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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. |
Find the locus of a point whose abscissaithree times its ordinate. |
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Answer» abscissa =x ordinate=y The locus is given by x=3y |
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| 2. |
Q- 1924 is divisible by 2 and ................a) 0 b) 7 c) 5 |
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Answer» Factors of 1924 1924/1 = 1924 gives remainder 0 and so are divisible by 1 1924/2 = 962 gives remainder 0 and so are divisible by 2 1924/4 = 481 gives remainder 0 and so are divisible by 4 1924/13 = 148 gives remainder 0 and so are divisible by 13 1924/26 = 74 gives remainder 0 and so are divisible by 26 1924/37 = 52 gives remainder 0 and so are divisible by 37 1924/52 = 37 gives remainder 0 and so are divisible by 52 1924/74 = 26 gives remainder 0 and so are divisible by 74 1924/148 = 13 gives remainder 0 and so are divisible by 148 1924/481 = 4 gives remainder 0 and so are divisible by 481 1924/962 = 2 gives remainder 0 and so are divisible by 962 1924/1924 = 1 gives remainder 0 and so are divisible by 1924 wrong answer |
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| 3. |
Find the length of the arc of a circle of dlameter 20 cm.,angle 36 at the centre.if the arc subtends anQ.2)Q.3) Convert the anglos a)44 b)-9r into degree. |
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Answer» thanks Convert the angel of 250° in degree (7π/12)C |
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| 4. |
44+4. |
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Answer» 48 is the correct answer |
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| 5. |
The diagonals of a rhomburhombuss are 16 cm and 12 cm. Find the length of each side of the |
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| 6. |
9. Write the natural numbers from 205 to 219. What fraction of them are odd numbers? |
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| 7. |
рео) - 7\"7 =4<Pn + 5┬л/: 19 |
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Answer» Like if you find it useful |
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| 8. |
92.One diagonal of a square is along the line 8x-15yone of its vertex is (1, 2). Then the equation of thethe square passing through this vertex, are |
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| 9. |
p ( का नए 2 भ्े cq %—Q’_S{\MQ’ |
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Answer» So-x^2+x-2= 0now x^2-x+2x^2-x+2x-2x(x-1)+2(x-1)(x+2)(x-1)= 0 |
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| 10. |
3.Length of the latus rectum of the parabola25 [(x-2)+(y-3)]-(3x-4y+7)2 is(A) 4(B) 2(C) 1/5(D )2/5 |
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| 11. |
The area of a quadrilateral is 120 cm2 and thediagonal is 20 cm. If the length of perpendicularsfrom one vertex is 7 cm, find the length ofperpendicular from the other vertex. |
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| 12. |
Q 19: Find the value of the following:cos220째 + cos 70째sin-31째 + sin- 5gosin'64+ cos 64째 sin 26 |
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Answer» cos^2 20+cos^270 + sin^264+ cos64 .sin26 sin^231+ sin^259 =cos^2(90-70) + cos^270 + sin^264+ cos64. sin(90-64) sin^2(90-59)+sin^259 =sin^270+cos^270 + (sin^264 +cos^264 ) cos259+sin259 = 1/1 +1 = 2 |
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| 13. |
find the value of p for which the root of the equation px(x-2)+6=0 ,are equal |
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Answer» then just solve the equation and then make it in the form of ax2 +bx + cthen by the formula of ✓b2-4ac =0find the value of pgot it??? Px2-2px+6=0 b2-4ac=0(given) (-2px)2-4(px2)(6)=0 4p2x2-4(6px2)=0 4p2x2-24px2=0 4p2x2=24px2 P2=24px2/4x2 p2=6p p=6p/p p=6 |
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| 14. |
Solve the following quadratic equations byfactorization method:(İ) 6√3x^2 + 7x=√3.(İİ) y^2 + 2√3y-9=0(iii) x^2-3√x + 6=0 |
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| 15. |
2.Show that any positive odd integer is of the form 6q+1, or 64 +3, or 6q + 5, where qsome integer |
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Answer» show that any positive odd integer is of the form 6q+1. or 6q+3. or 6q+5.where q is some integer |
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| 16. |
how that any positive odd integer is of the form 6q + 1, or 64 3,or 6q+5, where q isome integer |
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Answer» Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder. according to Euclid's division lemmaa=bq+r a=6q+r where , a=0,1,2,3,4,5then,a=6qora=6q+1ora=6q+2ora=6q+3ora=6q+4ora=6q+5 but here,a=6q+1 & a=6q+3 & a=6q+5 are odd. please like if you find it useful |
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| 17. |
Show that any positive odd integer is of the form 64+ 1, or 6q +3, or 6q + 5, where q issome integerhm |
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| 18. |
(a) 64(b) 46(c)66(d)444 Q 1924 is divisible by 2 and(a) 40(b) 7(c) 5(d) 3 |
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Answer» wrong answer |
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| 19. |
(iii) Prove that the normal chord to a parabola at the poiatwhose ordinate is equal to the abscissa subtends a right angle at the(M. U. 1969)focus. |
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| 20. |
(f) Write odd numbers between 100 and 160. |
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Answer» The odd numbers between 100 and 160 are:-101, 103,105, 106, 107, 109, 111 , 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133, 135, 137, 139, 141, 143,145,147 ,149,151,153, 155,157,159 |
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| 21. |
7m 20 +2m19. The sum of 2 consecutive odd numbers is 20, find the numbersf n numher by 10. Fin |
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Answer» Let two consecutive odd numbers are x and (x + 2) Given,x + (x +2) = 202x = 18x = 18/2 = 9 Two consecutive numbers are 9 and 11 |
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| 22. |
of the diagonals of a rhombus is 11.6 m. If area of the rhombuOnembus is 49.3 m2, find the length of theother diagonal |
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| 23. |
8. Find the condition such that one root of the equationx2+px+9 = 0 is 4 times the other. |
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| 24. |
[Find the equation of the parabola with vertex at (0, 0) and the focus at |
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Answer» focus of parabola is (0, a) . here a=2. so equation is x^2=4×2×y. x^2=8y |
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| 25. |
A straight line through the point A(3, 4) is such that its intercept between the axes isbisected at A. Its equation is(A) 3x-4y + 7-o(C) 3x + 4y 2515.(B) 4x+3y 24(D) x + y 7 |
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| 26. |
2.53. In A MNP, NQ is a bisector of Z N.If MN 5, PN 7 MQ 2.5 thenfind QP.Fig. 1.37 |
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Answer» Like my answer if you find it useful! |
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| 27. |
to.in the figure given below ,if <QPR-64° and <TRS-67° then find the valueof <QRP, when QP//RT.61+ |
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Answer» angle PRTwill be equal to 64 ° and the sum of ( both are alternative)angle PRT+angle QRP+ angle TRS = 180Angle QRP= 180°-(64°+67°)= 180-13149° Here as we know thatangle TRP =Angle QPR (alternate interior angles)Therefore angle TRS + PRT+PRQ=180 (linear pair)there for 67+64+ Angle PRQ =180ANGLE PRQ=180-131=49 The answer is 49 |
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| 28. |
In figure sides QP and RQ of ΔΡ0R are produced to points S and T respectivelyif LSPR-135 and LPOT-110, find LPRO135 |
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| 29. |
In Fig. 6.39, sides QP and RQ of ∆POR are produced to points S and T respectively.If angle SPR=135° and anglePQT=110°, find anglePRQ. |
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| 30. |
Typei)Find the equation of the parabola whose focus is (0,-2) and directrix is |
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Answer» wrong answer |
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| 31. |
The equation of the parabola whose focus is-4,5 and equation of directrix 2y-9 032 |
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| 32. |
160 If p-q+4q 2, when is the value of p when q 64?2 |
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Answer» Given, q = 64 Then,p = (64)^2/3 + 4(64)^-1/2 = (4)^3*2/3 + 4(8)^-2/2 = (4)^2 + 4(8)^-1 = 16 + 4/8 = 16 + 1/2 = 33/2 = 16.5 Therefore, value of p = 16.5 |
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| 33. |
Find the co-ordinates of vertex and the length of latus rectumof the parabola whose focus is (0, 0) and the directrix iline 2x + y = 1 . |
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| 34. |
Fig. 8.29resne mbus and P. Q. R and S are the mid-points of the sides AB, BC, CDrpectively. Show that the quadrilateral PQRS is a rectangle.3. ARC |
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| 35. |
sitine odd fIntaguwhene |
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| 36. |
0W that the quadrilateral PORS is a rhombus.26. Prove that the angle subtended by an arc at the centre is double the angle subtendedby it atany point on the remaining part of the circle. |
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| 37. |
if 2' is a root ofx^2-px + q-0 and p^2-4q then otherroot- |
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Answer» Thank you |
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| 38. |
Write the centre and radius of the circle (x+5)+(y-3)3(Score : 1Find the equation of the parabola whose vertex is (0,0), focus on x axis and passesthrough (2,3).(b)(Score: 2) |
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| 39. |
SECTION D10 The path of a train A is given by the equation 3x 4y -12 0 and path of another train Bis given by the equation 6x+8y- 48 0 Represent this situation graphicallyhis wife, one third of the remaining to his son and theto |
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| 40. |
)Find the equation of the ellipse, whose length of the major axis is 20 and foci are (0,±5). |
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Answer» The length of the major axis is 20,And the foci are on y-axis , the major axis is along the y-axis.The equation of ellipse is of the form :x²/a²+ y²/a²=1 a = semi major axis = 20/2 = 10c² = a²-b² gives 5² = 10²-b² i.e. b² = 75 Therefore the equation of the ellipse us : x²/75+y²/100 = 1 |
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| 41. |
3.In A MNP, NQ is a bisector of ZN.If MN = 5, PN = 7 MO = 2.5 thenfind QP>NFig. 1.37 |
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| 42. |
3. In A MNP, NQ is a bisec-tor of ZN. If MN = 5,PN = 7, MQ=2.5; thenfind QP. |
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| 43. |
(3) 2x2 -3x -5 |
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| 44. |
5-A departmental store announced a 30% discount on sports goods as a promotional sales initiative. After the,discount if a cricket kit costs 2100, find the original price of the kit.tudonto ot 0204 |
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Answer» Cricket kit cost after discount = 2100Original cost = 2100 + 2100*(30/100) = 2100 + 630 = 2730Original cost of kit is Rs 2730 |
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| 45. |
9४०० '|0६ ५४ + / 088० 09५05 (50Rigj BUS kit |
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| 46. |
LINUCan you construct a quadrilateral ARC whene AB = 3 cm, CD = 3cm, AD7.5 cm, AC = 8 cm andBD 4 cm? Justify your answer. |
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Answer» hello bhai 4 answer aayega ji ok |
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| 47. |
AB and AC are two chord of a circle whose radius is r. if p and q are the distance of chord AB and CD from the Central respectively and AB=2BC than prove that 4q*=p*+3r* |
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Answer» Given, AB and AC are two chords of a circle of radius rAgain, AB = 2AC and the distances of AB and AC from the center are p and q.Let AC = x, then AB = 2xWe know that the perpendicular drawn from the centre to the chord bisect the chord.So, CL = x/2 and AM = xNow, in ΔOLC,r2= q2+ (x/2)2=> r2= q2+ x2/4Again, in ΔOAM, 4q2= 4(r2- x2/4) (using 1) 4r2= 4r2- x2=> 4q2= 4r2- (r2- p2) (using 2)=> 4q2= 4r2- r2+ p2=> 4q2= 3r2+ p2 |
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| 48. |
Find the radius of the cin Ia In the adjoining figure, OD isperpendicular to the chord A of aeirele with centre O. If BC is a diameter,show that AC II CD and AC-2OD |
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| 49. |
し4.(a) Find the equation of the ellipse whose focus is (1,1), directrix is x + y + 1 = 0 andeccentricity = 1/3. |
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| 50. |
2. Find the equation of parabola with vertex (2,- 3) and directrix x -4y 48 0. |
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