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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 area Ul l/if one of its diagonals is 8 cm long, IIl7The floor of a building consists of 3000 tiles whichits diagonals areif the cost per m2 is 24.ombus shaped ande45 cm and 30 cm in length. Find the total cost of polishingnnts to buy a trapezium shaped field.to and twice |
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| 2. |
·guoン5: Eindth e--Peninarka.ctsliang le, eath--.vex hce (o,4), Co, a) and (3,c)ORand EcE 12 c n LADE8 Find LABC0018 |
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Answer» As AD/BD = 6/9 = 2/3 AE/EC = 8/12 = 2/3 So, AD/BD = AE/EC So, By converse BPT Theorem DE || BC , So, angle D = angle B angle ABC = 48° as Alternate angle are equal |
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| 3. |
Explain why\Delta \mathrm{ABC} \cong \triangle \mathrm{FED} |
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Answer» In ∆ ABC and ∆ FEDBC = DE ( given)angle ABC = angle FED ( 90°)angle BAC = angle EFD ( given)by AAS rule ∆ ABC and ∆ FED are congurent |
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| 4. |
. B et oumelc p;:togmegs}o“‘/;¢ y\j ८ ली (न € Selenft, o ¢e &% Joum ey |
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Answer» Let a be the First term, a3 be the third term, a5 be the 5th term and a7 be the 7th terma3 = 16a7 = a5 + 12 ............ (1)Let the common difference be "d"Common difference is equal in APSo,a7 = a5 + d + d = a5 + 2d............(2)From Equation (1) & (2)a5 + 12 = a5 + 2d2d = 12d = 6 From Given, we get that a3 = 16a3 = a + 2d = 16a + ( 2× 6 ) = 16 [ We know that d = 6 ]a + 12 = 16a = 4 So first term is 4 .... We can find AP by adding d continuouslySo,AP is 4, 10, 16, 22, 28....... |
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| 5. |
Eind d |
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Answer» thank you |
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| 6. |
Eind n13S30 |
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| 7. |
के जि et b |
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Answer» Let the angles of quadrilateral be x, 2x, 3x , 4x sum of all angles of quadrilateral = 360° x + 2x + 3x + 4x = 360° 10x = 360° x = 360/10 = 36° Ans. the four angles are :- x = 36°2x = 2 × 36 = 72°3x = 3 × 36 = 108°4x = 4 × 36 = 144° hit like if you find it useful |
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| 8. |
विस 9-1... ९059व ४ . S T Ol e e7. Prove that: tanB-secO0+1 1-sinकर के T रच i |
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| 9. |
15. IfTmATanB=-then prove that A+B=-4 |
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Answer» tan(A+B) = (tanA+tanB)/(1-tanA*tanB) = (5/6+1/11)/(1-5/6*1/11) = (61/11)/(66-5)/11 = 1 => (A+B) = tan-¹(1) = π/4 |
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| 10. |
1 \div [ 1 + 1 \div \{ 1 + 1 \div ( 1 + 1 \div 2 ) \} ] |
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| 11. |
AD and BC are equal andperpendiculars to a line segmentAB. Show that CD bisects AB. |
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| 12. |
Two tangents TP and TQ are drawn to a circle with centre O froman externarpoint T. Prove that 2PTQ-0 COFOOR |
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Answer» We know that, the lengths of tangents drawn from an external point to a circle are equal.∴ TP = TQIn ΔTPQ,TP = TQ⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)We know that, a tangent to a circle is perpendicular to the radius through the point of contact.OP ⊥ PT,∴ ∠OPT = 90º⇒ ∠OPQ + ∠TPQ = 90º⇒ ∠OPQ = 90º – ∠TPQ⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)From (1) and (2), we get∠PTQ = 2∠OPQ |
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| 13. |
The adjoining figure is made up of a triangle inscribed in a semi-circle inwhich AB= 20 cm, BC 15 cm and angle B is a right angle. Find the area ofshaded region. |
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Answer» Since ∆ABC is right angled at B , AC² = AB² + BC² = 15² + 20² = 25²AC = 25cm = 2r r = 12.5cmArea of semicircle : πr²/2 = 245.3 cm²Area of ∆ABC : 1/2 * 20 * 15 = 150 cm²Area of shaded region : 245.3 - 150 = 95.3 cm² In book, answer is 341.07 |
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| 14. |
बदन करी + कैरी et g aedबे हर) न अर + आर + अं न किक gle e |
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| 15. |
हर नि हे Leik-& A N G-Ax] FXv — Ket g |
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| 16. |
Two tangents TP and TQ are drawn to a circle with centre o from an external point T,then prove that PTQ-2 QPQ |
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| 17. |
Rahul scored 180 marks in the first test and 150 marks in the second test. The maximum marksin each test is 200. What is the decrease in his performance in percentage points?(A) 20%(B) 1596(D) 23% |
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| 18. |
1+iFit, the least positive integer m such that I ěŹti |
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| 19. |
&et b ey Ag Ag A i e i i e |
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Answer» (3a + 5b)/(3a - 5b) = 5 3a + 5b = 15a - 25b 30b = 10a a = 3b If you find this answer helpful then like it. |
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| 20. |
लि T रेडा ef et 2 ९९ १०1 g, U |
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| 21. |
) ioneThe point on the curve lz-5il -3 having the least positiveargument, is12 16(a) 55(d) (34) |
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| 22. |
et | Ooamndg |
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| 23. |
609*179 |
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Answer» 109011 is the right answer 109011 is the right answer. 109011 is the answeryou should multiply both numbers |
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| 24. |
1-609 1-tanbB e . aB |
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Answer» (Tan A/1-Cot A) +( Cot A / 1 - Tan A )= Sec A.Cosec A + 1 LHS = (sin A/cos A) / [1- (cos A/sin A)] + (cos A/sin A) / [1- (sin A/cos A)] = (sin A/cos A) / [(sin A - cos A)/sin A] + (cos A/sin A) / [(cos A - sin A) /cos A] = (cos^2 A) / [sin A(cos A - sin A)] - (sin^2 A) / [cos A(cos A - sin A)] = (cos^3 A - sin^3 A) / [(cos A * sin A) (cos A - sin A)] = (1/cos A * sin A) (sin^2 A + cos^2 A + sin A cos A) = (1/cos A * sin A) (1+ sin A cos A) = [1 / (cos A sin A) + (cos A sin A) / (cos A sin A)] = sec A * csc A + 1 = RHS it's really hard for me to understand |
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| 25. |
609 + 1 256 + 298 + 1 56 on suitable arrangement gives |
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Answer» First arrange number in proper arrangement 1256 + 609 + 298 + 156= 1865 + 454= 2319 |
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| 26. |
Prove that ; (2^a/2^b)^a+b * (2^b/2^c)^b-c * (2^c/2^a)^c+a =1 |
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Answer» 2^(a2-b2+b2-c2+c2-a2)=2^0 =1 |
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| 27. |
9. Pwo chords AB and CD of a circle with centre O, intersect at E. If20EA = <OED, Prove that AB = CD |
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| 28. |
7.) Chords AB and CD of a circle intersect eachother at point P such that AP CPShow that : AB CD |
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| 29. |
124. In a circle with centre O, chords AB and CDare parallel chords on opposite side of O. IfAB 20 cm, CD 48 cm and the distancebetween the chords is 34 cm, then the diameter(in cm) of the circle is:(1) 26(2) 39(3) 42(4) 52 |
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Answer» I hope this will help you |
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| 30. |
Simplify these expressions and find their valux = 4, a = -3 and b = -2:(i) 4x - 5- 7x + 8(ii) 3 (x + 2) + 5x - 7(iii) a +7+4(a - 5)(iv) 3(b + 2) + 56 - 7nce |
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Answer» 1)4*4-5-7*4+8=16-5-28+8-92)3(4+2)+5*4-7=18+20-7=31 here (1)4×4-5-7×4+8=16-5-28+8=9-36=-27 (ii)3(4+2)+5×4-7=12+6+20-7=18+13=31 hope this will help you like my answer and Mark it as best answer no 2 correct answer ha 1 )-92) 313)-284)-17 -9557) yiyituitt54679+568 1) 4(4)-5-7(4)+816-5-28+8-9 2) 3(4+2)+5(4)-73(6)+20-718+20-731 3) -3+7+4(-3-5)4+4(-8)-28 4) 3(-2+2)+5(-2)-73(0)-10-7-17 (1)4*4-5-7*4+8=16-5-28+8=24-32=(-8) 1)=12x-3,,(2)=-8×+9,,(3)=26,,(4)=2b+9 ans of all ? |
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| 31. |
In the given figure, O is the centre of a circle inwhich chords AB and CD intersect at P such thatPO bisects ZBPD. Prove that AB CD |
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| 32. |
,lu the depth of the canal.39. Find the area of a trapezium whose parallel sides are 11 m and 25 mlong, and the nonparallel sides are 15 m and 13 m long.40. The difference hetweetho lonahe |
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| 33. |
Je pia are zeros of polynomialEGO = 2*2-76 +5, find the valuof p2+9?. |
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Answer» value of p is 1and value of q is 5/2answer is 30/4 |
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| 34. |
2 For the least positive valu lk.ter which ar 2 tk 710=o has e alroots |
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| 35. |
\operatorname { sin } 24 ^ { \circ } + \operatorname { sin } 32 ^ { \circ } + \operatorname { sin } 204 ^ { \circ } + \operatorname { sin } 212 ^ { \circ } = |
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Answer» sin 24 + sin 32 + sin 204 + sin 212= sin 24 + sin 32 + sin (180+24) + sin (180 +32)= sin 24 + sin 32 - sin 24 - 2in 32=0 |
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| 36. |
\frac{\sin 32^{\circ}}{\cos 58^{\circ}} |
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Answer» sin 32°---------cos 58° sin 32° = sin(90-58) = cos 58° cos 58°-------- = 1cos 58° |
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| 37. |
(9()कीsin 65°cos 25° +cos 32°sin 58°— sin 28° sec 62° + cosec? 30° |
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| 38. |
Fill in the blanksa) sin 32° =_____ cosb) tan 40°*tan____=1 |
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Answer» a)sin32° * cos58°= 1b) cot 60* tan40° = 1 |
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| 39. |
tnicircleAărn |
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Answer» Area of the square=14²=296 cm² Two semicircles are inscribed in the square with its opposite sides as their respective diameters. Sum of area of the semicircles=2×(πr²/2)=πr²=3.14×7²=153.86 cm²≈154 cm² ∴Area of the shaded portion=Area of the square-Area of both the semicircles=( 296-154 ) cm²=142 cm² |
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| 40. |
32.) 4 (sin 30 + cost 609)-3 (cos 45-sin? 90. |
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| 41. |
Given A = {a, b, c), B-(2,3 } . Find the number ofrelations fromA to B. |
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Answer» the number of relation =n(A)×n(B)=3×2=6 |
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| 42. |
\frac { ( 7 ) ^ { 2 n + 3 } - ( 49 ) ^ { n + 2 } } { \{ ( 34 ) ^ { n + 1 } \} ^ { \frac { 2 } { 3 } } } |
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| 43. |
EXERCISE 1.2Represent these numbers on the number line.7/4. |
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| 44. |
(ii) (2x2 + 3x-5) by (3x -2) |
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| 45. |
5 \div 5 \div 5 \div 5 |
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Answer» solve is of number 5÷5÷5÷5 ans is 0•04 0.04 hoga question ka answer |
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| 46. |
80 \div 4 - 20 \times 2 \div 3 \text { of } 4 \div 4 + 30 |
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| 47. |
175 men can dig a canal 3150m long in 36 days. how many men are. required to dig a canal 3900m long in 24 days |
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| 48. |
(-153) \div[45 \div(-5)] |
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| 49. |
5 ^ { 7 } \times 5 ^ { 4 } \div 5 ^ { 8 } |
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Answer» thanks ( 7+4). 8 5. /5 11-8. 3 5. =5 |
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| 50. |
m \text { for which } 5 ^ { m } \div 5 ^ { - 3 } = 5 ^ { 5 } |
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Answer» using law of indiceshencea^m/a^n=a^m-nhence5^m-(-3)=5^3m+3=3m=0 5^m/5 ^3 = 5^5 === 5^m = 5^5+3 === 5^m = 5^8 ===. m =8 |
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