Explore topic-wise InterviewSolutions in Current Affairs.

(C) Telomerase

(A) RNA → DNA

Synthesis of DNA

(C) Synthesis of DNA

(C) It is not denatured at the melting temperature of DNA

Correct option (B) NAD⁺ 

(A) Relaxation of negatively supercoiled DNA

DNA ligase is  links the small pieces of DNA.

Correct Answer is: (c) Median 

Given: 

Weight of per parcel = 10.5 kg 

Formula Used: 

Total weight =  no. of quantity × weight in per quantity

Calculation: 

Total weight of 15 parcels = 15 × 10.5 = 157.5 kg 

∴ The total weight of 15 parcels is 157.5 kg

Concept used:

In statistics, a frequency distribution is a list, table or graph that displays the frequency of various outcomes in a sample. Each entry in the table contains the frequency or count of the occurrences of values within a particular group or interval

Calculation:

The number of students obtained less than 20 marks = 2 i.e (16, 18)

The number of students who obtained marks between 20 and 40 = 6 i.e (25, 30, 34, 21, 28, 33)

The number of students who obtained marks between 40 and 60 = 7 i.e (45, 56, 53,57, 47, 54, 45)

The number of students who obtained marks between 60 and 80 = 5 i.e (75, 65, 76, 78, 72)

The number of students who obtained marks between 80 and 100 = 5 i.e (87, 89, 82, 93, 91)

∴ The correct answer is

Given:

The length of the train is 450 meter

Concept:

Speed = Distance/Time

Calculation:

Let be assume length of the tunnel is p

⇒ 1 minute 12 seconds =(450 + p)/{90 × (5/18)}

⇒ 72 seconds =  = (450 + p)/25

⇒ p = 1350 m

∴ The required result will be 1350 m.

Given:
Speed of the train = 72 km/h
Length of the train = 240 m
Time to cross a tunnel = 1 minute

Concept used:
Distance = Speed × Time
km/h × 5/18 = m/sec
1 minute = 60 sec

Calculation:
Speed of the train = 72 km/h  
⇒ 72 × 5/18 = 20 m/sec
Time to cross the tunnel = 60 sec
Let the length of the tunnel be L.
(240 + L)/20 = 60
⇒ (240 + L) = 60 × 20
⇒ 240 + L = 1200
⇒ L = 1200 – 240
⇒ L = 960
∴ The length of the tunnel is 960 m.

Given :

length of train 1 = 200 m

Speed of train 1 = 90 km/hr = 25 m/s

Speed of train 2 = 108 km/hr = 30 m/s

Calculations :

 Let the length of tunnel be x m and length of the second train be L_T

According to question first train takes 24 sec to pass the tunnel so,

⇒ (200 + x)/25 = 24

⇒ x  = 400 m

∵  it is given that train 2 passes the tunnel in 30 s

⇒ ( L_T + x )/30 = 30

⇒ (L_T + 400)/30 = 30

⇒ L_T = 500 m

∴ The length of the second train is 500 m.

Given:

Length of train = 700 m

Speed of train = 72 km/hr

Train crossed the tunnel in 1 minute

Concept used:

To cross the tunnel train needs to travel = Length of train + length of tunnel

Speed × Time = Distance

Calculation:

Length of train = 700 m

Let the length of tunnel be x

Time taken to cross the tunnel = 1 minute = 60 sec.

Speed = 72 kmph = (72 × 5)/18 = 20 m/sec.

To cross the tunnel train needs to travel = Length of train + length of tunnel = 700 + x

Now,

Speed × Time = Distance

20 × 60 = 700 + x

⇒ x = 1200 – 700 = 500

∴ Length of tunnel = 500 m.

Given:

When 'p' is subtracted from each of 18, 15, 23 and 16, then the numbers so obtained in this order are in proportion.

Concept used:

Ratio and proportion

Calculation:

As per the question,

⇒ \(\frac{{18 - p}}{{15 - p}} = \frac{{23 - p}}{{16 - p}}\)

⇒ (18 - p) (16 - p) = (15 - p) ( 23 - p)

⇒ 288 - 34p + p² = 345 - 38p + p²

⇒ 4p = 57

⇒ p = 57/4

We need to find \(\left( {p - \frac{1}{4}} \right)\)

Given:

Length of train = 800 m

Speed of train = 78 km/h

Time to cross the tunnel = 1 minute = 60 seconds

Concept used:

1 km/h = 5/18 m/s

Formula used:

Speed = (Length of train + Length of tunnel)/Time

Calculations:

Speed = (Length of train + Length of tunnel)/Time

⇒ 78 × (5/18) = (800 + x)/60

⇒ 65/3 = (800 + x)/60

⇒ 65 × 20 = 800 + x

⇒ 1300 = 800 + x

⇒ x = 500 m

∴ The length of the tunnel is 500 m.

Given:

Speed of train = 132 kmph

Length of train = 110 meters

Length of platform = 165 meters

Formula used:

Speed = Distance/Time

Calculation:

To cross railway platform train has to travel distance equal to the sum of length of train and length of platform

Total distance = 165 + 110 = 275 meters

Speed in m/sec

⇒ 132 × (5/18) m/sec

Time = Distance/Speed

⇒ 275/(660/18)

⇒ (275 × 18)/660

⇒ 7.5 seconds

∴ Time taken by train to cross the platform is 7.5 seconds.

Given:

Time taken by the tap-1 to fill the tank = 2 hours

Time taken by the tap-2 to empty the tank = 3 hours

Calculation:

Let the total capacity of tank be 6 units.

Part of tank filled in 1 hour by the tap-1 = 6/2

⇒ 3 units

Part of tank emptied in 1 hour by tap-2 = 6/3

⇒ 2 units

Part of tank filled by both the taps together in 1 hour = 3 – 2

⇒ 1 unit

Time taken to fill the full tank = 6/1 = 6 hours

∴ In 6 hours will the tank be full when both the taps are open.

We have subtracted the 1 hour work of tap-1 and tap-2 while considering their together work because tap-2 is emptying the tank.

Given:

12 men complete one-fourth of a piece of work in 10 days, working 8 hours a day.

Concept used:

M₁ × D₁ × H₁/W1 = M₂ × D₂ × H₂/W₂ 

Where,

M → Number of men,

D → Number of days,

H → Number of hours,

W → Work

Calculation:

M1 × D1 × H1/W1 = M2 × D2 × H2/W2 

⇒ 12 × 10 × 8/(1/4) = 15 × D₂ × 6/1

⇒ D₂ = (12 × 10 × 8 × 4)/(15 × 6)

⇒ D₂ = 128/3 days

∴ 15 men would complete full piece of work in 128/3 days when they work 6 hours a day.  

Given:

Brazen can paint a wall in 12 days

Ballary can paint the wall in 21 days

Formula used:

If a person can complete the work in A days, then the person can complete 1/A work in 1 day.

Calculation:

Work completed by Brazen in 1 day = 1/12

Work completed by Ballary in 1 day = 1/21

Total work completed by them in 1 day = 1/12 + 1/21

⇒ Total work completed by them in 2 days = (7 + 4)/84

⇒ Total work completed by them in 2 days = 11/84

Total work completed by them in 7 × 2 days = 14 days = 77/84 

⇒ Remaining work = 1 – 77/84 = 7/84 = 1/12

Brazen can do 1/12 work in 1 day

Total time taken = 14 + 1 = 15 days

∴ Total time taken is 15 days

Given :

Mrinalini and Samyukta can paint half of the wall in 14 days

Mrinalini need 4 times more time than Samyukta 

Concept used :

Working efficiency is inversely proportional to time taken 

Total work = person's efficiency × Number of days he takes for work 

Calculations :

Half the work will be completed in 14 days 

Full work will be completed in 28 days 

Total work = 28 units 

Now,

Let the efficiency of Mrinalini be 'x' unit per day 

So the efficiency of Samyukta will be '4x'  per day 

Mrinalini and Samyukta per day work = x + 4x = 5x units 

So,

(5x) × 28 = 28 units 

⇒ 5x = 28 /28

⇒ x = 1/5 units 

Samyukta's per day work = 4x 

⇒ 4 × (1/5) = 4/5 units 

Total days of working of Samyukta = 28/(4/5) 

⇒ 35 days 

∴ Samyukta will complete the work in 35 days 

Given:

X and Y finish work together in 26 days.

X leaves after 10 days.

Y and Z finish the remaining work in 16 days.

The ratio of efficiency of X and Y = 3 ∶ 1

Concept used:

Work = Time × Efficiency

Calculation:

Total work done by X and Y = (1x + 3x) × 26 = 104x

Let the efficiency of Z be ‘z’.

Total work = (1x + 3x) × 10 + (1x + z) × 16

⇒ 104x = 40x + (1x + z) × 16

⇒ 1x + z = 4x

Or, efficiency of Z = 3x

Time is taken by Z to finish the work alone = 104x/3x = 34(2/3) days.

∴ Z will take 34(2/3) days to finish the work.

Given

Length of train = 150 m

Time taken to cross the pole = 12 sec

Formula used

Distance = speed × time

\(x \ m/sec = x \times {18 \over 5} km/hr\)

Calculation

Speed of train = distance/time

⇒ 150/12 = 25/2 m/sec

\(⇒ {25 \over 2} \ m/sec = {25 \over 2} \times {18 \over 5} km/hr\)

⇒ 45 km/hr

Given:

Harish is 50% more efficient than Rajesh.

Rajesh does 26% of work in 13 days.

Concept used:

Work done = Time × Efficiency

Calculation:

Number of days taken by Rajesh = 13 × 100/26 = 50 days

The ratio of efficiency of Harish to Rajesh = 3 ∶ 2

Time taken by both of them together to finish the work = 100/(3 +2) = 20 days

∴ Time taken by both to finish the work together is 20 days.

Given:

A is 50% more efficient than B.

Time is taken by A to finish the work = time taken by B – 15

Concept used:

Work = time × efficiency

When work done by A and B is the same, t ∝ 1/e; where ‘t’ is time and ‘e’ is efficiency.

Calculation:

The ratio of efficiency of A to B = 3 ∶ 2

The ratio of time of A to B = 2 ∶ 3

Then, time is taken by A to finish the task = time taken by B – 15

⇒ 2x = 3x – 15

⇒ x = 15

Time taken by A to finish work = 2x = 30 hrs

Then, time is taken by A to do double the work = 2 × 30 = 60 hours.

∴ Time is taken by A to finish double the work is 60 hours.

Given:

A and B alone can complete work in 9 days and 18 days respectively.

Calculations:

Let the work be completed in x days.

According to the question,

A worked for (x –3) days, while B worked for x days.

so (x - 3)/9 + x/18 = 1

⇒ (2x - 6 + x)/18 = 1

⇒ 3x - 6 = 18

⇒ 3x = 24

⇒ x = 8

∴ the correct answer is 8 days.

Let the present age of Prajakta be x

Then, the present age of her husband = x + 5

Now, According to question

The ratio between the age of Prajakta two year later and his husband one year later is 8 ∶ 9

⇒ (x + 2)/(x + 6) = 8/9

⇒ 9x + 18 = 8x + 48

⇒ x = 30

∴ Present age of Prajakta = 30 yrs

Present age of Prajakta's husband = 35 yrs

The ratio between the present age of younger son and his father is 1 ∶ 5

Let the age of the younger son and his father be x and 5x respectively

⇒ 5x = 35

⇒ x = 7

∴ Present age of younger son = 7 yrs

Now, the ratio between the present age of the elder son and the age of the younger son 1 year later is 2 ∶ 1

⇒ S/8 = 2/1

⇒ S = 16 

S → Present age of elder son

∴ Present age of elder son = 16 yrs

Age of Prajakta's father in law = 2 × 35 = 70 yrs

∴ Average age of the family = (30 + 35 + 7 + 16 + 70)/5 = 31.6 yrs

Given:

A piece of work is completed by 3 men and 4 boys in 7 days.

The same piece of work is completed by 5 men and 3 boys in 6 days

Formula used:

Work = Time × Efficiency

Calculation:

Let the efficiency of a man be M

And the efficiency of a boys be B

Now, using the first  given condition

⇒ Total work = (3M + 4B) × 7

⇒ Total work = 21M + 28B      ----(1)

Use the second given condition

⇒ Total work = (5M + 3W) × 6

⇒ Total work = 30M + 18W      ----(2)

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

⇒ 21M + 28B = 30M + 18W

⇒ 30M - 21M = 28B - 18W

⇒ 9M = 10B

⇒ M/B = 10/9

So, the ratio of efficiency of man and boy is 10 : 9

⇒ Time work = 21 × 10 + 28 × 9

⇒ Time work = 462

Four times of the total work is

⇒ 462 × 4 = 1848

The total efficiency of 1 man and 2 boys

⇒ 1 × 10 + 2 × 9

⇒ 10 + 18 = 28

⇒ Time = 1848/28 = 66 days

∴ 1 man and 2 boys will complete the four times of the same work in 66 days.

Given:

Time by which 2 men and 3 boys finish the work = 10 days

Time by which 3 men and 2 boys finish the work = 8 days

Concept used:

Work = time × efficiency

Calculation:

Let the efficiency of one man be 'm' and one boy be 'b'.

Then, (2m + 3b) × 10  = total work

Also, (3m + 2b) × 8 = total work

Equating the total work in both the cases, 

(2m + 3b) × 10  = (3m + 2b) × 8

⇒ 20m + 30b = 24m + 16b

⇒ 14b = 4m

⇒ 7b = 2m

Number of days in which (2m + b) can complete the work be D.

(2m + b) × D = (2m + 3b) × 10 = total work

Substituting 2m = 7b in the above equation,

⇒ 8b × D = 10b × 10

⇒ D = 100/8

⇒ D = 12 ½ days.

∴ The time taken to complete is 12 ½ days.

Given:

Let the present age of Rahul and mother be a years and b years respectively.

⇒ b = 3a

Then,

⇒ (a + 5) + (b + 5) = 70

⇒ a + b = 60

Solving,

a = 15 years and b = 45 years

∴ Required difference = 45 - 15 = 30 years.

Given:

12 men or 24 boys can do a work in 30 days.

Calculations:

Let the amount of work done by 1 man in 1 day be M.

and the amount of work done by 1 boy in 1 day be B.

according to the question,

12 × M × 30 = 24 × B × 30

so M ∶ B = 2 ∶ 1

so total amount of work = 24 × 1 × 30 = 24 × 30

now work done by 15 men and 6 boys in 1 day = 15 × 2 + 6 × 1 = 36

so the number of days taken by 15 men and 6 boys to complete the work = (24 × 30)/36 = 20 days 

∴ the correct answer is 20 days.

Given:

Three men can complete a piece of work in 6 days.

Formula used:

\(\frac{M_{1}\ × \ D_{1}}{W_{1}}\ =\ \frac{M_{2}\ × \ D_{2}}{W_{2}}\)

where M₁ = M₂ = number of man

D1  = D2 = number of days

W1 = W2 = part of work

Calculations: 

time taken = 6 days

part of work done in 1 day = 1/6

part work done in 2 days = (1/6) × 2 = 1/3

remaining work = 1 - 1/3 = 2/3

so ​​using above formula we get ​

\(\frac{3\ \times\ 2}{\frac{1}{3}}\ =\ \frac{6\ \times\ D_{2}}{\frac{2}{3}}\)

so D₂ = 2 days.

∴ the correct answer is 2 days.

As a society, we are yet to appreciate how crucial the work of the primary school teacher is, for society as a whole. Teaching at the primary level requires a specific kind of expertise which includes both knowledge and skills.

Following are some key actions required for making mathematics joyful for class while teaching Numbers—

• Participation
• Engagement
• Observations
• Making a hypothesis and verifying them
• Problem posing
• Experience with counting
• Problem-solving
• Visualization and representation
• Making connections
• Systematic reasoning
• Mathematical communication

Hence, we can conclude that to develop all the above characteristics a student is required to practice mathematics more and more. And also Numerals should be introduced only after the learners have experience with counting.

The pedagogy at the primary stage includes the following sequence of activities while dealing with different concepts, competencies, and skill—

• Children learn to enjoy mathematics rather than fear it.
• Children learn important mathematics—Mathematics is more than formula and mechanical procedures.
• Children see mathematics as something to talk about, to communicate through, to discuss among themselves, to work together on.
• Children pose and solve meaningful problems.
• Children use abstractions to perceive relationships, to see structures, to reason out things, to argue the truth or falsity of statements.
• Children understand the basic structure of Mathematics— Arithmetic, algebra, geometry, and trigonometry, the basic content areas of school Mathematics, all offer a methodology for abstraction, structuration, and generalization.
• Teachers engage every child in the class with the conviction that everyone can learn mathematics.

Analytic Method:

• It involves breaking up the conclusion into simpler steps and setting up relationships with what is given or known. It applies intuition and inductive reasoning. 
• It is a method of discovery. The solution or proof is arrived at through systematic reasoning.
• It proceeds from the conclusion to the hypothesis. 

If a teacher in Class VII asks her students to draw circles of different radii. She then asks them to calculate the ratio of circumference to diameter. The students find that this ratio is almost the same in the case of each circle (inductive reasoning and method of discovery).

Thus from the above-mentioned points, it is clear that the teacher is using the analytical approach of problem-solving.

• Inductive approach- 
  • Proceeds from the particular to the general; from the concrete to the abstract.
  • It takes care of the needs and interests of children. It is a developmental process.
• Deductive approach-
  • Proceeds from the general to the particular; from the abstract to the concrete.
  • The child is provided with information of facts, principles & theories. 
• Project method-
  • It is based on learning by doing.
  • Children learn through association, co-operation, activity etc. 

Given: 

A and B together can complete a work in 10 days

B working alone can complete the work in 15 days

Calculation:

A + B together can complete a work in a day = 1/10

B can complete a work in a day = 1/15

A can alone complete work in a day = 1/10 - 1/15 = 1/30

A can complete the total work in 30 days

∴ A can complete the total work in 30 days.

Given:

Speed of Motorboat in still water (x) = 18 km/h

Speed of River (y) = 12 km/h

Formula Used:

Speed Downstream = (x + y)

Calculations:

Speed of motorboat in still water (x) = 18 km/h

Speed of River (y) = 12 km/h

∴ Downstream Speed of the motorboat = (x + y)

⇒ 18 + 12

⇒ 30 km/h

Given∶

The ratio of average speed of onward to the the average speed of return journey is 7 ∶ 5.

The time taken in the whole journey of 490 km is 24 hours.

Formula used∶

Average speed = 2 x × y / (x + y)

Calculations∶

Let the average speed of the onward journey be x

Let the average speed of the return journey be y

Average speed = 2 x × y / (x + y)

According to the question∶

average speed of the onward journey 7x

average speed of the return journey 5x

Average speed = (2 × 7x × 5x) / (7x + 5x) = 490 / 24

⇒ x = 3.5

The average speed of the onward journey, 7x = 7 × 3.5

Given:

The difference between the speeds of A and B is 10 km/hr.

To travel a distance of 600 km, B takes 3 more hours than A.

Formula Used:

Distance = time x speed

Calculation:

Let speed of A be x km/hr and speed of B be (x – 10) km/hr

ATQ,

600/(x – 10) – 600/x = 3

⇒ x² – 10x – 2000 = 0

⇒ x = 50, - 40

But x cannot be - 40,

So, it will be 50

So, the speed of B = (50 – 10) km/hr = 40 km/hr

Time = (2400/40) hours = 60 hours

∴ B will take 60 hours.

Given:

Downstream distance = 24 km

Upstream distance = 24 km

Formula Used:

Speed = Distance/Time

D = B + S

U = B - S

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

Calculations:

24/D + 24/U = 14

⇒ 24/(B + S) + 24/(B - S) = 14

⇒ 24/(7 + S) + 24/(7 - S) = 14

⇒ [24 × (7 - S) + 24 × (7 + S)]/(7 + S) × (7 - S) = 14

⇒ 24 × 7 × 2/(49 - S²) = 14

⇒ 24 = 49 - S²

⇒ S² = 49 - 24 = 25

⇒ S = 5 km/hr

U = (B - S) 

⇒ U = (7 - 5) 

⇒ U = 2 km/hr

∴ The upstream speed is 2 km/hr.

Given:

Speed of River = 2 kmph

Distance upstream = 10 km

Time taken = 55 min

Formula Used:

Distance = T (x² – y²)/ 2x

Calculations:

Let Speed of motorboat in still water be x

Speed of stream (y) = 2 kmph

Time (T) = 55 min = 55/60 h

According to the formula,

Distance = T (x² – y²)/ 2x

⇒ 10 = 55/60 × (x² – y²)/ 2x

⇒ 240x = 11(x² – 4)

⇒ 11x² – 240x – 44 = 0

⇒ (x - 22) (11x + 2) = 0

Ignoring the negative value, we get

⇒ x = 22 kmph

∴ The speed of the motorboat in still water is 22 kmph.

Given:

Speed downstream is 3 times speed upstream.

In still water boat goes 50 m in 5 seconds.

Concept:

Speed downstream is sum of the speed of boat in still water and speed of stream

Speed upstream is difference of the speed of boat in still water and speed of stream.

To convert hours to minutes we have to multiply by 60

Formula:

Speed = distance/time

Time = distance/speed

Speed in m/s × 18/5 = speed in km/h

Calculation:

Speed of boat in still water = 50/5 = 10 m/s

Let the speed of stream be x m/s

Speed during downstream = (10 + x) km/h

Speed during upstream = (10 – x) km/h

3 × (10 – x) = (10 + x)

⇒ (10 – 3x) = (10 + x)

⇒ 3x + x = 30 – 10

⇒ 4x = 20

⇒ x = 5

Speed downstream = 10 + 5 = 15 m/s

Speed downstream = 15 × (18/5) = 54 km/h

Time to cover 45 km = 45/54 = (5/6) hours

Time taken to cover 45 km = (5/6)× 60 = 50 mins

∴ Time to cover the distance downstream is 50 minutes

Given:

Speed of boat in still water = 9 km/h

Downstream speed = 6 km/h + Upstream speed

Formula used:

Distance = Speed × Time

Calculation:

Let the speed of the stream be x km/h.

Downstream speed = (9 + x) km/h

Upstream speed = (9 – x) km/h

According to the question,

Downstream speed = 6 km/h + Upstream speed

⇒ (9 + x) = 6 + (9 – x)

⇒ 2x = 6

⇒ x = 3 km/h

Downstream speed = (9 + x) km/h

⇒ Downstream speed = 9 + 3

⇒ Downstream speed = 12 km/h

Time = 5 hours 20 min.

⇒ Time = (5 + 1/3) hours

⇒ Time = 16/3 hours

Required distance = Downstream speed × Time

⇒ Required distance = 12 × 16/3

⇒ Required distance = 64 km

∴ The distance covered by boat in 5 hours 20 min. downstream is 64 km.

Given:

Speed of car A = 72 km/h 

Speed of car B = 90 km/h 

If car B takes 2 hours lesser than car A for the journey,

Formula used:

Time =Distance/Speed 

Calculations:

Let the distance between the two cities be x km.

According to the question,

(x/72) – (x/90) = 2

⇒ (5x – 4x)/360 = 2

⇒ x = 720 km

∴ The total distance between both the cities is 720 km

Given:

Speed of boat in still water = 6 kmph

Speed of stream = 4 kmph

U/S + D/S = 8 hours

Formula used:

Speed of boat in still water is a kmph and speed of stream is b kmph

Speed D/S = (a + b) kmph

Speed U/S = (a – b) kmph

Calculations:

Speed in still water = 6 kmph

Speed of stream = 4 kmph

⇒ D_S speed = 10 kmph

⇒ U_S speed = 2 kmph

Assume distance to be D km

⇒ (D / 10) + (D / 2) = 8

⇒ 6D/10 = 8

⇒ D = 40/3 kms

∴ Distance covered is 40/3 kms