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Facts that Matter

•Celestial Objects:Objects, such as the stars, the planets, the moon and many other objects, in the sky are called celestial objects.

•Phases of the Moon:The various shapes of the bright part of the moon as seen during a month are called phases of moon.

The moon revolves round the earth. It also revolves round the sun along with the earth. The various stages of the moon during a month are shown in Fig. 17.1.

In position 1, the part of the moon facing earth does not receive light from the sun.

Hence it appears dark. The other part of the moon that is facing away from earth is lit by the sun light. This position of moon, when it appears dark on earth is known as new moon.

In position 2, 3 and 4, we see the moon not as a full disc but as a cresent. Position 2 of the moon is known as cresent moon, Position 3 as first quarter and position 4 as Gibbous (bright part is greater than a semicircle) moon.

In position 5, the Sun's rays fall directly on the part of the moon facing earth. In this position we see the moon as a full disc of light. This position of moon is known as full moon.

Ramesh read 25 pages while Suresh read 2/5 of 100 = 2 X 20 = 40 pages. So Ramesh read less.

give reasons

Number of pages he read in morning =45Number of pages he read in evening. =18 Therefore ,total pages he read altogether =45+18 =63

hi

36 is the correct answer of the given question

75 min is the best answer for this. sum

let(A+B)(A+B)fills 1 litre in 1 minutes then(A+B)(A+B)fills in 36 minutes = 36 litres

According to question

(A+B)(A+B)work only for 30 minutes then pipe filled by(A+B)(A+B)in 30 minutes is = 30 litres

remaining part =(36−30)=6(36−30)=6litres

6 litres part filled by A in =10 minutes

1 part filled by A=106106minutes

36 part filled by A =106×36106×3660 minutes

A+B=36A+B=36minutes

A=60 minutes

A's efficiency = 3 litres/minute

B's efficiency = 2 litres/minute.

B can alone fill the tank in

=T.CeffofB=1802=90T.CeffofB=1802=90minutes

75 minutes OK thankyou

90 minutes is the right answer of the following

75 minutes is the answer of this particular question

4 is the answer of these questions

percentage decrease = 75-25/75

= 50/75

= 0.66%

ALewis baseis an electron-pair donor. Because theLewis basedonates electrons, we can also say that it is a nucleophile, which is a substance that behaves as an electron donor.

So, when we look at anacid-basereaction, theLewis basedonates a pair of nonbonding electrons to the acid.

Lewis base is a substance that can donate an electron pair. those species that have a lone pair always form Lewis base so H^- is a lewis base as it possess a lone pair and can donate it.

This is equal to 75600 seconds

formula?

Multiply by 60....

Thanks sir

The process of farming organisms from the sea is called mariculture.

There are a total of 36 possibilities :

{ (1,1), (1,2), (1,3), (1,4), (1,5),(1,6) (2,1)(2,2)… …..(6,5) (6,6)}

The doublets among these are (1,1), (2,2),(3,3)(4,4),(5,5),(6,6)

So the probability of getting a doublet is :

No. Of doublets / total no. of outcomes =6/36=1/6

Probability = possible outcomes/ total outcomes

when a single die is thrown the number of outcomes is 6^1

when two dice are thrown the number of outcomes is 6^2

In throwing two dice the favorable cases of getting the sum as 7 are: (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3).

Therefore, the required probability is 6/36 or 1/6.

The dice quantity = 2(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)Probability= 1+6=7 2+5=7 3+4=7 4+3=7 5+2=7 6+1=7So,answer=6/36 1/6

Let the number of days be x.

Given that he has enough food to feed 20 animals in his cattle for 6 days.Later on, 10 more animals were added.

Therefore the number of days required to feed 30 animals:

20 * 6 = (20 + 10) * x

120 = 30 * x

x = 120/30

= 4.

Therefore the number of days = 4.

20*6=(20+10) *a120=30*aa=120/30a=4

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Consider all possible events and count them1 T2 T3 T4 T5 T6 T

1 H2 H3 H4 H5 H6 H

Here the likely events are 1 and total events are 12, which means the probability is 1/12

Probability = possible outcomes / total outcomes

Total outcomes = 6For getting number 5 or less than 5Possible outcomes = 5

Therefore, probability of getting a number 5 or less than 5 = 5/6

10000+10250+10790+9865+15350+10110/6

= 66365/6

= 11060.83

Loe the no of books she had initially = x

x/2 + 16 = 36

x/2 = 36-16

x/2 = 20

x = 40

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No. Of pages in the book=10×40=400 pages

Now, let the no. Of pages he should read be x.8×x=400x=400/8x=50 pages

what was the argument between the wind and the sun

Question 1:A die is rolled, find the probability that an even number is obtained.SolutionLet us first write thesample spaceS of the experiment.S = {1,2,3,4,5,6}Let E be theevent"an even number is obtained" and write it down.E = {2,4,6}We now use the formula of theclassicalprobability.P(E) = n(E) / n(S) = 3 / 6 = 1 / 2

Question 2:Two coins are tossed, find the probability that two heads are obtained.Note:Each coin has two possible outcomes H (heads) and T (Tails).SolutionThe sample space S is given by.S = {(H,T),(H,H),(T,H),(T,T)}Let E be the event "two heads are obtained".E = {(H,H)}We use the formula of the classical probability.P(E) = n(E) / n(S) = 1 / 4

Question 3:Which of these numbers cannot be a probability?a) -0.00001b) 0.5c) 1.001d) 0e) 1f) 20%A probability is always greater than or equal to 0 and less than or equal to 1, hence onlya)andc)above cannot represent probabilities: -0.00010 is less than 0 and 1.001 is greater than 1

Question 4:Two dice are rolled, find the probability that the sum isa) equal to 1b) equal to 4c) less than 13Solutiona) The sample space S of two dice is shown below.S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }Let E be the event "sum equal to 1". There are no outcomes which correspond to a sum equal to 1, henceP(E) = n(E) / n(S) = 0 / 36 = 0b) Three possible outcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence.P(E) = n(E) / n(S) = 3 / 36 = 1 / 12c) All possible outcomes, E = S, give a sum less than 13, hence.P(E) = n(E) / n(S) = 36 / 36 = 1

Question 5:A die is rolled and a coin is tossed, find the probability that the die shows an odd number and the coin shows a head.SolutionThe sample space S of the experiment described in question 5 is as followsS = { (1,H),(2,H),(3,H),(4,H),(5,H),(6,H)(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}Let E be the event "the die shows an odd number and the coin shows a head". Event E may be described as followsE={(1,H),(3,H),(5,H)}The probability P(E) is given byP(E) = n(E) / n(S) = 3 / 12 = 1 / 4

x² + 5x + 1/5 = 0

Discrimanat = b² - 4ac

5² - 4 × (5) × 1/5

25 - 4

21 > 0

Roots are real and distinct

thank you so much

-3/2 is the correct answer of the given question

how u find the value of x and y please tell it's urgent

total change is -1 m Or 1m backwards

original position is zero, 5m-6m=-1m backward from original position.

total change is -1 metre backwards

By applying the formula ,

M1 * D1 / W1 = M2 * D2 / W2

M-> MEN , D-> DAY , W-> WORK

therefore,

20 * 6 / 56 = 35 * 3 / W2

120 / 56 = 105 / W2

Answer -> W2 = 49.

धन्यवाद

7 percentage loss means (100-loss percentage)93 percentage is equal to = 6510and he wants to gain 8 percentagemeans (100+8) 108 percentagenow he wants to know the value of 108 percentage