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.

If x and y are two positive real number such that 4 x2 + y2-40 and xythe value of 2x + y?6thenfind

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2.

के ठल पत्ता (+व्पा तक 51% (1Hdanf) 4 cos (1 +cotA) o 1 बना एल:

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sinA(1+tanA)+cosA(1+cot)=sinA.cosecA+cosA.secA1+1=2

3.

Logo0s

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4.

1. If cosecA=-,5/4 then find the values of cotA, sintAand cos A

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5.

1+Cot?A &I लागू करके सिद्ध4 ‘” ) _ सर्वसमिका (05602 =: ... कीजिए-b CosA—SinA+1CosA + SinA—1 Cosec A+ CotA

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CosA-sinA+1/cosA+sinA-1=(cosA-sinA+1)(cosA+sinA+1)/(cosA+sinA-1)(cosA+sinA+1)=(cos²A-cosAsinA+cosA+cosAsinA-sin²A+sinA+cosA-sinA+1)/{(cosA+sinA)²-(1)²}=(cos²A-sin²A+2cosA+1)/(cos²A+2cosAsinA+sin²A-1)={cos²A+2cosA+(1-sin²A)}/(1+2cosAsinA-1) [∵, sin²A+cos²A=1]=(cos²A+2cosA+cos²A)/2cosAsinA=(2cos²A+2cosA)/2cosAsinA=2cosA(cosA+1)/2cosAsinA=(cosA+1)/sinA=cosA/sinA+1/sinA=cotA+cosecA=cosecA+cotA (Proved)

6.

1k 2l Cken]s L2y 1§ 30 0C k1 20 01 3l L 2 5 (1)ity 1 248 2yl bl b 8 ek 3} ()‘.’ I Gl bl 40 s Lk 20k S2Yln 1§ B213 e s S o9 4 ST RS Tey e —————————— e S

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7.

unite tuo sational oumbers betureand

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3/5= 0.62/3= 0.66no between 3/5 & 2/3 are= 0.62, 0.63, 0.64,0.65,0.658 etc

3/5=0.62/3=0=66 no between 3/5and 2/3

8.

The mass of a brick is 2 kg 750 g what is the total mass of 14 such bricks

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2.75 * 14= 38.5total mass of 14 bricks is 38kg 500g

1 block=2.75kg14 block=2.75×14=38.5

14* 2.750= 38.500= 38kg 500 g

9.

will snets came for a function. They were accommodated in 3halls with same number of guests in all the halls. How manyguests were there in each hall?

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No. Of guest = 65No. Of halls = 3Its not possible to divide 65 guest equally in 3 halls2 halls have 22 guests in each hall and third hall accommodate 21 guests

10.

en eny method

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differentiation of x³is 3x²

differentiation of x^n is nx^(n-1)

11.

1 - tan2 AIf3 cot A = 4, check whether 1 + tan A = cos2 A-sinA or no

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12.

PARALLELOGRAMS AND TRIANGLES163D, E and F are respectively the mid-points of the sides BC, CA and AB of a Δ ABC.Show that) BDEF is a parallelogram.(ii) ar (DEF)=-ar (ABC)4(i) ar (BDEF 2 ar(ABC)

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Given:ABC is a Triangle in which the midpoints of sides BC ,CA and AB are D, E and F.

To show:(i) BDEF is a parallelogram. (ii) ar (DEF) = 1/4ar (ABC)(iii) ar (BDEF) =1/2 ar (ABC)

Proof: i)Since E and F are the midpoints of AC and AB.

BC||FE & FE= ½ BC= BD

(By mid point theorem)

BD || FE & BD= FE

Similarly, BF||DE & BF= DE

Hence, BDEF is a parallelogram

.[A pair of opposite sides are equal and parallel]

(ii)Similarly, we can prove that FDCE & AFDE are also parallelograms.

Now, BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas.

∴ ar(ΔBDF) = ar(ΔDEF) — (i)

In Parallelogram AFDE

ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal) — (ii)

In Parallelogram FDCE

ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal) — (iii)

From (i), (ii) and (iii)

ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF).....(iv)

ar(ΔBDF) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC)

4ar(ΔDEF) = ar(ΔABC)(From eq iv)

ar(∆DEF) = 1/4 ar(∆ABC)........(v)

(iii)Area (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDF)ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF)

ar(parallelogram BDEF) = 2×ar(ΔDEF)(From eq iv)

ar(parallelogram BDEF) = 2× 1/4

ar(ΔABC)(From eq v)

ar(parallelogram BDEF) = 1/2 ar(ΔABC

13.

1 - tan2 A8. If3 cotA 4, check whether 1 + tan A Cos? A -sin'A or not.

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14.

. If3 cotA-4, check whether1+ tan AnsA- sin A or not

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thanks

15.

8. It3 cotA 4, check whether+ an C1 - tanAl + tan A

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(1-tan²A)/(1 + tan²A)=2cos²A -1

3cotA = 4 cotA = 4/3 tanA = 1/cotA = 3/4

LHS = (1-tan²A)/(1+ tan²A) = (1 - 9/16)/(1+ 9/16)= 7/25

RHS =2cos²A -1∵tanA = 3/4 ∴cosA = 4/5

= 2(4/5)² -1 = 32/25 -1 =7/25

LHS = RHS so, (1-tan²A)/(1 + tan²A)=2cos²A-1

as cos^2A-sin^2A=2cos^2A-1

16.

1 - tan2 A1 + tan AcOS A-sin'A or noti 3 cot A 4, check whether

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17.

8. Vikas can cover a distanceours on foot. How many km per hour doeshe walk?N1

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18.

uhh sidles-11.cenlee.DミandItyenve that

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19.

1-tan Acot A4,check whether+tanA -sin' A or not

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(1-tan²A)/(1 + tan²A)=2cos²A -1as sin^2a-cos^2a=2cos^2a-1=cos2a3cotA = 4 cotA = 4/3 tanA = 1/cotA = 3/4

LHS = (1-tan²A)/(1+ tan²A) = (1 - 9/16)/(1+ 9/16)= 7/25

RHS =2cos²A -1∵tanA = 3/4 ∴cosA = 4/5

= 2(4/5)² -1 = 32/25 -1 =7/25

LHS = RHS so, (1-tan²A)/(1 + tan²A)=2cos²A-1

20.

tan A1 + tanco A 4, check whether nA ' A - sin' A or not.

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21.

आकृति 6.19 में ए४॥ 80 और 07 ॥ 27 है। सिद्ध कीजिएBF _BE 4e EC

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22.

3. Vikas can cover a distanceAlcohours on foot. How many km per hour doeshe walk?

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23.

By what number shouldbe divided so that2ar4the quotient may be?27

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24.

Ny avIty=

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25.

(av) (x+2)'=x' -4

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Test it is OK let's see

26.

to getav

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5/12 is the correct answer of the given question

the answer is 10/24.

10/24 how is it correct =28-18/24 =10/24

27.

乀.av) 4li. + 80Find a quadraticits zeroes respec

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28.

If $ 3 \cot A=4, $ check whether $ \frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A $ or not.

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29.

30Lault Sivulu walk SU lila ley cover ut usuance cumpicuc Dupon12. Find the smallest length of a rope which can be measured in exact numberof times by three tapes measuring 1 m 20 cm, 75 cm and 1 m respectively.AIL_finetul noptilid

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30.

b.Simplity each ot the lol- the property:(ar 2707. Tell which property allo

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31.

The larger of two supplementary angles exceeds the smaller dy 58, then find

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32.

The larger of two supplementary angles exceeds the smaller by 18 degrees. Findthem.

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33.

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

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let the smaller be x so larger is x + 18 sum of supplementary angles is 180°x + x + 18 = 180°2x = 180° - 18°2x = 162°x = 162°/2x = 81°So, the angles are 81° and 81° + 18° = 99°

how to come 99

34.

) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

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35.

18) The larger of two supplementary angles exceeds the smaller by 20degrees. Find them

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36.

(i) The larger of two supplementary angles exceeds the smaller by 18 degrees Fndthem.

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37.

1 - tan2 ATlf3cot A-4, check whether I + tan 2A= cos' AsinA or not.

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38.

1- tan A8,1f3 cot A = 4, check whether I + tani A-coS A-sin A or not.

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39.

g. If 3 cot A 4, check whether tan? A1 + tan 2A= cos' A-sin?A or not.

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40.

००5 58 sin22° 5 cos 38° cosec 52°sin32° " cos 68° " \[3 (tan 18° tan 35° tan 60° tan 72° tan 55°)

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As cos 58°= cos(90-58)= cos32sin22°= cos(90-68)same cot38= cosec 52now

1+1-(1/√3*√3)2-1/35/3

41.

c. Clear screen (CS) command

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In computing,CLS(for clear screen) is a command used by the command line interpreters COMMAND.COM andCMD.EXEon DOS, FlexOS, OS/2, Microsoft Windows and ReactOS operating systems to clear the screen or console window of commands and any output generated by them.

42.

aroft dia he maike Asa, what is hisirait siss, t r Rs 60 a ess o Rs 1e0 is sulfteresd. Find itsAisa, fins sss netr dencostpriceSelli

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S.P.=600loss=100so C.P.=700so loss percentage=100/700*100=14.28%

43.

7. A shopkeeper sold an article at a gain of 5%. If he had sold it for? 16.50 less, he wouFind the cost price of the article.r lose by selli

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Let CP of the article = xat 5% gain, SP of article = x + (5/100)x = 1.05xat 5% loss, SP of article = x - (5/100)x = 0.95x

given that 1.05x - 0.95x = 16.5⇒ 0.1x = 16.5⇒ x = 16.5/0.1⇒ x = Rs. 165.

The cost price of the article is Rs. 165.

44.

324/5

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when we divide 324 by 5 Then solution is 64.8

45.

7. The price of a watch ist 50 more than twice the price of a gold ring. Let the price ofthe ring be x.(i) Express the price of the watch in terms of x.(ii) If the price of the watch is 208, find the price of the ring.

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46.

its original price.. The price of a sweater is increased by 8%. If its increased price is1566, findriginal price.

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Increase in price = 8%

Find the new selling price:Selling Price = 100 + 8 = 108%Selling Pirce = Rs 1566

Find the original price:108% = 15661% = 1566 ÷108 = Rs 14.50100% = 14.50 x 100 = Rs 1450

Answer: The original price is Rs 1450

47.

lungs

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Ans :- The lungs are the center of the respiratory (breathing) system. Every cell of the body needs oxygen to stay alive and healthy. Your body also needs to get rid of carbon dioxide. Your lungs are specially designed to exchange these gases every time you breathe in and out.

48.

2In the given figure, the diameter of two wheelshave measured 2 cm and 4 cm. Determine thelength of the belts AD and BC that pass aroundthe wheels, if it is given that belts cross each otherat right angles.

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49.

9, In the given figure, the diameter of two wheelshave measured 2 cm and 4 cm. Determine thelength of the belts AD and BC that pass aroundthe wheels, if it is given that belts cross each other2)at right angles.

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50.

ATHEMATICS:1. Find the value of (75)' by the shortcut method.buhich 324 must by multinte

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s = 75×75×75s = 421875

75×75×75so the answer is 421875

421875 the correct answer of the given question