^e = 1)

Again, taking log both sides,

log(log y) = log x^x

log(log y) = x log x

then differentiating w.r.t. x,

= ((d log(log y))/d log y) x ((d log y)/dy) x (dy/dx) 

= (xd log x)/dx + (log x) x (dx/dx)

= (1/log y) x (1/y) x (dy/dx) 

= (x) x (1/x) + log x

= (1dy/(y log y dx)) - 1 + log x 

= dy/dx = (1 + log x)y log y

= dy/dx =x^x e^{x^x}(1 + log x)

Answer is (d) 0

Answer is (a) 1

Answer is (d) vector(b.a)

Answer is (b) x = 2, y = - 8

Answer is (b) π/4

∫dx/(1 + sin x), x ∈ [0,π/2]

Let I = ∫dx/(1 + sin x), x ∈ [0,π/2]

∫(1 - sin x)dx/(1 + sin x)(1 - cos x), x ∈ [0,π/2]

∫(1 - sin x)dx/(1 - sin² x), x ∈ [0,π/2]

∫(1 - sin x)dx/(cos² x), x ∈ [0,π/2]

∫((1/cos² x) - (sin x/cos² x))dx, x ∈ [0,π/2]

∫(sec² x - tan x sec x)dx, x ∈ [0,π/2]

∫sec² x dx, x ∈ [0,π/2] - ∫tan x.sec x dx, x ∈ [0,π/2]

[tan x], x ∈ [0,π/2] - [sec x], ∈ [0,π/2]

((tan π/2) - tan 0) - ((sec π/2) - sec 0)

0 - (-1) 

= 1

A French philosopher Blaise Pascal invented “Pascaline” the first mechanical calculator.

10 times the loop will execute

3 types of iteration statements exist

Every loop has four elements that are used for different purposes.

These elements are:

1. Initialization expression

2. Test expression

3. Update expression

4. The body of the loop

1. Initialization expression(s) : 

The control variable(s) must be initialized before the control enters into loop. The initialization of the control variable takes place under the initialization expressions. The initialization expression is executed only once in the beginning of the loop.

2. Test Expression : 

The test expression is an expression or condition whose value decides whether the loop-body will be executed or not. If the expression evaluates to true (i.e., 1), the body of the loop is executed, otherwise the loop is terminated.

In an entry – controlled loop, the test – expression is evaluated before the entering into a loop whereas in an exit-controlled loop, the test – expression is evaluated before exit from the loop.

3. Update expression :

It is used to change the value of the loop variable. This statement is executed at the end of the loop after the body of the loop is executed. 

4. The body of the loop :

A statement or set of statements forms a body of the loop that are executed repetitively. In an entry – controlled loop, first the test-expression is evaluated and if it is nonzero, the body of the loop is executed otherwise the loop is terminated. In an exit – controlled loop, the body of the loop is executed first then the test – expression is evaluated. If the test – expression is true the body of the loop is repeated otherwise loop is terminated.

(a) Hat – Hat plays an important role in the development of the play. It stands for the city’s authority, ‘the Mayor’ as it was a part of the official uniform. When Dr. Stockmann put on the hat, he told his brother that the entire city was in his hand. He also added that with that supreme power he would throw him off the existing government. Dr. Stockmann saw the Mayor’s hat in the editor’s room and could realize that the Mayor was hiding in the next room, listening his conversation with the editor. This made him understand the Mayor’s plot of ruining his article.

(b) Stick – The stick at certain instances was a symbol of assertiveness, commitment, new ideas or sticking to the old things. The stick in Dr. Stockmann’s hand suggests his assertiveness, commitment and new ideas in the existing government with the support of everyone. But the stick in the Mayor’s hand, suggests that the whole situation was groomed by him to turn against Dr. Stockmann. The stick in the hand of the Mayor also proves his authority and command over other people.

(c) An envelope containing the letter – The envelope of the letter focuses over the issue of grievance which everyone keeps by hiding a secret. Dr. Stockmann took various efforts to expose the issue but finally Hovstad returned the envelope to him, which suggests how everyone tried to conceal the burning issue. The issue needed immediate attention, but was hidden from the public as the letter was hidden by the envelope.

India is a country of vast geographical diversity. Due to this, the vegetation found here also reflects variations on regional basis because of various controlling conditions. 

Indian forests can be classified as follows:

The distribution of forests on geographical basis can be explained as follows:

1. Evergreen Forests: Evergreen forests are found in the regions where there is more than 200 cm annual rainfall. In India, such type of forests in India are found in the western slope of Western Ghats, Andaman-Nicobar group of islands, Bengal, Assam, Meghalaya and Terai areas. 

2. Autumn Forests or Monsoon Forests: These type of forests shed their leaves in dry season. These type of forests are found in the lower part of Northern mountainous region, Vindhayachal and Satpura hills, Plateau of Chhota Nagpur, hilly area of Assam, southern part of Eastern Ghats and the leeward eastern regions of Western Ghats. 

3. Dry Forests: These type of forests are found at places where annual rainfall is 50-100 cm. Dry forests are mainly found in south – western Punjab, Haryana, Eastern Rajasthan and south-western Uttar Pradesh. 

4. Desert Forests: Desert forests are found in the areas where there is less than 50 cm rainfall annually. These are found in south – western Punjab, western Rajasthan, Gujarat and Madhya Pradesh.

5. Mountain Forests: These types of forests are found at Mahabaleshwar in Maharashtra and Pachmarhi in Madhya Pradesh. The height of these trees varies from 15 m to 18 m. The stenls of these forests have large circumference. In the Himalayan region of North India and hills of Assam, these type of forests are found at the height of more than 180 m. 

6. Tidal Forests: These type of forests are found at the mouth of peninsular rivers such as Mahanadi, Godavari, Krishna, Kaveri etc. and in the deltas of Ganga and Brahmaputra.