Explore topic-wise InterviewSolutions in Current Affairs.

Communalism is dangerous for National Integration.

1955 on the basis of language reorganization of states were done .

Udai Singh was ruler of Mewar.

(b)  Raigarh.

Akbar suppressed the rebellion of Abdullah Khan, the Governor of Malwa.

Shivaji's Agra Adventure was planned by Jai Singh.

(b) Prince Akbar

Shivaji sacked Surat and secured a large booty.

(a) Afzalkhan

The correct option is: (c) 4 3 2 1

Under the Mughal rule the judicial service was formed by Kazis.

Iraq is a land with widely different physical features. In its North-Eastern region, there are green plains, rivers with their banks full of trees and hills covered with all sorts of flowers. There is enough rain for cultivating different crops. Agriculture started here between BC 7000 and 6000. On the Northern side of Iraq there is a grassy plain. This place was suitable for grazing cattle. Naturally, the people here got a better life than those of mere agricultural farmers. 

In the Eastern side there were the tributaries of River Tigris and they offered excellent travel facilities towards the hilly regions. The southern side is a desert. It is in Iraq that the first writings and urban life started. The fertile silt deposited by Euphrates and Tigris Rivers made this area very suitable for cultivation and therefore it could nurture urban life.

(b) left fallow

Archaeological excavations were started, in Mesopotamia in the 1840s. Excavations in sites like Uruk and Mari continued for decades. The hundreds of buildings, statues, tombs, tools and seals in Mesopotamia and the thousands of written documents are useful sources in recreating Mesopotamian history.

During the Mughal rule the finance minister was known as Diwan.

During the Mughal rule, the police duties in the districts were entrusted to the officials known as Faujdar.

The word of 'Casta' of caste from Purtgali language

(b) Agriculture

Cattle herding and fishing.

Answer: statement is true

Indian people mainly depend upon Agriculture.

(d) All of these

 Rajput is a caste.

Answer: Prejudice

Balwant Roy Mehta samiti is associated with  Panchyati Raj

Sahjanand Saraswati is the father of peasant movement.

Vaishya is not a caste .

Vinoba Bhave is associated with Bhudan movement.

Institution is a market.

Jharkhand Mukti Morcha is related with  Tribal movement.

This answer is a sample

Skinfold measurements are also called as “fatfold thickness”.These measurements provide the information or data of the thickness of double folds of the skin and sub cutaneous adipose tissue at specific sites of the body. ln simple words skinfold provides the information about general fatness of the body.
The procedure of skinfold measurements is as under-
1. Triceps Skinfold - The arm of the subject or child should be hung loosely.Stand behind the subject and pull the vertical skinfold about 1/2 inch from the spot already marked.Keep the skinfold caliper perpendicular to the length of the fold centering the mark.Record the measurement to the nearest millimeter.
2. Sub scapula skinfold- After locating the marked point on the sub scapular region , pull a skin fold for about 34 inch above and keep the skin fold caliper perpendicular to the length of skinfold.Release the caliper and note the measurement to the nearest millimeter.

3. Suprailiac skinfold- The subject should stand straight with his feet together and arms relaxed.Pull a skinfold 34 inch above the marked point with the thumb and index finger.The skinfold caliper should be kept perpendicular to the
length of skin fold.Release the caliper and note the reading on the dial to the nearest of millimeter and record it.
4. Abdomen skinfold - After locating the already marked point, pull a horizontal skinfold to about 34 inch. Place the skinfold caliper perpendicular to the length of skinfold.Release the caliper and note the readingto the nearest of millimeter and record it.
5. Thigh skinfold — The person is made to stand with his weight on the left leg and right leg forward with knee slightly bent.Grip a skinfold on the already marked area on the mid thigh.Place the skinfold caliper and note the reading to the nearest of millimeter and record it.

Maximum value of sin θ   + cos θ  is √2

a²+b²+c²+d² = ab+bc+cd+da ) (a - b)²+(b - c)²+(c - d)²+(d - a)² = 0 ) a = b = c = d.  

Thus ABCD is a rhombus and  

[ABCD] = (1/2)(d₁d₂) .....(1)

where d₁ and d₂ are the lengths of the diagonals. Hence d₂ = (2[ABCD])/d₁ = 4units.

Correct option (C) greater than or equal to 2

Explanation:

sin²θ + cosec²θ  ≥ 2.

We count the number of 3-digit numbers with (i) at least one 5 and having no 3 and (ii) at least one 5 and having exactly one 3 separately. 

(i) Here we rst count the whole set and subtract the number of 3-digit numbers having no 5 from it. Since 3 is not there and 0 cannot be the rst digit, we can ll the rst digit in 8 ways. But we can ll the second and third digits in 9 ways(as 0 can be included). Thus we get 8 x 9 x 9 such numbers. If no 5 is there, then the number of such numbers is 7 x 8 x 8. Thus the number of 3-digit numbers not containing 3 and having at least one 5 is (8 x 9 x 9) - (7 x 8 x 8) = 8(81 - 56) = 200. 

(ii) If 3 is there as a digit, then it can be the rst digit or may be the second or third digit. Consider those numbers in which 3 is the rst digit. The number of such numbers having at least one 5 is (9 x 9) - (8 x 8) = 81 - 64 = 17. The number of 3-digit numbers in which the second digit is 3 and having at least one 5 is (8 - 9) - (7 x 8) = 16. Similarly, the number of 3-digit numbers in which the third digit is 3 and having at least one 5 is (8 - 9) - (7 x 8) = 16. Thus we get 17 + 16 + 16 = 49 such numbers. 

Therefore the number of 3-digit numbers having at most one 3 and at least one 5 is 200+49 = 249.

Note that P(n) - 2015 - (-2015) = P(n) divides P(P(n) - 2015) - P(-2015) for every  positive integer n. But P(n) divides P(P(n) - 2015) for every positive integer n. Therefore  P(n) divides P(-2015) for every positive integer n. Hence P(-2015) = 0.

Note  

In the original version of the problem the word 'non-constant' was missing. The  falsity of the statement was brought to the attention of the examiners by a  contestant who mentioned it in a remark at the end of a perfect solution to the  problem assuming that the polynomial is non-constant. Many students assumed  that the polynomial is non-constant and completed the solution. They deserved  full credit for doing so.

Given:

Length of sticks are 7.5 feet ad 3.25 feet

Formula used:

HCF of (A/B) and (C/D) = (HCF of A and C)/(LCM of B and D)

Calculation:

First convert decimal into fraction

⇒ 7.5 = 75/10 = 15/2

⇒ 3.25 = 325/100 = 13/4

Now, minimum length will be measured by HCF of {15/2, 13/4}

⇒ HCF = 1/4 = 0.25

∴ The minimum length can be measured by two stick is 0.25 foot.

1. Let f(x) = x³ - a²x + x + 2

∵ x - a is a factor

∴ f(a) = 0

⇒ a³ - a² + a + 2 = 0

⇒ a = -2

2. 3a² + 27b² + 5c² - 18ab - 30c + 237

= 3(a² + 9b² - 6ab) + 5(c² - 6c + 9) + 237 - 45

= 3(a - 3b)² + 5(c - 3)² + 192

∴ Smallest value of expression is 192 which occurs when a = 3b & c = 3.

3. P(2) = 3 & P(3) = 2

Let P(x) = x² + bx + c

∴ P(2) = 4 + 2b + c

P(3) = 9 + 3b + c

P(3) - P(2) = b + 5

⇒ b + 5 = 2 - 3 = -1

∴ b = -6

Also

4 - 12 + c = P(2)

c = 8 + P(2) = 8 + 3 = 11

∴ P(x) = x² - 6x + 11

Let the zeroes be a and b 

So, given a + b = 14 .....(1) 

Now, 

kx² + 7x - 5 

x = k, y = 7, z = -5 

From (1), 

a + b = 14

-y/x  = 14

-7/k = 14

14k = -7

k = -7/14

k = -1/2

Let f(x) = x³ - 6x² + 11x - 6 

Given: (x-1)(x-2) are the zeros of f(x). 

product of zeros = (x-1)(x-2) = x² - 2x - x + 2 = x² -3x + 2 

therefore, on dividing the f(x) with x² - 3x + 2 , we get, x-3 

Now, x - 3 = 0 

x = 3

Hence, the other zero is 3.

Sum of zeroes = 1  

Product of zeroes = 3  

So by formula of quadratic polynomial:-  

x² - (Sum of zeroes)x + (Product of zeroes) = 0

x² - 1x + 3 = 0

x² - x + 3 = 0

P(x) = x⁹ – 5x⁵ + 4x⁴ + 2x² + 1 

(i) The number of sign changes in P(x) is 2. The number of positive roots is atmost 2. 

(ii) P(-x) = -x⁹ + 5x⁵ + 4x⁴ + 2x² + 1. The number of sign changes in P(-x) is 1. The number of negative roots of P (x) is atmost 1. 

Since the difference of number of sign changes in P(-x) and number of negative zeros is even. P(x) has one negative root.

(iii) 0 is not the zero of the polynomial P(x). So the number of real roots is almost 3. 

∴ The number of imaginary roots at least 6.

Correct answer is: (b) K=10

3/6=5/k

⇒K=10

Correct answer is: (d) Four times

If the circumference of a circle increases from 2π to 4π then its area Four times the original area.

C₁/C₂=2πr/2πR

2π/4π=2πr/2πR

r/R=1/2

A₁/A2=πr² /πR²=(r/R)²=(1/2)²=1/4

A₂=4A₁

Correct answer is:  (d) \(\cfrac{a}{\sqrt{b^2-a^2}}\)
sin\(\theta\) = a/b 

H²=P²+B²

b²=a²+B²

B=\(\sqrt{b^2 -a^2}\)

tan\(\theta\) = P/B=a/\(\sqrt{b^2 -a^2}\)

Correct answer is: (a) 3

x + y = 2sin²\(\theta\)+2cos²\(\theta\)+1

=2(sin²\(\theta\)+ cos²\(\theta\))+1

=2+1=3

Correct answer is: (b) 44

2πr- r=37

r(2\(\times \cfrac{22}{7}\)-1)=37

r=37x7/37

r=7

circumference= 2\(\times \cfrac{22}{7}\) x7=44cm