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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. |
Evaluate:x+xlogx |
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Answer» this is wrong take x common from X +Xlogxthen put 1+logx = T then integrate it |
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| 2. |
l 30Which сor-5-64rational numbers is greater in thewhich of the two rationalliy(iv):36(ii) 2or O9n pair?2 33 4(v) ororIv) 777 |
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Answer» thanku😇😇 |
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| 3. |
integration of xlogx |
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| 4. |
1४) 2 [न (५४) 08.)कस 22 2. “ R |
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Answer» (- 8 )- (-5)= -3 minus minus is plus and plus minus is minus and when we sbtract -8+5 it becomes -3 as 8 is greater and has - minus sign |
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| 5. |
In the given figure, OACB is a quadrand of a circle withcentre O and radius 3.5 cm. If OD 2 cm, find the areaof the shaded region. |
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| 6. |
5. The ratio of income of a person to his savings is 10:1. If his savings of one year are6000, what is his income per month?1 |
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Answer» Let ratio of income:saving be 10x:xIf x=6000Income=10x=10*6000=60000 rs |
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| 7. |
the ratio of income of a person to his savings is 12:1.if his savings per year is 15000.find his monthly income? |
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Answer» I don't know this answer sorry his monthly income is 15000 because his is savings 15000 per year so multiply with 12 because his yearly savings in 1/12. 180000 divided by 12 to know monthly salary 180000/12=15000. his monthly income is 15000because his savings 15000 per year so multiply with 12 because his yearly savings in1/12. 18000 divided by 12 to know monthly salary 180000 |
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| 8. |
1) Damu works in the field of Raju. They have agreedto share the income obtained in the ratio of 4 : 3.Find the share of income received by each of them,if the income for one year is 21,000 |
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| 9. |
12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD-2 cm,find the area of the(6) quadrant OACB,(ii) shaded region. |
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| 10. |
6. In the adjoining figure, OACB is Aa quadrant of a circle with centreO and radius 3.5 cm. If OD 2 cmfind the area of the following:(i) quadrant OACE(ii) shaded region. |
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| 11. |
A streamer goes downstream and covers the distance between two ports in 5 hourswhile it covers the same distance upstream in 6 hours. If the speed of the stream is1 km/h, find the speed of the streamer in still water and the distance between two ports. |
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| 12. |
. A streamer goes downstream and covers the distance between two ports in 5 hourswhile it covers the same distance upstream in 6 hours. If the speed of the stream is1 km/h, find the speed of the streamer in still water and the distance between two ports |
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Answer» not clear please sent again |
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| 13. |
14. A boy was sent out for one rupee's worth of eggs. He broke 5 onhis way home, and his master had therefore to pay at the rate of 5paise more than the market price for 5. How many did the masterget for a rupee? |
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Answer» Total number of eggs initially bought= 10.Market price of each egg=Rs 1/10= 10 paise. Price that the master had to pay=( 10+5) paise=15 paise.Number of eggs that the master did get for a rupee=100 paise/15 paise=6 eggs |
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| 14. |
S. A streamer goes downstream and covers the distance between two ports in 5 hourswhile it covers the same distance upstream in 6 hours. If the speed of the stream is1 km/h, find the speed of the streamer in still water and the distance between two ports |
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| 15. |
15. The income of a man increase by 10%If his present annual income is 8,80(a) What will be his income 2 years(b) What was his income 1 year earli |
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Answer» 10% increases every year current us 880000 so after two years it will be 880000*1.1*1.1 = 1064800 income 1 year earlier 880000/1.1 = 800000 |
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| 16. |
between two consecutive years my income are in the ratio of 2:3 and my expenses in the ratio 5:9 if my income in the second year is 45000 and my expenses in the first year is 25000 then my total savings for the two years is |
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Answer» incomes are in 2:3 for second year income is 45000 so for first year income is 30000 expenses are 5:9 expense of first year is 25000 so expense of second year = 45000 so saving of first year = 30000 - 25000 = 5000 saving of second year = 45000 - 45000 = 0 so total savings of 2 years = 5000 + 0 = 5000 |
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| 17. |
Number of different words that can be formedby arranging the letters of the word'LEEKEEPER' such that no two consonantsare together is 6C2.k! then value of k is |
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Answer» it will be factorial 5 as all E can swipe to each other |
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| 18. |
In the given figure, OACB is a quadrand of a circle withcentre O and radius 3.5 cm. If OD 2 cm, find the areaof the shaded region.Ol |
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| 19. |
9.If shradha covers a distance of8 3 Kms, in one hour. Then find the distance she2 Hourscovers ine numb |
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Answer» in 1 hr. it travels 25/3 kms.in 12/5 hrs. it will travel (25/3)×(12/5)= 20 kms. wring answer answer is 30/15 |
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| 20. |
9. A motor car starts with the speed of 70km/hr with itsspeed increasing every two hours by 10 kmph. In howmany hours will it cover 345 kms? Ans 9/2 hrUCEED IJEE.J |
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Answer» The acceleration of motor car is. = 10/2 = 5km/hr² now using , newtons 2nd equation S =ut + 1/2*at²=> 345 = 70*t +1/2*5*t²=> 5t²+140t-690 = 0=> t²+28t-138 = 0=> t = [-28±√(28²+4*138)]/2=> t = (-28+26.8)/2 = 8.8/2 ≈ 9/2 hrs can u plz explain me as a aptitude student not like a science student plz you have to use this.. equation.. because speed will always.. vary here.. we can't use any direct aptitude , and basic science is also aptitude..👍 ok thanx |
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| 21. |
r 08. A car travels a distance of 550 km in 5 hours. What distance will the car coverHow many hours are needed to travel a distance of2750 kms? |
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Answer» For 2750 killometer = 2750/11= 25 hours |
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| 22. |
what is mid point theorm |
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Answer» The line segment connecting themidpointsof two sides of a triangle is parallel to the third side and is congruent to one half of the third side. |
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| 23. |
+_-3 =0A bird watcher on a cliff spotted a kingfisbĂŠr 7 m abeve him Earlier he had seen it Il m below. What wasThe kingfisher's change in altitude? |
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Answer» The bird watcher is able to see the kingfisher is 7 M above him Before he was seen at 11 M below of him The change in altitude of kingfisher is = 11 + 7 = 18 = 18 M the bird watcher is able to see the Kingfisher is 7m above him..Before he was seen at 11M below of himthe change in altitude of Kingfisher is =11+7=18=18Mso the answer is 18M |
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| 24. |
andlinejomingtheisthepoints(-1)and(4,-2).11. The slope of a line is double of the slope of another line. If tangent of the angle5,ordinasis thenerefore, equatiobetween them is,find the slopes of the lines.3 |
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| 25. |
A bird watcher on a cliff spotted a kingfisher 7 m above him. Earlier he had seen it 11 mbelow. What was the kingfisher's change in altitude? |
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Answer» Change in kingfisher's altitude= 7 - (-11)= 7 + 11= 18 m |
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| 26. |
अभी आपने देखा है कि08 को जिन्या मानकर एकया है कि 2013 है। एक परकार की सहायता से ) का -नकर एक चाप (arc) रखींचिए जो संख्या रेखा को बिन्दु Pएर काटता है। वp सख्या रेखा पर 2 के संगत होता है।वास्तविक संख्या रेखा पर 3 का स्थान निर्धारण कीजिए।न; आइए हम आकृति 1.7 को पुन: लें।उदाहरण ३ । वास्त|||3|||+-3-2-11 0 APQ2आकृति 1.8कक लंबाई वाले लंब: BD की रचना कीजिए (जैसा कि आकृति 1.8 में दिखाया30B पर एकक लंबाई वाले लंब। तब पाइथागोरस प्रमेय लागू करने पर, हमें OD= (2) +1 = 3 प्राप्त होता है।चरकार की सहायता से 0 को केन्द्र और OD को त्रिज्या मानकर एक चाप खींचिए जोसंख्या रेखा को बिन्दु Q पर काटता है। तब Q, 3 के संगत है।का स्थान निर्धारण करइसी प्रकार ॥ -1 का स्थान निर्धारण हो जाने के बाद आपसकते हैं, जहाँ एक धनात्मक पूर्णाक है। |
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Answer» what is the question please mention your question kindly Where is the Question?? |
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| 27. |
15. The income of a man increase by 10% every year.If his present annual income is 8,80,000(a) What will be his income 2 years later?(b) What was his income 1 year earlier ? |
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Answer» 10% increases every year current us 880000 so after two years it will be 880000*1.1*1.1 = 1064800 income 1 year earlier 880000/1.1 = 800000 |
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| 28. |
10. The diagonar ol od11. Find the height of a Cylinder whose radius is 7cm and the total surface area is 968 cm2mY 30 cm How many small cubes with side 6 cm |
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| 29. |
the form 3m or 3m+ 1 for some integer mHint: Letx be any positive integer theneach of these and show that they can beUse Euclid's division lemma to show th.The Fundamental Theorem of Arour earlier classes, you have seen thduct of its nrime factors. For instance |
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Answer» Thanks |
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| 30. |
Long Answer Questions49. Prove that if two lines intersect each other, then the vertically opposite angles are equal.INCERT Eors of rhe inter |
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Answer» Given two lines AB and CD intersect each other at the point O.To prove:∠1 =∠3 and∠2 =∠4 Proof:From the figure, ∠1 + ∠2 = 180° [Linear pair] → (1)∠2 + ∠3 = 180° [Linear pair] → (2)From (1) and (2), we get∠1 + ∠2 = ∠2 + ∠3∴ ∠1 = ∠3Similarly, we can prove ∠2 = ∠4 also. |
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| 31. |
Can u CRACK THE LOGIC ...???Question of lIT-B2222-T3 333- E4444- NThen5555-?Challenge all master brains! Explainwith logic9:52 PM |
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| 32. |
2 Diff.w.r.t.x, e^tog(logx) |
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| 33. |
A straight highway leads to the foot of a tower. A man standing at the top of theobserves a car at an angle of depression of 30°, which is approaching the foot of thetower with a uniform speed. Six seconds later, the angle of depression of thtower.to be 60°. Find the time taken by the car to reach the foot of the tower from this point. |
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Answer» Fuck uu madaechodddbhosadewala |
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| 34. |
* Qu. Find the Hict of 26,51, 91 ming fundamental theorm of arithmetica |
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| 35. |
In how many distinct permutations of the letter in MISSISSIPPI do the four I's not come together ? |
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| 36. |
Find the areofaq0tut U3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minghand in S minutes. |
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| 37. |
Find the least number which must be added tomber which must be added to 6203 to obtain a perfect square,Also find the square root of the number so obtained.Find the square TOG |
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| 38. |
simplify :(-225)*(-199)-(-225) |
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Answer» -225 X -199 - (-225)=> -225 X -199+225=> -225 X 26= -44,550 |
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| 39. |
10. If the distance of the point (r, y) from the point (a, 0) be a + x. Prove that y -A |
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| 40. |
Q8) integration w.r.t.xĹż cos2xdx |
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Answer» cos(2x) = cos^2(x) - sin^2(x). We also know the trig identity sin^2(x) + cos^2(x) = 1, so combining these we get the equation cos(2x) = 2cos^2(x) -1. Now we can rearrange this to give: cos^2(x) = (1+cos(2x))/2. This eventually gives us an answer of x/2 + sin(2x)/4 +c |
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| 41. |
Frequency927. A straight highway leads to the foot of a tower. A man standing at the top of the towerobserves a car at an angle of depression of 30°. which is approaching the foot of thetower with uniform speed. Ten seconds later, the angle of depression of the car isfound to be 60°. Find the time taken by the car to reach the foot of the tower from thispointORdh of eleation of a tower is 60°, and |
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| 42. |
igutive Marks : -1 In all other cases:0Ifnone of the bubbles is darkened.he cxample,it(A) (C) and (D) are all the correct options for a quention, darkening all thesehree will result in +4 marks; darkening only (A) and (D) will reault in +2 marks; and darkeningand (B) will result in-1 marks, as a wrong option is nlso darkenedThe sum of two positive integers-a' and 'b, is 528 and their H.C.F. is 33, then which of theAlowing cannot be the value of (a-b)(4) $96(8) 264(C) 6(D) 33 |
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Answer» a - b cannot be 6 as it is not a multiple of 33. All others are divisible by 33. |
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| 43. |
Iftwo positive integers A and B canexpressed as A=ab^2, B=a^3b whose a bare distinct prime numbers then find the LCMof A and B? |
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Answer» LCM is given by the product of prime numbers, each raised to its highest exponent value in prime factorisation. highest exponent of a is 3 and that of b is 2. Therefore, LCM = a³b² |
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| 44. |
distanll21. A police jeep is chasing a culprit going on a motorbil.at a speed of 72 kmkm/h, crossing thturning ten seconds later than the bike. Assuming ththey travel at constant speeds, how far from the turninorbike crosses a turningThe jeep follows it at a speed of 90will the jeep catch up with the bike?lumh overtakes another ca |
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| 45. |
R 1135 str 225 () 196 sht 3820(ii) 867 3255) 135 225 |
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Answer» 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. |
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| 46. |
135 225 |
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| 47. |
225 IFind the H.C.F. of two numbers 135 and 225 using Euclid division algorithm. |
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| 48. |
135 t 225 |
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Answer» (i) 135 और 225 225> 135 से, हम प्राप्त करने के लिए 225 और 135 पर विभाजन लेम्मा लागू करते हैं 225 = 135 × 1 + 90 शेष 90 remainder 0 के बाद से, हम प्राप्त करने के लिए विभाजन लेम्मा को 135 और 90 पर लागू करते हैं 135 = 90 × 1 + 45 हम नए डिविज़र 90 और नए शेष 45 पर विचार करते हैं, और प्राप्त करने के लिए विभाजन लेम्मा लागू करते हैं 90 = 2 × 45 + 0 चूंकि शेष शून्य है, इसलिए प्रक्रिया बंद हो जाती है। चूंकि इस स्तर पर विभाजक 45 है, इसलिए, 135 और 225 का एचसीएफ 45 है |
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| 49. |
02. A particle is moving in a straight line and given by the law s -af t bt + 6. It is known that the particle comes torest affer 4 seconds at a distance of 16 metres from the starting position. Then the retardation in m/sec- isa)-1(c)-2424 |
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| 50. |
The sum of distinct prime factors of 30 is |
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Answer» prime factors of 30=2*3*5hence sum is 2+3+5=10 |
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