Explore topic-wise InterviewSolutions in Current Affairs.

1. Transaction motive: It relates to the desire of the people to hold cash for the current transactions.

Mt = f (y)

Precautionary motive: It relates to the desire of the people to hold cash to meet unexpected or unforeseen expenditures.

Mp = f (y)

Speculative motive: It relates to the desire of the people to hold cash in order to take advantage of market movements regarding future changes in the price of bonds and securities in the capital market.

Ms = f (i)

Dadabhai Naoroji, William Digby, VKRV Rao and RC Desai made attempts to compute the per capita income in the colonial period.

The following are the two-fold motives behind the systematic deindustrialisation affected by the British:

1. Making India a Supplier of Raw Materials: The main motive of the British government was to make India a mere supplier of cheap raw materials to feed its own flourishing industrial base.

2. Making India a Market for Finished Goods: Another important objective of the British government was to use India as a virgin market to sell the finished goods produced by the British industries.

Given:

The number = 5224

Formula used:

Dividend = (Divisor × Quotient) + Remainder

Calculation:

⇒ 5224 = (43 × 121) + 21

⇒ The number which is subtracted from 5224 = 5224 - (43 × 121) = 5224 - 5203 = 21

∴ The required result will be 21.

Given: 

The divisor is 357 and the remainder is 219.

Calculation: 

Let the number is 'N'

According to question

N = 357k + 219

Now, this number is divided by 17

⇒ (357k + 219) ÷ 17

⇒ 357k/17 + 219/17

In, the part (357k/17) remainder is Zero, because 357 is divisible by 17

Now, we consider this part (219/17) 

Here, we found quotient is 12 and the remainder is 15

∴ The required value of the remainder is 15.

Given:

First number = 3

Next two greater prime numbers = 5 and 7

Concept used:

LCM will be the smallest number divisible by all of them

Calculation:

3 = 3 × 1

5 = 5 × 1

7 = 7 × 1

LCM will be the uncommon factors = 3 × 5 × 7

⇒ 105

∴ The smallest whole number divisible by 3, 5 and 7 is 105

Given:

Number given = 3

Next two greater prime numbers = 5 and 7

Concept used:

LCM of prime numbers are the product  of these numbers

Calculation:

Required number = 3 × 5 × 7

⇒ 105

∴ The required number is 105

Given

The sum of the squares of two positive numbers = 2426

The square root of one number = 7

Calculation

Let numbers be x and y

⇒ √x = 7

⇒ x = 49

According to question

⇒ x² + y² = 2426

⇒ (49)² + y2 = 2426

⇒ y2 = 2426 - 2401

⇒ y2 = 25

∴ The other number is 5

Given:

Five bells commence tolling together and toll at intervals of 9, 6, 4, 10 and 8 seconds respectively.

Concept used:

Least common multiplication 

Calculation:

A least common multiple of (9, 6, 4, 10 and 8)

We can write 9 = 3²

4 = 22 

6 = 2 × 3

8 = 23 

10 = 2 × 5

A least common multiple of (9, 6, 4, 10 and 8) = 23 × 3² × 5

⇒ 8 × 9 × 5

⇒ 360 sec

Bells ring together after every 360 sec 

Required number of times in 1 hour (60 × 60 seconds) = [(60 × 60)/360]

⇒ 10

But we have to add 1 because at starting all bells will be rung once a time after that the ring 10 times.

⇒ 10 + 1 

⇒ 11 times 

∴ The required number of times when they toll together is 11.

Given:

product of any two prime numbers except the smallest one

Calculation:

Prime numbers are 2, 3, 5, 7, ...

The smallest prime number is 2

So it can not be used in further calculation.

Product of any prime numbers,

Randomly picking from the above numbers,

⇒ 3 × 5 = 15

⇒ 11 × 9 = 99

and so on

Now we can say that the product will always be Odd number,

∴ The product will always be Odd number. 

Given:

5231405 × 99999.

Calculation:

5231405 × 99999.

⇒ 5231405 × (100000 – 1)

⇒ 523140500000 – 5231405

⇒ 523135268595

∴ The value of 5231405 × 99999 is 523135268595.

Given 

a + b = 27 

ab  = 182 

Calculation 

Let the 1st number be a and the 2nd number will be 27 – a 

Therefore, their product = a (27 – a)

It is given that the product of these numbers is 182 

Therefore a (27 – a) = 182 

⇒ a² – 27a + 182 = 0

⇒ a² – 13a – 14a + 182 = 0

⇒ a(a – 13) – 14(a – 13) = 0

⇒ (a – 13) (a – 14) = 0

⇒ a – 13 = 0 or a – 14 = 0 

⇒ a = 13 or a = 14

If first number = 13, then

⇒ Other number = 27 – 13 = 14 

If first number = 14, then

⇒ Other number 27 – 14 = 13 

∴ Therefore the numbers are 13 and 14.

Given:

4 bells beep at an interval of 4, 12, 20, and 25.minutes.

Beep together at 8 a.m.

Concept used:

Using LCM method

Calculation:

4 = 2 × 2

12 = 2 × 2 × 3

20 = 2 × 2 × 5

25 = 5 × 5

LCM of 4, 12, 20 and 25 = 2 × 2 × 3 × 5 × 5

⇒ 300 minutes 

⇒ 5 hour

So, 8 a.m. + 5 hour = 13 p.m. = 1 p.m.

∴ They will beep together again at 1 p.m.

Correct answer is: (c) 6:28 AM

LCM 0f 4,7,14=28

Bells will they ring together again at 6:28 AM

Given:

Three bells ring at the interval of 7, 10 and 12 seconds and they ring simultaneously at 2:00 pm.

Calculation:

Three bells rings at the interval of 7, 10 and 12 seconds and to calculate the time at which the bells ring again simultaneous we take the LCM of the given time intervals.

LCM of (7, 10 and 12) = 2 × 2 × 3 × 5 × 7 = 420 seconds = 7 minutes

∴ Three bells ring again after 7 minutes.

So, the time at which three bells ring again simultaneously = 2:00 pm + 7 min = 2:07 pm.

Concept used:

HCF = Highest common factor 

The greatest number which divides each of the two or more numbers is called HCF or Highest Common Factor.

Calculations:

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

H.C.F. of 1651 and 2032

1651 = 13 × 127 

2032 = 4 × 4 × 127

HCF of 1651 and 2032 = 127

∴ The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is 127.

The greatest 4 digit number is 9999,

Smallest 3 digit number is 100

Sum of 9999 + 100 = 10099

Concept:

 prime numbers are number that has only 2 factors: 1 and themselves. 

Calculation:

Smallest 3 digit number is 100 which is not a prime number.

101 is the next number which is divisible by 1 and 101

∴ 101 is the smallest prime number 

Given

Smallest two-digit prime number = 11

Concept

Remainder theorem

Dividend = Divisor × quotient + remainder 

A prime number is a number that is divisible by 1 and by itself only. For example 2,3, 5, etc.

Calculations

11 = 5 × 2 + Remainder

11 - 10 = Remainder

1 = Remainder

Concept:

Prime numbers are whole numbers greater than 1, that have only two factors i.e., 1 and the number itself

Calculation

Smallest three-digit prime number = 101

Greatest two-digit prime number = 97

Their difference = 101 - 97 = 4

Formulae Used:

(a + b)² = a² + 2ab + b²

(a - b)² = a2 - 2ab + b2

Calculation:

Let us consider a = 598, and b = 479

So, the numerator = a2 + 2ab + b² - (a2 - 2ab + b²) = 4ab

and the denominator = a × b = ab

∴ ? = 4ab/ab = 4

GIVEN:

2(4²² + 4²²)(3²² + 3²² + 3²²) = 12^{(p + 3)}

CALCULATIOON :

2(4²² + 4²²)(3²² + 3²² + 3²²) = 12^{(p + 3)}

⇒ 2 × 2 × 4²² × 3 × 3²² = 12^{(p + 3)}

⇒ 4²³ × 3²³ = 12^{(p + 3)}

⇒ 12²³ = 12^{(p + 3)}

On comparing –

⇒ 23 = p + 3

⇒ p = 20

According to question –

⇒ p/5 = 20/5 = 4

The value of the following is given by 100 (kilo ampere) x ( micro ampere) 100 milli ampere x 10 ampere  0.1 A.

Given:

Ratio of three numbers = 2 : 3 : 6

HCF = 45

Calculation: 

let the number be 2x, 3x, 6x

HCF is the common factor for all numbers

adding all the number = 2x + 3x + 6x = 11x 

∵ both the x and HCF is common in all the number

Substituting the value of HCF in x 

Sum of the three numbesr = 11 × 45 = 495

∴ The sum of the three number is 495

Given:

HCF = 13

LCM = 455

One number = 65

Concept used:

HCF × LCM = Product of the numbers

Calculation:

Let x be the second number.

13 × 455 = 65 × x

⇒ x = 13 × 455/65 = 91

∴ The second number is 91.

(d) Arrayed data

Given:

3 parallel straight lines

6 points = A1, A2, A3, A4, A5, A6

Calculation:

The maximum triangles would be in case of all 6 points are non-collinear. In such a case the number of triangles is ⁶C₃

⇒ (6 × 5 × 4) / (1 × 2 × 3)

⇒ 20

∴ Statement I is correct.

In another case, if the surveyor takes the position that A1 and A2 coincide on the 1^st line, similarly A3 and A4 coincide on the 2^nd line and A5 and A6 coincide on the 3^rd line. In such a case there will be 0 triangles.

Given:

For Statement I:

Cost price of 2 scarfs = Rs. 2200

Earns = Rs. 55 per scarf

Discount = 23%

For Statement II:

Marks up the C.P = 36%

Earns = Rs. 200

Calculation:

For Statement I:

The cost price of a shirt = Rs. 1100

Selling Price = 1100 + 55

⇒ Rs. 1155

Marked Price = 1155 / 77 × 100

⇒ Rs. 1500

Statement I is sufficient to answer the question.

For Statement II:

Let the cost price be x.

Marked Price = x × 136%

⇒ 1.36x%

Selling Price = x + 200

Statement I:

2X + 3 = 7 

2X = 7 - 3  

2X = 4 

X = 4 / 2

X = 2

The the value of X is known. 

So, statement 1 alone is sufficient to answer the question.

Statement II: 

3X + 4Y - 11 = 0 

Here the value of X and Y are not found.

So, statement 2 alone is not sufficient.

Thus X = 2, X is an integer number.

Hence, the correct answer is "Statement I alone is sufficient".

Diamagnetic ⇒ Na₂O, N₂O, SO₂, Cl₂O

Paramagnetic ⇒ KO₂, NO₂, ClO₂

Δ_r H = ∑B.E. (Reactants) – ∑B.E. (Products) 

= [B.E. (H₂) + B.E. (Br₂ )] – 2 B.E. (HBr) 

= 435 + 192 – 2 × 364 = – 101 kJ.

(b) negative 

(b) ΔH < 0, ΔS > 0 

(b) G = H – TS 

(a) spontaneous