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Given Data:

a, a + 2, a + 4 are prime numbers.

Concept Used:

3, 5, 7 are the only set of three prime numbers which are in arithmetic progression with common difference = 2

Calculation:

a, a + 2, a + 4, these three terms are in Arithmetic progression with common difference = 2.

Comparing a, a + 2, a + 4 with 3, 5, 7, we will get the value of a as 3

⇒ Number of possible values of a is 1 i.e. 3

∴ The number of values of a is 1.

Mistake Points There is a difference between the value of 'a' and the number of values of 'a'

Here the value of a = 3 but the number of values that 'a' can have is 1 because only 1 value satisfies for a i.e 3

Concept Used:

2-digit number × 2-digit number = 4-digit number.

To find the biggest 4-digit number which is a perfect square, we have to take the square of biggest 2-digit number.

Calculation:

2-digit biggest number = 99

99² = 9801

∴ The biggest 4-digit number which is a perfect square = 9801.

Concept Used:

To find the remainder of f(x), we have to put the value of x from x – a = 0

Calculation:

x – 1 = 0 ⇒ x = 1

f(x) = x¹³ + 1

putting x = 1,

f(1) = 1¹³ + 1 = 2

Remainder = 2

∴ When x¹³ + 1 is divided by x - 1, the remainder is 2.

Calculation:

According to the question,

A is a non-empty subset = A ⊆ C

B is a non-empty subset = B ⊆ C

By Distributive property,

 A ∪ (A ∩ B)  = (A ⋃ A) ∩ (A ⋃ B) = A

Given Data:

Rate = r = 5% per annum

Present population = 4410

Calculation:

Let population of village 2 years ago be x

After the first year, the population = x × (100 + 5)/100 = 21x/20

After second year, the population (Present population) = (21x/20) × (100 + 5)/100 = (21x/20) × (21/20) = 441x/400 = 4410(Given)

⇒ x = 4410 × 400/441 = 4000

∴ The population before 2 years = 4000.

5% = 1/20

For the first year,              20 → 21

For the Second Year,         20 → 21

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

                                         400 → 441

441 → 4410    (Given)

⇒ 400 → 4000

∴ The population before 2 years = 4000.

Formula Used:

 average = (sum of all observation)/ (number of observation)

Solution:

⇒ sum of all observation = 10 × 4.5 + 30 × 3.5 = 150

⇒ number of all observation = 40

⇒ average = 150/40

∴ average = 3.75

so, average = 3.75

Given:

side of cube = 14 cm.

∵ cylinder is cut from cube 

∴ r = 14/2 = 7 cm, h = 14 cm

Formula used:

 If side of a cube is a

∴ total surface area = 6 × a²

If the radius of a cylinder is r and height is h -

∴ total surface area = 2 × π × r × (r + h)

Solution:

total surface area of cylinder = 2 × (22/7) × 7 × (7 + 14)

⇒ total surface area of cylinder = 924 cm²

Formula used:

average = (sum of all observation)/(number of observation)

Concept used:

If we add or subtract anything in the given data, it will happen in the average also.

Solution:

there are 50 number which is subtracted from 53 and their mean (average) is - 3.5

∴ mean of given numbers = 53 - (-3.5) = 56.5

Formula Used:

If the total number of observation given is odd, then the formula to calculate the median is:

Median = {(n+1)/2}thterm

If the total number of observation is even, then the median formula is:

Median  = [(n/2)th term + {(n/2)+1}th]/2

Solution:

5, 9, 14, 15, x + 1, 2x - 13, 28, 30, 32, 34

∵ number of observation is even

∴ median = {(x + 1) + (2x - 13)}/2

⇒ 24 = (3x - 12)/2

⇒ 3x = 60

⇒ x = 20

∴ The value of x is 20

Formula Used:

  c1 × d1 = c2 × d2 

where c = number of cow and d = number of days

Given:

c1 = 44, d1 = 9 days, d2 = 12 days, c2 = x

∴ put these values in given formula:

⇒ 44 × 9 = 12 × x

⇒ x = 33

⇒ number of cows eating in 12 days is 33.

⇒ 11 cows less than the previous one