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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. |
In the given figure ,the side of square is 28 an and radius of each circle is half of the length of the side of the square where O and O' are centers of the circle .find the area of shaded region |
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| 2. |
.ygstarhotel in Delhi charges 10% sales tax on the price of the food taken. Mr Saxenaand his family had food for685. Find the total money he had to pay. |
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Answer» Tax charged on the price of the food taken=10%The family had food for RS.685So tax charged on this Rainy.685=10% of Rs.685=Rs. 68.5Therefore, final amount to be paid=Rs(685+68.5)=Rs753.5 |
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| 3. |
43.Ifin any binomial expansion a, b, c and d be the 6th, 7th, 8th and 9thtermsb2-ac 4arespectively, prove thatHOTS3cc--bd |
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| 4. |
N 3 i(Vi) 92%y>â 16 |
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Answer» 9x²y² - 16= (3xy)² - 4²= (3xy+4)(3xy -4) Please hit the like button if this helped you |
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| 5. |
A 3-star hotel in Delhi charges 10% sales tax on the price of the food taken. Mr Saxenaand his family had food for R 685. Find the total money he had to pay. |
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| 6. |
84. If HCFof 480 and 685 is expressed in the form 480X-4, mid the85* If d is HCF of 40 and 65, find the value of integers x and y which satisfy d 4065yCBSE 2014 |
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Answer» Given integers,40 and 65Applying Euclid's division lemma we get,65=40×1+25......................................1Applying Euclid's division lemma to 40 and 25 we get,40=25×1+15......................................2Applying Euclid's division lemma to 25 and 15 we get,25=15×1+10........................................3Applying Euclid's division lemma to 15 and 10 we get,15=10×1+5..........................................4Applying Euclid's division lemma to 10 and 5 we get,10=5×2+0...........................................5eq1,eq2,eq3,eq4 can also be written as,65-40×1=25..................................…..6 40-25×1=15........................................7 25-15×1=10.........................................8 15-10×1=5............................................9Eq9 we have,5=15-10×1By putting eq8 in above eq we get,5=15-(25-15×1)1 5=15-25×1+15×1 5=15×2-25×1By putting eq7 in above eq we get,5=(40-25×1)2-25×1 5=40×2-25×2-25×1 5=40×2-25×3By putting eq6 in above eq we get,5=40×2-(65-40×1)3 5=40×2-65×3+40×3 5=40×5-65×3 5=40(5)+65(-3)Given equation,d=40x+65yHere d is HCF.Therefore , 5=40x+65yOn comparing we get,x=5 and y=-3Hence value of x=5 and value of y=-3 |
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| 7. |
of the flat?taken. Mr Sazena2, A 3-star hotel in Delhi charges 10% sales tax on the price of the foodand his family had food for ? 685. Find the total money he had to pay |
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Answer» let x is the original amount of price of food then 685= x+ 10% of x = x + 10x÷100 =x+ 0.1x685 = 1.1x622=x |
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| 8. |
君1In figure, ΔFEC ΔGDB and L122. prove that ΔADE ~AABC.OTO |
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Answer» Given :∆FEC≅∆GBD , SO From CPCT WE GET, BD=CE----------------- ( 1 ) ALSO GIVEN :∠1 =∠2 , SO FROM BASE ANGLE THEOREM IN∆ADE WE GET AD =AE------------------------ ( 2 ) From equation 1 and 2 we get ADBD=AECE, So from converse of B.P.T. we get DE | | BC THEN , ∠1 =∠3 ( CORRESPONDING ANGLES AS DE | | BC AND AB IS TRANSVERSAL LINE ) AND ∠2 =∠4 ( CORRESPONDING ANGLES AS DE | | BC AND AC IS TRANSVERSAL LINE ) FROM ABOVE TWO EQUATIONS WE CAN SAY THAT : ∆ADE~∆ABC( ByAArule )( Hence proved) |
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| 9. |
excaudh36.timcoto. |
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Answer» the number is 12 |
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| 10. |
36. The general value of θ satisfying the equation 2 sin2e-3 sin θ-2 Ξ 0 is46π + (-1)"(?) |
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| 11. |
what is main source of energy |
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Answer» the main source of energy is sun |
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| 12. |
if D is the centers of and angle ADC is 150 find angle ABC |
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| 13. |
20. In a Δ ABC, AD bisects < A and <C > < B. Prove that < ADB > <ADC. |
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| 14. |
Expand (1 ++ x")" using binomial expansion. |
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| 15. |
Q. 6.A survey of 500 families was conducted to knovwtheir opinion about a particular detergent powIf 375 families liked the detergent powder and theremaining families disliked it, find the probabilitythat a family chosen at radom:Board Term IL, 2012, Set-01(i) likes the detergent powder(ii) does not like it. |
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| 16. |
17ě 92-522.232 132 |
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Answer» thank u |
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| 17. |
What should be added to 39.685 kg to get 400 hectogram? |
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Answer» 1 kg = 10 hectograms So 39.685 kg = 396.85 hectograms So 396.85 + 3.15 hectograms = 400 hectograms |
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| 18. |
927616. In the adjoining figure, find which lines are parallel. Give reasons.17. In tha fi |
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Answer» Lines l and m are parallel because180-88=92°which is equal to upper 92°It means corresponding angles |
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| 19. |
21)Find the particular solution of the Differential Equation(x + 1) (y -1) dx + (x-1) (y+1) dy 0Satisfying the condition x 2 when y 2 |
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| 20. |
Find the coefficient of at in the product (1 +2a) (2 - a)' using binomial |
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| 21. |
orrs 38.. The greatest terms are T37 and T1.Expand the following using binomial theo() (3x -2y)2. Compute the value of (V2+1) + (V2-13. Find the 11th term in the expansion of (1 |
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| 22. |
if you subtract 7from 5 times a number you get 17 |
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Answer» lets take the unknown number as xaccording to the question ,=5x-7=17=5x=17+7=5x=24=x=24/5=x=4.8therefore x = 4.8 let no be x, 5x-7=17; 5 x=17+7=24; 5x=24, x=24/5=4.8 |
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| 23. |
Diagonal of a quadrilateral ABCD bisects each other If <A = 35 determine <B |
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| 24. |
Diagonal AC of a quadrilateral ABCD, bisects angle A and angle C. Prove that triangle ABC is congruent to triangle ADC. |
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Answer» proves they are congruent |
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| 25. |
In a cyclic quadrilateral ABCD, the diagonalAC bisects the angle BCD. Prove that thediagonal BD is parallel to the tangent to thecircle at point A. |
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| 26. |
17. What should be added to 39.685 kg to get 400 hectogram? |
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Answer» 1 kg = 10 hectograms So 39.685 kg = 396.85 hectograms So 396.85 + 3.15 hectograms = 400 hectograms |
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| 27. |
17.What should be added to 39.685 kg to get 400 hectogram? |
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Answer» 1 kg = 10 hectograms So 39.685 kg = 396.85 hectograms So 396.85 + 3.15 hectograms = 400 hectograms |
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| 28. |
w10.What should be subtracted from 17 to get 11?1 |
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| 29. |
So, we get 3778688 = 92Example 17: Find the cube root of 592704 throughestimation |
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Answer» 84 is the right answer |
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| 30. |
2. LINEAR EQUATIONS IN ONE VARIABLE178. Solve the following problems by forming equations :( 1 ) We get i after subtracting from a fraction. Find the fraction. |
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Answer» 16/20 x/y-1/4=11/20x/y= 11/20+1/4x/y=16/20so 16/20 is the answer.... 16/20 is the right one |
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| 31. |
property of binomial theorem |
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| 32. |
1. Income and Expenditure Account is prepared on Cash Basis of Accounting. Is thestatement correct? |
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Answer» Income and expenditure account. The income and expenditure account is an account prepared bynon-trading concerns to ascertain surplus or deficit of income over expenditures for a particular period |
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| 33. |
te the following statements as an equation :(i) The sum of 4 times and 7 is 16 |
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Answer» equation will be4x +7 = 16 |
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| 34. |
determine the 4th term of (x/3+x/1)^5 |
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Answer» Explanation : 4th term = 5C3 (x/3)(x/3)(x/3)*(x/1)(x/1) = 10x*x*x*x*x/27 Answer : 10x*x*x*x*x/27 If you find this answer helpful then like it |
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| 35. |
that it lies on the bisecl12.100, AD丄CD and CBCD . If AQ = BP and DP = CQ, proveLDAQ = <CBP. |
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Answer» Given : AD⊥ CD and BC⊥ CDAQ = BP and DP = CQTo prove :∠ DAQ =∠ CBPProof :AD⊥ CD and BC⊥ CD∴∠ D =∠ C (each 90°)∵ DP = CQ (Given)Adding PQ to both sides. we getDP + PQ = PQ + CQ⇒ DQ + CPNow, in right angles ADQ and BPC∴ Hyp. AQ = Hyp. BPSide DQ = side CP∴Δ ADQ≡Δ BPC (Right angle hypotenuse side)∴∠ DAQ =∠CBP (Corresponding part of congruent triangles)Hence proved. |
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| 36. |
rationdued fractions as improperthe operatiete mixed fraat themall: Add 2 and 38 |
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Answer» 14/5 and 23/6 1.623441155622266245566667 1.623441155622266245566667 1.6234411556222.. is the right answer the correct add is |
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| 37. |
In the adjoining figureAB AD and CB CDProve that ΔABC ΔADC. |
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| 38. |
2. In fig AB and CDintersect each otherE such that DEBEand EA = ECShow that AD CB |
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Answer» In ∆ADE & ∆BCE,AE=CEDE=BEangle AED= angle BECBY SAS CRITERION,∆ADE congruent ∆BCEBY CPCT, AD=BCHENCE PROVED in ΔAED andΔBECDE=BE(given)EA=EC(given)ang(AED)=ang(BEC)[vert.opps angls]therefore by SAS congruenceΔAED≈ΔBECsoEA=EC(Cpct) |
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| 39. |
In a triangle ABC,AD is a median and 2 AD >AB+CB+CD |
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Answer» Thanks bro |
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| 40. |
ABCD is a quadrilateral such that diagonal AC bisects A and C prove AB=AD and CB=CD |
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| 41. |
9. In the given figure, diagonal AC of a quadrilateral ABCDbisects the angles A and C. Prove that AB AD and CB- CD |
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Answer» Given :- BAC = DACBCA = DCA TO PROVE :- AB = ADCB = CD PROOF :- IN TRIANGLE ABC AND ADCBAC = DACBCA = DCAAC = AC HENCE., TRIANGLE ABC CONGRUENT TRIANGLE ADC IMPLIES THAT., AB = ADCB = CD |
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| 42. |
In the figure below, diagonal AC of quadrilateral ABCD is bisector of ZC and A. Provethe igure below, diagonal AC ofthat AB AD and CB-CD |
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Answer» Given, AC is a bisector of ∠C and ∠ASo,∠CAB=∠DAC. --------(1)∠ACB=∠ACD. --------(2)Now, in triangles ABC and ADC∠CAB=∠DAC. (Given)∠ACB=∠ACD. (Given)AC=AC. (Common)So, by ASA triangle ABC is congruent to Triangle ADCSo by C.P.C.TAB=ADCB=CDHence proved. |
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| 43. |
)]02 पाक एड न V 995 _ ४ 9500 _ ऋ 39503 VY 395(छू 803 - के पाई) (के एष्ठ] + के 303 + [ ४1 |
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Answer» L.H.S = (1+cotA+tanA)(sinA-cosA)= (sinA+sinAcotA +tanAsinA-cosA -cosAcotA -tanAcosA)=(sinA+sinAcosA/sinA+tanAsinA-cosA-cosAcotA -cosAsinA/cosA)= (sinA+cosA+tanAsinA -cosA- cosAcotA -sinA)= tanAsinA -cosAcotA = sinAtanA-cotAcosA |
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| 44. |
6. किसी वस्तु का मूल्य ३ 350 है। यदिसेल्समैन इसे 14% के लाभ पर बेचे, तोवस्तु का विक्रय मूल्य है।| RRC ‘ग्रुप डी' दिल्ली 3 12.2013(a) १ 399(b) 395(C) 459(d) १ 364। |
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Answer» Option a is rightas 114 percent =we have to findwe are given value of 100 percentso 1 percent will be 3.5 percentagenow 114 percent will be 3.5 x 114399 114 प्रतिशत के रूप में = हमें खोजना होगाहमें 100 प्रतिशत मूल्य दिया जाता हैइसलिए 1 प्रतिशत 3.5 प्रतिशत होगाअब 114 प्रतिशत 3.5 x 114 होगा399 ans. (d) 364 crract ans. |
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| 45. |
KolkataWhat must be added to -35 to get 17?38 and -87. |
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| 46. |
6. किसी वस्तु का मूल्य ३ 350 है। यदिसेल्समैन इसे 14% के लाभ पर बेचे, तोवस्तु का विक्रय मूल्य है।RRC ‘ग्रुप डी' दिल्ली 3 12.2013(a) * 399(b) * 395(C) १ 459(d) १ 364 |
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Answer» 114 percentage = 3.5*114= 399Opti on A |
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| 47. |
35y²+13y-12 |
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Answer» 35y²+13y-12 the value of discriminant is ∆ = (13)²+4*35*12 = 1849 so, the value of y is = (-13±√1849)/70= (-13±43)/70= (-13+43)/70 , (-13-43)/70= 30/70 , -56/70= 3/7 , -4/5 |
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| 48. |
17. If the 2nd term of an AP is 8 and the 5thterm is 17, find its 19th term. [CBSE 2016]18The 4th te |
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Answer» 2nd term Second term (a2) = 8fifth term (a5) = 17 we know that general formula of a term in ap is a+(n-1)dtherefore, a2=a+d 8=a+da5=a+4d 17=a+4d solving the two equations simultaneously....we get....d=3therefore, a=5 thus the 19th term (a19) = a+18d = 5 + 18x3 = 5 + 54 = 59 |
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| 49. |
उ+ नB4y =13y, हे 0.10 |
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| 50. |
(vii) 13x+11y70; 11x+13y74t 4 and B are in |
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Answer» 13(13x+11y=7011(11x+13y=74 169x+143y=910121x+143y=81448x=96x=2 13y=74-22=52y=4 |
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