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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. |
V 3-v,1. If cos θ=and θ lies in Quadrant 111, find the values of all the other fivetrigonometric functions. |
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| 2. |
2. Factorise.\begin{array}{l}{4 p^{2}-9 q^{2}} \\ {16 x^{5}-144 x^{3}}\end{array} |
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| 3. |
ii) (3+ V2)(4+3- जŕĽ) |
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| 4. |
(11)1V2 +4/3 |
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| 5. |
Q3If cosx-s3 and π<X<ST, find the value of other five trigonometric functions and hence evaluate2:cos ecx cotx4534sec tan |
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| 6. |
03 f cosx-→ and兀<x<cosecx +cotxsecx tan x, find the value of other five trigonometric functions and hence evaluate45 14534 |
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Answer» Saale darthi pe boaj ha tu calculation bi nhi aati Chup chaap apne class ka question krr bas |
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| 7. |
If x=v2-4 and y= VI+,, find the value of x++y?+xy. |
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| 8. |
(iii) 2x + V2 = 3x - 4 - 3/2 |
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| 9. |
\frac { \operatorname { sin } x + \operatorname { sin } 3 x + \operatorname { sin } 5 x + \operatorname { sin } 7 x } { \operatorname { cos } x + \operatorname { cos } 3 x + \operatorname { cos } 5 x + \operatorname { cos } 7 x } = |
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Answer» numerator = sinx+sin3x+sin5x+sin7x= (sin7x+sinx)+(sin5x+sin3x)= 2sin(7x+x)/2.cos(7x-x)/2+2sin(5x+3x)/2co... [ sinC+sinD=2sin(C+D)/2cos(C-D)/2= 2sin4x.cos3x+2sin4x.cosx= 2sin4x[cos3x+cosx]dinominator = cosx+cos3x+cos5x+cos7x= (cos7x=cosx)+cos5x+cos3x)= 2cos(7x+x)/2.cos(7x-x)/2+2cos(5x+3x)/2.c... [cosC+cosD = 2cos(C+D)/2cos(C-D)/2= 2cos4x.cos3x+2cos4x.cosx= 2cos4x[cos3x+cosx]Numerator/Dinominator= 2sin4x[cos3x+cosx]/2cos4x[cos3x+cosx]= sin4x/cos4x= tan4x |
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| 10. |
\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x |
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| 11. |
Prove that\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x |
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Answer» 2 3 |
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| 12. |
4. A cone of metal of height 24 cm and radius of base 6 cm is melted and recast into a sphere. Find the[H.B.S.E. 2017 (Set-Cllradius of the sphere. |
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| 13. |
llu iy perimeter is 32 cm. Find its area.Cill aurd ItSA diagonal of a quadrilateral is 26 cm and the perpendiculars drawn to it from the oppovertices are 12.8 cm and 11.2 cm. Find the arca of the quadrilateral. |
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| 14. |
LLUS5 = 16 X 9 X 5 =720EXAMPLE 2 Find the HCF and LCM of 144, 180 and 192 by prime factorisation method.SOLUTION I Tsing the |
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Answer» This question had 192 too with 144 and 180. So here is the solution for that. Factors of 144 = 2×2×2×2×3×3 Factors of 180 = 2×2×3×3×5 Factors of 192 = 2×2×2×2×2×2×3 Common factors of 144, 180, 192 = {2,2,3} LCM (144,180,192) = 2×2×2×2×2×2×3×3×5 = 2880 So, HCF(144,180,192) = 2×2×3 = 12. |
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| 15. |
A sector of circle of radius 12cm has tge angle 120°.It is rolled up so that two bounding radii are joined together to form a cone.find the volume of the cone. |
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| 16. |
(a) A Rational NumoeWhich of the following is not an irrational number?(a) V24.(b) v3(c) v15(d) V4 |
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| 17. |
overali width in cm of soveral wide- screen televisions are 97 28 cm 98 cm48 L m and 97 94 cm. Express these numbers as rational numbers in the forma23range the widths in ascending order |
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Answer» Give, width of the wide screen television are 97.28 cm, 98.4/9 cm, 98.1/25 cm, and 97.94 cm => Width = 9728/100, 984/(9*10), 981/(25*10), 9794/100 => Width = 9728/100, 984/90, 981/250, and 9794/100 => Width = 9794/100, 9728/100, 984/90, (981*4)/(250*4) => Width = 9794/100, 9728/100, 984/90, 3924/1000 => Width = (9794*10)/(100*10), (9728*10)/(100*10), 984/90, 3924/1000 => Width = 97940/1000, 97280/1000, 984/90, 3924/1000 Width = 97940/1000, 97280/1000,3924/1000, 984/90 => Width = (97940*9)/(1000*9), (97280*9)/(1000*9), (3924*9)/(1000*9), (984*100)/(90*100) =>Width = 881460/9000, 875520/9000, 35316/9000, 98400/9000 = Width = 881460/9000, 875520/9000,98400/9000, 35316/9000 So, width in ascending order are: 97.94 cm, 97.28 cm, 98.4/9 cm, 98.1/25 cm |
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| 18. |
2.1muummaximlum VoV3ě y-xt tn.tteinn)", findol |
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| 19. |
The LCM of two numbers is 64699, HCF is 97and one of the numbers is 2231. Find theother number. |
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Answer» Let other number be n We knowProduct of LCM and HCF= Product of two numbers Then,64699 x 97 = 2231x n64699 x 97/2231 = nn = 29*97 = 2813 Therefore other number is 2813 |
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| 20. |
3,,In the figure below left side, I li n If L1-(2x + 50% 23-(x + 4y)o and L 5(5y+10. fndthe angles 41, 23 and 45.60°70°2 |
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Answer» Can you please provide the required figure. as they are talking about l and m lines and angles 1,3 and 5. |
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| 21. |
7.Find the value of x² +when x = 4+ V15 |
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| 22. |
LiFind the total cost of 12 chairs and 15 tables if the cost of 23 chairs and 13tables are 14720 and 11960 respectively.IA) 1570E) of these, 29.(C) 1560(D) R 1470ble School for Class VI is t 16920. What will |
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| 23. |
6) Simplify Toal-V15 |
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| 24. |
(g) If x = 4 - V15, find the value of (x +IX |
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Answer» 1/x=1/4-√15*4+√15/4+√15=4+√15/16-15=4+√15hence add both X+1/X=4-√15+4+√15=16 |
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| 25. |
Find the value of a and b.- a + b v15V3 |
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| 26. |
tan hou |
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Answer» The fractions having same same numerator and smaller denominator is larger as a whole. Hence, 3/4 > 3/6=> Rohit exercise for longer time. |
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| 27. |
oemliuLength of the fence of a trapeziurh sHapeuBC-48 m, CD 17 mand AD 40 m, find the3.find the area of this fieldC AB is perpendicular to the parallel sides AD and Bfielddrilateral shaped field is 24 m |
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| 28. |
xt find tht Value of2 x^{2}+x-1 |
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Answer» if x = 1, 2(1)² + 1 -1 = 2+1-1 = 2 |
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| 29. |
Shou -tht shabe s an odol positive integer ishole hamber m |
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| 30. |
4. A packet of sweets was distributed among 20 children and eagh of them received 4sweets. How many sweets will each child get, if the number of children is reduced by 4? |
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Answer» 20 children and each are getting 20 so total sweets=20×4=80children reduced by 4 so 16 children so each one receive 80/16=5 sweets |
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| 31. |
Pythagoras TheorenmIn a right angled triangle, the square of the hypotenuse is equal tothe squares of remaining two sides.Given : In Δ ABC, L'ARC-900To prove: AC ABBCConstruction :Draw perpendicular seg BD on side ACA-D-C. |
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| 32. |
Sood distributed 4.8 kg apples equally among 8 children.How much apples did each child get?Mona had bought 32.5 kg of sugar. She put it equally into 4 jars.How much sugar did she put in each jar?7. The cost of 16 chairs is ? 10,450.44. Find the cost of each chair.MATHS LAB activity UNDDIRIGHENMBERS ON THE SPIKE ABACUSENRICHMENT ACTIVITYYOU WILL NEEDabacus with 5-6 spikes, beads of different colours |
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| 33. |
3. A box contains 286 sweets to be distributed1 1 1among three children in the ratio :Find the number of sweets received by eachchild.4 |
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| 34. |
Length of the fence of a trapezium shaped field ABCD is 120 m. IfBC-48m, CD-17 m and AD 40 m, find the area of this field.SidAB is perpendicular to the parallel sides AD and BC. |
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| 35. |
ofatrapeipnadAD3. Length of the fenced the area of this fieldBC48 m,CD- 17 mand AD 40 m, find the area of this fc AB is perpendicular to the parallel sides AD and BC |
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| 36. |
Ex. 7 : Show that sin 10 sin 30 sin 50°- sin 7016S1(iv) - |
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Answer» Sin 10 Sin 30 Sin 50 sin 70⇒ (1/ 2) Sin 10 Sin 50 sin 70 multiply and divide with 2 Cos 10⇒ 2 Cos 10 Sin 10 Sin 50 sin 70/ (2 x 2 Cos 10) { 2SinA Cos A = Sin2A }⇒ Sin 20 Sin 50 sin 70 /(4 Cos 10) multiply and divide with 2⇒ 2 x Sin 20 Sin 50 sin (90 - 20) /(2 x 4 Cos 10)⇒ 2 Sin 20 Cos 20 Sin 50 / 8Cos 10⇒ Sin 40 Sin 50 / 8Cos 10 { 2SinA Cos A = Sin2A }Againmultiply and divide with 2.⇒ 2 x Sin 40 Sin ( 90 - 40) / 2 x 8Cos 10.⇒ 2 x Sin 40 Cos 40 / 2 x 8Cos ( 90 - 80)⇒ Sin 80 / 16Sin 80 = 1 / 16. |
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| 37. |
. Oc0s10° + sin 10०0810? - sin 10°........ |
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Answer» LHS = ( Cos 10 + Sin 10 ) / ( Cos 10 - Sin 10 ) { Multiply numerator and and denominator with 1 / Cos 10 } = [ 1 /Cos 10 ( Cos 10 + Sin 10 ) ] / [ 1/Cos 10 ( Cos 10 - Sin 10 ) ] = [ Cos10 / Cos10 + Sin10 / Cos10 ] / [ Cos10 / Cos10 - Sin10 / Cos10 ] = ( 1 + tan10 ) / ( 1 - tan 10 ) [ since sin 10 / cos 10 = tan 10 ] = ( tan 45 + tan 10 ) / ( 1 - tan 45 tan 10 ) [ since tan 45 = 1 ] = tan ( 45 + 10 ) [ since tan ( A + B ) = (tanA + tan B ) / (1 - tanAtanB)] = tan 55 = tan ( 90 - 35 ) = cot 35 [ since tan ( 90 - A ) = cot A ] |
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| 38. |
Rashid spent 35.75 for Maths books and 32.60 for Science bookamount spent by Rashid |
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| 39. |
35.75 for Maths book andäš32.60 for Science book. Find the totalRashid spentamount spent by Rashid.. |
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Answer» Amount spent for maths book = 35.75Amount spent for science book = 32.60Total amount spent = 35.75 + 32.60 = Rs 68.35 |
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| 40. |
\frac{\cos 10^{\circ}-\sin 10^{\circ}}{\cos 10^{\circ}+\sin 10^{\circ}}=\tan 35^{\circ} |
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| 41. |
3.Length of the fence of a trapezium shaped field ABCD is 120 m. IfBC-48 m, CD- 17 mand AD 40 m, find the area of this field. SideAB is perpendicular to the parallel sides AD and BC. |
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| 42. |
3. Length of the fence of a trapezium shaped field ABCD is 120BC-48 m, CD- 17 mand AD-40 m, find the area of this fieldAB is perpendicular to the parallel sides AD and BC |
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| 43. |
3.Length of the fence of a trapezium shaped field ABCD is 120 m.IfBC 48 m, CD 17 m and AD 40 m, find the area of this field. SideAB is perpendicular to the parallel sides AD and BC. |
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| 44. |
3. Length of the fence of a trapezium shaped field ABCD is 120 m.BC-48 m, CD- 17 m and AD 40 m, find the area of this field. SiAB is perpendicular to the parallel sides AD and BC. |
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| 45. |
Length of the fence of a trapezium shaped field ABCD is 120mBC-48 m, CD 17 m and AD 40 m, find the area of this field. S3.C AB is perpendicular to the parallel sides AD and BC. |
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| 46. |
3.Length of the fence of a trapezium shaped field ABCD is 120 m. IfBC -48 m, CD-17 mandAD-40 m, find the area of this field. SideB is perpendicular to the parallel sides AD and BC. |
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Answer» thank u sir |
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| 47. |
Length of the fence of a trapezium shaped field ABCD is 120 m. IfBC 48 m, CD 17 m and AD-40 m, find the area of this field. SideC ABis perpendicular to the parallel sides AD and BC.uboid |
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| 48. |
3. Length of the fence of a trapezium shaped field ABCD is 120 m.IfBC = 48 m, CD = 1 7 m and AD = 40 m, find the area of this field. SideAB is perpendicular to the parallel sides AD and BC |
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| 49. |
9. The length of the fence of a trapezium-shaped field ABCD is 130 mand side AB is perpendicular to each of the parallel sides ADand BC. If BC 54 m, CD 19 m and AD 42 m, find the area ofthe field. |
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| 50. |
If cos0+ sín9-y2cos θ then/2 cos θ thencosθ + sin θ1) v/2 sin θ3) 2sin 62) /2 sin θ4) -2sinnthenT.C |
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Answer» cosA+sinA=√2cosAsquaring both the sides =>(cosA+sinA)²=2cos²A=>cos²A+sin²A+2sinAcosA=2cos²A=>cos²A-2cos²A+2sinAcosA= -sin²A=> -cos²A+2sinAcosA= -sin²A=> cos²A-2sinAcosA=sin²Aadding sin²A on both the sides => cos²A+sin²A-2sinAcosA=2sin²A=> (cosA-sinA)²=2sin²A=> cosA-sinA=√2sinA |
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