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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. |
. Which is greater?e)0 33 or 3.300 G) 5.64 or 5.603or 0.4 (b) 0.07 or 0.02 (c) 3 or 0.8123 or 1.2 () 0.099 or 0.19 (g) 1.5 or 1.50 |
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| 2. |
bavztroed £12000 trom Py gad ot G % te ontwyears had ibdtiorw od thisrounded onualysimele inteetsumwhat ext31q oman t ω0uld i have to Pay |
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| 3. |
न गर्लग” eg 33. यदि (का 9-1, तो 9 का गन हो |
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Answer» tan ¢ = 1then, ¢=45° |
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| 4. |
DEisDEQ11. In the given figure, BA is perpendicular to AC,BFs EC. Show that ΔABC is congruent to ΔDEF.perpendicular to DF such that BAand |
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Answer» In ∆ ABC and ∆ DEFAB = DEBAC = FDE ( 90°)BC= EF ( as given BF = EC adding FC on both sides gives BC = =EC)So, by RHS rule, ∆ ABC and ∆ DEF |
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| 5. |
1. Solve the following pairs of equations by reducing them to a pair of linear equations:2 3ä¸+--=2(1) 2x , 3y2(ii)1 1 133x 2y 6 |
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| 6. |
33 Polynomia, x4+73 St 70c2+pata is enactlydivisible by at7x+12, then and thevalue þ and g |
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| 7. |
Bin the adjoining ligure, AC LAB DB I BAand AC OB show that no=B0 where dintheof AB and ED.point of intersectionsont |
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Answer» the old is inserted in the middle of a b and c d |
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| 8. |
-42 ab acCl14. Prove that babe 4a2b2c2ac be[Hint.Take a, b, c common from R1, R2, R3 andthen operate R1-Ri-R2] |
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| 9. |
\frac { - 36 } { 11 } \text { and } \frac { 49 } { 22 } \text { from the sum of } \frac { 33 } { 8 } \text { and } \frac { - 19 } { 4 } |
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| 10. |
e the following2x 3y1 1 133x 2y 6 |
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| 11. |
26. List out the bones of human hind limb (legs). |
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Answer» a)It includes Femur + Tibia and Fibula + Tarsals + Metatarsals + Phalanges b) Fovea capitis is depression in head of femur. c) Femur is longest and strongest bone of body. d) Femur is known as bone of thigh e) Greater trochenter, lesser trochenter 3rd trochonter are present in femur, of thigh and buttock muscles. |
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| 12. |
(A) 90(B) 135(C) 120A-5In the adjoining figure, AP and BP are angle bisectorsABCD. Then 2<APB =of ZA and B which meets at P on the2 |
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| 13. |
Dr ill the zeroes of the polynomial pV3 and 2-va.12.13two of terre |
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Answer» x^4-7x^3+9x^2+13x-4 //2+V3 (2+V3)^4-7(2+v3)^3+9(2+V3)^2+13(2+V3)-4=(16)+(9)+4V3-7(4+3+4V3)+9(4+3+4V3-)+26+13V3 -4=25+4V3-42+28V3+63+36V3+26+13V3-4==25-42+63+26+4v3+28v3+36v3+13V3=72+12V3 |
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| 14. |
1/2x+1/ 3y=21/3x+1/2y=13/6 |
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| 15. |
In the adjoining figure, ABCD is a l|gm whose diagonals AC and BD intersect atO. A line segment through O meets AB atP and DC at Q. Prove thatar(APQD) =ar(||gmABCD)/2. |
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| 16. |
Q.5 Tn the given figure, AC-RC. I.DCA = I.FCB and i PBC = I.FAC. Provethat △ DBC-△ LAC and DC-EC. |
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| 17. |
In the given figure, AC lI GD and AE I BF. Find the values of x, y and z115°85° |
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Answer» As AC || BD,115°+x=180° (Interior Angle)Therefore x=65°. x+y+86=180(Supplementary angle)y+65+85=180y+150=180y=30° As AE || BF,z+y=180(Interior Angle)z+30=180z=150° |
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| 18. |
1/2x +1/3y =2 ; 1/3x +1/2y = 13/6 |
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| 19. |
129. Solve for x and y:and y.52(2x+3y)127(3x-2y)(2x+3y(3x-2y) |
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| 20. |
5 In the figure, BC || DE. Find x. |
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Answer» X=120° is the right answer x=120 is the correct answer x=120° is the best answer |
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| 21. |
8. In the given figure (12.24), BA 1 AC and DE 1EF such that BA = DEand DF = BC. Prove that AC = EF. |
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| 22. |
33 and8-194and 20 from the surn of49e sum ot g36. Subtract the sum of |
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Answer» sum of (-36)/11 and 49/22 =-36/11 +49/22 =-23/22sum of 33/8 and -19/4=33/8+(-19/4)=-5/8subracting these sums=-5/8-(-23/22)=-5/8+23/22=37/88 |
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| 23. |
T o1 g 13 e 2 5 30 1035 हे है छह ते ४ Rie ) bl 2 कह ड् 2b b LSl | b 33 Il B कं यु |
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| 24. |
5. PMFind the solution of the equation 3x - 4y + 11 = 0; 7x +11y - 15 = 0.be solution of the pair of linear equation 3r5y-4- and 9x = 2y + |
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| 25. |
16. Draw a parallelogram ΔΑBCD in which AB-6.5cm AD-4.8cm and zBAD-700Measure its diagonals. |
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| 26. |
15. In the given figure,C is the midpoint of AB. IfZDCA = ZECB and ZDBC = ZEAC, provethat DC =EC. |
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| 27. |
The side BC of a AABC is produced, such that D ison ray BC. The bisector of ZA meets BC in L asshown in Fig. 12. Prove that ZABC + LACD -2LALC4 |
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Answer» Because Angle A is bisected. Therefore, Angle BAL is equal to Angle LAC.Let angle bal = x.Therefore lac =x.Triangle alc is a right angle triangle.In this angle alc = 90 °Therefore angle acl =acl + lac + |
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| 28. |
21. In the given figure, C is the mid-point of AB. If 4DCA LECB andLDBC-ZEAC, prove that DC-EC. |
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| 29. |
Ill limb12. In the figure 6.187, the side BC of AABC is producedto D. The bisector of ZA meets BC in L. Prove thatCDFig. 6.187 |
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| 30. |
1 in the adjoining figure, ABCD is a trapeziumin which AB DC and E is the midpoint ofAD. A line segment EFI AB meets BC at F.Show that Fis the midpoint of BC. |
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| 31. |
1. In the adjoining figure, ABCD is a trapeziumin which AB|| DC and E is the midpoint ofAD. A line segment EFll AB meets BC at F.Show that F is the midpoint of BC. |
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| 32. |
EXERCISE 9C1. In the adjoining figure, ABCD is a trapeziumDCin which AB|DC and E is the midpoint ofAD. A line segment EFll AB meets BC at F. EShow that F is the midpoint of BC. |
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| 33. |
VTATIEMAT8 cmCalculate the area of the designed region inFig. 12.34 common between the two quadrantsf circles of radius 8 cm each908 cm8 cn8 cmFig. 12.34 |
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| 34. |
8 cm16. Calculate the area of the designed region inFig. 12.34 common between the two quadrantsof circles of radius 8 cm each.908 cm908 cmFig. 12.34 |
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| 35. |
।1.24 क- सभाः द ध ध2. यटि बहुपट ४) के लिए 2-1-2 हो, तो बहुपदअवकल क्या होगा?।,सेभ देने |
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Answer» भाग 0 होगायहां पर हमने भागअल का उपयोग करते हुए कि आंसर दिया है |
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| 36. |
In figure, AC = AE, AB= AD and angleBAD=angleEAC. Showthat : BC=DE |
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Answer» Jaldi answer De di karo Mam |
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| 37. |
8.49, AC = AE, AB = AD and 4ADIn Fig.LEAC. Prove that BC = DE. |
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| 38. |
given figurE AC - AE,AB AD and BAD LEAC. Show that BC - DEOR8 |
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| 39. |
-15Solve the following pairs of equation by reducing them to a pairof linear equation.2x 3y1 1 133x 2y 6 |
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| 40. |
The solution of the linear equation 7 x-1=\frac{1}{6} is |
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| 41. |
09 503 + QE UIS (१द न्राइ्यु[क BIE bl % BBkl T |
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Answer» sin 30 degree= 1/2cos 60 degree= 1/2sin 30 degree+ cos 60 degree=(1/2) +(1/2)=2/2=1Therefore sin 30 degree+ cos 60 degree=1 |
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| 42. |
8 em16. Calculate the area of the designed region inFig. 12.34 common between the two quadrantsof circles of radius 8 cm each.8 cm90°8 cmFig. 12.34 |
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| 43. |
3816. Calculate the area of the designed region inFig. 12.34 common between the two quadrantsof circles of radius 8 cm each8 cm90°8 cmFie, 12.34 |
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| 44. |
MATHEMATI2388 cm16. Calculate the area of the designed region inFig. 12.34 common between the two quadrantsof circles of radius 8 em each.908 cm8 cm90°8 cmFig. 12.3412.5 Summary |
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| 45. |
8 em16. Calculate the area of the designed region inFig. 12.34 commos between the two quadrantsof circles of radius 8 cm each8 em908 cmFig. 12.34 |
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| 46. |
14. In figure (13.56), ABCD is a parallelogram in which LDAB 70and ZDBC 50°, compute LCDB and ZADB.70° |
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| 47. |
In the given figure, ΔΒΕΕ and ABAC are isosceles triangles and AEWhat can you say about ZADB and ZCEB? |
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| 48. |
The number of balls, each of radius1 cm that can be made from a solidsphere of radius 8 cm is |
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| 49. |
In AABC, AC> AB the bisector of LA meets BC at D. Prove that ZADB isan acute angle.. |
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| 50. |
In the adjacent figure, AC AE, AB ADand ZBAD ZEAC. Show thatBC = DE. |
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