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. |
1. Solve the following linear equatic(a) x+5=9 |
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Answer» x + 5=9 x=9/5Divide it x=1 x + 5=9 x=9/5Divide it x=1 x+5=9x=9-5x=4this is correct answer please like my answer x+5=9x=9-5×=4 is the correct answer x=4is the answer of the question X= 4 is the correct answer of the given question x+5=9x=9-4x=4the answer of this question is 4 x+5 =9=×=9-5=4 is your answer ×+5=9=×=9-5=4 is your answer x+5=9×=9-5×=4 . x+5=9x=9-5x=4 is correct ans x + 8=9x=9/5Divide it x=1 is the correct answer |
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| 2. |
ER.NEMob.982701CLASS- X(LINEAR EQUATICSolve the following system of equationcelemination method:YDS11.x=80, y = 30] |
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Answer» x/3 + y/4 = 11.......(1)5x/6 - y/3 = - 7......(2) Multiply eq(1) by 1/3 and eq(2) by 1/4 x/9 + y/12 = 11/3........(3)5x/24 - y/12 = - 7/4.....(4) Add eq(3) and eq(4)x/9 + 5x/24 = 11/3 - 7/4(8x + 15x)/72 = (44 - 21)/1223x/72 = 23/12x = 72/12 = 6 Put value of x in eq(1)6/3 + y/4 = 11y/4 = 11 - 2y = 9*4 = 36 Therefore,x = 6, y = 36 |
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| 3. |
Find the amount al ule Lil UL. JuFind the compound interest on 2500 for a year at 8% per annum com8% per annum compounded quarterly |
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Answer» A=p[ (1+r/n)^nt +1]= 2500 [( 1+8/4)^4×1+1]= [3^4 +1] ×2500=[ 81+1]×2500=82 ×2500Amoubt =205000 ( Given principle amt =2500Rsquaterly so n =41year so t=1r=8%) 200 repees is the answer interest calculated quarterly, rate becomes 1/4 times,rate = 8/4 = 2% time become 4 times, time = 1 x 4 = 4 Amount = p(1+R/100)^n = 2500(1+2/100)^4 = 2500(51/50)^4 = 2500(1.02)^4 = 2706.08 rs.. compound Interest = amount - principal = 2706.08 - 2500 = 206.08 rs.. (ANS) The correct answer of this question is :- 200 rupees. A=P(1+R/100)^n since c.i is calculated qurterly then n=4and, rate=8/4=2%so, A=2500(1+2/100)^4A=2500(1+2/100)^4A=2500×(102/100)^4A=2706.08 rs.hence C.I.=A-P =2706.08-2500 =206.08 rs. ans. Given,Principal=₹2500Rate=8%p.a.Time=1 yearThen,=Amount=P(1+R/100)power 2=2500(1/1 + 8/100) power 2=2500(108/100)power 2. {by taking LCM)=2500×108×108/100×100=₹2916A=₹2916CI=A-P=2916-2500=₹416 Rs 206.08 is the compound interest 2500*1*8/1002500*8/10020000/100200 The answer is 8% That should be answered by the formulas I=ptr/100=2500×1×8/100=200compound interest is 2500+200=2700 R = 8%May this help you 200 is the correct answer. please like as best C. I=2706.08 is a correct answer 500×8÷100 equals to 200 answers are given why should i give hiiihow are u dear 😂 😂 😍 ❤ ❤ 💍 The correct answer is r=8%. |
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| 4. |
vis/83fachorise --isfachooise ul-2042 +0ai |
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Answer» Please specify question that needs to be answered |
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| 5. |
Simplify:(6--34257+115 |
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Answer» 19/3-14/5+22/15= 5 is the right answer 5 is the answer of this question 5 is correct answer ok 5 correct answer for this question so, 1st u need to multiple and then add the fraction - 6=1/3 i.e, 19/3 do it and u will get ans 5 😊 5 is currently answer answer is 5check it 5 is the correct answer 5 is the correct answer =19/3-14/5+20/15=19×5-14×3+22×1/15=95-42+22/15=75/15=3. is the correct answer to your question 5 is the right answer =19/3-14/5+22/15=19×5-14×3+22×1/15=95-42+22/15=75/15=5 is the correct answer to your question 5 is the answer for this easy question 31/15 is the correct,right,perfect answer |
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| 6. |
4. Simplify6) (a-b) |
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Answer» wrong answer |
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| 7. |
Simply Fixation Questions |
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Answer» "Fixation" is a concept that was originated by Sigmund Freud to denote the persistence of anachronistic sexual traits. The term subsequently came to denote object relationships with attachments to people or things in general persisting from childhood into adult life |
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| 8. |
(ě¨)-25(ç˝°-5, given that x+1 x* 22x +3x-314. Solve for x2x-32x +3 |
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| 9. |
simply 65+(-13)+(-72)+(-8) |
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Answer» 65-13-72-8 = 65 - 93 = 28 simply 65/(-13)/(-72)/(-8) |
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| 10. |
एक s ek eL e |
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Answer» (i) 135 and 225Since 225 > 135, we apply the division lemma to 225 and 135 to obtain225 = 135 × 1 + 90Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain135 = 90 × 1 + 45We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain90 = 2 × 45 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 45,Therefore, the HCF of 135 and 225 is 45. (ii) 196 and 38220Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain38220 = 196 × 195 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 196,Therefore, HCF of 196 and 38220 is 196. (iii) 867 and 255Since 867 > 255, we apply the division lemma to 867 and 255 to obtain867 = 255 × 3 + 102Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain 255 = 102 × 2 + 51We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain102 = 51 × 2 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51. Like my answer if you find it useful! |
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| 11. |
e Lotk hRehlie Lo b12 Lk pib 80 2P212 ०९ 919 ४. 2७ [१ '€डै कयु डाक b L2k4 कप Uy 2k bl DAIAD T) stz 5६1 |)ildOH L lielkers DEylliak] TR ledib | MK kK ‘D1 ADH Lk O€I e 0T “pie |
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| 12. |
L 0e siem\( - 2 हानि [ |
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Answer» thnx |
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| 13. |
rnd theo8t8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, theangle of elevation of the top of the statue is 60° and from the same point the angle ofelevation of the top of the pedestal is 45. Find the height of the pedestal,1S |
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| 14. |
In fig., ABCD is a cyclic quadrilatreral. If ZBAC-50° andfind ABCD60 °50° |
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Answer» As ABCD is a cyclic quadrilateralthen in cyclic quadrilateralsum of opp.angles is equal to 180 degree angle bac +angle bcd =18050 + angle bcd =180angle bcd = 130 |
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| 15. |
Find the vea of the shaedion in givens fi whe14 cm.ABCD Sasquare of sie |
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Answer» area of square = (side)^2 = (14)^2 sqcm = 196 sqcmarea of 4 circle = 4 × πr^2 =4×(22/7)(7/2)^2 =154 sqcmshaded area = (196-154)sqcm = 42 sqcm |
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| 16. |
PRACTICE1. In Fig. AB is a 6 m highpole and CD is a ladderinclined at an angle of60째 to the horizontaland reches up to apoint D of pole. IfAD 2.54 m, find thelength of the ladder.(use 3 1.73) |
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| 17. |
6in Fig. 510, ACBD. then prove that ABCDFig. 5.10 |
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| 18. |
-2 5 13EXAMPLE 3 The points A (4,5, 10), B (2, 3, 4) and C (1, 2, -1) are three vertices of a parallelogramABCD. Find vector and cartesian equations for the sides AB and BC and find the coordinates of D[CBSE 2010] |
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Answer» Thanks a lot |
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| 19. |
Practice QuestioSet 1Like & Unlike Terms1. Find the factors of the terms.B. 7abretbor the given pair of terA. 8xy |
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Answer» A. 8 xy Factors are 2³ × x × y 2 , 4, 8, x, y, 2x, ... |
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| 20. |
6. Simplify6. Simply (4) |
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| 21. |
What lernof tarpaulin 3 m wide will be required to make conical tent of height 8 mand base radius 6 m? Assume that the extra length of material that will be required forstitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14). |
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| 22. |
of tarpaulin 3 m wide will be required to make conical tent of height 8Assume that the extra length of material that will be required fs. What lengthand base radius 6 m?stitching margins and wastage in cutting is approximately 20 cm |
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| 23. |
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π-3. 14). |
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| 24. |
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 mand base radius 6 m? Assume that the extra length of material that will be required forstitching margins and wastage in cutting is approximately 20 cm (Use 3.14).s |
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| 25. |
TnApQ R,ăźE and Fare the mid pents..at sig |
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Answer» If you like the solution, Please give it a 👍 |
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| 26. |
— | <te कि13 LI kb |
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Answer» 1/2 |
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| 27. |
The radius of the base of a cone is increasing at the rate of 3 cm/min and the altitude is decreasin24cmithe rate of 4 cm/min. The rate of change of lateral surface when the radius-7 and altitude of these |
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| 28. |
and reaches Bangalore at s20a men esues Chennel at l030 pum on Saturday and reachesRnd sie dhuration of the joumey |
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| 29. |
7DUD) Blast - UutC) Blow31.Simplify ((33) - 2 x (2²-3668)(2+)-2x 3-4x4-2B) 9A) 723c) 71 |
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Answer» Answer = c) (7*9+1)/9 [ (3³)-² * (2²)-³ ] / [ (2^4)-³ * 3^-4 * 4-² ]>>>>> { a^-m = 1/a^m }= [ (2^4)³ * 3^4 * 4² ] / [ (3³)² * (2²)³ ]>>>>> { (a^m)^n = a^(m*n) }= [ 2^(4*3) * 3^4 * (2²)² ] / [ 3^(3*2) * 2^(2*3) ]= [ 2^12 * 3^4 * 2^4 ] / [ 3^6 * 2^6 ]>>>>> { a^m * a^n = a^(m+n) }= [ 2^(12+4) * 3^4 ] / [ 2^6 * 3^6 ]= [ 2^16 * 3^4 ] / [ 2^6 * 3^6 ]= ( 2^16 / 2^6 ) * ( 3^4 / 3^6 )>>>>> { a^m / a^n = a^(m-n) }= 2^(16-6) / 3^(4-6)= 2^10 / 3^(-2)>>>>> { a^(-m) = 1/ a^m }= 2^10 * 3²= 1024 * 9= 9216 simplify is answer 64/9 a) 65 b) 65 c) 64 it my answer answer is C) 7(1/9) (3^3)-2 × (2^2)^-3/ (2^4)-2 × 3^-4× ( 4)^-2= 3^-6 × 2^-6/ 2^-8 ×3 ^-4 ×2^-4= 3^-6+4 × 2^-6+8+4= 3^-2 × 2^6= 64/9option C is right C is the correct answer option b is correct answer option A is correct answer =c)(7*9+1)/9 answer is a righr answer =9216 is the correct answer answer c 7\19 simplify but beautifully questions option A is right The answer is 7.99999... i.e. C option A is the answer option c is right ans answer is 64/9 option is third ans -- c (7*9+1)/9। . answer = c is Wright answer simplify is answer is 64/9 option c is the right answer. Answer is 64/9 because of 2 power 6/3power 2 64/9 is the right answer to this question. C is the correct answer for haw long had the seagul been alone C is the correct option Bruh! the correct answer is c 65/9-a 65|/7-b 64/9-c (A =65),(B=65), (C=64) later by some problems C is the right answer answer=c) (7*9+1)/9 is the best tha correct answer is (a) option c Answer is-64/9.This is correct. B is the right answer C is correct option 64/9 A option is the correct answer hu na |
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| 30. |
कि- e LT केN oy )) LI |
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Answer» (cosecA - sinA)(secA - cosA)(tanA + cotA) =1 LHS = (cosecA - sinA)(secA - cosA)(tanA + cotA) =(1/sinA - sinA)(1/cosA - cosA)(tanA + 1/tanA) =(1 - sin2A / sinA)(1 - cos2A / cosA)(tan2A + 1 /tanA) =(cos2A / sinA)(sin2A / cosA)( sec2A / tanA) =cosA X sinA (1/cos2A / sinA /cosA) =cosA X sinA (1 / cosA sinA) = cosA sinA X 1 /cosA sinA = 1 |
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| 31. |
Deifine-like decimal, like firection, papelunlike decimal, with three example1:0 |
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Answer» no ans hjskdhdhhdhsbsjdjehdbdbsbebhebdbdbdbdbdbddbd |
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| 32. |
. EpE- AP EAS + टू दल o% ५८.s e |
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Answer» 2√2+5√3+√2-√3=3√2+4√3 |
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| 33. |
to find UUTPractice1. Complete the following table to dray(II) x - y = 4(1) x + y = 3x+y=3x 3 Oyol 5 3(x, y) (3, 0) 0 (0, 3).2. Solve the following simultaneous equo(1) x + y = 6 x - y = 4.13 |
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Answer» Bhai accept it as best please because I am your fan X=3, 2, 0; y=0, 5,3 (X,Y)=2,5 |
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| 34. |
In fig., ABCD and PQRC are rectangles and Q is the mid-point ofAC. Prove that () DP = PC, (ii) PR =AC/2 |
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Answer» part 1 Part 2 |
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| 35. |
| 19. ज्ञात करें : [z-eas |
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Answer» 1/3e^x^3+c is answer |
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| 36. |
andisapproAmasoeasestitching marginswastage in cutting10. AJoker's cap is in the form of a right circular cone of base radius 7 cm and height 27 cm.Find the area of the sheet required to make 10 such caps. |
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Answer» this is ..... height =27 cm not 24 Radius = 7cmheight = 27 cm l =√r² +h²l =√7² + 27²l =√49 + 729l = √778l = 27.89 Curved surface area of one conical cap =πrl= (22/7)×(7)×27.89= 613.638 cm² 6136.38 is the answer |
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| 37. |
\int\left[\frac{3}{y-2}+\frac{4}{(y-2)^{2}}\right] d y=3 \int \frac{d y}{y-2}+4 \int \frac{d y}{(y-2)^{2}} |
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| 38. |
\int _ { 0 } ^ { 4 } \int _ { 0 } ^ { 2 \sqrt { z } } \int _ { 0 } ^ { \sqrt { \left( 4 z - x ^ { 2 } \right) } d z d x d y } |
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| 39. |
Exercise 1.1l. Using appropriate properti3 52 5 62. Write the e |
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Answer» please like my answer if you find it useful thanks |
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| 40. |
LI INII E INTEGRALS (BOARDLemPrDSETEVALUATE THE FOLLOWING INTEGRALSdx(N.C.E.R.T PROBLEN |
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| 41. |
on sig length ofole is ro assabendwhat is the prea ofAPP 2 |
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| 42. |
A vender bought 9 sweets for ? 1. At what rateshould he sell it to gain 80%? |
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Answer» He has bought 9 sweet for 1 rupee it means 1 sweet for 1/9 rupee so if he want to sent it for 80% profit then, selling price=buying price +80% profit Selling price= 1/9 +1/9*80/100 selling price=1/9+0.8/9 =(1+0.8)/9 =1.8/9 =0.2 rupee =20 paisaeach It means he has to sold all his sweet for1.8rupee |
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| 43. |
x. y are two numbers with mean proporticn 4 and third properti nalvalue of x. y.32.Find |
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| 44. |
aluate the following integrals: |
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Answer» Kise ne answer ni ditaa naa Chl koi naa |
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| 45. |
The propertiPropertyta pety)ThesameSimplify03- |
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| 46. |
A vender bought toffees at 6 for a rupee.How many for a rupee must he sell togain 20% is |
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| 47. |
ind the foHowing integrals |
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| 48. |
Find the following integrals:ĐĽ |
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| 49. |
how to post anwer in this app |
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Answer» Touch on type your answer.. And then post your answer touch on bottom buttonand you can type answerthen touch on blue button ANSWER touch on bottom buttonand you can type answerthen touch on blue button ANSWER |
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| 50. |
I. Evaluate the following integrals. |
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Answer» ∫e²ˣdx=e²ˣ/2+cthis is the solution |
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