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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. |
Write the nth term of the sequence1. 3,7,11,15,.....2. 4,8,12,16,.....3. 7, 10, 13,16,.....4. -2,-4,-6,-8,......5. ⅔, ⁴/³, ⁶/³, ⁸/³,.....6. ¼,⅖,³/⁶,⁴/⁷,.......7. 1,2,4,8,.....8. 2,1,0,-1,......9. -5,-9,-13,-17,....10. 12,7,2,-3,...... |
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Answer» Used formula is an = a + (n - 1)d 1. 4n - 1 2. 4n 3. 3n + 4 4. -2n 5. 2n/3 6. n/(n+3) 7. 2n-1 8. 3 - n 9. -4n - 1 10. -5n + 17 |
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| 2. |
What is first five natural number |
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Answer» The first five natural numbers are 1, 2, 3, 4, and 5. |
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| 3. |
Find the square of:(i) \( (a+4 b) \) |
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Answer» by using formula:(a+b)² = a² + b² + 2ab we get: a² + (4b)² + 2(a)(4b) =a² + 16b² + 8ab (a+4b)2 = (a)2 +2(a)(4b) + (4b)2 => a2 +8ab + 16b2 |
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| 4. |
lcm of 10,15,20 |
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Answer» First, we divide each by 2 if the number divides by 2 then we write the quotient otherwise we write the same number in the next line. Next, we will divide the numbers by 2 and the same procedure of writing is carried out. Again, next we divide by 3 and the same as above and again we divide by 5 and hence we obtain 1 in the last row. This is the end of the division procedure. We have to divide till we get 1 in the next row. |
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| 5. |
Subtract (– 540) from the sum of ( – 238) , 436 and ( – 301 ) |
| Answer» Answer: - 643 | |
| 6. |
For a GP, if `(m+n)^(th)` term is p and `(m-n)^(th)` term is q, then `m^(th)` term is ……. .A. pqB. `sqrt(pq)`C. `p/q`D. `q/p` |
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Answer» Correct Answer - B In any GP if `(m+n)^(th)` term i.e., `a_(m+n)=p` and `(m-n)^(th)` term i.e., `a_(m-n)=q` then `a_(m)=sqrt(pq)` |
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| 7. |
The equation of normal to the curve y = `log_(e)` x at the point P( 1, 0) is ..... .A. 2x+y=2B. x-2y=1C. x-y=1D. x+y=1 |
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Answer» Correct Answer - D We have, y`=log_(e)xrArr(dy)/(dx)=1/x` `therefore` Slope of normal at P(1,0) `=-1/((dy//dx)_((1,0)))` `=(-1)/1`=-1 `therefore` Equation of normal at P(1,0) `rArry-0=(-1)(x-1)` `rArry=-x+1` `rArrx+y=1` |
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| 8. |
If A non-singular matrix such that `(A-2l)(A-4l)=0` then A=`8A^(-1)=` . .A. IB. 0C. 3ID. 6I |
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Answer» Correct Answer - D We have `absAne0` `rArrA^(-1)` is exists Since,(A-2I)(A-4I)=0 `rArrA^(2)-A(4I)-2I(A)+8IdotI=0` `rArrA^(2)-4AI-2AI+8I=0` `rArrA^(2)-6AI+8I=0` `rArrA^(2)-6A+8I=0` On pre multiply both sides by `A^(-1)`, we get `A^(-1)A^(2)-6A^(-1)A+8A^(-1)I=0` `rArrIA-6I+8A^(-1)`=0 `rArrA-6I+8A^(-1)=0` `rArrA+8A^(-1)=6I` |
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| 9. |
Deagrams which are used to compara the net deviation of related varibles with respect to time and location are :A. deviation bar diagramsB. simple bar diagramsC. multiple bar diagramsD. pie diagrams |
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Answer» Correct Answer - a A. |
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| 10. |
Which of the following statements is correct ?A. Bars may be vertical or horizontalB. Breadth of the bars remains the sameC. All bars are based on some common base lineD. All of these |
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Answer» Correct Answer - d D. |
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| 11. |
If G(3, -5, r) is centroid of triangle ABC where A(7, - 8, 1), B(p, q, 5) and C(q + 1, 5p, 0) are vertices of a triangle then values of p, q, r are respectively ...A. `6, 5, 4`B. `-4,5,4`C. `-3,4,3`D. `-2,3,2` |
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Answer» Correct Answer - D Here,`((7+p+q+1)/3,(-8+q+5p)/3,(1+5+0)/3)` `rArr((8+p+q)/3,(-8+5p+q)/3,2)=(3,-5,r)` `rArr(8+5+q)/3=3,(-8+5p+q)/3=-5,2=r` `rArr8+p+q=9,-8+5p+q=-15,r=2` `rArrp+q=1,5p+q=-7,r=2` `rArrp=-2,q=3,r=2` |
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| 12. |
`int1/((x^(2)+1)^(2))`dx=…A. `tan^(-1)x-1/(2x(x^(2)+1))+c`B. `1/2tan^(-1)x+x/(2(x^(2)+1))+c`C. `tan^(-1)x+1/(x^(2)+1)+c`D. `tan^(-1)x+1/(2(x^(2)+1))+c` |
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Answer» Correct Answer - B `int1/(x^(2)+1)dx=1/(x^(2)+1)` `int1dx-int(d/(dx)(1/(x^(2)+1))int1dx)dx` `=x/(x^(2)+1)-int(-2x)/((x^(2)+1)^(2))xdx` (integration by parts) `=x/(x^(2)+1)+2int(x^(2)+1-1)/((x^(2)+1)^(2))dx` `=x/(x^(2)+1)+2intx/((x^(2)+1))dx-2int(dx)/((x^(2)+1)^(2))` .....(i) And also we know that `int(dx)/(x^(2)+1)=tan^(-1)(x)` From Eqs. (i) and (ii), we get `tan^(-1)x=x/(x^(2)+1)+2tan^(-1)x-2int(dx)/((x^(2)+1)^(2))` `rArr2int(dx)/((x^(2)+1)^(2))=tan^(-1)x+x/(x^(2)+1)` `rArrint(dx)/((x^(2)+1)^(2))=1/2tan^(-1)x+x/(2(x^(2)+1))=c` |
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| 13. |
If `theta=(17pi)/3` then, `tantheta-cottheta`= …A. `1/(2sqrt3)`B. `(-1)/(2sqrt3)`C. `2/(sqrt3)`D. `-(2)/(sqrt3)` |
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Answer» Correct Answer - D We have, `theta=(17pi)/3` `thereforetantheta-cottheta=tan((17pi)/3)-cot((17pi)/3))` `=tan(5pi+(2pi)/3)-cot(5pi+(2pi)/3)` `=tan(2pi)/3-cot(2pi)/3` `=tan(pi-pi/3)-cot(pi-pi/3)` `=-tanpi//3+cotpi//3` `=-sqrt3+1/sqrt3)=(-3+1)/(sqrt3)=(-2)/(sqrt3)` |
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| 14. |
If `A=[(1+2i,i),(-I,1-2i)]`, where `i=sqrt(-1)`, then `A(adjA)=`. . .A. `-2l`B. 2lC. 5 lD. 4l |
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Answer» Correct Answer - D We have, `A = [{:(1 +2i, i),(-I,1-2i):}]` `:. |A| = |{:(1 + 2i, i),(-I,1-2i):}| = (1 + 2i) (1 - 2i) + i^(2)` `= 1 - (2i)^(2) + (-1) = 1 + 4 - 1 = 4` `:. A` (adj A) = |A| I = 4I |
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| 15. |
If `x^(2) = e^(x-y) `, then `(dy)/(dx)` at x = 1 is ……..A. eB. 1C. 0D. `-1` |
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Answer» Correct Answer - C Given have, `x^(y) = e^(x - y)` taking log on both sides, we get y logx = (x - y) log e = (x - y) When x = 1, then (log 1) = (1 - y) `implies y = 1` On differentiating both sides, `y((1)/(x)) + log x. (dy)/(dx) = 1 - (dy)/(dx)` `implies (dy)/(dx) (logx + 1) = 1 - (y)/(x)` `implies (dy)/(dx) (logx + 1) (x - y)/(x)` `implies (dy)/(dx) = ((x - y))/(x(logx + 1))` When x =1, then `((dy)/(dx)) = (1 -1)/(1(log1 +1)) = 0` |
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| 16. |
In the case of a periodic function possessing half-wave symmetry, which Fourier coefficient is zero? (a) an (b) bn (c) a0 (d) none of above. |
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Answer» Correct option (c) a0 |
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| 17. |
Which of the following periodic function possesses even symmetry? (a) cos3t (b) sin t (c) t . cos 50 t (d) (t + t2 + t5). |
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Answer» Correct option (a) cos3t |
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| 18. |
A periodic function has zero average value over a cycle and its Fourier series consists of only odd cosine terms. What is the symmetry possessed by this function. (a) even (b) odd (c) even quarter-wave (d) odd quarter-wave |
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Answer» (c) even quarter-wave |
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| 19. |
If the Fourier coefficient bn of a periodic function is zero, then it must possess ______ symmetry. (a) even (b) even quarter-wave (c) odd (d) either (a) and (b). |
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Answer» (d) either (a) and (b). |
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| 20. |
Find k that value of the function in the interval is continuousf(x)= k cos x, x<=0 2x -k, x>0 |
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Answer» f(x) \(=\begin{cases}k\,cos\,x;&x\leq0\\2^x-k;&x>0\end{cases}\) f(0-) \(=\underset{x\rightarrow0}{Lim}\) k cos x = k cos 0 = k f(0+) \(=\underset{x\rightarrow0}{Lim}\) 2x - k = 20 - k = 1 - k ∵ f(x) is continuous at x = 0 ∴ f(0-) = f(0+) ⇒ k = 1 - k ⇒ 2k = 1 ⇒ k = 1/2. |
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| 21. |
In a Fourier series expansion of a periodic function, the coefficient a0 represents its (a) net area per cycle(b) d.c value(c) average value over half cycle (d) average a.c value per cycle |
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Answer» (b) d.c value |
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| 22. |
A given function f (t) can be represented by a Fourier series if it(a) is periodic (b) is single valued (c) has a finite number of maxima and minima in any one period(d) all of the above |
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Answer» (d) all of the above. |
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| 23. |
limf(x) =x+x2 +x3 -3/x-1 where x tends to 1 |
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Answer» \(\underset{x\rightarrow 1}{lim}\) f(x) = \(\underset{x\rightarrow 1}{lim}\) \(\frac{x+x^2+x^3-3}{x-1}\) (0/0 type) = \(\underset{x\rightarrow 1}{lim}\) \(\frac{x^3+x^2+x-3}{x-1}\) = \(\underset{x\rightarrow 1}{lim}\) \(\frac{(x-1)(x^2+2x+3)}{x-1}\) = \(\underset{x\rightarrow 1}{lim}\) (x2+2x+3) = 12 + 2 x 1 + 3 (By taking limit) = 1+2+3 = 6 |
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| 24. |
If the average value of a periodic function over one period is zero and it consists of only odd harmonics then it must be possessing _______ symmetry. (a) half-wave (b) even quarter-wave (c) odd quarter-wave (d) odd. |
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Answer» (a) half-wave |
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| 25. |
If in the Fourier series of a periodic function, the coefficient a0 = 0 and an = 0, then it must be having _______ symmetry. (a) odd (b) odd quarter-wave (c) even (d) either (a) or (b). |
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Answer» (d) either (a) or (b). |
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| 26. |
Evaluate: lim (x→0) ((cos 9x – cos 7x)/(cos 3x – cos5x)) |
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Answer» \(\lim\limits_{x \to 10} \) \(\cfrac{cos9x-cos7x}{cos3x-cos5x}\) (\(\frac{0}{0}\)type) = \(\lim\limits_{x \to 10} \) \(\cfrac{-9sin9x +7sin7x}{-3sin3x+5sinx5x}\) (By applying D.L.H Rule) = \(\lim\limits_{x \to 10} \) \(\cfrac{-9\frac{sin9x}{9x}\times9x+7\frac{sin7x}{7x}\times7x}{-3\frac{sin3x}{3x}\times3x+5\frac{sin5x}{5x}\times5x}\) = \(\lim\limits_{x \to 10} \) \(\cfrac{x\Big(-81\frac{sin9x}{9x}+49\frac{sin7x}{7x}\Big)}{x\Big(-9\frac{sin3x}{3x}+25\frac{sin5x}{5x}\Big)}\) = \(\cfrac{-81\lim\limits_{x \to 10}\frac{sin9x}{9x}+49\lim\limits_{x \to10}\frac{sin7x}{7x}}{-9\lim\limits_{x \to 10}\frac{sin3x}{3x}+25\lim\limits_{x \to 10}\frac{sin5x}{5x}}\) = \(\frac{-81+49}{-91+25}\) = \(\frac{-32}{16}\) = -2 (∵\(\lim\limits_{x \to 10}\) \(\frac{sinax}{ax}\) = 1) |
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| 27. |
When the negative half-cycle of a complex waveform is reversed, it becomes identical to its positive half-cycle. This feature indicates that the complex waveform is composed of (a) fundamental (b) odd harmonics (c) even harmonics (d) both (a) and (b) (e) both (a) and (c) |
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Answer» (d) both (a) and (b) |
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| 28. |
A periodic waveform possessing half-wave symmetry has no (a) even harmonics (b) odd harmonics (c) sine terms (d) cosine terms |
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Answer» (a) even harmonics |
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| 29. |
An unbalanced system of 3-phase voltages having RYB sequence actually consists of (a) a positive-sequence component (b) a negative-sequence component (c) a zero-sequence component (d) all of the above. |
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Answer» (d) all of the above. |
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| 30. |
Given J-tube has 2.4 mL of air at a pressure of 1 atm. On adding mercury, volume of air is reduced to 1.9 mL as shown. Difference in the level of mercury in two columns is: A. (a)700 mmB. (b)200 mmC. (c )900 mmD. (d)760 mm |
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Answer» Correct Answer - B Atmospheric pressure `=76` mmHg `implies P_(1)V_(1)=P_(2)V_(2)` `P_(2)=(P_(1)V_(1))/V_(2)=(760xx2.4)/1.9=960` mmHg `:.` Difference in column `=960-760=200` mmHg |
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| 31. |
Why is Transformer Rated in KVA, not in KW. |
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Answer» Copper losses ( I²R) depends on Current which passing through transformer winding while Iron Losses or Core Losses or Insulation Losses depends on Voltage. So the Cu Losses depend on the rated current of the load so the load type will determine the power factor, that is why the rating of Transformer in kVA, and not in kW. |
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| 32. |
Concerning Tc-99m: a. It is the daughter nucleus of Mo-99 and is produced following the emission of gamma rays b. Tc-99m and Tc-99 differ in their half-lives c. It emits gamma rays with 140 keV d. Pertechnetate (TcO4-) is used in imaging to localize Meckel’s diverticulum e. It emits beta particles |
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Answer» a. False. The parent compound of Tc-99m is Mo-99, but it is produced following the emission of a negative beta particle. Tc-99m subsequently emits gamma rays to form its isomer Tc-99. b. True. Tc-99m and Tc-99 are isomers and differ in their half-lives and energy states. They have the same atomic and mass numbers. c. True. Tc-99m emits gamma rays with 140 keV, which are suitable for imaging. d. True. TcO4- is trapped in the thyroid gland and the salivary glands, but is also taken up by the gastric mucosa or by ectopic gastric mucosa (Meckel’s diverticulum). e. False. Tc-99m results in the emission of pure gamma rays. |
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| 33. |
Which of the following are true regarding the sensitivity of collimators? a. It is a measure of the fraction of total gamma rays falling on the collimator that pass through the holes to the crystalb. Increasing the number of holes increases the sensitivity c. Increasing the sensitivity reduces the patient dose d. Narrower holes result in increased sensitivity e. Longer holes reduce sensitivity |
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Answer» a. True. b. True. Increasing the number of holes results in a higher fraction of gamma rays passing through the collimator. c. True. The higher the sensitivity, i.e. the higher the number of gamma rays passing through the collimator to form the image, the lower the amount of radio nuclide needed and hence the lower the patient’s dose, but this is usually at the expense of the resolution. d. False. Wider holes allow a higher fraction of gamma rays to pass through. e. True. Longer holes result in a narrower angle of acceptance resulting in more gamma rays being stopped by the collimator. |
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| 34. |
Regarding the effective half-life of a radionuclide: a. It takes into account the biological and physical half-lives b. It decreases as a result of renal failure c. The dose received by an organ is proportional to the effective half-life d. It differs among individuals for a given radio pharmaceutical e. It is generally increased in patients with liver failure |
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Answer» a. True. The effective half-life is calculated from both the biological and physical half-lives. b. False. The effective half-life may not depend on renal excretion e.g. Kr-81m, and, even if it does, renal failure results in an increased biological half-life and therefore an increased effective half-life. c. True. The dose to an organ increases in proportion to the effective half-life. d. True. This is mainly due to the fact that the biological half-life varies among individuals according to their disease state and physiological factors. e. False. This will only be the case in those radio pharmaceuticals that are excreted by the biliary system. |
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| 35. |
Desirable properties of a radionuclide for imaging include: a. High levels of cross-reactivity with blood proteins b. A half-life measured in hours, not minutes or days c. It decays to a stable daughter d. Low activity per unit volume e. Emission of beta particles |
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Answer» a. False. This would result in high blood pool activity, which may reduce uptake in the target organ. b. True. The half-life should ideally be a few hours, roughly equal to the time from injection to scanning. c. True. Decay should be to a stable daughter in order to minimize the dose to the patient. d. False. A radionuclide should have a high specific activity, i.e. high activity per unit volume. e. False. Beta particles act as secondary electrons and deposit an unnecessary dose in patients. |
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| 36. |
∫tan2 x/2 dx = ?(a) tan x/2 - x + c (b) tan x/2 + x + c (c) 2tan x/2 + x + c (d) 2tan x/2 - x + c |
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Answer» (d) 2tan x/2 - x + c |
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| 37. |
Company A sells goods of Rs. 5,00,000 in the third quarter of the Fiscal year 2020-21. The cost price and additional expenses of goods are accounted to be Rs. 3,25,000. What will be the profit after tax if 25% of tax is levied?1. Rs. 1,31,2502. Rs. 1,75,0003. Rs. 1,40,2504. Rs. 1,30,750 |
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Answer» Correct Answer - Option 1 : Rs. 1,31,250 Given: S.P. = Rs. 500,000 C.P. = Rs. 325,000 Tax% = 25% Formula used: Profit = S.P. – C.P. Profit after tax = Profit × (1 – (Tax ÷ 100)) Calculations: Calculating profit earned ⇒ Profit = 5,00,000 – 3,25,000 ⇒ Profit = 1,75,000 ⇒ Profit after tax = 1,75,000 × (1 – (25 ÷ 100)) ⇒ Profit after tax = 1,75,000 × 3 ÷ 4 ⇒ Profit after tax = 43,750 × 3 ⇒ Profit after tax = 1,31,250 ∴ Profit after tax is Rs. 1,31,250 |
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| 38. |
Expenses paid but not accounted as expenses means ________. |
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Answer» expenses paid in advance but services yet to be enjoyed An accrued expense is a liability that represents an expense that has been recognized but not yet paid. Not every transaction requires an immediate exchange of cash for goods and services. Sometimes, especially when there is a prolonged history of ongoing transactions between two parties, formal invoicing and payment requirements can occur after the expense associated with the transaction has been recognized. Accured Expense |
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| 39. |
The scaler product of the vectors 5i + j - 3 k and 3i - 4j + 7k is (a) 10 (b) –10 (c) 15 (d) –15 |
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Answer» Correct option: (b) –10 |
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| 40. |
The scalar product of 5i + j - 3k and 3i - 4j + 7k is (a) 10(b) -10(c) 15(d) -15 |
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Answer» Answer is (b) -10 |
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| 41. |
Write down the number of significant digits in a)123.3. b)2004.4 c )400 d)0.0003 |
| Answer» (a) 4 (b) 5 (c) 1 (d) 4 | |
| 42. |
What is product B in the given reaction ? |
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Answer» The correct option is (ii) Nitrobenzene
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| 43. |
Ambidentate ligands like NO2- and SCN- are : a) unidentate b) didentate c) polydentate d) has variable denticity |
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Answer» The correct answer is: A |
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| 44. |
What is difference between an emulsion and a gel? |
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Answer»
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| 45. |
At a distance of `5 cm and 10 cm` outward from the surface of a uniformly charged solid sphere, the potentials are `100 V and 75 V`, repectively. Then.A. potential at its surface is `150 V`B. the charge on the sphere is `(5//3)xx10^(-10)C`C. the electric field on the surface is `1500 V//m`D. the electric potential at its centre is `225 V` |
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Answer» Correct Answer - A::C::D Potential at 5 cm from surface`=(KQ)/(R+5)=100` Potential at `10` cm from surface `=(KQ)/(R+10)=75 rArr R=10` cm. `:.` Potential at surface `=(KQ)/(R)=(100xx15)/10=150 V` Electric field on surface`=(KQ)/R^(2)=(100xx15Vxxcm)/(100 cm^(2))` `=1500 V//m` |
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| 46. |
Bill ________ play tomorrow. A) going to B) can C) is going D) can to |
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Answer» Correct option is B) can |
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| 47. |
At the post office he asked ________ . A) stamps B) some stamps C) four stamps D) for four stamps |
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Answer» Correct option is D) for four stamps |
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| 48. |
The girl ________ house he visited was Elizabeth. A) of which B) of whom C) which D) whose |
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Answer» Correct option is D) whose |
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| 49. |
The actor ____ house we visited was Robert Redford. A) which B) whose C) who D) of whom |
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Answer» Correct option is B) whose |
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| 50. |
Answer the following questions according to the text.Elizabeth parked the car and then went into the busy station to meet Jane who was going to spend the weekend with her. Elizabeth’s friends often came for the weekend, but there was something a little different about Jane’s visit. She and Jane hadn’t seen each other for 15 years. While at the university, they had shared a flat together for 2 years but then each had married and Jane and her husband had lived abroad a great deal. Elizabeth began to ask herself. ‘Will we even be able to recognize each other after all these years?” Right then she saw Jane walking towards her and smiling straight at her with the old unforgettable smile quite unchanged. 1. The first thing Elizabeth noticed about Jane when they met at the station was _____. A) that she had become quite old. B) how well she was looking. C) that she had not forgotten how to smile. D) her smile which had not changed. E) that she had already been married. 2. Elizabeth’s friendship with Jane _____. A) went back to their university years.B) lasted only for two years when they were at the university. C) ended soon after they had each married. D) has always been envied by all their friends. E) found its best expression in their weekend visits to each other.3. Elizabeth is worried _____. A) because Jane’s husband may have changed a great deal after all these years. B) in case she won’t be able to find a place to park the car. C) about Jane and her husband who are going to live abroad. D) in case Jane will change her mind about coming for the weekend. E) in case after a separation of so many years they may both fail to recognize each other. |
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Answer» 1 D) her smile which had not changed. 2 A) went back to their university years 3 E) in case after a separation of so many years they may both fail to recognize each other. |
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