Explore topic-wise InterviewSolutions in Current Affairs.

(A) Saccharomyces 

(a) the value of a scalar

(c) a vector

(d) the magnitude of a vector.

Explanation:

The value of a scalar, a vector and the magnitude of a vector do not depend on a given set of coordinate axes with different orientation. However, components of a vector depend on the orientation of the axes.

Given:

Alpha and Beta can finish the task in 18 days; Beta and Gama can do it in 24 days while Gama and Alpha can finish it in 36 days.

Calculation:

Time taken by (Alpha + Beta) to finish the work = 18 days

So, (Alpha + Beta)’s 1 day work = 1/18     ----(1)

Time taken by (Beta + Gama) to finish the task = 24 days

So, (Beta = Gama)’s 1 day work = 1/24     ----(2)

Time taken by (Gama + Alpha) to finish the work = 36 days

So, (Gama + Beta)’s 1 day work = 1/36      ----(3)

On adding eq. (1), (2) and (3), we have

2(Alpha + Beta + Gama)’s 1 day work = 1/18 + 1/24 + 1/36 = (4 + 3 + 2)/72 = 9/72 = 1/8

Also, (Alpha + Beta + Gama)’s 1 day work = 1/(2 × 8) = 1/16

So, Alpha, Beta, Gama together can make a sports car in 16 days.

Now, Alpha’s 1 day’s work = {(Alpha + Beta + Gama)’s 1 day’s work – (Beta + Gama)’s 1 day’s work} = 1/16 – 1/24 = (3 - 2)/48 = 1/48

Hence, Alpha alone can make the car in 48 days.

Beta 1 day’s work = {(Alpha + Beta + Gama)’s 1 day’s work – (Gama + Alpha)’s 1 day’s work}

= 1/16 – 1/36 = (9 - 4)/144 = 5/144

Hence, Beta alone can make the sports car in 144/5 days.

Gama’s 1 day’s work = {(Alpha + Beta + Gama)’s 1 day’s work – (Alpha + Beta)’s 1 day’s work}

= 1/16 – 1/18 = (9 - 8)/144 = 1/144

Hence, Gama alone can make the sports car in 144 days.

∴ Alpha, Beta and Gama alone can finish the task in 48 days, 144/5 days and 144 days respectively.

Correct answer is Chaugan

Given:

1 man + 1 woman + 1 boy can finish work in 5 days

1 woman can finish work in 10 days

1 boy can finish work in 16 days

Calculation:

Let the time taken by 1 man be ‘x’ days.

⇒ (1/x) + (1/10) + (1/16) = 1/5

⇒ 1/x = (1/5) – (1/10) – (1/16)

⇒ 1/x = 3/80

⇒ x = 80/3

Given:

Number of days taken to finish a piece of work together = 60 days

Calculation:

Let the number of days taken by Mamta be x

Efficiency of Mamta = 1/x units

Number of days taken by Geeta x + (100/100) × x = 2x

⇒ Efficiency of Geeta = 1/2x

According to the question,

Mamta and Geeta together can finish the work in 60 days

Now,

\( \Rightarrow \;\frac{1}{x}\; + \;\frac{1}{{2x}} = \;\frac{1}{{60}}\)

\( \Rightarrow \;\frac{{2 + 1}}{{2x}} = \;\frac{1}{{60}}\)

\( \Rightarrow \;\frac{3}{{2x}} = \;\frac{1}{{60}}\)

⇒ x = 90

∴ Mamta will complete her work alone in 90 days.

GIVEN:

A toffee costs Rs. 1 however, a discount of 4% is allowed on all the toffees purchased after 1000 toffees.

CONCEPT:
Basic concept of discount

FORMULA USED:

Selling price = Marked price × [1 – (Discount/100)]

CALCULATION:

Total cost = (1 × 1000) + (1 – 1 × 0.04) × 3000

⇒ Total cost = 1000 + 2880 = Rs. 3880

Given:

A can do a piece of work = 24 days

B can do a piece of work = 12 days

Concept used:

Total work = LCM

Formula used:

Efficiency = (Total work)/(Total time)

Both can do same in a day = total efficiency ÷ total work

Calculations:

Total work = LCM = 24

Total work completed in one day = 1 + 2 = 3

Work done by A and B together in one day = 3/24 = (1/8)^th of total work

∴ A and B can do (1/8)^th of total work in one day together

Given:

A can finish a task in 45 days

B can finish the same task in 30 days

(A + B) work together for 3 days.

Concept used:

If a person does work in 'n' days, then one day work will be (1/n) part of the total work.

Calculation:

A's one day work = 1/45 part of the total work

B's one day work = 1/30 part of the total work

Worked by both for one day's work = (1/45) + (1/30)

⇒ (2 + 3)/90

⇒ 5/90

⇒ 1/18 part of the total work.

Now Worked by them for 3 days = (1/18) × 3

⇒ 1/6 part of the total work

Now remaining work = 1 - (1/6)

⇒ (6 - 1)/6

⇒ 5/6 part of the total work

But B completes a work = 30 days

The remaining work (5/6) part will be completed by B = (30) × (5/6)

⇒ 5 × 5

⇒ 25 days.

∴ The remaining work, B will finish in 25 days.

Given:

A, B, and C takes 20 days, 15days, and 12 days respectively.

B and C worked for 3 days.

Formula Used:

Total work is the LCM of time taken

Efficiency = Total work/Time taken

Calculation:

Total units of work = LCM of 20, 15, and 12

⇒ Total units of work = 60 units

Efficiency = Total work/Time taken

Efficiency_A = 60units/20days

⇒ EfficiencyA = 3 units/day

Efficiency_B = 60units/15days

⇒ Efficiency_B = 4 units/day

Efficiency_C = 60units/12days

⇒ Efficiency_C= 5 units/day

B and C worked for 3 days.

Total units of work completed by them,

⇒ Units of work completed = (B + C) × 3 days

⇒ Units of work completed = (4 + 5) × 3

⇒ Units of work completed = 9 × 3

⇒ Units of work completed = 27 units

Remaining units of work = 60 – 27

⇒ Remaining units of work = 33 units

33 units of work completed by A in the starting days.

⇒ Number of Days taken A = 33/3

⇒ Number of Days taken A = 11 days

∴ The number of days A worked for is 11 days. 

Given:

Cost Price (CP) of the first variety of tea = Rs.18 per kg

Cost Price of the second variety of tea = Rs.20 per kg

Ratio of adding the two varieties = 5 ∶ 3

Selling Price (SP) of the mixture of two varieties = Rs.21 per kg

Concept Used:

Profit% = [(SP – CP)/CP] × 100

Calculation:

Let the quantity of first variety added be 5x kg, and

The quantity of second variety added be 3x kg

So, the total Cost Price = (18 × 5x) + (20 × 3x) = Rs.150x

Also, the total quantity of the sold mixture = 5x + 3x = 8x kg

Hence, the total Selling Price = 21 × 8x = Rs.168x 

So, the overall Profit% = [(168x – 150x)/150x] × 100 = 12%

∴ The overall Profit Percentage is 12%

Given :

The ratio of x and y in alloy A = 5 : 2

The ratio of x and y in alloy B = 3 : 4

The ratio of A and B in alloy C = 4 : 5

To find :

Percentage of metal x in alloy C

Calculation :

Let the Quantity of metal x in alloy C be x

⇒ Quantity of metal x in alloy A = 5/7

⇒ Quantity of metal y in alloy A = 2/7

⇒ Quantity of metal x in alloy B = 3/7

⇒ Quantity of metal y in alloy B = 4/7

A.T.Q.,

⇒ The ratio of x and y in alloy C \( = \;\frac{{\left( {\frac{5}{7}} \right) × 4 + \left( {\frac{3}{7}} \right) × 5}}{{\left( {\frac{2}{7}} \right) × 4 + \left( {\frac{4}{7}} \right) × 5}}\)

⇒ The ratio of x and y in alloy C \( = \;\frac{{\frac{{20}}{7} + \frac{{15}}{7}}}{{\frac{8}{7} + \frac{{20}}{7}\;}}\)

⇒ The ratio of x and y in alloy C = 35/28

⇒ Quantity of x in alloy C = 35/63

⇒ Quantity of x in alloy C = 5/9

⇒ Percentage of x in alloy C = (5/9) × 100

∴ The percentage of x in alloy C is \(55 \frac{5}{9}\)%.

Solution:

Let the speed of boat in still water be u km/hr and the speed of the stream is v km/hr.

Given:                            

u : v = 5 : 2

Formula Used:

Speed while travelling downstream = u + v

Speed while travelling upstream = u - v           

Calculations:

Let u = 5x and v = 2x

Speed while travelling downstream = 5x + 2x = 7x

Speed while travelling upstream = u-v = 5x - 2x = 3x

Time(upstream) = 42/3x and  Time(downstream) = 42/7x

It is given that it takes 4 hour more to travel 42 km upstream than to travel same distance downstream.

⇒ 42/3x = 42/7x +4

⇒ 14/x = 6/x + 4

⇒ 8/x = 4

⇒ x = 2

Speed in still water = u = 5x

⇒ 5 × 2 = 10 km/hr 

Given :

Length of train = L_T = 200 m

Length of bridge B = L_B = 400 m

Calculations:

Let the speed of train = S_T

And Length of bridge A = L_A

Formula used :

Speed of train = (length of train + length of bridge) /time taken

S_T  = (L_T + L_A )/15      ----(1)

Also , S_T = (L_T + L_B )/30      ----(2)

On equating equation (1) and (2) we get

⇒ (L_T + L_A )/15 = (L_T + L_B )/30

⇒ (200 + L_A)/15 = (200 + 400)/30

⇒ L_A = 100 m

Given:

Speed of the train = 60 km/h

Length of the train = 125 m

Time to cross a bridge = 30 sec 

Concept used:

Distance = Speed × Time

km/h × 5/18 = m/sec

Calculation:

Speed of the train = 60 km/h  

⇒ 60 × 5/18 = 50/3 m/sec

Let the length of a bridge be L.

(125 + L)/(50/3) = 30

⇒ (125 + L) = (30 × 50)/3

⇒ 125 + L = 500

⇒ L = 500 – 125

⇒ L = 375

∴ The length of bridge is 375 m.

Given:

Speed of the boat in still water = 13 km/h 

Speed of the stream = 4 km/h 

Formulas used:

Downstream (Speed) = Speed of the boat + Speed of the current 

Time = Distance/Speed 

Calculation:

Time = 68/(13 + 4) 

⇒ 68/17 = 4 hours 

∴ The required time = 4 hours 

Given:

A travels a distance double than that travelled by B.

The speed of B is double than the speed of A.

If B increases his speed by 5 km/hr, the ratio of time taken by A and B is 14 : 3.

Formula Used:

Distance = time x speed

Calculation:

Let the distance A travels be 2d and B travels d

Speed of A is x and speed of B be 2x

ATQ,

14/3 = 2d/x × (2x + 5)/d

⇒ x = 15

⇒ 2x = 30

Time taken by B = (270/30) hours = 9 hours

∴ B will take 9 hours to travel 270 km.

Given:

Speed = 48 km/hr

Time = 4 hour 20 min = 13/3 hours

Formula used:

Distance = Speed × Time

Calculation:

Distance between station A and B = 48 × (13/3) km = 208 km

∴ The distance between station A and B is 208 km

Given:

Three person are walking from A to B.

Their speed are in the ratio = 4 : 3 : 5.

Formula used:

Time = Distance/Speed 

Calculation:

Time = 1/4 : 1/3 : 1/5 

Let L C M of 4, 3, and 5 = 60

Time (1/4)× 60 : (1/3)× 60 : (1/5)× 60

∴ The time ratio will be 15 : 20 : 12.

As we know, overall average speed of the bus = (Total distance covered)/(Total time taken)

= (84 × 11/3 + 66 × 16/3 + 110)/(11/3 + 16/3 + 2)

= (28 × 11 + 22 × 16 + 110)/(27/3 + 2)

= 11 × (28 + 32 + 10)/11 = 70 km/h

Given

Smith travels at  speeds of 10 km/hr, 12 km/hr, 15 km/hr, 20 km/hr along sides of the square

Formula used

Average speed =\(\frac{{total\;distance\;}}{{total\;time}}\)

Calculation

Let assume that side of square be x

⇒ Total distance = perimeter of square

⇒ 4x

⇒ Total time = \(\frac{x}{{10}}+\frac{x}{{12}}+\;\frac{x}{{15}}+\frac{x}{{20}}\)

⇒ \(\frac{{18x}}{{60}}\)

⇒ Average speed = total distance / total time

⇒  \(\frac{{4x}}{{\frac{{18x}}{{60}}}}\)

⇒ \(\frac{{40}}{3}\;km/hr\)

⇒ \(13\frac{{\;\;1\;\;}}{3}\;km/hr\)

Formula Used :

Speed = Distance/Time

Calculations :

Initial distance = d₁ = speed₁ × time ₁

⇒ d₁= 60 × 2 = 120km

Now if the distance is doubled then speed will be = (2 × 120)/2 km/hr

⇒ 120 km/hr

Percentage increased = (120 – 60)/60 × 100 %

⇒ 100% 

∴ The percentage increase is 100% 

Let 3d be total distance of journey.

As we know, overall average speed of the car = (Total distance covered)/(Total time taken)

= 3d/(d/78 + d/84 + d/90) = 3/{(14 × 15 + 13 × 15 + 14 × 13)/( 6 × 13 × 14 × 15)}

= 3 × 16780/(210 + 195 + 182) = 3 × 16780/587 = 3 × 27.9 = 83.7 km/h

Given:

Speed of Saurabh = 10 km/h

Speed of Shivam = 2 km/h

Concept used:

Speed = Distance/time

Distance = Speed × time

Calculations:

Time taken by Saurabh to cover the distance = Distance/speed

⇒ 50/10 = 5

The speed of Shivam = 2 km/h

Distance travel by Shivam in one hour = 2 km

Remaining distance = 50 – 2 = 48 km

Remaining hours for Shivam to travels the remaining distance = 4 hours

Speed needs to travel 48 km in 4 hours = 48/4 = 12 km/h

Speed which Shivam should increase from second hour = 12 – 2 = 10 km/h

Given:

Total distance travelled = 78km

Time = 3 hours 30 minutes = 3 + 30/60 = 7/2 hours

Formula:

Speed = Distance/time

Calculation:

Let speed of Vaibhav and speed of Vignesh be a kmph and (91 - a) kmph respectively.

⇒ 78/a + 78/(91 - a) = 7/2

⇒ 7098 - 78a + 78a = 7/2 × (91a - a²) 

⇒ a² - 91a + 2028 = 0

⇒ a2 - 52a - 39a + 2028 = 0

⇒ a(a - 52) - 39(a - 52) = 0

⇒ (a - 52)(a - 39) = 0

a = 52 or a = 39

Vaibhav's speed is more than Vignesh, so a = 52kmph

∴ Speed of Vaibhav is 52kmph.

Correct answer is Rajamundhry

Explanation:

Statement Either P marries Q or X marries Y
The given statement explains any of the actions or both the actions follow.

Option 1 follows the statement because one action follows (X marries Y).
Option 2 follows the statement because one action follows (P marries Q).
Option 3 follows the statement because both actions follow (P marries Q and X marries Y).
Option 4 does not follow the statement because any of the actions follow (Neither P marries Q nor X marries Y).

Let the experiment is throwing a fair die.

 And the event E is getting a number greater than 2 on the upper face of die.

Total number of outcomes = n(S) = 6. 

Numbers on die which are greater than 2 are 3, 4, 5 and 6 . 

Therefore, total number of outcomes favourable to event E is n(E) = 4.

Now, probability of getting a number greater than 2 = \(\frac{Total \,outcomes\, which\, faurable\, to \, event\, E}{totatl number\, of\, outcomes}\)

\(\frac{n(E)}{n(S)} = \frac{4}{6} = \frac{2}{3} = 0.67\)

Hence, the probability of throwing a number greater than 2 with a fair die is 0.67.

(A) Meandering

Meandering (as adjective) = proceeding in a convoluted or undirected fashion

The correct answer is Moon Express.

• On July 20, 2016, Moon Express became the first company to receive U.S. government approval to send a robotic spacecraft beyond traditional Earth orbit and to the Moon.
• This was the first time in history that any government signatory to the Outer Space Treaty exercised its rights and obligations to formally authorize and supervise a commercial entity to fly a mission beyond Earth orbit.
• This historic ruling is a breakthrough U.S. policy decision supporting our commercial lunar exploration and discovery and heralding a new era of expanding space enterprise.

• Moon Express is an American privately held early-stage company formed by a group of Silicon Valley and space entrepreneurs.
  • It had the goal of winning the $30 million Google Lunar X Prize, and of ultimately mining the Moon for natural resources of economic value. 
• EXPEDITION ONE: LUNAR SCOUT
  • The Lunar Scout expedition will be the first commercial voyage to the Moon.
  • This historic expedition will demonstrate the cost-effectiveness of entrepreneurial approaches to space exploration, carrying a diverse manifest of payloads including the International Lunar Observatory, “MoonLight” by the INFN National Laboratories of Frascati and the University of Maryland, and a Celestis memorial flight.

Correct option (C) was allowed to

Given:

A boat goes 5 km upstream and 7.5 km downstream in 45 minutes

Also, it goes 5 km downstream and 2.5 km upstream in 25 minutes

Concept used:

Upstream speed = Speed of the boat in still water - Speed of flow of the stream

Downstream speed = Speed of the boat in still water + Speed of flow of the stream

Calculation:

Let the Downstream and Upstream speed be 'd' km/hr and 'u' km/hr respectively

 As per the question,

5/u + 7.5/d = 45/60     -----eq^-n(1)

2.5/u + 5/d = 25/60     -----eq^-n(2)

Multiplying eq^-n(2) by 2 and then subtracting eq^-n(1) from (2), we get

\({2.5 \over d}\) = \({5 \over 60}\)

⇒ d = 30

By putting the value of 'd' in any of the eq^-n, we get the value of 'u'

⇒ u = 10

Now

Time is taken to travel 6 km upstream

= 6/10 hour or 36 minutes

∴ The time required to travel 6 km upstream is 36 minutes

Given:

A boat covers a certain distance downstream in 1/2 hour, while it comes back in 3/2 hours.

The speed of the stream is 5 km/hr.

Concept used:

Speed = \(\frac{{Distan ce}}{{Time}}\)

Calculation:

Let the speed of the boat be x km/hr

Let the speed of the current be y km/hr

\(\frac{{x + 5}}{{x - 5}} = 3\)

x + 5 = 3x - 15

2x = 20

x = 10 km/hr

GIVEN:

 Speed of boat in still water : Speed of current = 5x : 1x

Distance to be travelled = 300 km.

FORMULA USED:

Speed = Distance covered/Time taken

CALCULATION:

300/6x = T

300/4x = T + 1

Solving,

x = 25

Speed of boat in still water = 5x = 125 kmph

GIVEN:

Three statements.

CALCULATION:

We know that:

For all-natural values of ‘a’ and ‘n’ :

When aⁿ is divided by (a – 1), then remainder will always be 1.

When aⁿ is divided by (a + 1), then remainder-

• When ‘n’ is even, then remainder will always be 1.
• When ‘n’ is odd, the remainder will always be ‘a’.

A:

Remainder when 34⁶ is divided by (34 – 1) = 1

B:

Remainder when 34⁸ is divided by (34 + 1) = 1

C:

Remainder when 34⁵ is divided by (34 + 1) = 34

Hence, only A is TRUE. 

Given:

Distance = 48 km

Downstream Time = 20 h

Upstream Time = 4 hrs more than Downstream

Formula Used:

Speed of boat in still water = ½ (Speed Downstream + Speed of Upstream)

Calculations:

Speed of Downstream = 48/20 = 2.4 km/h

Speed of Upstream = 48/24 = 2 km/h

∴ Speed of boat in still water = ½ (Speed Downstream + Speed of Upstream)

⇒ (1/2) × (2.4 + 2)

⇒ (1/2) × 4.4

⇒ 2.2 km/h

The speed of the boat is 2.2 km/h.

Calculations:

Let the speed of boat in still water be s m/sec

As the object covers 100m in 50sec, so the distance traveled by the object will be with the help of the speed of the stream only.

∴ Speed of stream = 100/50

⇒ 2 m/s

Given that,

⇒ (s – 2) = (s + 2) – 50 (s + 2)/100

⇒ (s – 2) = (s + 2)(1 – 1/2)

⇒ s = 6 m/sec

⇒ s = (6 × 18)/5

⇒ s = 108/5 km/h

∴ Distance covered by the boat in still water in 5 hours

⇒ 108 × 5/5

⇒ 108 km

The distance the boat can cover in still water in 5 hours is 108 km.

Given:

Time take by boat to cover 154 km = 7 hours

Speed of stream = 4 km/h

Concept used:

Upstream speed = Downstream speed – (2 × speed of stream)

Calculation:

Downstream speed of the boat = 154/7

⇒ 22 km/h

Upstream speed of boat = 22 – (2 × 4)

⇒ 22 – 8

⇒ 14 km/h

Time is taken by boat to cover the same distance upstream = 154/14

⇒ 11 hours

Given:

Downstream distance = 25 km

Time taken = 5 hours

Speed of Boat : Speed of current = 3 : 2

Formula Used:

Speed = Distance/Time

D = B + S

U = B - S

where, D → Downstream speed, U → Upstream speed, B → Speed of boat in still water, S → Speed of stream.

Calculations:

Let the speed of boat and speed of stream be 3x and 2x respectively.

Speed = Distance/Time

⇒ D = 25/5

⇒ D = 5 km/hr

Now, D = B + S

⇒ 5 = 3x + 2x

⇒ 5 = 5x

⇒ x = 1

B = 3x = 3 × 1

⇒ B = 3 km/hr

Similarly, 

S = 2x = 2 × 1

⇒ S = 2 km/hr

Now, U = B - S

⇒ U = 3 - 2 = 1 km/hr

Speed = Distance/Time

⇒ 1 = 25/Time

⇒ Time = 25 hours

∴ The time taken by ship to cover 25 km upstream is 25 hours.

CONCEPT:

Dividend = (Quotient × Divisor) + Remainder

CALCULATION:

Suppose divisor is k, then, x = 23k + 8

According to the question –

⇒ (4x + 15) = [4(23k + 8) + 15]

⇒ 92k + 32 + 15

⇒ 92k + 47

When (92k + 47) is divided by 23, then, 92k is completely divisible by 23 and leaves 1 as a remainder when 47 is divided by 23.

⇒ Rem[(92k + 47)/23] = Rem(47/23) = 1

Given:

The ratio of the speed of the boat in still water to the speed of the stream = 16 : 5

Distance upstream = 16.5 km

Time upstream = 45 minutes

Distance = 17.5 km

Calculations:

Let the speed of the boat in still water be 16x.

Speed of Stream be 5x.

Upstream speed = 16x – 5x = 11x

Speed = Distance/Time

⇒ 11x = 16.5/45 × 60

⇒ 11x = 22

⇒ x = 2

Speed of the boat in still water = 32 km/h

Speed of stream = 10 km/h

Downstream speed = 32 + 10 = 42 km/h

Distance = 17.5

Time = 17.5/42 × 60

⇒ 5/12 × 60

⇒ 25 minutes

The time taken by the boat to cover the distance of 17.5 km downstream is 25 minutes.