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Correct answer is (d) Spine layout

a) Individual to district

Sales Performance can be evaluated on the basis of Individual to district.

I will include the following columns in the visitors’ register: 

A. Date 

B. Time in 

C. Time out 

D. Name of the visitor 

E. Address of the visitor 

F. Phone No of the visitor 

G. Purpose of the visit 

H. Name of the person whom the visitor wants to meet 

I. Entry pass No issued to visitor 

J. Visitor’s signature 

K. Security supervisor/ officer signature

Correct option: c) Display window

Buying performance may be evaluated on the basis of Net sales.

Correct option: d) Make payment

Cytoplasmic inclusions are cytoplasmic molecular aggregates, such as pigments, organic polymers and crystals. They are not considered cell organelles.

Fat droplets and glycogen granules are examples of cytoplasmic inclusions.

(d) Abscissic acid

Porin is an abundant polypeptide present in bacterial cell walls. It helps in the diffusion of solutes.

C.P of each article be Rs 1 

C.P of 21 articles = Rs.18 ,S.P of 18 articles = Rs 21.  

Gain% = [(3/18)*100]% = 50/3%

(B) are perpendicular to each other

Correct option:

(A) 0.3 

Given:

The investment ratio of A, B, and C = 2: 3: 5

Profit ratio = 3: 4: 2

Concept used:

Time of investment ∝ Profit/investment 

Calculation:

Let time periods investments be T1, T2, T3

The ratio of investments is A, B, and C = 2: 3:  5 

The profits in the ratio = 3: 4: 2

Time period T1, T2, T3 = p1/x1: p2/x2: p3/x3

⇒ T1,T2, and T3 = 3/2, 4/3, 2/5

⇒ T1 ∶ T2 ∶ T3 = 45 ∶ 40 ∶ 12 

∴ Ratio of investment is 45 ∶ 40 ∶ 12.

(SP of 33m)-(CP of 33m)=Gain=SP of 11m 

SP of 22m = CP of 33m 

Let CP of each metre be Re.1 , 

Then, CP of 22m= Rs.22,SP of 22m=Rs.33. 

Gain%=[(11/22)*100]%=50%

Formula used

Average speed = 2xy/(x + y) where x and y are speed

Calculation

Speed from Agartala to Udaipur = x = 40 km/hr

Speed from Udaipur to Agartala = y = 60 km/hr

Average speed = (2 × 40 × 60)/(40 + 60)

⇒  4800/100 = 48 km/hr

Given:

Speed of Rahul from A to B = 3 m/s

Speed of Rahul from B to A = 2 m/s

Speed of Sahil from A to B = 4 m/s

Speed of Sahil from B to A = 1 m/s

Distance between A and B = 1200 m

Formula used:

Speed = Distance/Time

Calculations:

Time taken by Rahul to complete whole journey = (1200/3) + (1200/2)

⇒ 400 + 600

⇒ 1000 sec

Time taken by Sahil to complete whole journey = (1200/4) + (1200/1)

⇒ 300 + 1200

⇒ 1500 sec

∴ The time taken by Sahil and Rahul respectively is 1500 seconds and 1000 seconds.

Let the original speed of Mridank and Mrehul be x km/hr 

They walked for 2 hrs when Mridank ijured his legs

∴ Distance travelled by both of them = Speed × Time

⇒ Distance = x × 2 = 2x km

∴ Remaining distance = 125 – 2x km

Now, speed of Mridank = x × (60/100) = 3x/5

∴ According to question

(125 – 2x)/(3x/5) – {(125 –2x)/x} = 2

⇒ (125 – 10x – 75 + 6x)/3x = 2

⇒ 50 – 4x = 6x

⇒ x = 5

∴ Speed of Mridank and Mrehul = 5 km/hr

∴ Distance at which Mridank injured his legs = 5 × 2 = 10 km

Correct answer is: (a) 3/4

Possible outcomes are (HH), (HT), (TH), (TT)

Favorable outcomes(at the most one head) are (HT), (TH), (TT)

So probability of getting at the most one head =3/4

Correct answer is: (b) 90ᵒ

2sin²β – cos²β = 2

Then 2 sin²β – (1- sin²β) = 2

3 sin²β =3 or sin²β =1

β is 90ᵒ

52 + 38 – 88 = 2

Correct option  (A) a decrease in the threshold voltage

Given:

A and B can do a work in 60 days and 75 days respectively

Both A and B work together for 10 days. 

Calculation:

Work done by A in 1 day = 1/60

Work done by B in 1 day = 1/75

Work done by both (A + B) in 1 day = (1/60 + 1/75)

⇒ (5 + 4)/300

⇒ 9/300 = 3/100

Now, Work done by both (A + B) in 10 days = [10 × (3/100)] = 3/10

Remaining work = [1 – (3/10)] = 7/10

Work done by A in 6 day = [6 × (1/60)] = 1/10

Now, Remaining work = (7/10 – 1/10) 

⇒ (7 – 1)/10 = (6/10) days

According to the question:

(A + C) × 6/10 = 24

⇒ (A + C) = (24 × 10)/6 = 40 days

Work done by both (A + C) in 1 days = 1/40

Work done by C in 1 day = (1/40 – 1/60) 

⇒ (3 – 2)/120 = 1/120

∴ Total time work done by c  in 120 days.

Given:

P can do a piece of work = 20 days

Q can do a piece of work = 40 days

P leaves that work 10 days before completion

Concept used:

Total work = LCM

Formula used:

Efficiency = Work/Time

Calculations:

Total work = LCM = 40

Let the work finishes in x days

So, Time in which P works = (x – 10)

Work done by P in (x – 10) days = 2 × (x – 10)

Work done by Q in x days = 1 × x = x

Now,

⇒ 2(x – 10) + x = 40

⇒ 2x – 20 + x = 40

⇒ 3x = 60

⇒ x = 20 days

∴ Time in which the total work has completed is 20 days

Given:

Ratio of the age of Seema and Reema = 2 : 3

Concept used:

The ratio a : b shows a/b 

Where,

Antecedent = a

Consequent = b

Calculation:

Let the present age of Seema be 2x

Let the present age of Reema be 3x

3x – 2x = 6

⇒ x = 6

⇒ Present age of Seema = 2 × 6 = 12

⇒ Present age of Reema = 3 × 6 = 18

After six years age of Seema = 12 + 6 = 18

After six years age of Reema = 18 + 6 = 24

⇒ The ratio of Seema and Reema after six years = 18 : 24 = 3 : 4

∴ The ratio of age of Seema and Reema after six years is 3 : 4. 

Given:

\(\frac{1}{{x\;}} + \frac{1}{9}\; = \frac{1}{{27}}\)

a2 + p2 = q2

Calculation:

In the 1st condition,

\(\frac{1}{{x\;}} + \frac{1}{9}\; = \frac{1}{{27}}\)

⇒ \(\;\frac{{9 + x}}{{9x}}\) = \(\frac{1}{{27}}\)

⇒ 9x = 27(9 + x)

⇒ x = 3(9 + x)

⇒ x = 27 + 3x 

⇒ -2x = 27 

⇒ x = -13.5

In the second statement, 'x' is not given.

⇒ We cannot find 'x' using the second statement.

Hence 'x' can find using 1st statement.

Given:

P and Q can complete work in 6 days.

Q and R can complete it in 12 days.

P and R took 15 days to complete the same piece of work. 

R works every fourth day.

Concepts used:

Efficiency = Work/Time

Time = Work/Efficiency

Calculation:

1-day work of P and Q = 1/6

1-day work of Q and R = 1/12

1-day work of P and R = 1/15

1 days work of P, Q and R = 1/2[(1/6) + (1/12) + (1/15)]

⇒ 19/120

Work done by P and Q in 3 days = 3/6 = 1/2

Work done by P, Q and R till fourth day = 1/2 + 19/120 = 79/120 

Remaining work = 1 – (79/120) = 41/120

Time = Work/Efficiency

Time taken by P and Q to complete the remaining work = (41/120)/(1/6) = 2.05 days

Total time taken by P, Q and R = 4 + 2.05 days = 6.05 days

∴ P, Q and R complete the work together when P and Q are assisted R on every fourth day in 6.05 days.

Given:

A can complete a piece of work in 10 days and B can complete it in 20 days.

A starts working and B works on every alternate day.

Before 5 days of the completion of work, B left the work.

Concepts used:

Efficiency = Work/Time

Calculation:

A can complete a piece of work in 10 days and B can complete it in 20 days.

Efficiency = Work/Time

1-day work of A = 1/10

1-day work of B = 1/20

1 day work of A and B = 1/10 + 1/20 = 3/20

A and B started working together but before 5 days of the completion of work, B left the work.

5 days work of A = 5/10 = 1/2

Remaining work = 1 – (1/2) = 1/2

A and B work together on every alternate day.

2 days work of A and B = 1-day work of A + 1-day work of A and B

⇒ 1/10 + 3/20

⇒ 5/20 = 1/4

1/4 work is done by A and B in 2 days.

⇒ 1 full piece of work is done by A and B in 8 days.

⇒ 1/2 work is done by A and B in  4 days.

Total time taken to complete work = 5 days + 4 days = 9 days.

∴ A and B complete piece of work in 9 days if B assists A on every alternate day and if B left work before 5 days of completion of work.

President of the function and Principal of the college, Prof. Dayananda Saraswathi, PTA Vice President Mrs Neena Gupta, special invitees, parents, staff and students.

Annual day is a day of joy. It is a day that records the happenings of one whole year. It is also a day that showcases the talents of the students. So, there is no second thought about the day being the most significant one. Matching the significance of the day in knowledge and experience Is our chief guest Mr Vedamurthy. True to his name, he is the very fountainhead of knowledge. Coupled with his knowledge is his fine flow of the English language as he is an M.A. in English. He is of KAS 2008 batch.

After having served as lecturer In Chitradurga, he now works as Assistant Commissioner of Food and Civil Supplies Department, Bellary. He has already endeared himself to the people of Bellary because of his amiable and helpful nature. His contributions to the field of education are tremendous. He has started a Guidance Centre which helps students avail of Government scholarships and Government-funded research projects. Today he will address us on the topic ‘All that you have is now. Let’s hear it for Mr Vedamurthy, our esteemed chief guest.

GIVEN:

Three statements.

CONCEPT:

Mensuration

FORMULA USED:

Area of circle = πR²

Area of square = a²

CALCULATION:

A:

Diameter of circle = Side of square = 14 cm

Radius of circle = 14 / 2 = 7 cm

Maximum possible area of a circle = (22 / 7) × 7²

= 154 cm²

B:

Area of a path = Area of park with path - Area of park without path

= (7 + 2 + 2)² - 7²

= 121 - 49

= 72 cm²

C:

Maximum number of square tickets = Area of large piece of paper / Area of one ticket

= (18 × 21) / 1.5²

= 12 × 14

= 168

Given:

A pump can fill a pot in 2 hours.

With leakage, the pot can be filled in \(2\frac{1}{3}\) hours

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Work done in 1 hour by the pump = 1/2

Work done in 1 hour by leakage and pump = 3/7

Work done by the leakage in 1 hour = (1/2) – (3/7)

⇒ (7 – 6)/14

⇒ 1/14

∴ The leakage can empty the pot in 14 hours.

Given:

Two pipes X and Y can fill a container in \(37\frac{1}{2}\) minutes and 45 minutes respectively

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Pipe X fills the container in \(\frac{{75}}{2}\) minutes.

Part of the container filled by X in 30 minutes

⇒ (2/75) × 30

⇒ 4/5

Remaining part = 1 – (4/5)

⇒ 1/5

Now, 1 part is filled by pipe Y in 45 minutes

1/5 part is filled in = 45 × 1/5

⇒ 9 minutes

∴ The pipe Y should be turned off after 9 minutes.

Statement : “If you know the answer, then you don’t guess.”

Option 1) If you don’t know the answer, then you guess. → This is totally converse of statement given so will not be logically statement.   

Option 2) You don’t know the answer, or you don’t guess. → This statement will be logically correct

Option 3) You know the answer, and you guess. → As per the statement it is given about when you donot guess nothing.    

Option 4) If you don’t guess, then you know the answer. → This is totally converse of statement given so will not be logically statement.   

Hence, Option 2 is correct answer.

Correct option is D) take

Securities Markets is a place where buyers and sellers of securities can enter into transactions to purchase and sell shares, bonds, debentures etc.

Correct option is D) ticket

National Exchange for Automated Trading (NEAT)

Initial Public Offering

Hydro power plants are of prime importance as about 25 per cent of our energy requirement in India is met by hydro Power Plants. 

(i) A high rise dam is constructed at a suitable place on the river to obstruct the flow of water and thereby, collect water in larger reservoirs. Due to rise in water level the kinetic energy of flowing water is transformed into potential energy of stored water. 

(ii) The water from the high level in the dam is carried through sluice gates and pipes to the turbine of electric generator, which is litted ar the bottom of the dam. Due to flowing water, turbine is rotated at a fast rate and hydel electricity is produced. 

(iii) A hydro power plant converts the potential energy of falling/stored water into electricity.

Correct option is (c) Capital Expenditure ₹2,50,000; Revenue Expenditure ₹80,000; Deferred Revenue Expenditure ₹1,00,000 

Correct option is (c) ₹5,000 will be subtracted 

Correct option is: (A) 2