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.

1a. There are three events A, B, C one of which must happen and only one can happen ata time. If the odds are 8 to 3 against A, 5 to 2 against B; find the odds against C.

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2.

nd her loss percentage.n buys Diwali cards at850 per 100 cards and sells then at? 136 a 1mand his gain per cent.

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Rs of one card 850/100=8.5RsRs of 10 cards = 10*8.5=85RsProfit Rs = 136-85=51Rsprofit of 100 cards = 51*10 = 510gain percent = 510/850*100 = 60 %

3.

The dimensions of a rectangular box are in the ratio of 2: 3:4 and the difference between the costof covering it with a cloth at the rate of 10 and ? 11 per mbox13.isき1,300. Find the dimensions of thend 1 5 m deen is to be tiled. If each tile is

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Let common multiple of 2:3:4 be x=> Breadth = 2x=> Length = 3x=> Height = 4x

Surface area = 2(2x*3x + 3x*4x + 4x*2x) = 2(6x²+12x²+8x²) = 52x² cm²

As per question,=> Rs. [11(52x²)-10(52x²)] = Rs. 1300=> 1(52x²) = 1300=> x² = 1300/52 = 25=> x =5length= 2*5= 10cmbreadth= 3*5= 15height = 4*5= 20

4.

kg of rice at t 36 per kg and 25 kg of rice at 32 per kg. He mixed the twond sold the mixture at 38 per kg. FInd his gain per cent in the wholetransaction

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Cost of 1kg rice = ₹36Cost of 20kg rice = ₹36 × 20 = ₹720 Cost of 1kg rice = ₹32Cost of 25kg rice = ₹32 × 25 = ₹800 Total weight = 45kg Total cost = ₹1520Selling price of 45kg rice = ₹38 × 45 = ₹1710 Gain = 1710 - 1520 = 190 Gain% = (190/1520) × 100 = 12.5%

5.

ed for348000, its value depreciates at 10% per annum during the first year andA car is purchasat 20% per annum during the second year.9.What will be its value after 2 years?

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6.

Identify the mode of nutrition where sunlight is used along withCO2 and water to prepare food

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The mode of nutrition here is autotrophic nutrition and this type of autotrophic nutrition is photoautotrophs which undergo photosynthesis

7.

2. Additive inverse of - 3 - 4i is(a) - 3+4i,(b) +3+4i.(c) 3+4i.(d) +3-4i.

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c is the correct ans explanation : additive inverse is the no. which on adding gives 0 for eg : additive inverse of -5 is 5 as -5+5 would give 0 now in this case 3+4i on addition with -3-4i would 0 hope u understood

8.

\frac \frac 53 2 \times 26.5 \times \frac 106 4 - \frac 33 2 \times \frac 66 4 \times 16.5 \frac 53 2 \times \frac 106 4 %2B \frac 106 4 \times 16.5 %2B \frac 66 4 \times 16.5

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1st one ans is 91101.656

2nd one answer is 14025.129

9.

A ball falls vertically after being dropped.The ball falls a distance d metres in a time of t seconds.d is directly proportional to the square of tThe ball falls 20 metres in a time of 2 seconds.a)Find a formula for d in terms of t.

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d is proportional to the square of tSo we can write:

(1) . . d = K * t^2 . . where K is some constant.

At the moment we don't know the value of this constant. But we do know that the ball has fallen 20 metres after 2 seconds. This info will give us the value of K.

Substituting d=20 and t=2 into equation (1), we get

20 = K * 2^220 = K * 4K = 5

So, using the value 5 for K, equation (1) becomes:

d = 5t^2 . . (d in metres, t in seconds)

This is the formula for d in terms of t

10.

1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 +

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11.

\frac { 4 ^ { n + 4 } - 5 \times 4 ^ { n + 2 } } / { 4 ^ { n + 1 } times 11 }

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12.

Directions (Q. Nos. 24-25) If f(-x) = -f(x), then f(x) isf(x) is evenodd function and if f(-x) = f(x), thenfunction.2[*f(x), if f (x) is evenAlso, I flo) dx=112*f(x), if f (x) is odd- The value of the integral, sin’x dx is(d) 1-TC(a) o(b)cC/2

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By integral value I understand that you are asking the geometrical meaning of an integral of a function. values in the integer are integral value. Like1,2,3 ,4,5etc are integer values. The Anti Derivative of any function/Number/Variable is it's Integral value.

13.

f(I)+f(3) + f(x)f(21+f(-) + f(l/2)32=35, then x =2x-1, if x>1If f(x)=İx2 +1, ifand if-1x1

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14.

Let f be differentiable for all x, if f (1) =-2 and f' (x)2 2 for all x e [1,6], then(A) f (6) < 8(C) f (6)·25(B) f (6) 2 8(D) f (6) s5

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Hence, option (B) is correct.

15.

There are 12 boys and 6 girls in how many ways A commity of 5 can be form. Such that it is contain 2 boys and 3 girls

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16.

There are 12 boys and 6 girls in how many ways A committee of 5 can be form. Such that it is contain 2 boys and 3 girls

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12C2 × 6C3 = 12×11/2 × 6×5×4/3×2 = 6×11×5×4 = 120×11 = 1320

17.

\left(x^{3}+4 x^{2}+16 x+61\right) \div(x-4)

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x*x*x-4x*x+8x*x-32x+48x-192+131 =x*x(x-4)+8x(x-4)+48(x-4)+131 =(x*x+8x+48)(x-4)+131 so q=x*x+8x+48 reminder = 131

18.

4. 5, 6, 16, 2, 244, 1245(A) 28 (B) 55 (C) 57 (D) 61 (E) ofthese

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19.

8.Given below are the marks of 7 studentsin a science test.6, 7, 8,7,9,9,10Use the data to answer the following questions.A. What is the range of the data?B. Find the mean of the marks obtained by the7 students.

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Writing the numbers in increasing order

6,7,7,8,9,9,10

A. Range : 6-10

B. Mean = 6+7+7+8+9+9+10/7

= 56/7

= 8

Like my answer if you find it useful!

20.

a = \frac { 1 } { 2 + \sqrt { 3 } } \text { and } b = \frac { 1 } { 2 - \sqrt { 3 } } \text { then the value of } 2 a ^ { 2 } + 3 a b - 2 b ^ { 2 }

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21.

(i) In how many ways can 6 boys and 4 girls be seated in a round tableso that two girls never be seated together.

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Let us treat two girls as a single entity(for sake of ‘togetherness’)

Now these two girls can be selected from 4 girls in 4c2 ways.

Now,total no of members =6 boys+3 girls (2 girls has been counted as one).

We know that n members can be seated in a circle in (n-1)! Ways.

So 9 members can be seated in 8! Ways.

And the two girls who were treated as one could exchange their position in 2! Ways.

So total no of ways so that 2 girls be seated together= [(8!×2!×4c2)]

So,required answer is 9!-(8!×2!×4c2)

22.

n(S)1113. 6 boys and 6 girls sit in a row randomly, find the probabilityEx.t all the 6 girls sit together.

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23.

Example 14. In how many ways 8 boys and 6 girls can be arranged in a row sothat no two girls are together?[R.G.T.U. 2007

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One boys one girlb- g-b-g-b-g-b-g-b-g-b-g

two boys then one boy one girl b-b-g-b-g-b-g-b-g-b-g-b-g-b

one boys one girl b-g b-g

one boys one girlb- g b- g

24.

allsEx. 13. 6 boys and 6 girls sit in a row randomly, find the probability[L. I. T.79]that all the 6 girls sit together.

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25.

Find the actual lower class limits, upper classlimits and the mid-values of the classes10- 19, 20 - 29, 30-39 and 40 - 49.

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26.

Find the sum of the lower limit of the median class and the upper limit of the modalclass :Classes10-2020--3030-4040-5060 - 7050-60I 7FrequencyI 519

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27.

(1) UUUUUU2If 4x+i (3x-y) = 3+i(-6), where x and y are real numbers, then find the values of xand y.Seatt

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By comparison😊

we compare the real part and the imaginary parts so we got this value

4x= 3 so x= 3/43x-y= -63(3/4)-y= -69/4+6= y33/4= y

thanks

28.

14, k का मान ज्ञात कीजिए यदि बिन्दुk, 2-2 k ),(-k+1,2 k) तथा(-4-k, 6-2 k)संरेख हैं।

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29.

owiing qlestions: (Any One)04From the top of a lighthouse, 100 m above sea level, the angle of depression of aship, sailing towards it, changes from 30 to 450. Determine the distance trathe ship during the period of observation.i)velled by

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30.

26. A ship sails out to an island at the rate of 15 km/hr and sails back to thestarting point at 10 km/hr. Find the average sailing speed for the wholejourney

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31.

Exam. 15.01.2017) (IInd Sitting)43. Three Science classes A, B andC take a Life Science test. Theaverage score of class A is 83.The average score of class B is76. The average score of class Cis 85. The average score of classA and B is 79 and average scoreof class B and C is 81. Then theaverage score of classes A, B andC is

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32.

The points A(1,-2), B(2,3), C (k,2) and D(-4,-3) are the vertices of a parallelogram. Findthe value of k.ORFind the value of k for which the points (3k-1,k-2), (k,k-7) and (k-1,-k-2) are collinear

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33.

r,s and t are real numbers and r v s, then show that the roots of the equation(r-s)x2+7r+ s)x-3(r-s)-0 are real and unequal.

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34.

30. Find the value (s) of k for which the points (3k -1,k-2),(k, k -7) and (k -1,-k-2)CBSE 2014]are collinear

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35.

3 2i7 +4iFind real numbers A and B, if A + iB

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36.

43)If sin θ =, then find cose

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37.

Find the distance between the points (a cos θ, a sin e) and (-a sin θ, a cose).

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38.

2 keagecR) — ((atBLeed) =P st R (aBecAr 20 2)

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LHS = (tanA + cosecB)^2 - (cotB-secA)^2 = (tan^2A + 2tanAcosecB + cosec^2B) - (cot^2B - 2cotBsecA + sec^2A) = tan^2A + cosec^2B + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA

Substituting (tan^2A = sec^2A - 1) and (cosec^2B = 1 + cot^2B) in the above step: (sec^2A - 1) + (1 + cot^2B) + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA = 2tanAcosecB + 2cotBsecA = 2(tanAcosecB + cotBsecA)

RHS = 2tanAcotB(cosecA + secB) = 2tanAcotB((1 / sinA) + (1 / cosB)) = 2(tanA / sinA)cotB + 2tanA(cot B / cosB) = 2(sinA / (cosAsinB))cotB + 2tanA(cosB / (sinBcosB)) = 2(1 / cosA)cotB + 2 tanA(1 / sinB) = 2secAcotB + 2tanAcosecB = 2(secAcotB + tanAcosecB)

39.

6. If a- cose + i sin e, then find the value of(1- a)

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40.

For any two vecioShow that the points A(-2i +3j +5k), B(i+2j +3 k) and C (7i k)are collinear.

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41.

6, Find real numbers x and y if (x-iy) (3+5i) is the conjugate of-6-24i.

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42.

: REAL NUMBERS:1.The LCM of two numbers is 760 and their product is 6080. Find their HCFd LcE of hers to he 378 and 18 respectively?

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LCM*HCF=product of two numbershenceHCF=6080/760=8hence HCF is 8

43.

. As observed from top of light house, 100m high above the sea levdepression of a ship sailing directly towards it, changes from 30 to 60distance travelled by the ship during the period of observation.ind the

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thank you

44.

Q22. From the top of a vertical tower, the angles of depression of two cars, in thesame straight line with the base of the tower at an instant are found to be 30° and60°. If the cars are 100m apart and are on the same side of the tower, find theheight of the tower.

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45.

Real NumbersV Find the HF of 13.5 and 345 wing Euclid'sdivision lemma.

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By using Euclid's division algorithm,

Since 345 > 135, we've to divide 345 by 135.

By usingEuclid's division lemma,

345 = 135 * 2 + 75

Since, r ≠ 0, we've to divide 135 by 75 to get,

By usingEuclid's division lemma,

135 = 75 * 1 + 60

Since r ≠ 0, we've to divide 75 by 60 to get,

By usingEuclid's division lemma,

75 = 60 * 1 + 15

Since, r ≠ 0, we've to divide 60 by 15 to get,

By usingEuclid's division lemma,

60 = 15 * 4 + 0

Since, r = 0, the divisor of the last step will be the divisor of the given two numbers.

Therefore, 15 is the HCF of 135 and 345.

46.

3 \times 8+6 \times 11+9 \times 14+\ldots

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47.

\frac { 3 \times 7 \times 11 ^ { 8 } } { 21 \times 11 ^ { 3 } }

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48.

118/517 (Basement), Kaushalpuri, NealIn the given figure. ABCD is a quadrilateral. A line through D, parallel to AC, MeetsProduced in PProve that: ar (A ABC)- ar (quad. ABCD)

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There is something wrong in the question.

In the given figure triangle ABC is a part of quadrilateral.

So area of ∆ABC will always be less than area of quadrilateral ABCD.

49.

\frac { 3 \times 7 ^ { 2 } \times 11 ^ { 8 } } { 21 \times 11 ^ { 3 } }

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3×7^2×11^8/21×11^33×7^2×11^8=3.15×10^1021×11^3=279513×7^2×11^8/21×11^3=3.15×10^10/27951=1126972.201=1.126×10^6

50.

If cosec O=18+! FindBo Sin Ox cose

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