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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. |
8. Problem: Find the equation of the circle passing through (4, 1), (6, 5) and having theCentre on the line 4x + y-16 0. |
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| 2. |
Find the co-ordinates of a point p on line x + y--13, nearest to the circle x2 + y2 + 4x + 6y-5-(B) (- 15, 2) |
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Answer» The line through the nearest point must pass through center of circle and must be perpendicular to given line. Therefore, equation of line perpendicular to x+y+13=0 will be y-x+λ =0. It will be passing through center of circle which is (-4/2, -6/2) = ( -2,-3) So, (-3)-(-2)+λ=0 which gives λ= 1.Hence , equation of line will be y-x+1=0. Now, intersecting point of x+y+13=0 and y-x+1=0 will give the required point.Solving for x and y , x=-6 and y=-7So, the point is (-6,-7). |
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| 3. |
Set of values of m for which a chord of slope m of the circle x^2 +y^2 4parabola y^2=4x is |
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Answer» y^2= 16x comparing this with y^2= 4ax we get focus of this parabola is at (a,0) = (4,0) now focal chord : y = m(x-4) ..................1 m is slope of this chord.... this chord is tangent of (x-6)^2+ y^2= 2 this is equation of circle ... for circle x^2+y^2= a^2 if line is tangent then its equation is given by y = mx +(-)a(1+m^2) for (x-6)^2+ y^2= 2 tangent will be y = m(x-6) +(-)a(1+m^2)^1/2 y = mx - 6m +(-)(2+2m^2)^1/2 ..................2 eq 2 & eq 1 are same line so -4m = -6m +(-)(2+2m^2)^1/2 (equating intercepts) 2m = (2+2m^2)^1/2 m^2= 1 m = +(-) 1therefore possible values for m are +1 & -1 |
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| 4. |
(i) Exterior angle of a regular polygon having n sides is more than thatof the polygon having n2 sides by 50°. Find the number of the sidesof each polygon. |
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Answer» Exterior angle of a regular polygon having n side = 360/n Exterior angle of a regular polygon having n² side = 360/n² 360/n-360/n^2=50360n-360/n^2=505n^2-36n+36=0(5n-6)(n-6)=0n=6 |
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| 5. |
Question: Divide 3x2 - x3 - 3x + 5 by x - 1and verify the division algorithm. |
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Answer» There is a mistake in question 3x^2-x^3-3x+5 also ,-x^3+3x^2-3x+5x^2+x-1 ) -x^3+3x^2-3x+5 ( -x+4 - x^3 -x^2+x ________________ 4x^2-4x+5 4x^2 - 4x-4 _______________ -8x+9 clearly , remainder = -8+9 quotient = -x+4 using division algorithm , a = b*q+r = (x^2+x-1)(-x+4)-8+9 =-x^3+3x^2-3x+5 also , 3x^2-x^3-3x+5 ( dividend ) |
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| 6. |
2.Divide (3 - x + 2x) by (2 - x) and verify the division algorithm |
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| 7. |
What can you say about the angle sum of a convex polygon with number of sides ?a) 7(b) 8(c)10(d) n |
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Answer» The angle sum of a convex polygon ofnsides is (n−2) × 180º.So the angle sum of the convex polygons having number of sides as above will be as follows.(a) (7 − 2) × 180º = 900°(b) (8 − 2) × 180º = 1080°(c) (10 − 2) × 180º = 1440°(d) (n− 2) × 180° |
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| 8. |
6. If the common difference of an A-P is-6, find a,-47 tOisthe centre of a circde, PQ is a chord and the tanges |
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| 9. |
4. Fill in the missing1010 |
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Answer» 4/10 ................ |
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| 10. |
Q.4 Find the missing terl,13,,3sinAcalculate cosA and tanA. |
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Answer» Sin a =perpendicular/hypotenuse=3/4then,base =√( 4²-3²)=√7 then cosa=base/hypotenuse= √7/4 and tana =perpendicular/base=3/√7 this app not working 😠 |
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| 11. |
Given that the vertices A, B, C of a quadrilateral ABCD lie on a circle.Also ZA +ZC 180°, then prove that the vertex D also lie on the same circle.2. |
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Answer» Given. A+C=180° and A, B, C Lie on a circle To prove. D lies on the same circle as A, B, C. Soln. -- In a quadrilateral, if the sum of opposite angles is 180°, then it is a cyclic quadrilateral. In quad. ABCD, A + C=180°,Therefore, ABCD is a cyclic quad. In a cyclic quad., the vertices lie on the same circle. Thus, D also lies on the same circle..Hence, proved! |
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| 12. |
36. In fig OPQR is a rhombus whose three vertices P, Q, R lie on a circle of radius S cm.Find the area of the shaded region.23-O+323 |
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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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| 13. |
parabola y-=4X.2. Find the area of the circle 4x 4yhichis interio to the parabola x2 -4y.2. |
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| 14. |
11. The focus of a parabola is (1, 5) and its directrix is the straight line x + y + 2 = 0 Findthe vertex and the latus rectum of the parabola |
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| 15. |
iv)Anequilateraltriangle is inscribed in the parabola y - 4ax, such that one vertex ofthis triangle coincides with the vertex of the parabola. Then find the side length ofthis triangle. |
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| 16. |
16·A) Divide f (x) = 4x^4-3x^3-2x^2+x-7 by g(x) x-1 find remainderand quotient. Also verify using division algorithm. |
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| 17. |
11. An arch is in the form of a parabola with its axisvertical. The arch is 10m height & 5m wide at the base.How wide is it 2m from the vertex of the parabola?INCERT] |
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| 18. |
2. What is the sum of the interior angles of a polygonc. 10 sidesliieular polygon5 sidesb. 8 sides |
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Answer» Sum of Interior Angles = (n-2) × 180° a. 540 degree b. 1080 degree c. 1440 degree |
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| 19. |
Divide f (x) = 4x^4-3x^3-2x^2+x-7 by g (x)=x-1 find remainderand quotient. Also verify using division algorithm. |
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| 20. |
Simple-intsust matelor At the end, 4yoh,Loh utim in |
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| 21. |
Divide p) . 5x-3x+62-7by 9 (x) = 1-2* by using longdivision, and find its remainder |
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Answer» p(x)=5x^3-3x^2+6x-7; g(x)=1-2×;; x=-1/2;; p(x)=5(-1/2)^3-3(-1/2)^2+6(-1/2)-7=-5/8+3/4-6/2-7= 5+6-12-56/8=-1-56/8=-57/8 |
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| 22. |
Simplify and Find value of missing variable\frac{1}{3}\left(-\frac{7}{4} k+1\right)-\frac{10}{3} k=-\frac{.13}{8} |
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Answer» thanks handwriting cannot be understood |
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| 23. |
|| boll%, 1}}}z Male এ ৮। lon Jay le]>'1lf. If/a24 hole |
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| 24. |
5. Determine if 23100 is divisible by 33.'lon |
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Answer» a number is divisible by 3, if it's sum is divisible by 32+3+1+0+0= 6and 6 is divisible by 3so number is divisible by 3 |
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| 25. |
91. DGJ, FIJ, NKH, JMP, ?(2) LOR(4) LOM(1) LON(3) LOQ |
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Answer» LOR is the correct answer |
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| 26. |
The parallel sides of a trapezium are 15 Im and 10 m long and its non-parll sidesare 8 m and 7 m long. Find the area of the trapezium. |
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Answer» Like if you find it useful |
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| 27. |
16. The numerator of a fraction is 6 more than the denominator. If 5 is subtracted from the denominator,the fraction becomes Find the fraction. |
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Answer» Let the numerator be xand denominator ve (x+6)After adding 2,New numeraror = (x+2)New denominator = (x+6+2)= (x+8)New Fraction= 1/2(x+2)/(x+8)= 1/22(x+2) = 1(x+8)2x+4 = x+82x-x = 8-4x= 4Hence,nunerator = x= 4denominator = (x+6)= (4+6) = 10Original fraction= 4/10 or 2/5 2/5 is the best answer.. |
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| 28. |
5. A fraction becomes, if 2 is added to botlhthe numerator and denominator. If 3 is added |to both the numerator and the denominator,it becomes. Find the fraction. [HSLC 16] |
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| 29. |
The numerator of a fraction is 4 less than the denominator. If 1 is added to boththe numerator and the denominator, the fraction becomes . Find the fraction.Let the denominator be x. |
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| 30. |
chairprice of each.The denominator of a fraction is 3 more than the numerator If2 is added to the numeraior a5 is added to the denominator, the fraction becomesFind the fraction |
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Answer» but the answer is 4/7 Sorry. Mistook 5 for 6. This is the correct solution. Apologies for the error. thnkiu u |
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| 31. |
and remainder.90 सेमी की लम्बाई वाली एक लड़की बल्ब लगे एक खंभे केआधार से परे 1.2 मी/से की चाल से चल रही है। यदि बल्बभूमि से 3.6 मी की ऊँचाई पर है, तो 4 सेकण्ड बाद उस लड़की3की छाया की लम्बाई ज्ञात कीजिए।A girl of height 90 cm is walking away from thebase of a lamp-post at a speed of 1.2 m/s. If thelamp is 3.6 m above the ground, find the lengthof her shadow after 4 seconds.P. T,0.=03/(Set: A) |
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| 32. |
ion2.a) On what sum of money will the diference between the compound interest and simple imest for 2 years be equal to? 25, tf the rate of interest charged for both is 5% pa? |
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Answer» Difference=PR^2/10000PR 25=P×5×5/10000P = 10000 app try kar raha tha kya??? |
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| 33. |
of uleu1. Govind borrowsat 10% simplevests the moneySincsecondm.indest. He immediately invests theborrowed at 10% compoundcompounded half-yearly. How mydoes Govind gain in one year ?ompound intereslow much moneyof |
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Answer» P=18000rsR=10%T=1 yearS.i. =P *R*T/10018000*10*1/100=1800rsCompounded halfyearlyP =18000 rsT=2 half yearsR=5%A=p[1+r/100]^t=18000[1+5/100]^2=18000[21/20]^2=18000*21*21/400=45*441=19845rsC.i.=19845-18000=1845rsMoney he gained =1845-1800=45rs P=18000rsR=10%T=1 yearS.i. =P *R*T/10018000*10*1/100=1800rsCompounded halfyearlyP =18000 rsT=2 half yearsR=5%A=p[1+r/100]^t=18000[1+5/100]^2=18000[21/20]^2=18000*21*21/400=45*441=19845rsC.i.=19845-18000=1845rsMoney he gained =1845-1800=45rs |
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| 34. |
6.A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areasof the corresponding minor and major segments of the circle(Use n-3.14 and 3-1.73) |
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| 35. |
8 x + 6 y + 3 = 03 \pi / ( \text { and } ) 4 x + 3 y + 1 = 0 |
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Answer» ii) 8x+6y= -3 .....(1 4x+3y = -1.......(2so multiply second eq by 2 and subtract it from eq 1we get 0= 1 which is absurd so there is no solution for this pair of equation. tell me what I solve in this question |
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| 36. |
more interest and by how4.y,wnopaysmuch'?1borrowed? 12,000 from Jamshed at 6% per annum sinple interHad I borrowed this sum at 6% per annum compound interest,wwould I have to pay?for 2 years.what extra amount |
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| 37. |
2. Use induction to prove that n^3 -7n +3, is divisible by 3, for all natural numbers n |
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| 38. |
ABC is a right-angledLB-90, LA = 30° and20" In the adjoining figure,triangle in whichAC = 20 cm.Find (i) BC, (ii) AB. |
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Answer» I Dont know sorry sorry |
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| 39. |
In the figure 7.48, square ABCD isinscribed in the sector A-PCQ. Theradius of sector C-BXD is 20 cm.Complete the following activityto find the area of shaded region.P.Fig. 7.48 |
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| 40. |
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areasof the corresponding minor and major segments of the circle.(Use n-3.14 and 3-1.73)6. |
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| 41. |
From the given figure alongside find the differencebetween the perimeter of the square and thecircumference of the circle inscribed in the squarc.8. |
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| 42. |
A room is 12.5 in long and 8 m wide. A squarc carpet of side 8 Im is laid on its floor. Find thearea of the floor which is not carpeted. |
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| 43. |
Use principle of mathematical induction to prove that1.2 +2.3 + 3.4 + ....+ n(n+1) = n(n + 1) (n +2),VnEN |
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| 44. |
Use mathematical induction to prove:If n is any odd positive integer then n(n^2 - 1)is divisible by 24. |
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| 45. |
23. If points (7,-2).(5.1),and (3,k) are collinear then find the value of k.24. Find the value of 4sin 60'+3tan-300-8sin 45'-cos'45"11. |
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| 46. |
1 2(3 1 (7 11If, then write the value of k. |
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| 47. |
Show that the points (12,8), (-2, 6) and (6,0) form a right angled triangle. Also find the areatriangle. |
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| 48. |
27. Find the greatest number that will divide 445, 572 and 699, leaving remainders 4, 5, 6respectively.test number that wil divide 445, 572 and 699, leaving remainders 4.5.6 |
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Answer» Since the respective remainders of 445, 572 and 699 are 4, 5 and 6, we have to find the number which exactly divides (445-4), (572-5) and (696-6). So, the required number is the HCF of 441, 567 and 693.Firstly, we will find the HCF of 441 and 567. ∴ HCF = 63 Now, we will find the HCF of 63 and 693. ∴ HCF = 63 Hence, the required number is 63. |
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| 49. |
find the greatest number that will divide 508 635 762 leaving remainders 4 5 6 |
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Answer» Subtract 4 , 5 and 6 respectively from 508, 635 and 762 we get the number 504, 630 and 756. Let a = 630 and b = 504 Euclid division algorithm 630 = 504 × 1 + 126 504 = 126 × 4 + 0 HCF (630,504) = 126 Find HCF of 756 and 126 756 = 126 × 6 + 0 HCF (756,126) = 126 Thus HCF of 504, 630 and 756 = 126 Hence the number is 126 |
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| 50. |
lf the three linesbx+ by+-0and cr + c2 y+1oat tay+ 1 0,urrent, show that at least two of the three constants a, b,c are equal.are |
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