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A mathematical determinant whose first row consists of n functions of x and whose following rows consist of the successive derivatives of these same functions with respect to x.

Concept :

The binary number system with only two independent digits, 0 and 1, is a base-2 number system. All larger binary numbers are represented in terms of ‘0’ and ‘1’.

The decimal number is equal to the sum of binary digits (Dn) times their power of 2 (2n)

Decimal = --- + D2 × 22 + D1 × 21 + D0 × 20 

In order to convert the binary fractions to decimal numbers, we use negative power of 2 to the right of the binary point.

= D-1 2-1 + D-2 2-2 + ---

Calculation:

Given binary number is: 1001.0010

The decimal equivalent of the integer part will be:

(Decimal)_Integer = (1) × 23 + (0) × 22 + (0) × 21 + (1) × 20

= 8 + 1 = 9

The fractional part will be:

(Decimal)Fractional = (0) × 2-1 + (0) × 2-2 + (1) × 2-3 + (0) × 2-4

= 2^-3 = 0.125

∴ The resultant decimal equivalent is:

(1001.0010)₂ = (9.125)₁₀

Given:

The ratio of the monthly income of Radhey and Ranu = 12 ∶ 9

The ratio of their expenditure = 11 ∶ 7

Saving of each of them = Rs. 18,000

Formula used

Income = Expenditure + Saving

Calculations:

Let the income of Radhey and Ranu be 12x and 9x

Expenditure of Radhey = 12x – 18000

Expenditure of Ranu = 9x – 18000

According to the question,

⇒ (12x – 18000)/(9x – 18000) = 11/7

⇒ 84x – 126000 = 99x – 198000

⇒ 15x = 72000

⇒ x = 4800

The total income of Radhey and Ranu = 12x + 9x = 21 × 4800 = Rs. 1,00,800

∴ The total income of Radhey and Ranu is Rs. 1,00,800

Given-   

Ratio of earning of Rohit and Ravi = 5 : 4

Ratio of their expenditure = 6 : 5 

Ratio of their Savings = 3 : 2

Concept Used-

Income  = Expenditure + Savings

Calculation-

Let the earnings of Rohit and Ravi be 5x and 4x and their expenditure be 6y and 5y

Savings of Rohit = 5x - 6y

Savings of Ravi =  4x - 5y

According to Question-

5x - 6y + 4x - 5y =  5000

⇒ 9x - 11y = 5000  ---------(i)

(5x - 6y)/(4x - 5y) = 3/2

⇒ 10x - 12y = 12x - 15y

⇒ 2x - 3y = 0 -------(ii) 

Solving Equation i and ii we get x = 3000 and y = 2000

Monthly earning of Rohit = 5 × 3000

⇒ 15000

∴ Half Yearly earning of Rohit = 6 × 15000

⇒ 90000

Concept:

Decimal to Binary:

To convert the decimal number into binary follow the below steps:

• Take the decimal number as a dividend.
• Divide this number by 2 (2 is the base of binary so divisor here).
• Store the remainder in an array (it will be either 0 or 1 because of divisor 2).
• Repeat the above two steps until the number is greater than zero.
• Print the array in reverse order (which will be the equivalent binary number of a given decimal number).

Calculation:

Given number is 151.75

Converting the 151 into binary

Step 1: Divide (151)₁₀ successively by 2 until the quotient is 0:

151/2 = 75, remainder is 1
75/2 = 37, remainder is 1
37/2 = 18, remainder is 1
18/2 = 9, remainder is 0
9/2 = 4, remainder is 1
4/2 = 2, remainder is 0
2/2 = 1, remainder is 0
1/2 = 0, remainder is 1

Step 2: Read from the bottom (MSB) to top (LSB) as 10010111. This is the binary equivalent of decimal number 151 (Answer).

Converting the 0.75 into binary we get 0.11

Now the total conversion is:

(151.75)₁₀ = (10010111.11)₂

Calculation:

\(\rm \frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} +...+\frac{1}{{n \times (n+1)}} \)

\(\rm = \frac{{2\; - \;1}}{{1 \times 2}} + \frac{{3\; - \;2}}{{2 \times 3}} + \frac{{4\; - \;3}}{{3 \times 4}} +... + \frac{{(n+1)\; - \;n}}{{n \times (n+1)}}\)

\(\rm = \frac{1}{1} - \frac{1}{2} + \;\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} +... + \frac{1}{n} - \frac{1}{n+1}\)

\(\rm = 1 - \frac {1}{n+1}\)

\(\rm = \frac {n+1 -1}{n+1}\)

\(\rm = \frac {n}{n+1}\)

Given:

Efficiency of A is 80% more than B and B can finish a work in 56 days.

Concept:

This is a typical problem of time & work.

Time required to complete a work is inversely proportional to the efficiency of the worker.

Formula used:

Time required = Total work / Rate of work done

Calculations:

Efficiency of A is 80% more than B.

So, ratio of rate of work done between A and B = 180 : 100 = 9 : 5

Let’s assume, A works 9x work per day and B works 5x work per day.

Total volume of the given work = 56 × 5x = 280x

If they work together, rate of work = (9x + 5x)/day = 14x works/day

Time required = 280x/14x = 20 days

Given:

Speed of boat in still water = 11 kmph

Speed of stream = 2 kmph

Distance = 117 km

Formula used:

Downstream speed = Speed of boat + Speed of stream

Upstream speed = Speed of boat – Speed of stream

Speed = Distance/Time

Calculations:

Let the downstream speed and upstream speed be d and u respectively

Downstream speed

⇒ d = 11 + 2 = 13 kmph

Upstream speed

⇒ u = 11 – 2 = 9 kmph

Time taken in going upstream and downstream

⇒ (117/9) + (117/13)

⇒ 13 + 9

⇒ 22 hours

Given:

Ratio of the present ages of Abhik and Anuja is 3 : 2.

After 15 years ratio of ages of Abhik and Ankan will be 9 : 5.

After 15 years ratio of ages of Anuja and Ankan will be 7 : 5

Calculation:

Let us assume present ages of Abhik, Anuja and Ankan be 3x, 2x and y respectively.

ATQ,

\(\frac{{3x\; + \;15}}{{y\; + \;15}} = \;\frac{9}{5}\)

⇒ 15x + 75 = 9y + 135

⇒ 15x – 9y = 60      ----(1)

And,

\(\frac{{2x\; + \;15}}{{y\; + \;15}} = \;\frac{7}{5}\)

⇒ 10x + 75 = 7y + 105

⇒ 10x – 7y = 30       ----(2)

Solving (1) & (2), y = 10

Present age of Ankan = 10 years.

Given:

Average age of 5 members of the family is 30 years and second son is 12 years old at present.

Concept:

This is an age related question and can be solved just by simple calculations.

Calculation:

Total age of the family at present = 5 × 30 = 150 years

When second son was born, number of members in the family = 3 [since the youngest son was not born at that time]

And, 12 years ago their sum of ages = (150 – 4 × 12) = 102 years

Their average age at that time = 102/3 = 34 years

Students often makes mistakenly takes number of family members as 4 when calculating. Beware of that mistake.

Given:

Downstream speed = d = 15 kmph

Upstream speed = u = 8 kmph

Formula used:

Speed of boat in still water = (d + u)/2

Speed of stream = (d – u)/2

Calculations:

Let the speed of boat in still water be B and the speed of stream be S

Speed of boat in still water

⇒ B = (15 + 8)/2

⇒ B = 23/2

⇒ B = 11.5 kmph

Speed of stream

⇒ S = (15 – 8)/2

⇒ S = 7/2

⇒ S = 3.5 kmph

∴ The speed of boat in still water is 11.5 kmph and the speed of stream is 3.5 kmph

Given:

Initial mixture = 3 : 5

Final mixture = 3 : 9

Calculation:

Let’s, calculate this sum in terms of parts.

⇒ Initial mixture = 3 + 5

⇒ Initial mixture = 8 parts

⇒ Final mixture = 3 + 9

⇒ Final mixture = 12 parts

2 liters were replaced while forming final mixture

⇒ 12 – 8 = 2 L

⇒ 4 Parts = 2 L

⇒ 1 Part = 1/2 L

⇒ 8 Parts = 8 × (1/2)

⇒ Initial Quantity = 4 L + 2L (∵ 2 L was replaced)

⇒ Quantity of olives = (3/8) × 6

∴ Quantity of olives = 2.25 L

Given:

Initial quantity of mixture = 164 liters

Acid : Water = 21 : 20

Concept:

When replacing the mixture with water, quantity of acid will only decrease whereas quantity of water will first decrease and then increase.

Calculations:

Initial quantity of acid = 164 × 21/41 = 84 liters

Initial quantity of water = 164 – 84 = 80 liters

Let us assume the volume of mixture that has to be replaced by 10 liters of water = 41x

[Note how we are choosing the values so that we can avoid complicated fractions]

In 41x liters mixture,

Quantity of acid = 21x liters

Quantity of water = 20x liters

∴ Final quantity of acid = (84 – 21x) liters

∴ Final quantity of water = (80 – 20x + 10) liters = (90 - 20x) liters

ATQ,

(84 – 21x)/ (90 - 20x) = ¼

⇒ 336 – 84x = 90 – 20x

⇒ x = 246/64 = 123/32

∴ Quantity of mixture to be replaced = 123/32 × 41 = 157.6 liters

Correct option: (2) Angle strain

Explanation:

Angle strain is found in cyclic compounds.

Given:

Two numbers are respectively 40% and 80% more than a third number

Calculations:

Let the third number be 100x

So, First number = 100x × (140/100) = 140x

Second number = 100x × 180/100 = 180x

Required ratio = 140x ∶ 180x

⇒ 7 ∶ 9

∴ Required ratio is 7 ∶ 9

Given:

40/100) × x = (15/100) × y

Calculation:

40/100) × x = (15/100) × y

⇒ x/y = 15/40

⇒ x/y = 3/8

⇒ x ∶ y = 3p∶ 8p

⇒ (y - x)/(y + x) = (8p - 3p)/(8p + 3p)

⇒ (y - x)/(y + x) = 5p/11p

∴ (y - x) ∶ (y + x) = 5∶ 11

The Correct option is 2 i.e. 5∶11

Given:

The ratio between two numbers = 3 : 4

Formula used:

If a number is increased by y%, then the new number will be [(100 + y)/100] × a number

Calculation:

Let the first number be 3x and the second number be 4x

According to the question, the first number increased by 40%

Now new first number = [(100 + 40)/100] × 3x

⇒ New first number = (140/100) × 3x

⇒ New first number = 1.4 × 3x

⇒ New first number = 4.2x

And the second number is increased by 50%

Now new second number = [(100 + 50)/100] × 4x

⇒ New second number = (150/100) × 4x

⇒ New second number = 1.5 × 4x

⇒ New second number = 6x

The ratio of the first number and the second number = 4.2x/6x

⇒ The ratio of the first number and the second number = 42/60

⇒ The ratio of the first number and the second number = 7/10

∴ The ratio of the first number and the second number is 7/10.

Given:

The number A is 50% more than the number C.

The number B is 40% more than the number C.

Concept Used:

 The increased value = Value × (100 + percentage)/100

Calculation:

Let the value of C be x.

Then , the value of A = (100 + 50)/100 × x = 3x/2

The value of B = (100 + 40)/100 × x = 7x/5

Then, the ratio of A and B = 3x/2 : 7x/5

The ratio of A and B = 15 : 14

∴ The ratio of the number A and B is 15 : 14. 

Given:

Seats of IT, mechanical and civil in a college are in ratio 4 : 4 : 5

Calculation:

Let the seats of IT, mechanical and civil be 4x, 4x and 5x

After increment seats of IT will be = 4x × 120/100 = 4.8x

After increment seats of Mechanical will be = 4x × 150/100 = 6x

After increment seats of civil will be = 5x × 120/100 = 6x

⇒ required ratio = 4.8x : 6x : 6x

⇒ After solving ratio = 4 : 5 : 5

Jowai headquarter town

As charge +Q is at the center of the circle, 

Therefore,

V_A = V_C 

If V_B is potential at point B, 

Then,

V_B = V_A = V_B - V_C 

∴ W_AB = Q(V_B - V_A) = W₁ 

W_BC = Q(V_B - V_C) = W₂ 

∴ Numerically, 

W₁ = W₂

Absorption transitions of Balmer series will be those which start from second energy-level(n=2). But because at ordinary temperatures almost all the atoms remain in their lowest energy level (n=1), hence absorption transitions can start only from n=1 (not from n=2,3,4....levels). Thus, only Lyman series is found in the absorption spectrum of Hydrogen atom whereas in the emission spectrum, all the series are found.

Correct option is (C) A classical atom based on Rutherford's model is doomed to collapse.

According to Rutherford, e– revolves around nucleus in circular orbit. Thus e– is always accelerating (centripetal acceleration). An accelerating change emits EM radiation and thus e– should loose energy and finally should collapse in the nucleus.

Correct option is (3) 4 MHz

Band width = upper frequency - lower frequency = 10 - 6 = 4MH_z

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

Since E=0, therefore the potential V inside the surface is constant. Because there is no potential difference between any two points inside the conductor, the electrostatic potential is constant throughout the volume of the conductor.

d) Both a and c

c) On effector cell membranes of glandular and smooth muscle cells

b) Benztropine

d)  Ventricle

Correct option is B) for

Correct option is  D) advertisement