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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. |
The area of a rectangle is 108 cm2. Ifits lengthratio of its breadth to its length is12 cm, the(a) 4: 3(c) 3: 4(b) 12:108 |
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Answer» Area of rectangle = L×BGiven L = 12108 = 12 × BB = 108/12 = 9So ratio of breadth to LengthSo L : B = 9 : 12 = 3 : 4C) Option is correct. |
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| 2. |
108 |
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| 3. |
() 6x-108 |
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Answer» 6x -10 = 8.=> 6x = 18=> X = 3. X/3 -7 = 4=> x/3 = 11=> X = 33. Please hit the like button if this helped you |
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| 4. |
108/2 |
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Answer» 108÷2=54 |
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| 5. |
(e)\frac{1}{3} \cdot \frac{1}{9}, \frac{1}{27}、 |
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| 6. |
5 \sqrt 3 %2B 2 \sqrt 27 %2B \frac 1 \sqrt 3 |
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Answer» 5√3 + 2√27 + 1/√3 5√3 + 6√3 + 1/√3 × (√3/√3) 11√3 + √3/3 34√3/3 |
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| 7. |
k Prove that a diameter AB of a circle bisects all those chords which are parallel to thetangent at the point A.INCERT EXEMPLAR |
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| 8. |
x^2 %2B 1/(x^2)=27 |
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| 9. |
1 \cdot \overline 27 |
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| 10. |
2 \cdot ( \frac { 2 } { 3 } ) ^ { 15 } \div ( \frac { 2 } { 3 } ) ^ { k } = \frac { 8 } { 27 } |
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Answer» 8/27=2^3/3^3on the LHS2/3)^15-khence equate the exponents15-k=3k=12 thanks |
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| 11. |
A regular hexagon has a side 84 A regular hexagon has a side 8 cm. Find the area of hexagon.area of hexagonidor as 4 m. 6 m and 8 m |
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| 12. |
A reqular hexagon has a side 8 cm. Find the area of hexagon. |
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| 13. |
9.0.69hord of a circle with centre O, AOC is a diameter and AT is the tangent at Aa1 AB is awn in Fig. 8.70. Prove that <BAT = <ACB.asINCERT EXEMPLAR] |
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Answer» Let ∠ACB = x and ∠BAT = y.A tangent makes an angle of 90 degrees with the radius of a circle,so we know that ∠OAB + y = 900……..(1)The angle in a semi-circle is 90, so ∠CBA = 900. ∠CBA + ∠OAB + ∠ACB = 1800 (Angle sum property of a triangle) Therefore, 90 + ∠OAB + x = 1800So, ∠OAB + x = 900………….(2)But OAB + y = 900, Therefore, ∠OAB + y = ∠OAB + x ………….[From (1) and (2)]x = y.Hence ∠ACB = ∠BAT. |
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| 14. |
\frac{3.27^{x+1}+27.3^{3 x}}{3.3^{3 x+2}-\frac{1}{3} .27^{x+1}} |
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| 15. |
3^(-5*x %2B 2)/27^(1/3)=81 |
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Answer» your answer is x= -5/3 3^(-5x+2) / 27⅓=813^(-5x+2) / 3³⅓=3⁴3^(-5x+2) -1=3⁴ since base are equal, (-5x+2)-1=4-5x+1=4-5x=4-1x=-3/5 answer is = –3/5.thanks |
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| 16. |
3^(x - 1)=1/27 |
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Answer» Like if you find it useful |
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| 17. |
o~1o-1हरwn |
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Answer» 8/15 + 4/15 = 1 2 / 1 5 -4/5 is the right answer. 8 / 15 + 4 / 15 = - 1 2 / 15 -12/15 is the right answer -12/15 is your answer correct answer is 12/15 8/15 + 4/15= 1 2 / 1 5 -12/15 is the correct answer -(8/15)-(4/15)=-12/15 =-4/5 ans. -4/5 is the correct answer of it. 8/-15 +4/-15=(8+4)/-15=12/-15=4/-5 answer _12/15 is correct answer (-8-4)/15-12/15is correct answer -8/15 + -4/15=-8/15-4/15= -12/15 -4/5 is the correct ans ok -4/5 is the correct answer -4/5 is the correct answer |
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| 18. |
wn~ |2ilन] |
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Answer» q=3/2 is right answer q= 3/2 is the correct answer of the given q=3/2 is the right answer |
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| 19. |
3 ^ { x - 1 } = \frac { 1 } { 27 } |
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Answer» 3^(x-1)=1/27=1/(3^3)=3^(-3)so x-1=-3so x=-3+1=-2 |
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| 20. |
z?叶teornt«o(Yf.thegume?fi.rs tη tenm S+Wn |
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Answer» Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)We can consider the 9th term as the m th term.Let’s consider the ratio these two AP’s m th terms as am: a’m→(2)Recall the nth term of AP formula, an= a + (n – 1)dHence equation (2) becomes,am: a’m= a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getam: a’m= [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1: S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].now substitute the value of m as 9so the answer becomes120/95 |
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| 21. |
| + छा है. (] -tan A ) ¢ wn’—-——T— — - . A(x) + oot IA ) 1 -om A ) |
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Answer» L.H.S1+tan^2A/1+cot^2A=(1+sin^2A/cos2A)/(1+cos^2A/sin^2A)=((cos^2A+sin^2A)/cos^2A)/(sin^2A+cos^2A)/sin^2A=(1/cos^2A)/(1/sin^2A)=1/cos^2A*sin^2A/1=sin^2A/cos^2A=tan^2A M.H.S(1-tanA/1-cotA)^2=(1+tan^2A-2tanA)/(1+cot^2-2cotA)=(sec^2A-2tanA)/(cosec^2A-2cosA/sinA)=(sec^2A-2*sinA/cos A)/(cosec^2A-2cosA/sinA)=(1/cos^2A-2sinA/cosA)/(1/sin^2A-2cosA/sinA)=[(1-2sinAcosA)/cos^2A]/[(1-2cosAsinA)sin^2A]=(1-2sinAcosA)/cos^2A*sin^2A/(1-2sinAcosA)=sin^2A/cos^2A=tan^2A R.H.S=tan^2A Hence L.H.S=M.H.S=R.H.S thanks |
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| 22. |
: W_,._ ‘7 Mjflv&w’_‘cte ol a WWWMW Fend 4/Wn "/ 1.सह 8L% s ANt<L |
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| 23. |
12. Given S = !! [2a + (n = 1 wn, find d when S-800,n = 16, and a = 5, |
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Answer» thank you mod neethu |
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| 24. |
R T T AT,‘r e s ‘wn w1 e 30 मकरe pades 4§§05 ot 10 penda Gहूo |
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| 25. |
50. Which one of the following can befolded to form the given solid?(2)WN(4) |
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Answer» 1) is the correct answer 1) is the correct answer 1) is correct answer 1) the correct good answer a is the right answer |
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| 26. |
\frac { 3 \times ( 27 ) ^ { n + 1 } + 9 \times 3 ^ { ( 3 n - 1 ) } } { 8 \times 3 ^ { 3 n } - 5 \times ( 27 ) ^ { n } } |
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| 27. |
10. The rainfall in mm) in a city onus of a certain week was recorded as follows:DayRainfall22 2221Monday0.0Tuesday12.2Wednesday2.1Thursday0.0Friday20.5Saturday5.3Sunday1.0O Find the range of the rainfall in the above data() Find the mean rainfall for the week.i) On how many days was the rainfall less than the mean rainfall ? |
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| 28. |
8 The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:Mon Tue Wed Thurs Fri Sat SunRainfall .0 12.2 2. 0. 2055.5 1.0(in mm)ind the range of the rainfall in the above data.Findthe mean rainfall for the week.(ii) On how many days was the rainfall less than the mean rainfall. |
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| 29. |
III LIETED LIHULLULL.10. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:DayRainfall(in mm)Monday0.0Tuesday12.2Wednesday2.1Thursday0.0Friday20.5Saturday5.3Sunday1.0(i) Find the range of the rainfall in the above data.(ii) Find the mean rainfall for the week.(iii) On how many days was the rainfall less than the mean rainfall ? |
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| 30. |
\frac{2^{3} \times 3^{3} \times 5}{2^{2} \times 3^{2}} |
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| 31. |
\frac{3 \times 27^{n+1}+9 \times 3^{3 n-1}}{8 \times 3^{3 n}-5 \times 27^{n}} |
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Answer» Part 1 Part 2 |
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| 32. |
\frac { 3 \times 27 ^ { n + 1 } + 9 \times 3 ^ { 3 n - 1 } } { 8 \times 3 ^ { 3 n } - 5 \times 27 ^ { n } } |
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| 33. |
\frac{9^{n} \times 3^{2} \times 3^{n}-(27)^{n}}{3^{3 m} \times 2^{3}}=3^{-3}, \text { prove that }(m-n)=1 |
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| 34. |
\left( \frac { 2 ^ { 6 } \times 3 ^ { 4 } } { 6 ^ { 3 } } \right) ^ { 2 } \times \left( \frac { 3 ^ { 3 } \times 2 ^ { 5 } } { 3 \times 8 } \right) ^ { 3 } |
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| 35. |
\frac { 5 ^ { 2 } \times 5 ^ { 6 } } { 5 ^ { 3 } } \times 3 ^ { 2 } \times 3 ^ { 3 } |
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| 36. |
\frac { ( 5 ) ^ { 2 n + 3 } - ( 25 ) ^ { n + 2 } } { [ ( 125 ) ^ { n + 1 } ] ^ { 2 / 3 } } |
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| 37. |
23. The bookshop of a particular scnool has 10 dozen chemistry books, 8 dozen physicsbooks, 10 dozenrespectively. Find the total amount the bookshop will receive from selling all the boksusing matrix algebra.economic books. Their selling price are Rs 80, Rs 60, and Rs 40 each |
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| 38. |
If 1, ω, ω2 are the cube roots of unity, then prove the following :3 .3 |
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| 39. |
.Ammu's father bought some note books of the same price. Total price is Rs.360.Ifthe price of each book were less by 2 rupees, we would got 2 books more. Find thenumber of books bought. Find the cost of one book. |
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| 40. |
18. A boy has 3 library tickets and 8 books of his interest in the library, Of themse &bedoes not want to borrow Chemistry part II, unless Chemistry part I is also becmowedIn how many ways can he choose the three books to be borrowed ? |
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Answer» The first option is he chooses any of the books other then chemistry part 2 = 7C3 = 35 The second option is he selects both the chemistry books and any other book from the remaining options = (2C2 * 6C1) = 6 So he can choose in a total of 41 ways. |
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| 41. |
If the side of a square is tripled, how many times the perimeter of thefirst square will that of the new square be? |
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Answer» Let side of 1st square be Aso perimeter will be 4A in New SquareSide = 3APerimeter = 4×new side= 12A Therefore new perimeter is 3 times the Original perimeter |
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| 42. |
In how many ways can one select a cricket team of cleven from 17 players inwhich only 5 players can bowl ifeach cricket team of Ilmast include exactly â |
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| 43. |
1. How many numbers between 400and 500 are exactly divisible by 12,15 and 20?(a) One(c) Three(b) Two(d) of these |
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Answer» if a number is divisible by 12 15 and 20 then it is divisible by LCM of these 3 LCM of 12 15 and 20 can be computed by factorisation 12 = 3 * 2 * 215 =3 * 520 = 5 * 2 * 2 si\o LCM = 3 * 2*2* 5 (higest power of each prime in each number)= 60 so the number is divisible by 60 now 2 multiles of 60 *60*7 and 60*8 are between 400 and 500 so number of numbers = 2 right thanks |
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| 44. |
1. Compare the fractions:57(i)and812 |
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| 45. |
1. Compare the fractions:7(i(i)and8 12 |
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Answer» 5/8 is greater beacause when we divide 5/8 we get 0.625and when we divide 12/7 we get 0.54 0.62 > 0.54 |
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| 46. |
How does Plasmodium reproduce. Is this method sexual orasexual ? |
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Answer» The protozoan Plasmodium has a complex life cycle in which asexual multiplication in the vertebrate host alternates with an obligate sexual reproduction in the anopheline mosquito |
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| 47. |
Q3.Find the modulus of the following3+i |
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Answer» (3+i+3i+3i^2)/(3-i)-(3-i-3i+3i^2)/(3+i)=(3+4i-3)/(3-i)-(3-4i-3)/(3+i)=4i/(3-i)+4i/(3+i)=(12i+4i^2+12i-4i^2)/(9-i^2)=(24i-4+4)/(9+1)=24i/10 |
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| 48. |
Compare Fractions withSame DenominatorCompare. write <, >, or =4. 67,0음824 |
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Answer» 4.<7.>10>13.<this is the answer thank you |
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| 49. |
The first ionisation energy of lithiumwill be46.46. |
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Answer» The value of the ionisation energy (2370 kJ mol-1) of lithium is much higher than hydrogen, because the nucleus now has 2 protons attracting the electrons instead of 1. Lithium is 1s^22s^1. Its outer electron is in the second energy level, much more distant from the nucleus. |
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| 50. |
38. Hydrogen has high ionisation energythan alkali metals, due to its |
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Answer» The atomic number of hydrogen is 1. It means that hydrogen has just 1 electron and an electronic configuration of K 1. This electron is close to the nucleus and it has an high electrostatic force with respect to the nucleus. So, a lot of energy is required in order to pull off that 1 valence electron. Where as, any other element after Hydrogen has valence electrons more distant from its and the valence electrons are not pulled by the nucleus with a greater force and the energy levels to pull them off from their valence shells are low. |
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