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. |
A sum amounts tosum.23762 in 2 years at 9% per annum, compounded annually. Find the4. |
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| 2. |
13. A certain sum invested at 4% per annumcompounded semi-annually amounts to*7803 at the end of one year. Then, thesum is |
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| 3. |
Date .Page3. 2 +ĺ¸1521 |
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Answer» (2/3 + 7/15) - (-1/21 + - 3/7)= [(10 + 7)/15] - [(-1 - 9)/21] = (17/15) - (-10/21) = (17*7 + 50)/105= (117 + 50)/105= 167/105 |
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| 4. |
\int \sqrt{1+\sin x} d x=\begin{array}{l}{-2 \sqrt{1-\sin x}+c} \\ {\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)} \\ {\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)} \\ {2 \sqrt{1-\sin x}+c}\end{array} |
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Answer» page 2 |
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| 5. |
7. (a) 3025 = 2? 52 x11”? |
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Answer» 11^2 is correct answer. |
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| 6. |
\frac { ( 1 + \operatorname { sin } x ) ^ { 2 } + ( 1 - \operatorname { sin } x ) ^ { 2 } } { 2 \operatorname { cos } ^ { 2 } x } = \frac { 1 + \operatorname { sin } x } { \operatorname { cos } ^ { 2 } x } |
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Answer» bahi answer galat hai |
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| 7. |
solve : x11 + x25 =2 |
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Answer» 1/(x +1) + 2/(x + 5) = 1/22(x + 5 + 2x + 2) = (x + 1)(x + 5)2(3x + 7) = (x^2 + 6x + 5)6x + 14 = x^2 + 6x + 5x^2 - 9 = 0x^2 - (3)^2 = 0 (x - 3)(x + 3) = 0x = - 3, 3 |
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| 8. |
24. What will be unit digit in368x7° x11'l x9684?(1) 1(2) 2(3)3(4) 4 |
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Answer» 4 should be i the answer |
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| 9. |
कह बी o QR €9 - X11 + cupmind कि |
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Answer» Put in X=-2hence40(-2)^2+11(-2)-6340*4-22-63=160-22-63=75 is the remainder |
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| 10. |
यास करें।12.45 X 749.70 X 120.83 X11 |
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Answer» 12.45×7=87.1549.70×12=596.40.83×11=9.13 plz accept as best 12.45x7=87.1549.70x12=596.40.83x11=9.13 87.15596.49.13 is the right answer 12.45 ×7= 87.1549.70×12= 596.4 0.83×11= 9.13 us the answer a) 87.15b) 596.4c) 9.13 1:87.15 2:596.4 3:9.13 |
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| 11. |
DatePage.s mudeě¤.3 |
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Answer» the points are marked in blue and the centre (0,0)no. of points inside is 9. alternatively , you can put all these points un the equation x²+y² , and show them as < 9. |
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| 12. |
Date.PageX6003 25 x 600 |
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Answer» =325×600=195000 answer. =325×600=195000 answer 195000 is a correct answer 325×699=195090 is the right answer 325×600325×6×1001950×100195000. 325×600195000 is the right answer |
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| 13. |
DatePage No.*+55a2+1=3 |
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Answer» Given,x + 1/x = 5 Squaring both sides(x + 1/x)^2 = 25x^2 + 1/x^2 + 2*x*1/x = 25x^2 + 1/x^2 = 25 - 2x^2 + 1/x^2 = 23 |
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| 14. |
DATEचाकूलेट के कुल हिस्से - 12राजी का हिस्सासगधा का हिस्सा 112X14शीला का हिस्सा 4x721-2मज का हिस्सा न -(3+1+12-9मंजू चौकलेट काभाग खाएगी। |
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Answer» this answer is worng and answer is -9 yes the answer is right....total chocolate= 12manju chocolate=33/12=1/4 |
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| 15. |
ePolynomialsGraphs-²+2x-15 |
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Answer» x²+5x-3x-15X(X+5)-3(X+5)(X+5)(x-3) answer |
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| 16. |
BOSSPage No.DEDate: | |-دا (3) +12+ی |
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Answer» it will be 9.5hope this helps u |
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| 17. |
29. The angle of elevation of the top of a tower 30 m high from the foot of another tower inthe same plane is 60° and the angle of elevation of the top of the second tower from thefoot of the first tower is 30. Find the distance between the two towers and also, the heightof the other tower. |
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Answer» From which book you have posted this questionplease tell us so that we can provide you faster answer |
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| 18. |
Sketch the graphs(i) y 2 sin x |
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| 19. |
classmateDate66Pagefind-12cast 2 16.80orangesOptoranges! |
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| 20. |
A TV tower stands vertically on a bankof a canal. From a point on the otherbank directly opposite the tower, theangle of elevation of the top of thetower is 60°. From another point 20 maway from this point on the line joingthis point to the foot of the tower, theangle.of elevation of the top of thetower is 30° (see Fig. 9.12). Find theheight of the tower and the width ofthe cana.11. |
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| 21. |
= . el e A eÂŽ (xii) (sin A +cosec A+ (cos A +sec A)2=7 + tan? A + cot? A1 |
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| 22. |
A TV tower stands vertically on a bankof a canal. From a point on the otherbank directly opposite the tower, theangle of elevation of the top of thetower is 60°. From another point 20 maway from this point on the line joingthis point to the foot of the tower, theangle of elevation of the top of thetower is 30° (see Fig. 9.12). Find theheight of the tower and the width ofthe canal.II. |
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| 23. |
1. Write the following in the ex() (a +2b+c)22-5p)(iv) (m + 2n(vii) (ab + bc + ca)(x) (x+2y + 4z)x112(xii) (2x+3y+ 22) |
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| 24. |
ATV tower stands vertically on a bankof a canal. From a point on the otherbank directly opposite the tower, theangle of elevation of the top of thetower is 60° From another point 20 maway from this point on the line joingthis point to the foot of the tower, theangle of elevation of the top of thetower is 30° (see Fig. 9.12). Find theheight of the tower and the width ofthe canal. |
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| 25. |
11.ATV tower stands vertically on a bankof a canal. From a point on the otherbank directly opposite the tower, theangle of elevation of the top of thetower is 60°. From another point 20 maway from this point on the line joingthis point to the foot of the tower, theangle of elevation of the top of thetower is 30° (see Fig. 9.12). Find theheight of the tower and the width ofthe canal |
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| 26. |
Page:Date:3 |
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Answer» let sonus age be x4x-3=3x+74x-3x=7+3x=10 years let sonu age be a A.T.G.S4×a-3=3×a+74a-3 = 3a+74a-3a=7+3 a=10Sonu's age=a=10yeara |
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| 27. |
Seeta sells a dining set to Neeta for? 6000 and gains 20%. For how much should stsell it to increase her profit by another 5%? |
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Answer» let the original price be xx+20%×x=6000120x/100=6000x=6000×100/120x=5000To gain profit of 25%5000+25%×50005000+12506250So seeta had to sell the dining set at 6250 to gain profit of more 5% |
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| 28. |
\frac{x y}{25 x^{2}-y^{2}}-\frac{2 x^{2} y}{10 x^{2} y+2 x y^{2}} |
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| 29. |
DatePage3 |
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Answer» derivative= (-sinx-cos x)at x= 2π/3-(sin2π/3+cos 2π/3)-(cos30+sin30)-(√3/2+1/2)-(√3+1/2)please like the solution 👍 ✔️👍 |
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| 30. |
A TV tower stands vertically on a bankof a canal. From a point on the otherbank directly opposite the tower, theangle of elevation of the top of thetower is 60°. From another point 20 maway from this point on the line joingthis point to the foot of the tower, theangle of elevation of the top of thetower is 30° (see Fig. 9.12). Find theheight of the tower and the width ofcanal. |
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| 31. |
Date12 32. |
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| 32. |
(xii) $2.ă18, ://8¡432. |
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Answer» From which book have you taken this question? Please tell us so that we can provide you faster answer. |
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| 33. |
4(xii) 2x2++, , |
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Answer» 2x^2 + 7/2x + 3/4 = 08x^2 + 14x + 3 = 08x^2 + 12x + 2x + 3 = 04x ( 2x + 3) + 1(2x + 3) = 0(4x + 1)(2x + 3) = 0x = - 1/4, - 3/2 |
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| 34. |
21xii)x11 |
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| 35. |
31m11 â 0(x+h)"-x3(xii)x1l |
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| 36. |
(xii) 12.18,./18 ,%2, |
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Answer» a1=√2,a2=√8= 2√2a3=√18=3√2a4=√32=4√2_________________________________common difference a2-a1= 2√2-√2=√2so d=√2 I want these to be answer |
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| 37. |
tbook for Class XIIPART ISin7S |
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Answer» (cos15*cos75)-(sin75*sin15)=(sin(90-15)*sin(90-75))-(sin75*sin15)=(sin75*sin15)-(sin75*sin15)=0 |
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| 38. |
36. A man sold his old dining table set at a lossof 20%. Ifhe had sold it for Rs 800.00 more,he would have received a profit of 5%. Thecost price of the Dining table set is |
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Answer» 25% = 8001 % = 800/25= 32100%(cp) = 32 × 100= 3200 will be the answer |
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| 39. |
&35Byo3 G |
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Answer» 17/5+18/5=35/5=7 ans |
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| 40. |
. ीध जपप्तु पकद रे ० 1294 मूडo3 L % 2o ,.'¥ q . ;- |
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Answer» In order to prove that the sum of angles of a triangle is 180, you must know the theorems of angles of a triangle. We know that, alternate interior angles are of equal magnitude. This'll help us get the answer. Let us assume a triangle ABC. Now we have to substitute the angles. ∠PAB + ∠BAC + ∠CAQ = 180°. where PQ line parallel to side BC touching vertex A. ∠PAB = ∠ABC & ∠CAQ = ∠ACB BECAUSE alternate interior angles are congruent.. Hence we get, ∠ABC + ∠BAC + ∠ACB = 180° [Proved] |
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| 41. |
कक5 783x 783X 783+ 217x217x217 _,रन 8. o3 783 783x217+217x217 )8, |
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| 42. |
) Is 10xH6 + (-2)| = 10 x 6 + 10 x (-2)? |
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| 43. |
The shadow of tower, when the angle of elevation of the sun is 45o is foube 10 m longer than when it was 60o. Find the height of tower.s. |
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Answer» Let AB be the tower with height h.Let AC and AD be the shadows when elevation of sun are 60 degrees and 45 degrees.As per given, CD=10mlet us assume CA=xIn triangle ACB,tan60°=opposite side /adjacent side√3=h/AC√3=h/xx=h/√3 ------ equation (1) In traingleDAB, tan45°=AB/AD=h/(AC+DC)1=h/(x+10)x+10=h-----equation(2) By substituting the value of x in equation 2 we get: h/√3+10=h h-h/√3=10 h√3-h=10√3 h(√3-1)=10√3 h=10√3/√3-1 Rationalizing factor is√3+1 h=10√3(√3+1)/[(√3-1)x(√3+1)] h=10√3(√3+1)/(3-1) h=10√3(√3+1)/2 h=5√3(√3+1) m h=5(3+√3) =15+5*√3 =15+5*1.732 =15+8.660 =23.66 m ∴ Height of tower is 23.66m hit like if you find it useful |
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| 44. |
020. The shadow of a tower, when the angle of elevation of the Sun is 45° isfound to be 10 metres longer than when it was 60°. Find the height of the tower. |
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Answer» thank you |
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| 45. |
The shadow of a tower standing on level ground is found to be 45 m longer when Sun'saltitude is 30°, then when it was 60. Find the height of the tower.5 |
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| 46. |
tten ned the value sf χ。nt |
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| 47. |
दे. हि Oz pod_dns volis Sf - T Cc) Qउड़ान के रह. |
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Answer» sinA = 12/13cosA = √(1- sin^2 A) = √(1- 144/169) = 5/13 (sinA - cos^2 A)/(2sinA - cosA)= (12/13 - 25/169)/(24/13 - 5/13)=(131/169)/(19/13)=131/247 |
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| 48. |
(xii) If20 and sf = 100, then what is the value of x? |
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| 49. |
26.The shadow of a tower standing on a level ground is found to be 30 m longer when thesun's altitude is 30° than when it is 60. Find the height of the tower |
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| 50. |
Example 5: The shadow of a tower standinga level ground is found to be 40 m longerwhen the Sun's altitude is 30° than when it is60. Find the height of the tower. |
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Answer» thanks |
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