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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. |
(25^(2*n - 7)*5^(2*(n %2B 6)))/125^(2*n) |
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| 2. |
lgebra)Ex.5. Find the value of' a' for which the rootsundβ of the equation x^2 -6x + a = 0, satisfy therelation 3 α + 2 β = 20 |
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| 3. |
find the time in which Rs550 will amount to Rs 605 at 4% per annum simple interest.(Bilaspur, 2014) [Ans. 2years] |
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Answer» SI=P×R×T/100605-550=550×4×T/100;55=55×4×T/10;T=10/4=2.5yr |
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| 4. |
Verify Rolle's theorem for the following functions:f(x) = x* in (-1,1) |
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Answer» for solving this type of question you need to learn the basic. |
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| 5. |
Problem seto1. Select the appropriate alternative.(1) In A ABC and A PQR, in a oneto one correspondenceABBC CAQR PR PO- then(A) A POR ~ A ABC(B) A PQR ~ A CAB(C) A CBA ~ A POR(D) A BCA - A PORRLFig. 1.67 |
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Answer» triangle PQR ~ triangle CAB |
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| 6. |
PA.9. In the given figure ABIICD.HLOAB=124° ,124。<OCD = 136°, then <AOC =?(b) 90° o(d) 110(a) 80°138 |
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| 7. |
The area of a rectangle is 360 cm2, which is equal to 90% of the area of a square. What is each sideof the square?(a) 20 cm(b) 15 cm(c) 40 cm(d) 25 cm |
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| 8. |
equal area.25. The base BC of AABC is divided at D such that BD DC. Prove thear(ABD) =ar(-ABC).idee |
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| 9. |
Example 10: In the following figurerectangle in which the segments AP and AQ are drawnas shown. Find the length of (AP + AQ).ABCD is a30°30°60 cm |
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Answer» In right triangle APB30° + 90° + <PAB = 180°<PAB = 180° - 120° = 60° Then, cos 60° = B/H = AB/AP 1/2 = 60/APAP = 60*2 = 120 cm In right triangle AQDsin 30° = P/H = AD/AQ1/2 = 30/AQAQ = 30*2 = 60 cm Therefore, length of (AP + AQ) = 120 + 60= 180 cm |
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| 10. |
Find the domain of the following functionsf(x, y)-25-x2 -y |
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Answer» for domain the term under the root should always be > 0 so, 25 -x² -y² > 0=> x²+y² < 25 so, this is an equation of circle having radius 5 units and all the points lying inside it.. will be in the domain from x = -5 to 5 and. y = -5 to 5 |
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| 11. |
4 x (5 6) 4x5+4x65 ) |
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Answer» LHS=4×(5+6)=4×11=44 RHS=4×5+4×6=20+24=44 LHS=RHSHence, Verified |
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| 12. |
In a constituency of the voters had voted for candidate A whereas had votedfor candidate B. Find the rational number of the voters who had voted for others. |
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Answer» Answer:Rational number of voters who had voted for other = 1/10Step-by-step explanation:Sincelet total number of voters = 100Then 19/25 of total voters = (19/25)×100 = 1900/25 = 767/50 of total voters = (7/50)×100 = 700/25 = 14So Candidate A have voters = 76and Candidate B has voters = 14Remaining voters = 100 - 76 - 14 = 10SO voters of other candidates = 10And fraction of voters of other candidate = 10/100 = 1/10ThusRational number of voters who had voted for other = 1/10 1----10 answerplzzz like Rational number=1/10; Voters=100; (19/25)×100=1900/100=76,; 7/50 of total voters=(7/50)×100=700/25=14, candidate A=76; B candidate B=14; remaining voters=100-76-10=10; fraction of voters other candidates=1/10; Rational number of voters had voted for other=1/10 1/10 is the correct answer of the given question |
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| 13. |
9. In the given figure, AB and CD arestraight lines through the centre O of acircle. If <AOC= 80° and LCDE=40。find (i) 4DCE, (i) ZABC.80° |
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| 14. |
If sec θ-x +-, prove that: sec θ + tan θ-2x or4x2x |
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| 15. |
LAOBShow tIn the adjoining figure, AOB is a straight line.Find the value of Hence, find AOCZCOD and ZBOD(3x+7)09-13. In the |
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Answer» The sum of angles on a line is 180.=> (3x+7)+(2x-19)+X = 180=> 6x - 12 = 180=> 6x = 192=> X = 32. Angle AOC = 3x +7 = 103. Angle DOB = X = 32. Angle COD = 2x - 19 = 45. Please hit the like button if this helped you. |
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| 16. |
4x(2x-1) = 2x-1 |
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| 17. |
4x (2x-1) = 2x-1 |
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| 18. |
Evaluate 105106 using appropriate identit |
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Answer» step by step explanation = 105*106100(5+6)100(11)1100 100 (5+6)100(11)1100 1100 is the answer for that question BANK of each other in my life in a couple of years of experience and knowledge of how Eirik (100+5) .(100+6)x=a x=b. x=a+b a=5 100(5+6)100(11)=1100 b=6 1100 is the best answer. |
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| 19. |
9. In parallelogram ABCD, wo points P and Q areBQtaken on diagonal BD such that DP(see Fig. 8.20) Show that) AP CO(iv) AQ CP(v) APCQ is a parallelogramFig. 8.20aR und CO are |
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| 20. |
Q22: In the given figure, a circle touches the side BC of AABC at P and touches AB and AC producedat Q and R respectively. If AQ 5 cm, find the perimeter of ΔABC. |
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Answer» AQ = AR = lengths of tangents from point A to the ex-circle of triangle ABC.BQ = BP = lengths tangents from B onto the excircle.CR = CP = lengths of tangents from C onto the excircle. AQ = AB + BQ = AB + BP AR = AQ = AC + CR = AC + CPAQ + AR = 2 AQ = AB + BP + PC + CA = perimeter of triangle ABC So perimeter = 2 * 5 cm = 10 cm Like my answer if you find it useful! |
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| 21. |
10 triangles ofequal area25. The base BC of AABC is divided at D such that BDar(AABD) = a (4ABC) anDC. Prove that15. A mediaAD is aPD is16. CO isAis |
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| 22. |
\frac { 2 } { 3 } \times \frac { 3 } { 5 } + \frac { 5 } { 2 } - \frac { 3 } { 5 } \times \frac { 1 } { 6 } |
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Answer» 2/3*(3/5) +(5/2)-(3/5)×(1/6)= 2/5 +5/2 -1/15= (12+75-2)/30= 85/30= 17/6 |
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Q 22: In the given figure, a circle touches the side BC of AABC at P and touches AB and AC producedat Q and R respectively. If AQ 5 cm, find the perimeter of AABC |
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Answer» Like if you find it useful |
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| 24. |
A circle touches all the four sides of a quadrilateral ABCD, Prove that |
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Answer» Like my answer if you find it useful |
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| 25. |
-\frac{2}{3} x \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6} |
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Answer» by BODMAS rule-2/5+5/2-1/10=(-2×2+5×5-1)/10=(-4+25-1)/10=20/10=2 |
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| 26. |
If a circle touches all the four sides of aparallelogram, the parallelogram is a rhombus. |
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Answer» Let ABCD be a parallelogram which circumscribes the circle. AP = AS [Tangents drawn from an external point to a circle are equal in length] BP =BQ [Tangents drawn from an external point to a circle are equal in length] CR= CQ [Tangents drawn from an external point to a circle are equal in length] DR = DS [Tangents drawn from an external point to a circle are equal in length] Consider, (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)AB + CD = AD + BC But AB= CD and BC = AD [Opposite sides of parallelogram ABCD] AB + CD =AD + BC Hence 2AB = 2BC Therefore, AB= BC Similarly, we get AB= DA and DA = CD Thus, ABCD is a rhombus. |
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| 27. |
मान ज्ञात कीजिए --\frac{5}{2}-\frac{3}{5} \times \frac{7}{2}+\frac{3}{5} \times\left(\frac{-2}{3}\right) |
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| 28. |
- \frac { 2 } { 3 } \times \frac { 3 } { 5 } + \frac { 5 } { 2 } - \frac { 3 } { 5 } \times \frac { 1 } { 6 } |
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| 29. |
Using appropriate properties-\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6} |
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| 30. |
Using appropriate properties find:-\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6} |
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| 31. |
17. The quadrilateral formed by joining the mid points of consecutive sides of a trapezium will be a:a) Parallelogramb) Rhombusc) Trapeziumd) Square |
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Answer» The answer is Rhombus |
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| 32. |
2) A circle touches all sides of a parallelogram. So the parallelogram must be a(A) rectangle(B) rhombus(C) square(D) trapezium |
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Answer» Answer: B)RhombusExplanation:Let ABCD be a parallelogram which circumscribes the circle.AP = AS [Tangents drawn from an external point to a circle are equal in length] BP =BQ [Tangents drawn from an external point to a circle are equal in length] CR= CQ [Tangents drawn from an external point to a circle are equal in length] DR = DS [Tangents drawn from an external point to a circle are equal in length] Consider, (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)AB + CD = AD + BCBut AB= CD and BC = AD [Opposite sides of parallelogram ABCD] AB + CD =AD + BCHence 2AB = 2BC Therefore, AB= BC Similarly, we get AB= DA and DA = CDThus, ABCD is a rhombus. |
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| 33. |
Q1. Write the formulae of(0) Area of parallelogram(ii) Area of trapezium |
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Answer» i) Area of Parallelogram = Base × Height ii) Area of Trapezium = ( a + b) × h/2 a = base 1 b = base 2 h = Distance between parallel sides area of. parallelogram ,base ×height |
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| 34. |
हनन 85 21४ ४ 1205 हु 1| 22Ul Ble 1855 0§ Sakk १२५ ०२९ 0द 1७: हि 21% कि (8) |
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Answer» 20m/s20*18/5= 52km/hour |
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| 35. |
EX1.Find using appropriate propertiesa) XX |
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| 36. |
Maths.Write each properties of trapezium,.parallelogram,rectangle, rhombus, square. |
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| 37. |
What is the measure of angle between the bisectors of adjacent angles in linear paiPoint P is on the hicocto |
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Answer» let the angle be xthen x+x=180°2x=180°x=90° |
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| 38. |
1 .â(29) = 2022 (27) |
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Answer» Please hit the like button if this helped you Thank u very much |
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| 39. |
IHolidayHome workMathsfindusingpropertiesEX(-97+3x/915 125 112azAdditiveinverse of |
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| 40. |
\left(\frac{1}{4}-\frac{1}{6}\right) \text { of }\left(\frac{2}{3}-\frac{5}{12}\right) \times\left(\frac{5}{8}-\frac{7}{12}\right) |
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Answer» (2/3-5/12)×(5/8-7/12)=(8-5)/12 × (15-14)/24=3/12×1/24=1/4×1/24=1/96(1/4-1/6)=(3-2)/12=1/121/12 of 1/96=1/(12×96)=1/1152 Thnx |
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| 41. |
1. Complete the following factor trees and writedown the prime factors of each number90 85 150800 |
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| 42. |
Answer thequiWhat is a parallel? Write three features of parallels. |
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Answer» Parallel lines are the lineshaving the same distance continuously between them. features of parallel line are;-1) they never intersect each other2)The distance between the parallel lines will be same at all points.3)Slope of parallel lines are same.4)The angles formed when two parallel lines are cut by another line called transversal are same. |
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| 43. |
olve the following pairs of equationsequations:byreducingthemtoapaiAd)1+1=2.1.1-13In |
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Answer» hit like if you find it useful |
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| 44. |
13) Find the value of x and y which satisfy thefollowing equations (x,yeR)+ 2y)+ (2x -3y)i +4i 5 |
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Answer» thanks dude |
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| 45. |
solve :3x +2x = ??? |
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Answer» 3x + 2x = (3+2)x = 5x is the best answer 5x is the best answer 🏆🏆🏆 5x is the correct answer |
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| 46. |
Solve 3x+2-11 |
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Answer» 3x = 11-23x = 9 x = 3So the value of x is 3. |
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| 47. |
6. Find the common factors of the following:(a). 8, 24, 32(c) 35, 15(e) 24, 36(b) 64, 76(d) 42,36(f) 11, 33, 887. Make factor trees of each of the given numbers.(a) 4800(b) 540(c) 630(d) 7500 |
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Answer» a-8b-4c-5d-6e-12f-11 -8-4-5-6-12-11is the right answer a)8= 2×4 =2×2×224=2×12 =2×2×6 =2×2×2×332=2×16 =2×2×8 =2×2×2×4 =2×2×2×2×2common factors = 2×2×2=8 b)64=2×32 =2×2×16 =2×2×2×8 =2×2×2×2×4 =2×2×2×2×2×276=2×38 =2×2×19common factors = 2×2 = 4 c) 35=5×715=3×5Common factors = 5d)42=2×21 =2×3×736=2×18 =2×2×9 =2×2×3×3common factors = 2×3 = 6 e)24=2×12 =2×2×6 =2×2×2×336=2×18 =2×2×9 =2×2×3×3common factors = 2×2×3 = 12 f)11=1×1133=3×1188=2×44 =2×2×22 =2×2×2×11comman factors = 11 |
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| 48. |
rections: Solve each problem. Show all work using MOactor the equation below to solve for when e)-.Factor the equation below to solve for x when f(x) o.f(x)2x2 + 9x-35a) x = 2.5 or x = 7d) x = 2.5 or x=-7 |
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| 49. |
\int \frac{\cos 2 x-\cos 2 x}{\cos x-\cos \alpha} d x |
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Answer» thanks |
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| 50. |
5.Rationalise the denominators of the following:()17-J65+2(iv)523- S5(17+35202 +3137-336.If both a and hare |
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