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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. |
Find angle in radians between the hands of a clock at 7:20 pm. |
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| 2. |
CD and GH are respectively the bisectors of angle ACB and angle EGF such that D and H lie on sides AB and FE of triangle ABC and triangle EFG respectively. if triangle ABC is similar to triangle FEG , show that; triangle DCB is similar to triangle HGE |
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| 3. |
Fill in the blanks:(i) A triangle has(i) The sum of the angles of a triangle is(iii) The sides of a scalene triangle are of(iv) Each angle of an equilateral triangle measures(v) The angles opposite to equal sides of an isosceles triangle are(vi) The sum of the lengths of the sides of a triangle is called itssides,angles and ....vertices.lengths. |
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| 4. |
The displacement (in metres) of a particle varies with time (in seconds) ass(t) = 3t2-5t + 2Q2.I.II.III.Find the velocity of the particle at all instances when its displacement becomes zero.Find the acceleration of the particle when the velocity is 1 m/s.Find the displacement of the particle when it comes to rest momentarily. |
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Answer» given s(t) = 3t²-5t+2 = 0.... v(t) = ds/dt = 6t-5 1) for s(t) to be zero..=> 3t²-3t-2t+2 = 0=> 3t(t-1)-2(t-1) = 0=> (t-1)(3t-2) = 0=> t = 1, 2/3 so, the Velocity when displacement is 0V1 = 6(1)-5 = 1m/s and V2 = 6(2/3) -5 = -1m/s 2)a = dv/dt = 6 m/s²... always.. 3) when particle is at rest , velocity is 0so 6t-5 = 0 => t = 5/6 so, displacement is = 3(5/6)²-5(5/6)+2 = 2.08. -4.16+2 = 4.08-4.16 = 0.08 m |
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| 5. |
7. Coffee costing 250 per kg was mixed with chicory costing 75 per kg in the ratio 5:2 for acertain blend. If the mixture was sold at 230 per kg, find the gain or loss per cent.Hint. Let 5 kg of coffee be mixed with 2 kg of chicory. |
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| 6. |
How many kilograms of tea at 50 per kg should bemixed with 35 kg of tea costing 60 per kg so as to sellthe mixture at 57 per kg without gaining or losinganything in the transaction |
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Answer» Let the amount of ₹50 tea be ‘x’ so the equation becomes 50x + 60*35 = 57*(x + 35) solve this for ‘x’ 7x = 3*35 or x= 15kg so you will need to mix15kgof the ₹50 tea |
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| 7. |
transacton.offee costing 250 per kg was mixed with chicory cosung ? 75 per kg in the ratio ocertain blend. If the mixture was sold at 230 per kg, find the gain or loss per cent.7. C |
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| 8. |
Q15) From the given figure, find the values of angles x&y,given that AB //CD50%i 3o |
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Answer» x + 50 = 180 so x = 130 now x and y are same due to parallel lines so y = x = 130 |
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| 9. |
17 p - 2 = 49 |
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Answer» Ans :- 17p - 2 = 4917p = 49 + 217p = 51p = 51/17p = 3 thank you so much |
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| 10. |
17*p - 2=49 |
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Answer» 17p-2=4917p=51p=51/17=3 17p=51p=51÷17p=3this is how it will be solved |
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| 11. |
\frac{-3 x-28}{x^{2}-49}=\frac{3}{17} |
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Answer» please show how you factorise |
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| 12. |
\frac { x ^ { 2 } - 3 x - 28 } { x ^ { 2 } - 49 } = \frac { 3 } { 17 } , x x \pm 7 |
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Answer» Thanks a lot Samita |
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| 13. |
12. If 5 is a zero of x² + px + 8, then find p and other zero. |
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| 14. |
Polynomials12. Zero of the polynomial p(x)-2-5x is22 |
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Answer» p(x) = 2-5x2-5x=0x=2/5 x=-5/2(d) one is the correct answer sorry, (c) one is the correct answer that is -2/5 |
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| 15. |
9. The sums of n terms of two arithmetic progressions are in the ratio5n +4: In + 6. Find the ratio of their 18th terms.11 |
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| 16. |
in 3o |
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| 17. |
6.28 X 3o 25 |
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| 18. |
Find the value of t in each of the following cases: 4t=52^2 -40^2 |
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Answer» 4t=1104t=276 is the value of t no this is not right answer ohh I am sorry this is right answer |
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| 19. |
\left. \begin{array} { l } { 6 x + 5 y = 52 } \\ { 2 x + 2 y = 26 } \end{array} \right. |
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| 20. |
-ीन्फबुक 52 2/ VingJoun of eagonsts |
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| 21. |
x2-3x-28 3x-49 172 |
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Answer» x*x-3x-28 = (x-7)(x+4) x*x-49 = (x-7)(x+7) x*x-3x-28 / x*x-49 = (x-7)(x+4)/(x+7)(x-7) =x+4 / x+7 so x+4 / x+7 =3/17 17x+68=3x+21 14x=-47 so x=-47/14 Thanks |
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| 22. |
FACtoise6x2- 52-2) |
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| 23. |
0o¢ (| 0के 3o |
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Answer» that is the answer |
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| 24. |
किःd=guish w) 3o 2ye B |
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| 25. |
e et 29, 20 8 फिट बा 2A8 = 3O |
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Answer» Like if you find it useful |
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| 26. |
How many words can be formed changing theorder of the letters of the word TEG?18. |
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Answer» there are 4 letter and they distinct sopossible permutations is 4! = 24 |
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| 27. |
6.Find iāland lbl.if (à + b).(a-b)-8and I äl=81b1 |
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Answer» Given~|a|=8|b| As we know, |a|=a |b|=b Then we can write a=8b……………(1) And there is,(a+b)(a-b)=8 Solving, a(a-b)+b(a-b)=8 a^2-b^2=8 From 1 (8b)^2-b^2=8 64b^2-b^2=8 63b^2=8 b=8/63hencea=8*8/63=64/63 |
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| 28. |
2. By what number should 3-3 be multiplied to obtain 42 |
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Answer» 1/27*x = 4x= 27*4=10 8 let the required number be X x × 3^-3 = 4x × 1/3^3 =4x × 1/27 = 4x = 4×27x = 108 is the answer of the question. 1/27×x=4x=27×4=108 answer. 108 is the correct answer of the given question 108 is the best answer |
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| 29. |
24. If a, b, c are all nonzero and a +b+c= 0, prove thatàb= 3.bc "ca* ab - 3. |
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| 30. |
Im |
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| 31. |
387आगेप्रश्नमालासूत्र एकाधिकेन पूर्वेण द्वारा योग कीजिए।9 87656 3 2 17895 2 260543 |
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Answer» 312047 is the correct answer 312047 is rite answer |
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| 32. |
9(2-3x)-17 (5x-2)=10 |
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Answer» 9(2 - 3x) - 17(5x - 2) = 10 18 - 27x - 85x + 34 = 10 112x = 44 x = 44/112 = 22/56 18-27x-85x+34=10 -27x-85x =10-34-18 -112x =-42 x =-42/122 x =22/56 |
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| 33. |
(33/52)*((2*(2/17))*(7*(2/9))) |
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| 34. |
\frac{3}{4}+7 \frac{2}{17}-9 \frac{15}{15} |
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| 35. |
(17/3)/((2*(4/9))) |
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Answer» 78/11 is the correct answer of the given question 78/11 is the right answer. |
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| 36. |
2. By what number should 3be multiplied to obtain 4? |
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| 37. |
tegim Teama |
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Answer» ∫ (e^x - 1) / (e^x + 1) dx let u = e^x ; du = e^x dx = ∫ (u - 1) / (u (u + 1)) du = ∫ (2u - (u + 1)) / (u (u + 1)) du = 2 ∫ du / (u + 1) - ∫ du / u = 2 ln(u + 1) - ln(u) + C = 2 ln(e^x + 1) - x + C |
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| 38. |
Im-m |
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| 39. |
71/m[3o? + (3")2 ]* = 81 at, at m=? |
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Answer» 6 is the correct answer 6 is the right answer. plz like my answer 6 is correct answer for question |
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| 40. |
Hadiu 6 cm Theat peint T Find Jangth TP |
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Answer» Given radius, OP = OQ = 5 cmLength of chord, PQ = 4 cmOT ⊥ PQ,∴ PM = MQ =4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]In right ΔOPM,OP²= PM²+ OM²⇒ 52= 42+ OM²⇒ OM²= 25 – 16 = 9Hence OM = 3cmIn right ΔPTM,PT²= TM²+ PM²→(1)∠OPT = 90º [Radius is perpendicular to tangent at point of contact]In right ΔOPT,OT2²= PT²+ OP²→(2)From equations (1) and (2), we getOT²= (TM²+ PM²) + OP2²⇒ (TM + OM)²= (TM²+ PM²) + OP²⇒ TM²+ OM²+ 2 × TM × OM = TM²+ PM²+ OP²⇒ OM²+ 2 × TM × OM = PM2+ OP²⇒ 32+ 2 × TM × 3 = 42+ 52⇒ 9 + 6TM = 16 + 25⇒ 6TM = 32⇒ TM =32/6 = 16/3Equation (1) becomes,PT²= TM²+ PM² = (16/3)2+ 42 =(256/9) + 16 = (256 + 144)/9 = (400/9) = (20/3)2⇒PT = 20/3∴the length of tangent PT is(20/3) cm |
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| 41. |
By what number should 6be divided to obtain 42? |
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| 42. |
17. By what number should 6 be divided to obtain? |
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Answer» 56/9*3/14= 4/3 answer |
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| 43. |
\left. \begin{array} { l } { \text { add } } \\ { 7 \frac { 3 } { 8 } + 5 \frac { 1 } { 2 } + 3 \frac { 5 } { 2 } } \end{array} \right. |
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| 44. |
7. By what number should 6 - be divided to obtain 4 |
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Answer» 8/7 is the answer pls like |
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| 45. |
17. By what number should 6 be divided to obtain 4? |
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| 46. |
- By what number should 6 2 be divided to obtain 4? |
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| 47. |
7 on twice of a number, we get 49Find the number. |
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Answer» Thank you So much |
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| 48. |
-4910. Which rational number has to be multiplied with 64 to get the product |
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| 49. |
The least number that must be addedto 1300 so as to get a perfect square is[A] 49[B] 59 |
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Answer» B) is correct answer |
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| 50. |
17. By what number should 6 2 be divided to obtain 4 |
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Answer» thanks |
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