Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

(a) Prove geometrically that cos (A + B) cos A cos B-sin Asin BHence find cos 75°

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cos75°=cos(45°+30°)=cos45°cos30°-sin30°sin45° = (1/√2)(√3/2)-(1/2)(1/√2)=(√3-1)/2√2

2.

5 naprimary school there shall be 3 teachers to 60 students. If there are 400 studentsenmiled in the school how many teachers should be there inthe school in the same ratio?

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3.

In a primary school there shall be 3 teachers to 60 students. If there are 400 studentsenrolled in the school, how many teachers should be there in the school in the same ratio?

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given that in a primary school has the answer wiilbe 9 teachers

If there are 3 teachers for 60 students,Therefore for 20 students 1 teacher will be needed.

For 400 students ,teachers required=400/20 =20 teachers

20 is the right answer of the following

60 student =3 teacher1 teacher=20 student400student=400/20=20 teacher

20 teacher is the right answer

if 3 teacher for 60 students said that 1 teacher = 60/3 = 20 students so that in 400 students = 400/20 = 20. teachers

20 teachers is the right answer

primary school have 9 teacher

4.

49(a). Prove geometrically thatHence find cos 75cos (A + B)=cos Acos B-sin AsinB

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5.

angle and sin θ-cos θ, find the value of θ and hence, find the value ofθ is an acute2tan2 θ + sin2 θ-1.12

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6.

6Find the value of:(i) sin 75°

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7.

x+4y=45 and 3x+6y=75 hence,find the value x and y for mx+ny=86 from the value n is 56.

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x + 4y = 45 3x + 6y = 75 3x + 12y - 3x - 6y = 35 - 75 6y = 60 y = 10 x + 40 = 45 x = 5 mx + ny = 86 5m + 56×10 = 86 5m = 86 - 560 = -474 m = -94.8

8.

sn - 6 and sin - 10. find the values710, find the valuesof cos A and cos B. Hence using the formulacos (A + B) = cos A cos B-sin A sin B, show thatBoard Term-1, 2012, Set-55]A+B=45°.

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SinA =1/√5cosA=√(1-sin²A) (sin²A+cos²A=1)=√1-1/√5² (1-cos²A=sin²A=√1-1/5. (cosA=√1-sin²A) =√5-1/5=√4/5cosA=2/√5sinB =1/√10cosB=√1-sin²B=√1-1/√10²=√1-1/10=√10-1/10=√9/10cosB=3/√10Cos(A+B)=cosA. cosB-sinA. sinB=2/√5 x 3/√10 - 1/√5 x 1/√10=6/√50-1/√50=5/√50=5/5√2=1/√2cos(A+B)=1/√2Cos(A+B)=Cos 45°A+B=45°

9.

HowDoes raincause

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Rain is formed during a process which is usually known as the water cycle. The water cycle involves a number of steps, including evaporation, cloud formation, precipitation (rain), relocation, and then evaporation again.

Evaporation occurs when liquid water is vaporised into water vapour, allowing it to become a part of the atmosphere. Heat from the sun acts on the water molecules in liquid water, increasing their energy. These particles then ‘break away’ from the liquid, becoming airborne. When there is high humidity it simply means that there is a high concentration of water molecules in the air: a lot of evaporation has occurred.

When the amount of water vapour in the air reaches a level where no more water can evaporate, we say that the air is fully saturated, or that the humidity is at 100%. If the conditions then change so that the air can hold less water vapour (for example, if it gets colder or the barometric pressure decreases), then some of it will begin to condense and form liquid water again. To begin with, only tiny particles of water will be formed which are too light for gravity to pull them back to the ground. If this happens near ground level, fog or mist will be formed. However, if it happens in the upper atmosphere, we will see collections of these water particles as clouds.

Inside a cloud, the tiny water particles move around, bumping into each other. When they hit each other, they can combine to form a larger droplet. Once this has happened enough times, a droplet of water which is heavy enough to fall to the ground is formed. The form of this fall – known as precipitation – depends on atmospheric conditions. It may be snow, rain, sleet, hail, etc. Once precipitation, or rainfall, has occurred, the liquid water will move around the earth’s surface, usually towards the ocean, until it is evaporated again and the cycle restarts.

10.

MENTAL MATHS CORNERMark True' or Talse for the following statements1 x+3is a factor of x-2.3x+61 The factor of sy-+ bs - ay are (y-b a).s a-1is a factor of 2+Sx2 x-2 is a factor of -8xy2xr 3x 1 is divided by x 1, then the remainder is zero(1 - 3x)

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11.

(2) Wirite rationalizing factor of y8Write rationalizing factor of /8.

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Rationalising means making an surd into rational number Heere √8=√4 ×√2√8=2√2=2√2rationlising means converting into rational here 2√2 is not a rational so, converting by multiplying 2√2 by √2 Then we get ,4 as number Here, 4 is a rational number so, rationalising factor of √8 is √2

12.

I. a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is5. Then calculate the least prime factor of (a + b)

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13.

A toy merchant allows 25% discount on the marked price otmeprofit of 20%. If he gains 360 at the sale of one toy, find the man-Hint : First

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pls mark this Ans as best😊😊

14.

5cycle merchant allows 25% discount on the marked price of the cycles and still makes a prof20%. If he earns a profit of 360 on the sale of one cycle, find the marked price of the cycle.Hint: First find the C.P. of the cycle.]

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15.

p(x) = 4x² + 12x + 5

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16.

Tourny OR 192 km-From aeown g taken- 2 hour's more ey SeatroMy、assenrainMore, pǐnd dhe speed othe laster and

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17.

OATHl'sthansaction5. A cycle merchant allows 25% discount on the marked priceof the cycles and still makes a profit of20% If he earns a profit of t 360 on the sale of one cycle, findHint: First find the C efthodlethe marked price of

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Profit = 20% = 360 Rs

So, Cost Price = 360 / 20 * 100 = 1800 Rs

And Sales Price = 1800 + 360 = 2160 Rs

Discount = 25% of marked price

So Sales Price = (100 - 25) % = 75% of marked price

=> 2160 = 75% of marked price

=> marked price = 2160 / 75 * 100 = 2880 Rs

18.

A merchant changed his tradediscount from 25% to 15%. Thiswould increase selling price by:(2) 6.36(3) 13 %(4) 169%33

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bhaiya thank you...

19.

15·The circumference of the base of a right circular cylinder is 132 cm. Find the volume of the cylinder, if itsheight is 22 cm.

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mam formula type answer dejia

ghanks

20.

the strength of a school is 2500 .out of this,600 are in pre_primary section ,700 in middle school,and the remaining are in high school ,and in high school .Find the percentage of students in each section

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the strength of a school is 2500 .out of this,600 are in pre_primary section ,700 in middle school,and the remaining are in high school ,and in high school .Find the percentage of students in each section?600 in pre primary,700 in middle and remaining in high school.so students in high school=(2500-(600+700))=1200so percentage of pre primary students=(600/2500)*100=24%percentage of middle school students=(700/2500)*100=28%percentage of high school students=(1200/2500)*100=48%

21.

Find the distance betweem the points(i) A(2-3) and B-6, 3)

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22.

tensile strenght determination formula

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23.

7. A certain school has 300 students, 142 of whom are boys. It has 30 teachers. 1uwhom are men. What percent of the total number of students and teachers inschool is female?

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24.

10.A certain school has 300 students, 142 of whom are boys.It has 30 teachers, 12 of whom are men.What percent of the total number of students and teachers in the school are female?

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25.

the zeros of the polynomials f(x)=4x^-12x+9

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thank you sir

26.

as 30 teachers, 12and teachers into complete at this point, IUW TUIS17. A certain school has 300 students, 142 of whom are boys. It has 30 +whom are men. What percent of the total number of students and toschool is female?

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27.

There are 56 teachers in a school out of them, 42 are lady teachers and the rest are male teachers, find the ratio of:(a) male teachers to the lady teachers.(b) lady teachers to the male teachers.(c) male teachers to total number of teachers

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no. of male teachers = 56-42 = 14. so a) 14:42 = 1:3b) 42:14 = 3:1c) 14:56 = 1:4

28.

Draw the net of a hexagonal prism

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29.

(iv) Places A and B are 100 km apart on a highway, One car starts from A and anothfrom B at the same time. If the cars travel in the same direction at different speedey meet in 5 hours. If they travel towards each other, they meet in 1 hour. Whare the speeds of the two cars?

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thanks

30.

The following observation have been arranged in ascending order. If the median of the data is 23.5,find the value of x.12, 16, 17, 19, x, x+3,27, 37, 38,40

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31.

+Cisequalto2-(12x-(f(x) + 64x) = 0 is twice of other, then find the maximum value of theIf one root of the equation tfunction f(x), where x eR.

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Given quadratic equation: t² − 12xt − (f(x) + 64x)=0roots are: (12x±√(144x²+4f(x)+64×4x))/2±2√(36x²+f(x)+64x)/2 =6x±√(36x²+f(x)+64x)as one root is twice the other, or f(x)=-32x²-64xHere, f(x) is a quadratic equation with D ≧ 0 and aSo the maximum value at x = -b/2a = 64/(2×-32) = -1∴ The maximum value of f(x )= f(-1) = -32(-1)² - 64(-1) = 32 (Ans)

32.

.The length of a line segment is 13 units and the coordinates of one end point are (-6,1). If the ordinate of theother end point is-1, find the abscissa of the other end point.

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Let AB be the line segmentTherefore, A(6,-7) and B will be (x,-1) as y-coordinate is called ordinate and x-coordinate is called abscissa.By applying Distance Formula,AB²=(x2-x1)²+(y2-y1)²13²=(x-6)²+(-1+7)²169=x²-12x+36+36⇒x²-12x+72=169⇒x²-12x-97=0

Like my answer if you find it useful!

33.

Find the common factor 24, 36

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24= 2*2*2*336= 2*2*3*3comman factor = 2*2*312

34.

Find the common factor of 2x, 3x2.4.

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2x = 2*x3x^2 = 3*x*x4 = 2*2

Since there is no common factor visible.

Therefore, 1 is the only common factor of the given terms.

35.

Of the five villages P, Q, R, S andT situated close to each other, P is to the Westof Q, R is to the South of P, T is to the North ofQ and S is to the East of T. Then, in whatdirection R is with respect to S?

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36.

bracket open 3 + root 3 bracket close bracket open 3 minus root 3 bracket close

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37.

If the expressions x2-11x + a and x2-14x+2a have a common factor anda 0, then find the common factor

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38.

-1: If A= {x/x is a factor of 6}B = {x / x is a factor of 8}find the A-B and B-A

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A-B=3,6B-A=4,8 correct answer

A-B=3,6B-A=4,8is the best answer

39.

Ex. 1: If A= {x / x is a factor of 6}B = {x / x is a factor of 8}find the A-B and B-A

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A-B=3,6B-A=4,8 Answer.

40.

find the common factor 12,24,36

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12 is the right answer

41.

903.3. Int0.33. Define internal energy(1 markpossesses a fixed quantity of

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Ans :- Internal energy is defined as the energy associated with the random, disordered motion of molecules. It is separated in scale from the macroscopic ordered energy associated with moving objects; it refers to the invisible microscopic energy on the atomic and molecular scale.

42.

10. A field in the form of a parallelogram has base15 dam and altitude 8 dam. Find the cost ofwatering the field at 10 P per square metre

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Area of paralelogram=b*h.=1 dam = 10 metres so,b=15*10=150.and h=8*10=80.Hence,80*150=12000.m^2.now,cost of watering=area *10=12000*10=120000paise=1200rupees.

43.

(- 7/8) + 8/9

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(-7/8)+(8/9)=[-7×9 + 8×8]/8×9=1/72

44.

ली LA b |44. स0 श्रे0 के प्रथम 7 पदों का योग 63 और अगले 7 पदों का योग 161 है इसका>४वां पद ज्ञात कीजिए।सन कक ey

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Sn= ( n / 2) [ 2a + ( n -1)d sum of the first 7 terms of an A.P is 63 i. e S7= 63.( 7 / 2) [ 2a + 6d ] = 63

2a + 6d = 18 --------(1)

Sum of its next 7 terms = 161.

Sum of first 14 terms = sum of first 7 terms + sum of next 7 terms.14 पदों का योगफल बराबर होगा 7 पदों का योगफल + 7 पदों का योगफल

S14 = 63 + 161 = 224

( 14 / 2) [ 2a + 13d ] = 224.

7 [ 2a + 13d ] = 224.

⇒ [ 2a + 13d ] = 32 -------92)

BySolving equation (1) and (2) we obtainसॉल्व करने पर प्राप्त होगा

d = 2

a = 3.

t28= a + ( 28 - 1) d

t28= 3 + ( 28 - 1) 2

t28= 57.

28th term= 57.

45.

Example 24 In a school there are 20 teachers who teach mathematics or phthese, 12 teach mathematics and 4 teach both physics and mathematics, HcTeach physics?

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46.

2.उत्तर 3:(a) 578 x 161

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578*161=93058 is the right answer

578×161=93058therefore 93058 is the right answer

47.

(D)SIn a class, 30 students passed in Maths,20 passedin Physics and 8 passed in both. How manystudents failed in both if the class has 45students?

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48.

The HGF and 1.C. M of TWO Nymbol are 17 and1229 respectively. If one of the numbersis 459. Find the other.

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I don't know of this answer

49.

Two sides of a triangle are 7 units and 10 units.Which of the following can be the length of thethird side?

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Thanks

Please post the options also.

50.

2 \frac{\cos 67^{\circ}}{\sin 23^{\circ}}-\frac{\tan 40^{\circ}}{\cot 50^{\circ}}+\cos 0

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