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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. |
cos 20°+cos 70°Find the value of sin59+ sin2 31° |
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Answer» Value ofcos^2 20 + cos^2 70/sin^2 59 + sin^2 31 We knowsin^2 (90 - x) = cos^2 xcos^2 (90 - x) = sin^2 xSin^2 x + Cos^2 x = 1 Then,sin^2 (90 - 20) + cos^2 70/sin^2 59 + cos^2 (90 - 31) = sin^2 70 + cos^2 70/ sin^2 59 + cos^2 59 = 1/1 = 1 |
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| 2. |
\operatorname { cos } 70 ^ { \circ } \operatorname { cos } 10 ^ { \circ } + \operatorname { sin } 70 ^ { \circ } \operatorname { sin } 10 ^ { \circ } = \frac { 1 } { 2 } |
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| 3. |
+ 2 t re-o)+ z = 0 have equal roots?ora16. The angry Arjun carried some arows for fighting with Bheeshm with half the arrows, he cutdown the arrows thrown by Bheeshm on him and with six other arrows he killed the rathdriver of Bheeshm. With one arrow each he knocked down respectively the rath, flag and thebow of Bheeshm. Finally, with one more then four times the square root of arrows the laidBheeshm unconscious on an arrow bed(i) Find the total number of arrows Arjun had(Ăźi) Which mathematical concept is used in abuve problem?(iii) Which value should be promoted in this problem? |
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Answer» Let the Arjun carry x no of arrows.He used x/2 arrows to cut down Bheeshm arrows.He used 6 arrows to kill the rath driver and another 3 for rath,flag and bow.He use 4√x +1 arrows to put Bheeshm on arrow bed.From above the we have x/2+6+3+4√x+1=x ⇒ x +20 +8√x= 2x ⇒ x-8√x-20=0 ⇒ √x(√x-10) + 2(√x-10)=0 ⇒ (√x+2)(√x-10)=0 ⇒ √x=10 0r x=100 ( neglecting -tive value)Arjun had 100 arrows |
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| 4. |
lim _ x arrow 3 x %2B 3 |
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Answer» Lim x tends to 3thenx+3=3+3= 6 |
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| 5. |
lim _ x arrow - 2 \frac x ^ 5 %2B 32 x %2B 2 |
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| 6. |
8. In the figure the arrowhead segments are parallel. findthe value of x and y59060o |
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Answer» y is 59 as Alternate angle equal. x is 60 as corresponding angle are equal so x+y = 59+60 = 119 |
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| 7. |
cos 70°cos10°+sin70°sin10°=1/2 |
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| 8. |
7. Find the simple interst on8600 from 12 Octob.2005 to 6 March 2006 at 8% per annum |
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Answer» Days from 12 October 2005 to 6 March 2006 is 145 days P= 8600R% = 8%T= 145 daysSi = prt/100= 8600×8×145/100 = 99760 answer is wrong👻 |
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| 9. |
7. Find the simple interst on 8600 from 12 October2005 to 6 March 2006 at 8% per annum. |
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| 10. |
\cos 20^{\circ}+\cos 40^{\circ}+\cos 140^{\circ}+\cos 160^{\circ} |
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| 11. |
Find the circumcentre of the triangle formed by the points0a, 3), (0, -2), 3,1) [March 2018AP, May 2016Ts, 2006] |
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Answer» LetA(1,3),B(0,-2),C(-3,1).Slope ofAB=(-2–3)/(0–1) =-5/-1=5. Mid point of AB ( (1+0)/2,(3–2)/2 ) = (1/2,1/2). Equation of perpendicular bisector of AB is … y - 1/2 = (-1/5)(x - 1/2) => x + 5y -3=0 ……(1) Slope of BC=( 1+2)/(-3–0) =3/-3= -1 Mid point of BC ( (0–3)/2, (-2+1)/2 ) = (-3/2,-1/2) Equation of perpendicular bisector of B is……. y - (-1/2) =1{x-(-3/2)} =>y + 1/2 = 1(x + 3/2) ie x-y+1 = 0…..(2) Solving (1) and (2) we get x=-1/3 and y = 2/3. So the circumcenre is (-1/3, 2/3) do u another method |
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| 12. |
. Find the circumcentre of the triangle formed by the points1,3), (0, -2), (- 3, 1) [March 2018AP, May 2016TS, 2006] |
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Answer» LetA(1,3),B(0,-2),C(-3,1).Slope ofAB=(-2–3)/(0–1) =-5/-1=5. Mid point of AB ( (1+0)/2,(3–2)/2 ) = (1/2,1/2). Equation of perpendicular bisector of AB is … y - 1/2 = (-1/5)(x - 1/2) => x + 5y -3=0 ……(1) Slope of BC=( 1+2)/(-3–0) =3/-3= -1 Mid point of BC ( (0–3)/2, (-2+1)/2 ) = (-3/2,-1/2) Equation of perpendicular bisector of B is……. y - (-1/2) =1{x-(-3/2)} =>y + 1/2 = 1(x + 3/2) ie x-y+1 = 0…..(2) Solving (1) and (2) we get x=-1/3 and y = 2/3. So the circumcenre is (-1/3, 2/3) |
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| 13. |
. Prove that the points (3, 0) (6, 4) and (-1,3) are the vertices of a right angled isosceles triangle.(March, 201e |
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| 14. |
\cos 20 ^ { \circ } \cos 40 ^ { \circ } \cos 60 ^ { \circ } \cos 80 ^ { \circ } = \frac { 1 } { 16 } |
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Answer» Cos20°cos40°cos60°cos80°=(1/2)(2cos20°cos40°)(1/2)cos80° [∵,cos60°=1/2]=(1/4)[cos(20°+40°)+cos(20°-40°)]cos80°=(1/4)(cos60°+cos20°)cos80°=(1/4)(cos60°cos80°+cos20°cos80°)=(1/4)(1/2)cos80°+(1/4)cos20°cos80°=(1/8)cos80°+(1/4)(1/2)(2cos20°cos80°)=(1/8)cos80°+(1/8)[cos(20°+80°)+cos(20°-80°)]=(1/8)cos80°+(1/8)(cos100°+cos60°)=(1/8)cos80°+(1/8)cos100°+(1/8)cos60°=(1/8)(cos80°+cos100°)+(1/8)×(1/2)=(1/8)[{2cos(80°+100°)/2}{cos(80°-100°)/2}]+(1/16)=(1/8)(2cos90°cos10°)+(1/16)=0+(1/16) [cos90°=0]=1/16 (proved) |
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| 15. |
A bus leaves nandland at 10am if it is travelling at the speed of 45km/h at what time will it reach rajapur which is 270km away? |
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| 16. |
EXFind the areas of the following polygons:1.8 cmcmD←ㅡㅡㅡㅡㅡ12cm>25 m611 m5.5 m10 m |
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| 17. |
\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}=\frac{1}{16} |
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| 18. |
Try to open my locked suitcase which has the biggest 5 digit odd number as thpassword comprising the digits 7, 5, 4, 3 and 8. Find the password. |
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| 19. |
lit _ x arrow 0 \frac \sin 5 x %2B \sin 3 x \sin 6 x - \sin 2 x |
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| 20. |
\frac \cos 20 ^ \circ - \cos 70 ^ \circ \cos 20 ^ \circ %2B \cos 70 ^ \circ |
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| 21. |
: (8600 : coscef â 1gy =â""o) coscef + 1 |
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Answer» like if you find it useful |
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| 22. |
\begin{array}{l}{\text { (i) } \frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}} \\ {\text { (ii) } \sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}}\end{array} |
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| 23. |
xample 9: Selvi's house has an overhead tank in the shape of a cylinder. Thisis filled by pumping water from a sump (an underground tank) which is in theshape of a cuboid. The sump has dimensions 1.57 m Ă 1.44 m Ă 95cm. Theoverhead tank has its radius 60 cm and height 95 cm. Find the height of the waterleft in the sump after the overhead tank has been completely filled with waterfrom the sump which had been full. Compare the capacity of the tank with that ofthe sump. (Use3.14 |
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| 24. |
5. Cost price 8600,transport charges? 250,porterage 150,selling price 10000 |
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| 25. |
Find the simple interst on 8600 from 12 October2005 to 6 March 2006 at 8% per annum |
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Answer» Time period is = 20+30+31+31+28+6 = 141 days = 141/365 = 0.4 years now S.I = P*R*T/100 = (8600*8*0.4)/100 = 275.2 Rs |
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| 26. |
24. Selvi's house has an overhead tank in the shake ofa cylinder. This is filled by pumping water from asump (an underground tank) which is in the shapeof a cuboid. The sump has dimensions 1.57 m x1.44 m .95 m. The overhead tank has its radius60 cm and height 95 cm. Find the height of thewater left in the sump after the overhead tank hasbeen completely filled with water from the sumpwhich had been full. Compare the capacity of thetank with the of the sump. (Use3.14). |
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Answer» Dimensions of cylindrical tank are — Radius (r₁) = 60 cm = 0.60 m Height (h₁) = 95 cm = 0.95 m ∴ Volume of Tank = πr²h = 3.14 × 0.6 × 0.6 × 0.95 V1= 1.074 m³ The dimensions of cuboid shape sump are — Length (l) = 1.57 m Breadth (b) = 1.44 m Height (h) = 0.95 m ∴ Volume of the sump =l.b.h. = 1.57 × 1.44 × 0.95 m³ V2= 2.148 m³ When the tank is completely filled from the sump, remained volume of the sump = V2– V1 = 2.148 – 1.074 = 1.074 m³ Now the length and breadth will remain same for the sump but height of water level will decrease due to decrease in water level. ∴ Volume of the water in the sump =l × b×h₂ ⇒ 1.978 = 1.57 × 1.44 ×h₂h₂= 0.475m Hence, the height of water, left in the tank = 0.475 m. and the Volumes of tank and sump are V₁= 1.074 m³&V₂= 2.148 m³respectively v₁/v₂₌1.074/2.148= 1/2 |
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| 27. |
A bus leaves nandgaon at 10 am if it is travelling at a speed of 45km/h at what time will it reach rajapur which is 270km away? |
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Answer» Speed=distance/time45=270/timetime=270/45time=6 It will take 6 hours which means it will reach at 4pm |
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| 28. |
\frac \operatorname sin 30 ^ \circ %2B \operatorname tan 45 ^ \circ - \operatorname cosec 60 ^ \circ \operatorname sec 30 ^ \circ %2B \operatorname cos 60 ^ \circ %2B \operatorname cot 45 ^ \circ |
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Answer» (1/2 + 1 - 2/√3)/(2/√3 + 1/2 + 1)=(3/2 - 2/√3)/(3/2 + 2/√3)=(3√3 - 4)/(3√3 + 4) |
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| 29. |
x^2 %2B 10*x %2B 60 |
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| 30. |
cos(x)=sin(30^circ)*sin(60^circ) %2B cos(30^circ)*cos(60^circ) |
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| 31. |
(-60^circ*cosec %2B sin(30^circ) %2B tan(45^circ))/(cos(60^circ) %2B sec(30^circ) %2B cot(45^circ)) |
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| 32. |
tan 0 _ 0ŕĽ06o Tâtang = ! +1ano +cotd| |
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Answer» hit like if you find it useful |
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| 33. |
Prove that: के हक 1cosec®—cot® sin® sin® cosec® + cotd’ |
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Answer» LHS=1/(cosecA-cotA) -1/sinA =(cosecA+cotA)/cosec^2A-cot^2A) -cosecA =cosecA+cotA-cosecA =cotA RHS=1/sinA-1/(cosecA+cotA) =cosecA-(cosecA-cotA)/(cosec^2A-cot^2A) =cosecA-(cosecA-cotA) =cotA LHS=RHS hit like if you find it useful |
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| 34. |
4. tanf + cotd=A sec8 cosec 08609.00860 9 8. 560 0+ 00860 900860 08600 |
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Answer» Let theta = A Then,tan A + cot A= sin A/cos A + cos A/sin A= (sin^2 A + cos^2 A)/cosA.sinA= 1/cosA.sinA= secA.cosecA (A) is correct option |
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| 35. |
Nandini's house is -- km from her school. She walked some distance and then took a bus for 1 km to10reach the school. How far did she walk. |
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Answer» Answers Distance from Nandini’s house to school =9/10 km Distance covered by bus =1/2 km Distance she walked = distance from house to school – distance she covered by bus. =9/10−1/2 =9/10−5/10 =4/10 =2/5 km Distance walked by Nandini =2/5 km walk = 9/10-1/2 = 9-5/10 =4/10 = 0.4 km = total distance from house to achool is 9/10 km she walked for x km she travelled through bus for 1/2km toatl =walk + bus 9/10 =x +1/2x= 9/10-1/2x = 9-5 / 10 x=4/10 =2/5 km ×= 2/5 km 2/5 is the correct answer of the given question 2/5 km is the best answer |
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| 36. |
22. If e is the eccentricity of the earth's orbit, the ratio of maximumdistances (perihelion and aphelion) of the planet is(A) 0.(B) 1.(C) ite(D) LTE |
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Answer» option c is the correct answer of the given question C is the best answer |
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| 37. |
-72 %2B 35 %2B 30 %2B 60 |
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Answer» 53 30+60+35-72125-72=53 |
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| 38. |
7. Nandini's house is to km from her school. She walked some distanee and then10took a bus forkm to reach the school. How far did she walk? |
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| 39. |
\frac \operatorname cos 30 ^ \circ %2B \operatorname sin 60 ^ \circ 1 %2B \operatorname cos 60 ^ \circ %2B \operatorname sin 30 ^ \circ = \frac \sqrt 3 2 |
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Answer» cos 30 + sin 60 /(1 + cos 60 + sin 30) = sqrt(3)/2 + sqrt(3)/2/ (1 + 1/2 + 1/2) = 2sqrt(3)/2 / (1 + 1) = sqrt(3)/2 |
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| 40. |
Find the distance of a point P(x, y) from the origin.In an ADi |
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Answer» Ty |
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| 41. |
28. A manufacturer involves twelve children in colouring pen stands all over excludingbase which are in the shape of a cylinder made of wood of thickness 2 cm. Theinner radius of the cylinder is 4 cm and its height is 14 cm. Find the area they hadto paint if 50 pen stands were given to them for painting. |
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Answer» Find the area painted as follows: Find the outer surface area of the cylindrical pencil stand.Area of the outer cylinder surface is: Perimeter of the cylinder× height of the cylinder πD × H = 22/7 × 12× 14 = 528 cm² Area of the inner cylindrical part: (assuming that the base is 2cm thick as well)The inner height would be 12 cm πD × H= 22/7 × 8× 12 = 301.632cm² The area of the top part painted:Area of the larger circular surface - smaller surface πR² -πr² = 3.142×6² - 3.142× 4² = 113.112 - 50.272 = 62.84cm² Add the area to find the total area painted: 528 cm² + 62.84cm² +301.632cm² = 892.472cm² The area painted in 1 stand is 892.472 cm² The area of 50 such stands is 892.472 cm²× 50 = 44623.60 cm² hit like if you find it useful answer is wrong correct answer is 347600/7 cm^2 |
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| 42. |
17. Prove that 2225 cosecA + cotdCoSA+sinA-1 |
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| 43. |
A man ,Jaflf.s,f 4and Larni A a’dxSalany wor T)g\::(;jz‘-f{i“( 1०Afar रद 222८ |
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Answer» Increment of every year is =1800-1500= 300 10-4 =6 year. :. increment in 6yrs=Rs. 3001. ? 300*1/6 = 50 every year 50 RS is the increment.1) starting salary = 50*4 = 200 increment in 4 years .so 1500-200= 1300 /- 1st year salary2)anuual increment is 50/- RS. |
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| 44. |
What is drag ? |
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Answer» air resistance is known as drag the frictional force exerted by fluids on an object is known as drag. In fluid dynamics, drag is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid. This can exist between two fluid layers or a fluid and a solid surface. In fluid dynamics, drag is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid. This can exist between two fluid layers or a fluid and a solid surface I called drag. In fluid dynamics, drag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid.[1] This can exist between two fluid layers (or surfaces) or a fluid and a solid surface. Unlike other resistive forces, such as dry friction, which are nearly independent of velocity, drag forces depend on velocity.[2][3] Drag force is proportional to the velocity for a laminar flow and the squared velocity for a turbulent flow. Even though the ultimate cause of a drag is viscous friction, the turbulent drag is independent of viscosity.[4] |
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| 45. |
11. In Δ ABC and Δ DEF, AB-DE,ABIDE, BC-EFand BC l| EF. Vertices A, B and C are joined tovertices D, E and F respectively (see Fig. 8.22).Show that0 quadrilateral ABED is a parallelogram(i) quadrilateral BEFC is a parallelogram(m) ADI CFand AD-C(iv) quadrilateral ACFD is a parallelogram(v) AC-DF(vi) Δ ABC Δ DEFFig. 8.22 |
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Answer» thankyou |
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| 46. |
( 3 x - 3 x ^ { 2 } - 12 + x ^ { 4 } + x ^ { 3 } ) \div \left( 2 + x ^ { 2 } \right) |
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Answer» thanks |
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| 47. |
3 ( x ^ { 2 } + 1 / x ^ { 2 } ) - 16 ( x + 1 / x ) + 26 = 0 |
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Answer» thankyou soooooo....much..... |
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| 48. |
-12*x^3*y^2 %2B 60*(x^5*(y^3*z^2)) %2B 36*(x^3*y^4) |
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Answer» 12 12x³y²(3y²+5x²yz²-1) |
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| 49. |
(276) Perimeter of a rectangular garden is 360 m and its area is 8000 m2. Find the nlength of the garden and also find itsbreadth. (The length is greater than breadth)dounstream and cames hack the |
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| 50. |
logbase10 8000 = 3 (l + logbase10 2) |
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