Explore topic-wise InterviewSolutions in .

GIVEN

The five-digit number 136xy is divisible by 3, 7, and 11.

CALCULATION:

Firstly, we evaluate the LCM of (3, 7 and 11) = 3 × 7 × 11 = 231

The largest possible value of 136xy is 13699.

When we divide 13699 by 231, we obtain 70 as the remainder.

So, required number = 13699 – 70 = 13629

∴ x = 2 and y = 9

According to question –

⇒ x² + y² + 5y/3 = 2² + 9² + 5(9)/3

⇒ 4 + 81 + 15

⇒ 100

∴ The correct answer is 100.

Given:

The given polynomials are (x² – bx + 8) and (x² – 11x + 3a) are divisible by (x + 5)

Concept Used:

According to factor theorem If (x - a) is factor of P(x) then P(a) = 0

Calculation:

Let the polynomials are P(x) = (x² – bx + 8) and G(x) = (x² – 11x + 3a). If (x + 5) is the factor of polynomials. Then according to the factor theorem

∴ P(-5) = 0

Now, put x = -5 in P(x)

So, 33 + 10b = 0

⇒ b = -33/10

And also G(-5) = 0

Now, put x = -5 in G(x)

So, 80 + 3a = 0

⇒ a = -80/3

So, product of both is positive and both the value is in fraction

Hence, option (2) is correct

Formula for sum S of  n interior angles  of n sided polygon is

S = (n−2) × 180˚

Since polygon is regular all its interior angles is same, sum of them is 135 degree times n

So we have S = 135n˚

As per formula,

S = (n−2) × 180 = 135n˚

180(n−2) = 135n

180n − 360 = 135n

45n = 360

n = 360/45 = 8

Number of sides is 8.

\(\left|\begin{array}{ccc} b+c & a+b & a \\ c+a & b+c & b \\ a+b & c+a & c \end{array}\right|\)

\(= \) \((b + c) \begin{bmatrix}b + c&b\\c + a&c\end{bmatrix} - (a + b)\begin{bmatrix}c + a&b\\a + b&c\end{bmatrix} + a\begin{bmatrix}c + a&b + c\\a + b&a + c\end{bmatrix}\)

\( = (b + c)[(bc + c^2) - (bc + ab)] - [(a + b)(c^2 - b^2 + ac - ab)] + a[(a^2 + 2ac + c^2 - (ab + ac + b^2 + bc)]\)

\(= (b + c)(c^2 - ab) - [(a + b)(c^2 - b^2 + ac - ab)] + a(a^2 - b^2 + c^2 + ac - ab - bc)\)

\( = bc^2 - bc^2 - ab^2 - ab^2 + ab^2 + ab^2 - ac^2 + ac^2 - a^2c + a^2c + a^2b - a^2b + a^3 + b^3 + c^3 - 3abc\)

\( = a^3 + b^3 + c^3 - 3abc\)

Option (1) is correct. 

Given:

P is two years older than Q who is twice as old as R.

The total of the ages of P, Q and R be 52

Calculations:

Let R's age be x years.

So, Q's age = 2x years.

P's age = (2x + 2) years.

According to the question

⇒ (2x + 2) + 2x + x = 52

⇒ 5x = 50

⇒ x = 10

Q's age = 2x = 20 years

∴ Q’s age is 20 years

(i) P(x < 30)  = P\((\frac{x-\mu}\sigma<\frac{30-25}5)\)

 = P\((\frac{x-\mu}\sigma<1)\)

 = 0.8413 (∵ \((\frac{x-\mu}\sigma∼N(0,1))\)

(ii) P(x > 18) = 1 - P(x \(\leq18\))

 = 1 - P\((\frac{x-\mu}\sigma\leq\frac{18-25}5)\)

 = 1 - P\((\frac{x-\mu}\sigma\leq1.4)\)

 = 1 - (P(x\(\leq\) 0) - P(-14 \(\leq\) x \(\leq\) 0))

 = 1 - P(x \(\leq\) 0) + P(0 \(\leq\) x \(\leq\) 1.4)

 = 1 - P (x \(\leq\) 0) + (P(x \(\leq\)) - P(x \(\leq\)0))

 = 1 - 2P(x \(\leq\) 0) + P(x \(\leq\) 1.4)

 = 1 - 2 x 0.5 + 0.9192

 = 0.9192

(iii) P(25 < x < 30) = P(0 < \(\frac{x-\mu}{\sigma}<1\))

 = P(x \(\leq\) 1) - P(x \(\leq\) 0)

 = 0.8413 - 0.5

 = 0.3413

Given:

mean of four numbers = 22

mean of three smallest numbers = 19

data range 15

Formula used:

Mean = Sum of observations/Number of observations

Data range = Largest number – Smallest number

Calculation:

Sum of all numbers = 22 × 4 = 88

Sum of three smallest numbers = 19 × 3 = 57

Largest number = 88 – 57 = 31

Smallest number = 31 – 15 = 16

Sum of three largest number = 88 – 16 = 72 

Mean = 72/3

⇒ 24

∴ The mean of three largest number is 24

Given:

Observations = a, a + 3, a + 6, a + 9 and a + 12.

Mean = 10

Formula used: 

Mean(a) = sum of observations/number of observations

Calculation:

According to the question, 

Mean(a) = sum of observations/number of observations

⇒ (a + a+ 3 + a + 6 + a + 9 + a + 12)/5 = 10

⇒ (5a + 30)/5 = 10

⇒ 5(a + 6)/5 = 10

⇒ a + 6 = 10

⇒ a = 4

∴  a is equal to 4.

Given:

mean of four numbers = 110

largest number = 135

Formula used:

Mean = Sum of observations/Number of observation

Range of set = largest number – smallest number

Calculation:

Sum of all numbers = 440

Sum of numbers other than the largest = 440 – 135 = 305

for the value of the range to be maximum other two numbers should be 134 and 133 as all numbers are different.

Smallest number = 305 – (134 + 133) = 38

Range = 135 – 38 = 97

∴ The range of the set of these four numbers is 97

Given:

Mode of data = 7

Mean of data = 8

Formula Used:

Mode = (3 × Median) - (2 × Mean)

Calculation:

7 = (3 × Median) - (2 × 8)

⇒ 3 × Median = 16 + 7

⇒ Median = 23/3

∴ The Median of the data is 23/3

Given:

The given data,

12, 11, 11, 13, 15, 16, 20, 12, 13, 14, 11

Concept used:

Mode is the most repeated value in a data

Calculations:

Values repeated in the given data,

⇒ 12 = 2 times

⇒ 11 = 3 times

⇒ 13 = 2 times

The most repeated term,

Mode = 11

∴ The mode of the following data is 11

Given:

Data = (1, 5, 9, 2, 2, 3, x, a, 9)

Arithmetic mean = 5

Formula Used:

Arithmetic mean = Sum of terms/Number of terms

Calculations:

Number of terms = 9

Sum of terms = 1 + 5 + 9 + 2 + 2 + 3 + x + a + 9

⇒ Sum of terms = 31 + a + x

Arithmetic mean = Sum of terms/Number of terms

⇒ 5 = (31 + a + x)/9

⇒ (31 + a + x) = 5 × 9

⇒ (31 + a + x) = 45

⇒ a + x = 45 - 31 = 14

Arithmetic mean of a and x = (a + x)/2

⇒ Arithmetic mean of a and x = 14/2

⇒ Arithmetic mean of a and x = 7

∴ The arithmetic mean of a and x is 7.

Given:

Arithmetic mean of family's height = 135 cm

Number of members = 3

Height of new member = 119 cm

Formula Used:

Arithmetic mean = Sum of heights/Number of members

Calculations:

Number of members = 3

Arithmetic mean = Sum of heights/Number of members

⇒ 135 = Sum of heights/3

⇒ 135 × 3 = Sum of heights

⇒ Sum of heights = 405 cm

Height of new member = 119 cm

Sum of heights of all members = 405 cm + 119 cm

Arithmetic mean of heights = 524/4

⇒ 

∴ The new arithmetic mean of heights of all the members of the family is 131 cm.

Arithmetic mean or average of number 2¹⁰ and 2²⁰

⇒ Average = (2¹⁰ + 2²⁰)/2

⇒ Average = 2¹⁰/2 + 2²⁰/2

⇒ Average = 2⁹ + 2¹⁹      

∴ The arithmetic mean of 210 and 220 is 29 + 219      

Given:

The number are 20, 20, 21, 21, 22, 22, x, 23, 23, 24, 25, 27

The median of ascending order = 22.5

Formula used:

When N is even

Median = [(N/2 + 1)^th + N/2^th]/2

Calculation

Median = (6^th + 7^th) ÷ 2 = (22 + x) ÷ 2

⇒ 22.5 = (22 + x) ÷ 2

⇒ 45 = 22 + x

⇒ x = 23 

∴ The required value of x is 23.

Given:

Mean = 5

Median = 6

Formula used:

Mode = 3Median - 2Mean

Calculations:

Mode = 3Median - 2Mean

⇒ Mode = 3 × 6 - 2 × 5

⇒ Mode = 8

∴ The mode is 8.

Given:

S = {2, 5, 7, 15, 21, 23, 32}

Formula Used:

Arithmetic mean = (Sum of terms)/(Number of terms)

Calculations:

S = {2, 5, 7, 15, 21, 23, 32}

Number of terms = 7

Arithmetic mean = (Sum of terms)/(Number of terms)

Sum of terms = {2 + 5 + 7 + 15 + 21 + 23 + 32} = 105

⇒ Arithmetic mean of S = 105/7 = 15

∴ The arithmetic mean of S is 15.

Given:

Mean of 27, 26, 17, 25, 43, 19, 37, 10

Formula used:

Mean = Sum of observations/Number of observations

Calculation:

Sum of observations = (27 + 26 + 17 + 25 + 43 + 19 + 37 + 10)

⇒ 25.5

∴ The mean of given set value is 25.5

Here multiplication operation has more priority than addition. hence 

1. a = 5 + 15 = 20

2. Here 9 is an interger 

‘9’ is a character 

“9” is a string

(I) def COUNTLINES(): 

file=open('abc.txt','r') 

lines = file.readlines() 

count=0 

for w in lines: 

if w[0]=="A" or w[0]=="a": 

count=count+1 

print(“Total lines “,count) 

file.close()

Local Area Network

It is a Hub/Switch

1. Twisted pair cables and coaxial cables. 

2. Twisted Pair Wire : 

Two copper wires individually insulated, twisted around each other and covered by a PVC. There are two types of twisted pair wire. They are UTP and STP. It is very cheap and easy to install.

Coaxial Cable: 

A sturdy copper wire(conductor) is insulated by plastic. This is covered just like a mesh by a conductor , which in turn is enclosed in an protective plastic coating. Compared to twisted pair wire it is more expensive, less flexible and more difficult to install. But it is more reliable and carry for higher data rates.

There is no risk and gets a fixed salary.

1. Fused bulb 

2. Defective switch 

3. Defective Circuit of warning lamp

1. Storing the waste oil properly for collection and recycling 

2. Store waste oil in a safe place 

3. The use of expert services 

4. Do not pour it down the drain or the septic system 

5. Use the appropriate heater if burning the oil as a disposal means 

6. Use waste oil disposal approved equipment 

7. Dispose of everything associated with the used oil

Plants, Animals.

Carbon brush

Concept:

 A resultant force is the force (magnitude and direction) obtained when two or more forces are combined.

Calculations:

Given, ABCD is a quadrilateral. Forces \(\overrightarrow{AB}, \overrightarrow{CB}, \overrightarrow{CD}\) and \(\overrightarrow{DA}\) act along its sides.

⇒\(\vec {AB} = \vec b - \vec a\)

⇒\(\vec {CB} = \vec b - \vec c\)

⇒\(\vec {CD} = \vec d - \vec c\)

⇒\(\vec {DA} = \vec a - \vec d\)

 A resultant force is the force (magnitude and direction) obtained when two or more forces are combined.

Resultant Force =  \(\overrightarrow{AB} + \overrightarrow{CB} + \overrightarrow{CD}+\overrightarrow{DA}\) 

⇒Resultant Force = \(\rm \vec b - \vec a + \vec b - \vec c + \vec d - \vec c + \vec a - \vec d\)

⇒Resultant Force = \(\rm 2 \vec b - 2 \vec c\)

⇒Resultant Force = \(\rm 2 (\vec b - \vec c)\)

⇒Resultant Force = \(\rm 2\overrightarrow {CB}\)

(a) Amplify weak signals 

The correct answer is: (c) 3.

Given:

Case I

SP = Rs. 1000

Case II

SP = Rs. 1800

Formula used:

P = S.P. – C.P.

L = C.P. – S.P.

P% = (P/C.P.) × 100

L% = (L/C.P.) × 100

Where,

P → Profit 

L → Loss

SP → Selling price

CP → Cost price

Calculations:

Let the C.P. be x.

According to question 

{(x – 1000)/x} × 100 = {(1800 – x)/x} × 100

⇒ x(x – 1000) = x(1800 – x)

⇒ 2x = 2800

⇒ x = 1400

So, loss = 1400 – 1000 = 400

L% = 100 × (400/1400) = (200/7)%

∴ The loss% is (200/7)%.

(d) All of the above

(b) Optocoupler 

Given:

A, B, and C starts a business with an investment of Rs. 75,000, Rs. 90,000, and Rs. 1,05,000 respectively.

After a year B gets Rs. 12,000 as the profit.

Formula used:

The ratio of profit sharing = Ratio of (Investment × Time)

Calculations:

Since the time period is the same for all three, the ratio of profit sharing will depend upon the ratio of investment only.

The ratio in which profit will be shared between A, B and C

⇒ 75,000 ∶ 90,000 ∶ 1,05,000

⇒ 5 ∶ 6 ∶ 7

Suppose total Profit = Rs. x

⇒ 12,000 = x × 6/18

⇒ x = Rs. 36,000

Given:

Discount % (D1) = 20%, Profit % (P₁) = 60%

Discount % (D₂) = 30%, Profit % (P2) = ?

Formula used:

S.P = M.P × (100 – D%)/100 

C.P = S.P × 100/(100 + P%) 

P = S.P – C.P 

P% = (P/C.P) × 100

Where,

S.P → Selling price 

C.P → Cost price 

M.P → Marked price 

D% → Discount% 

P% → Profit% 

Calculations:

Let the M.P of the product be 100x 

So, S.P of the product = 100x × (100 – 20)/100 = 80x 

C.P of the product = 80x × 100/(100 + 60) =50x 

If the discount of the product is 30% 

Now, the S.P of the product = 100x × (100 – 30)/100 = 70x 

Profit = 70x – 50x = 20x 

Profit% = (20x/50x) × 100 = 40%

∴ The required profit% is 40%

(a) Tunnel diode 

Given:

Total investment = 75500

Profit = 45300

Concept used:

Person’s Profit = Investment × Time period

Calculation:

Let x be the investment of C

C = x      ----(1)

B = (x + 4500)      ----(2)

A = (x + 4500 + 3500)      ----(3)

We know that,

x + (x + 4500) + (x + 4500 + 3500) = 75500

⇒ x + (x + 4500) + (x + 8000) = 75500

⇒ 3x + 12500 = 75500

⇒ 3x = 63000

x = 21000

Put the value of x in equation 1, 2, and 3.

A = 29000

B = 25500

C = 21000

Ratio of Profit = 29000 : 25500 : 21000

⇒ 58 : 51 : 42

A’s share = [58/(58 + 51 + 42)] × 45300

∴ Share of A in the business is Rs. 17400.

(b) Schottky diode

Given:

Investment of A  = Rs. 92,500

Investment of B = Rs. 1,12,500

B's share in profit = Rs. 9,000

Concept used:

If time of investment is equal for A and B then, ratio of profit earned by A and B is same as the ratio of investment done.

Calculations:

Ratio of profit = 92,500 : 1,12,500

⇒ 37 : 45

Let total profit be y.

B's share in the profit = {45/(45 + 37)} × y

⇒ (45/82) × y = 9,000

⇒ y = 7,38,000/45

⇒ y = 16,400

∴ The total profit earned by them together is Rs. 16,400

Given:

Amount invested by A =Rs. 92,500

Amount invested by B = Rs. 1,12,500

B's share in profit = Rs. 9,000

Concept used:

Divide the total profit according to investment ratio.

Calculation:

Let the profit earned by A be Rs. X

According to the question,

(92500/112500) = (X/9000)

⇒ X = (92500 × 9000)/112500

⇒ X =  37 × 200

⇒ X = 7400

∴ The profit earned by A is Rs. 7400

Given:

X's investment = Rs. 19,000

Y's investment = Rs. 17,000

Total profit = Rs. 10,800

Concept used:

Divide the total profit according to investment ratio.

Calculations:

The total investment of X in one year = 19000 × 12 

The total investment of Y in one year = 17000 × 12 

The ratio of equivalent capitals of X and Y = (19000 × 12) ∶ (17000 × 12) = 19 ∶ 17

Thus X’s share = 10800 × 19/36 = Rs. 5,700

∴ The share of X is Rs. 5,700.

(b) 20 µsec 

that the time constant RC should satisfy the following condition

1/fc << RC < 1/fm

= 1/ 1 MHz << RC < 1/ 2KHz

= 1µs < RC < 0.5 ms

(c) Load resistor

The correct answer is: (b) High

Given:

Expenditure on food = 30%

Expenditure on room rent = 10%

Ram saves 25% from remaining amount

After saving Ram invest in Mutual funds and life insurance policy in ratio of 5 : 4 respectively and difference is Rs.2,500

Concept:

If A is x% of B

Then, x% = (A/B) × 100

Calculation:

Let, Ram’s monthly income = Rs.100x

Expenditure on food = (30/100) × 100x

⇒ Rs.30x

Expenditure on room rent = (10/100) × 100x

⇒ Rs.10x

Remaining amount after expend on room rent = 100x – 30x – 10x

⇒ Rs.60x

Saving amount of Ram = (25/100) × 60x

⇒ Rs.15x

Remaining amount after saving Rs.15x = 60x – 15x

⇒ Rs.45x

Now, Rs.45x invest in Mutual funds and life Insurance policy in ratio of 5 : 4

Invest in Mutual funds = (5/9) × 45x

⇒ Rs.25x

Invest in Insurance policy = (4/9) × 45x

⇒ Rs.20x

Amount Invested in Mutual funds – Amount Invested in Insurance policy = Rs.2500

25x – 20x = 2500

⇒ 5x = 2500

⇒ x = 500

Ram’s monthly income = 100 × 500

⇒ Rs.50,000

∴ Monthly income of Ram is Rs.50,000.

Given:

Investment of Pawan = Rs 18,000

Investment of Raman = Rs 25,200

Profit earned by both at the end of one year = Rs 8,640

Concept used:

Investment × time of investment ≡ Profit

Calculation:

Ratio of Profit of Pawan and Raman = 18,000 × 1 ∶ 25,200 ∶ 1 = 5 ∶ 7

Let the profit received by Pawan and Raman be 5x and 7x respectively.

Total profit, 8640 = 5x + 7x

⇒ 12x = 8640

Or, x = 720

Share of Pawan, 5x = Rs 3,600

∴ Share of Pawan in the profit is Rs 3,600.