Explore topic-wise InterviewSolutions in Current Affairs.

(D) Statement-1 is False, Statement-2 is True.

Answer is (c) Statement I is true, but Statement II is false.

Answer is (b) Both the statement are true, but statement II is not the correct explanation of Statement I.

Answer is (c) Statement I is true, but Statement II is false.

Answer is (b) Both the statement are true, but statement II is not the correct explanation of Statement I.

Answer is (b) 1/13

Answer is (b) 90°

Answer is (c) π/4

Answer is (b) log|sin x + cos x| + c

Answer is (b) 2

Answer is (a) 1/15

Answer is (a) a unit matrix

Answer is (b) [(1,1),(1,0)]

Answer (a) [(1,-2),(-2,1)]

Answer is (d) (0,0,1)

Answer is (c) symmetric

Given, f(x) = (3 - x³)^1/3

L.H.S. = fof(x)

= f[f(x)]

= f{(3 - x³)^1/3}

= 3 - {(3 - x³)^1/3}^1/3

= [3 - 3 + x³]^1/3 = (x³)^1/3

= x

Answer is (c) ((x/5) - 4)

L.H.S. = sec²(tan^-1 2) + cosec²(cot^-1 3)

Let tan^-1 2 = α

tan α = 2

∵ sec² α = 1 + tan² α

1 + 2² = 1 + 4 = 5

sec² α = 5

Let again cot^-1 3 = β

cot β = 3

∴ sec² α + cosec² β

5 + 1 + cot² β

= 5 + 1 + (3)²

= 5 + 1 + 9 = 15 = R.H.S.

Answer is (b) -π/3

L.H.S. = tan^-1 √x

= (1/2) (2 tan^-1 √x)

= (1/2) x cos^-1 ((1 - (√x)²)/(1 + (√x)²))

= (1/2) cos^-1 ((1 - x)/(1 + x)) 

= R.H.S

I = ∫x².e^{x^3} dx

= (1/3)∫e^z dz = (1/3) e² + c

put, x³ = z

on diff, 3x² dx = dz

x² dx = dz/3

= (1/3) e^{x^3} + c

Answer is (c) -1

Answer (d) 1

Answer is (a) 1

Answer is (b) [(-1,0),(0,-1)]

Answer is (b) 0

Answer is (a) 0

\(\begin{bmatrix}x&8\\3&3\end{bmatrix}=0\)

⇒ 3x - 24 = 0

⇒ x = 24/3 = 8

Answer is (b) 8

Answer is (a) -9

1. n = 2 + 8 = 10.

2. DAUGHTER this word has 8 different letters. A, U, E are the vowels. Treat these 3 as one unit, then there are 6 units and can be permuted in 6! ways. The above vowels can be permuted in 3! ways. Hence the total number of words is 6! × 3! = 4320.

Hence the number of different at-least 3 flag signals = 120 + 360 + 720 + 720 = 1920. 

Concept used :

Mode of a given group of numbers is the value that appears most often in the given data group 

Calulations :

For a group 4, 5, 5, 6, 4, 3, 2, 2, 5, 1

Mode of the group = 5 (5 came 3 times in the given group)

∴ 5 is the mode of the given numbers.

Given:

5 × 4Pr = 6 × 5Pr – 1

Formula used:

P (n, r) = n!/(n –r) !

Where, The number of possible arrangements for are objects from a set of n object.

Calculation:

5 × 4Pr = 6 × 5Pr – 1

⇒ 5 × [4!/(4 – r)!] =  6 × [5!/{5 – (r – 1)!}]

⇒ 5 × [4!/(4 – r)!] = (6 × 5 × 4!)/(6 – r)!

⇒ (5 × 1)/(4 – r)! = (6 × 5)/(6 – r) × (5 – r) × (4 – r)!

⇒ 1 = 6/(30 – 11r + r²)

⇒ r² – 11r + 30 = 6

⇒ r2 – 11r + 30 – 6 = 0

⇒ r2 – 11r + 24 = 0

⇒ r2 – 8r – 3r + 24 = 0

⇒ r(r – 8) – 3(r – 3) = 0

⇒ (r – 8) × (r – 3) = 0

If, (r – 8) = 0

⇒ r = 8

Either, (r – 3) = 0

⇒ r = 3

In expression n = 4 or 5

⇒ n ≤ 5

⇒ n ≠ 8

⇒ n = 3

∴ The value of r is 3.

The correct answer is the Frequency.

• Fr equency:
  • The total number of frequencies observed in observation of data.
• Mean:
  • The "mean" is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers
• Median:
  • The "median" is the "middle" value in the list of numbers
• Range: 
  • List numbers are just the difference between the largest and smallest values.

Given:

Mean = 38

Range = 31

The middle number is 20 less than the sum of the other two numbers

Formula used:

Mean = (x + y + z)/3

Range = biggest term – smallest term

Calculation:

Let the three numbers is x, y, and z.

Mean = (x + y + z)/3

⇒ 38 = (x + y + z)/3

⇒ x + y + z = 114      ….(i)

According to question,

Middle term = (first term + last term) – 20

⇒ y = (x + z) – 20

Value of y putting in equation (i)

⇒ x + z + (x + z) – 20 = 114

⇒ 2(x + z) =134

⇒ x + z = 67      ….(ii)

According to question

Rang = biggest term – smallest term

⇒ z – x = 31      ….(iii)

add the equation of (ii) and (iii)

⇒ 2z = 98

⇒ z = 49

Pattern of the series is:

78 + 2 = 80

80 + 3 = 83

83 + 5 = 88

88 + 7 = 95

95 + 11 = 106

Therefore wrong number in the series is 104.

So, starting with 104, the series will be 104, 106, 109, 114.

∴ The Third term in the 2nd series is 109.

There is a possibility with the logic.

Here logic used is prime no. 2,3,5, 7 ,11 But difference given is 2,3,5,7,9

So one can think it is an odd no series and can consider 1st term as the wrong term to make the difference, 1 which is odd

But there is no option for 83 which will be third term in the 2nd series if starting from 79.

Pattern of the series is:

9 × 2 + 1 = 19

19 × 3 + 1 = 58

58 × 4 + 1 = 233

233 × 5 + 1 = 1166

1166 × 6 + 1 = 6997

Therefore wrong number in the series is 229. 

So, starting with 229, the series will be 229, 459, 1378, 5513.

∴ The Second term in the 2nd series is 459.

Given:

Let initial amount A and B have before shopping be Rs. 10a and Rs. 13a respectively.

Calculation:

The amount left with A after shopping = 350 × 2/7 = Rs. 100

The amount left with B after shopping = 350 - 100 = Rs. 250

Let amount spent on shopping by A and B each be Rs. m 

⇒ 10a - 100 = m

⇒ 13a - 250 = m

Then,

⇒ 10a - 100 = 13a - 250

⇒ a = 50

∴ The amount spent by A on shopping = 500 - 100 = Rs. 400

Formula used:

SI = (P × R × T)/100

Amount = P × (1 + R/100)^t

CI = P × [(1 + R/100)t- 1]

where, SI → Simple interest, CI → Comp[und interest

P → Principal, R → Rate of interest, T → Time period

Calculations:

Statement I:

Let the principal amount be P.

CI at 25% for 2 years =  P × [(1 + 25/100)2- 1]      ----(1)

SI at 5% rate for 1 year = (P × 5 × 1)/100      ----(2)

Now, (1) - (2) = Rs. 250

⇒ 11P/16 - P/20 = Rs. 250

⇒ 51P/80 = Rs. 250

⇒ P = Rs. 20000/51

So, statement I is sufficient to answer the question.

Statement II:

Let the principal be P.

⇒ (3P + 2) = P × (1 + 3.5/100)t

Since we have 1 equation 2 variables

So it cannot be solved

So, statement II is not sufficient to answer the question.

Statement III:

CI at 8% annually = P × [(1 + 8/100)t- 1]

CI at 8% semi-annually = P × [(1 + (8//2)100)t- 1]

P × [(1 + 8/100)t- 1] -  P × [(1 + (8//2)100)t- 1] = Rs. 279

Since we have 1 equation 2 variables

So it cannot be solved

So, statement III is not sufficient to answer the question.

∴ The data in statement I alone is sufficient to answer the question, while the data in statement II and statement III alone are not sufficient to answer the question.

Statement I:

Given:

Added quantity of water = 20 l

Ratio of alcohol and water after addition of water = 8 : 3

Calculation:

Let x be the initial quantity of water

Equating the quantity of water

{(x + 12)/(120 + 12)} = 3/(8 + 3)

⇒ (x + 12)/132 = 3/11

⇒ (x + 12)/12 = 3

⇒ x + 12 = 36

⇒ x = 36 – 12 = 24

Quantity of water in 120 liters mixture is 24 l

Quantity of alcohol in 120 liters mixture is 120 – 24 = 96 l

∴ Statement I alone is sufficient to answer the question.

Statement II:

Given:

Quantity of mixture replaced with alcohol = 20 l

The ratio of alcohol and water after replacement = 5 : 1

Calculation:

Let the quantity of alcohol be x

⇒ Quantity of alcohol after replacement = (x/120) × 100 = 5x/6

⇒ Quantity of water after replacement = 100 – (5x/6)

Equating with given ratio

{(5x/6) + 20}/{100 – (5x/6)} = 5/1

⇒ {(5x/6) + 20} = 500 – (25x/6)

⇒ (30x/6) = 480

⇒ x = 480/5 = 96

Quantity of alcohol in 120 liters mixture is 120 – 24 = 96 l

The quantity of water in 120 liters mixture is 24 l

∴ Statement II alone is sufficient to answer the question.

∴ Either statement I alone or statement II alone is sufficient to answer the question.

Given:

A can complete a work in 80 min and B can complete the same work in 40 min. 

Calculation:

 The efficiency of A = 1/80 per minute

 The efficiency of B = 1/40 per minute

 Time taken by both A, B to complete the work = ( 1/A + 1/B )

 ( 1/80 + 1/40 ) = ( 1 + 2 )/80 = 3/80

⇒ Time Taken = 80/3 

∴ Time Taken = 26 minutes 40 seconds

Given:

A's 1 hour's work = 1/10

B's 1 hour's work = 1/15

Calculation:

(A + B)'s 1 hour's work = 1/10 + 1/15 = 1/6

(A + B)'s 4 hours work  = 4 × 1/6 = 2/3

Remaining work after 4 hours = 1 - 2/3 = 1/3

Now, Time taken to complete 1/3 of work is 12 hours

So, time taken to complete 1 work = (12 × 3) hours

⇒ 36 hours

Then,

(A + B + C) 1 hour's work = 1/10 + 1/15 + 1/36 = 35/180

⇒ 7/36

∴ If all worked together, then work will be completed in 36/7 hours

Given:

A can do a work in 6 days, B can do the same work in 8 days

And C can do the same work in 12 days

Formula Used:

Efficiency ∝ 1/Time taken

Calculation:

The efficiency of A = 1/6 per day

The efficiency of B = 1/8 per day

 The efficiency of C = 1/12 per day

 A, B, C can do the same work in = ( 1/6 + 1/8 + 1/12 )

⇒ ( 4 + 3 + 2 )/24 = 9/24

∴ Time taken to complete work is 8/3 days.

Formula used:

Simple Interest = (P × R × T)/100

Where, P = Principal

R = Rate of interest per annum

T = Time

Statement (I):

Simple interest = 10%

Time = 5 years

∴ The statement (I) alone is not sufficient to answer the question.

Statement (II):

Rakesh return 5% of the total amount

Return amount he received Rs. 240.

5% of total amount = Rs. 240 

∴ The statement (I) alone is not sufficient to answer the question.

∴ Both the statements I and II together are needed to answer the question.