Formula used:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

Calculation:

Let,

We can see that, when 1 < x < 2,

Value of x - 1 is positive

Hence, option 3 is correct.

Calculation:

Given, x³ + 1 > x2 + x 

⇒ x³ - x² - x + 1 > 0

⇒ x² (x - 1) - 1 (x - 1) > 0

⇒ (x² - 1) (x - 1) > 0

if (x² - 1) > 0 which gives x = +1

and if x - 1 > 0 it gives x = 1

Thus we can say that x > -1.

Given:

x = (634)²³ – (276)³⁹ + (263)³⁴

Concept used:

The unit digit of the power of a number repeats itself every fourth time.

Calculation:

Unit digit of x = Unit digit of (634)²³ – (276)³⁹ + (263)³⁴

(634)²³ – (276)³⁹ + (263)³⁴

⇒ (634)^{4 × 5 + 3 –} (276)4 ^{× 9 + 3 +} (263)^{4 × 8 + 2}

⇒ Unit digit will be determined by unit digit of (634)³ – (276)³ + (263)²

Unit digit of (634)³ = 4

Unit digit of (276)³ = 6

Unit digit of (263)² = 9

Hence, unit digit will be 4 - 6 + 9 → 13 - 6 = 7

∴ The unit digit of x is 7.

Solution:-

⇒ Sufficient

S2: y = 4

⇒ Insufficient

Hence, option 1 is correct.

Given:

a0 = 5

Statement I

a0 = 5

a₁ = 3.a_1-1 =  3.a₀ = 3 × 5 = 15 

a₂ = 3.a₂-1 =  3.a₁ = 3 × 15 = 45

a₃ = 3.a₃-1 =  3.a₂ = 3 × 45 = 135

a₄ = 3.a₄-1 =  3.a₃ = 3 × 135 = 405

a₅ = 3.a₅-1 =  3.a₄ = 3 × 405 = 1215

a₆ = 3.a₆-1 =  3.a₅ = 3 × 1215 = 3645

a₇ = 3.a₇-1 =  3.a₆ = 3 × 3645 = 10935

∴  a0 + a1 + _ _ _ _ _ + a7 = 5 + 15 + 45 + 135 + 405 + 1215 + 3645 + 10935 = 16400

Alternatively

5, 15, 45, 135, 405, 1215, 3645, 10935 are in Geometric Progression

Common Ratio (r) = 3, a = 5, n = 8 

Sum formula in GP = a (rⁿ -1)/r-1 = 5 × ( 3⁸ -1)/ 3 - 1 = 5 × (2187 - 1)/3 -1 = 10930/2 = 16400

Statement II

an > 0, where it is implied that sum of the series will be greater than 0. Evidently, the result cannot be determined.

∴ Statement I alone can give the answer to the question.

Given:

Following observation is given along with frequency

Formula Used:

Median = l + ((N/2) - F)/f) × h

Where,

l is the lower limit of median class

f is the frequency of median class

h is the class size of median class

F is cumulative frequency of the class preceding the median class

And N = ∑f

Calculation:

To calculate the median first we need to calculate N and also need to calculate the cumulative frequency of the given observation

Formula Used:

Median = l + (((N/2) - F)/f) × h

Where,

l is the lower limit of median class

f is the frequency of median class

h is the class size of median class

F is the cumulative frequency of the class preceding the median class

And N = ∑f

Calculation:

To calculate the median first we need to calculate N and also need to calculate the cumulative frequency of the given observation

∴ N/2 = 70/2 = 35

The median of given data is 48 which lies in the median class 40 - 60

Now, we can find the value of f, F, h and l from the table formed above

∴ f = 20, F = (5 + x), N/2 = 70/2 = 35, l = 40 and h = 20

Put all the values in the formula of median

∴ 48 = 40 + (35 – (5 + x))/20) × 20

⇒ x = 22

Now, 45 + x + y = 70

∴ x + y = 25

⇒ y = 3

Hence, option (4) is correct

Given:

The values of 1, 2, 3, ....., n with respectively frequencies x, 2x, 3x, ...., nx 

Concept used:

Where f_i and x_i denote the observations and respective frequencies respectively.

Calculation:

∴ The mean of the values of 1, 2, 3, ....., n with respective frequencies x, 2x, 3x, ...., nx is 

Given:

The given values =  a + 4, a – 3.5, a – 2.5, a – 3, a – 2, a + 0.5, a + 5 and a – 0.5

Concept used:

If n is odd

Median = [(n + 1)/2]^th observations

If n is even

Median = [(n/2)^th + (n/2 + 1)^th observations]/2

Calculation:

a + 4, a – 3.5, a – 2.5, a – 3, a – 2, a + 0.5, a + 5 and a – 0.5

Arrange the data in ascending order

⇒ a – 3.5, a – 3, a – 2.5, a – 2, a – 0.5, a + 0.5, a + 4, a + 5

Here, the n is 8, which is even

Median =  [(n/2)th + (n/2 + 1)th observations]/2

⇒ [(8/2) + (8/2 + 1)/2] term

⇒ 4^th + 5^th term

⇒ [(a – 2 + a – 0.5)/2]

⇒ [(2a – 2.5)/2]

⇒ a – 1.25

∴ The median of the data set is a – 1.25

Formula Used:

Where l = lower limit of median class,

n = number of observations,

h = class size, f = frequency of median class,

cf = cumulative frequency of class preceding the median class.

Calculation: 

From the given table we have formed the above table of cumulative frequency to find the median.

In the table, n = 150 + x + y = 150 + 78 = 228

which will lie between 42 + x and 107 + x,

therefore the median class: 40 - 50

l = 40, f = 65, cf = 42+x