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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. |
A peacock is siting on a pillar 9 m high. A snake at a distance of 27 m from the pilming to a hole at the base of the pillar. Seeing the snake, the peacock pounces uIf their speeds are equal, then find at what distance from the hole is snake caught |
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Answer» Let AB be the pillar of height 9 meter. The peacock is sitting at point A on the pillar and B is the foot of the pillar. (.. AB = 9)Let C be the position of the snake which is at 27 meters from B. (.. BC = 27 and ABC = 90) As the speed of the snake and of the peacock is same they will travel the same distance in the same time...Now take a point D on BC that is equidistant from A and C. (Please note that snake is moving towards the pillar and hence B-D-C)Hence by condition 1 AD = DC = y (say)Take BD = x Now consider triangle ABD which is a right angled triangleUsing Pythagorus theorem (AB + BD = AD).. 9 + x = y9 = y - x = (y-x)(y+x)81/(y+x) = (y-x)y+x = BC = 27Hence 81/27 = (y-x) = 3y - x = 3y + x = 27Adding gives 2y = 30 or y = 15x = 27 - y = 12 Thus the snake is caught at a distance of x meters or 12 meters from the hole. |
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| 2. |
29. A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m awayfrom the bottom of the pillar, a snake is coming to its hole at the base of the pillarSeeing the snake the peacock pounces on it. If their speeds are equal, at what distance |
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| 3. |
A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snakethe peacock pounces on it. If their speeds are equal, at what distance from the hole is thesnake caught? |
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| 4. |
29. A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m awayfrom the bottom of the pillar, a snake is coming to its hole at the base of the pillarSeeing the snake the peacock pounces on it. If their speeds are equal, at what distancefrom the hole is the snake caught? |
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Answer» Let the distance covered by the peacock be AC = x mHence the distance covered by the snake is DC = x mIn right ΔABC , by Pythagoras theorem we haveAC2= AB2+ BC2 x2= 92+ (27 – x)2⇒ x2= 81 + 729 – 54x + x2⇒54x = 810∴x = 15 |
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| 5. |
Datobhaight JineХ"30° |
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Answer» sum of all these angles = 180° ...( linear pair) => 30+42+18 +(2x-48)+(3x-12) = 180=> 90+5x-60 = 180=> 5x = 180+60-90=> x = 150/5 = 30° |
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| 6. |
\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h} \text { gui } f(x)=2 x^{2}+3 |
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Answer» f(2+h)=2(h^2+4h+4)+3=2h^2+8h+11f(2)=11lim h->0 (2h^2+8h+11-11)/h=lim h->0 2h+8=8 |
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| 7. |
29. A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m awayfrom the bottom of the pillar, a snake is coming to its hole at the base of the pillarSeeing the snake the peacock pounces on it. If their speeds are equal, at what distancefrom the hole is the snake caught ? |
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Answer» Let the distance covered by the peacock be AC = x m Hence the distance covered by the snake is DC = x m In right ΔABC, by Pythagoras theorem we have AC2= AB2+ BC^2x2= 92+ (27 – x)2⇒ x2= 81 + 729 – 54x + x2⇒54x = 810∴x = 15please share your valuable feedbackby liking the solution 👍 |
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| 8. |
4. Prove dh of the linear equation x +2y -8. At which point(s) does the graph cut thex-axis and y-axis. |
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| 9. |
In the figure AB and CD are two diameters of a circle (withto each other and OD is thearea of the shaded region.16.eter of the smaller circle. If OA -7 cm, find thethe sm |
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| 10. |
How many tiles each 25 cm long and 12 cm wide will be required to lay a path 12.5and 4.8 m wide |
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Answer» your are nice |
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| 11. |
wing pax.110°yd |
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Answer» sum of angles S and P Should be 180° ( parallelogram property) so. angle p = x = 180° -110° = 70° now opposite angles of parallelogram are equal so, y = 110° |
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| 12. |
WingIIDthe Nalue |
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Answer» thanks a lot |
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| 13. |
2 2 | ( जद दान दि 0.0. ) 2 Lo 1004 चलकी... D o e £ ey fn %Rके गए (४ 9 N . D L//(rJ,rw)r 20e o |
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Answer» A+ 13 d = 2(a+7d) a + 13d= 2a+ 14d -a -d = 0 a+ d = 0 = a2 ---- --------> 1a + 5d = -8 -----> 22-1 a +5d = -8a +. d = 0------------------a = 2D = -2 a 20 = 2+ 19*-2 = -36 S20 = 10(2+(-36)) = -340 |
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| 14. |
ही हक 21 ey ही कक Lo L123 234 345 n(n+1)(n+2)2 n(n+3)An+Nn+2) |
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| 15. |
Golve the inequalitu51 tuy soleăš22,92 |
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Answer» Here is a similar question- |
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| 16. |
c)SimfM7ĺşdecSe CJolNat232 |
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Answer» Question you have submitted is incomplete. Please post a complete question. what is your exact question. can you tell me how can i learn it write from o to 4 like this give rootover and divide by 2 next then sin table will look like thiscos table will be in opposite order of sin table then tan table will be sin/cos then cot will be 1/tansec will be 1/coscosec will be 1/sin now,you get it or not?????? how to write tan and and cot sec and cosec pls tell you are good teacher 1 min wait.............. ok sir or mam like my post this is for tan this is for cosec... thanks are you a boy or girl and from where good night this is for cot... kk girl yarr....photo se pata nahi chal rahaa kha se ho didi mai pta odishaa see....samjh mein agaya naa.... ha samjh aa gya |
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| 17. |
s&3 zaos of a poly 2+27-1-2x auand - thatferid all the gemsid all the seal |
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| 18. |
(n)thealcu9. In Fig. 12.27, AB and CD are two diameters of acircle (with centre O) perpendicular to each otherand OD is the diameter of the smaller circle. IfOA 7 cm, find the area of the shaded region. |
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| 19. |
A peacock is sitting on a pillar 9 metre high. Asnake at a distance of 27 metre from the pillar iscomming to a hole at the base of the piller. Seeingthe snake, the peacock pounces upon it. If theirspeeds are equal, then find au what distance fromthe hole is snake caught?are 2(i) 27The(2 Do all t |
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Answer» Thank you |
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| 20. |
3 ^ { - h } = ( \frac { 1 } { 27 } ) ^ { ( 1 - 2 h ) } |
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Answer» Thanks |
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| 21. |
23. Find the least number which should be subtracted from 27583 so that the differencu no Ut April 2013exactly divisible by 35. |
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Answer» quotient = 788remainder = 3 quotient 788 & remainder is 3 it is the correct answer. |
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| 22. |
Prove that ACD = ABC + DEC |
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| 23. |
5170 (9реж-рен9)(05 [190 + рен) |
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| 24. |
95x+95x=190 then x=? |
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Answer» x=1 is the right answer |
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| 25. |
LCM of 100 and 190 |
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Answer» 100 = 2×2×5×5 190 = 2×5×19 LCM = 2×2×5×5×19 = 1900 From which book have you taken this question? Please tell us so that we can provide you faster answer |
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| 26. |
HCF of 100 and 190 |
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| 27. |
wing complennember |
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| 28. |
In the fig, APB& COD are semícircle BSD are semicircle of diameter 7cm each, while ARCand BSD are semicircles of diameter 14 cm each. Find the perimeter of the shaded region(T-22/7)(Ans :66 cm)凸 |
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| 29. |
In Fig. 12.31, a square OABC is inscribed in a quadrant OPBO IfOA 20 cm, find thearea of the shaded region. (Use Ď = 3.14)21cm7cmFig. 12.32neFig. 12.31Frodii 21 em and 7 cm |
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Answer» 😮 |
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| 30. |
golveey wing lo |
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Answer» logx=3/4log(4/3)logx=3/4(log4-log3)logx=3/4(0.12)logx=0.094X=e^0.094X=1.0985 |
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| 31. |
Find the H.C.F. of 95 and 190 |
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Answer» First find prime factors of 95 and 190 95 = 5*19190 = 2*5*19 HCF is highest common factor. For given numbers highest common factor is 95 190=95×2+0so the HCF is 95 |
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| 32. |
8. In the adjoining figure, ABand CD are two diametersof a cirele, with centre O,perpendicular to each other pand OD is the diameter of thesmaller circle. If OA = 7 cm,find the area of the shaded region. |
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| 33. |
7/ A chord of a circle of radius 12 cm subtends anangle of 120° at the centre. Find the area of thecorresponding segment of the circle.(Use Tt-3.14 and /3 1.73) |
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Answer» Area of a triangle = 1/2 a × b × sin( angle between a and b) |
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| 34. |
7. A chord of a circle of radius 12 cm subtends anangle of 120° at the centre. Find the area of thecorresponding segment of the circle.(UseTt 3.14 and 31.73) |
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| 35. |
6. A chord of a circle of radius 15 cm subtends an angle of 600 at the centre. Find the aroof the corresponding minor and major segments of the circle.(Use Tt 3.14 and 3 -1.73)7. A chord of a circle of radius 12 cm subtends anangle of 120° at the centre. Find the area of thecorresponding segment of the circle.(Use π = 3.14 and V3 = 1.73) |
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| 36. |
-Jul 14 a1. Express 0.0000073 in the standard form.Axnlain Parthe |
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Answer» 0.0000073=7.3*10^-6 wri |
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| 37. |
\text { for gp }\sqrt { 2 } , 2,2 \sqrt { 2 } , \ldots . .15 ^ { \text { th } } \text { terms then } T _ { 15 } = |
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| 38. |
2. Form the pair of linear equatioรกs i the tulu(if they exist) by the elimination methodwe add I to the numerator and substract I from the denominator, a fraction reduces5 if we only add 1 to the denominator What is the fraction?later Nuri will be twice as |
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Answer» The fraction : x/y Add 1 to numerator and subtract 1 from denominator,gives us 1: (x + 1) / (y - 1) =1 x+1=y-1 x+2=y [i] Add 1 to denominator,gives us 1/2: x/(y+1)=1/2 2x=y+1 [ii] Substitute [i] in [ii]: 2x=(x+2)+1 2x=x+3 x=3. Substitute x=3 in [i]: 3+2=y y=5. The original fraction : 3/5. Like my answer if you find it useful! |
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| 39. |
ofinA bag contains 150 kg wheat. A truck can carry 883 bagstrucks carry?wheat |
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Answer» total weight = 150*883 = 132450 kg |
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| 40. |
15. The (p + q) th and (p-q)th terms of a G.P are m and n respectively. Prove that its pth termisvmandqth term is m(m) |
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| 41. |
DatePageRepnesent the neal numben given by2212 Ď<5 or the number line-i |
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Answer» The box contains the real numbers which satisfy the above condition. |
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| 42. |
Ernnple 9 : Show that () . 2353535âŚ-0.235 can be expressed in the formwhere p and q are integers and0 |
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| 43. |
Q.4. Formm the pair of linear equatioorfollowing problems and find their solutions(if they exist) by any algebraic method:(a) A part of monthly hostel charges is26fixed and the remaining depends onthe number of days one has taken foodin the mess. When a student A takesfood for 20 days she has to pay 1000:as hostel charges whereas a studentB, who takes food for 26 days, pays :1180 as hostel charges. Find the fixedcharges and the cost of food per day.23or |
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| 44. |
(-190)*(-55)+(-190)*(-45) |
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Answer» (-190)×(-55)+(-190)×(-45)=10450+8550=19000 is the answer |
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| 45. |
13. Jyothika secured 35 marks out of 50 in English au 27 H JULHuthamatics In which subject did she get more marks and how much?1.What |
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Answer» Marks in English=35Marks in Maths. =27Now, 35-27 =7Therefore , Marks she secured more in is English by 8 marks |
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| 46. |
CUETESI PAPLITA. 1. What are integers? Write all integers from -5 to 5.2. In each of the pairs given below, find the larger integer.(i) 0.-3(11) -4, -6(iii) -99, 9 |
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| 47. |
29. In a circle OA = 14cm OB = 7cm14 cm7 cmFind the area of the shaded portion. |
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Answer» OA =14 OB=7THEN ROUND SHAPE=OA-OB=14-7=7 area of OA-area of OB=πR^2-πr^2 R=14 cm,r=7 cm Area of shaded portion is 616-154=462 cm×cm ta OA=14 OB=7 THEN ROUND SHAPE =OA-OB=14-7=7 area of shaded region=π(R²-r²)π(14²-7²)π(14+7)(14-7)22/7×21×722×21462cm² OA=14 OB=7 SO,SHAPE =OA _OB= 14_7 =7 the correct answer is the 7 R=14 ,r= 7 cm is the correct answer correct area of shaded portion = area of whole circle -area of small unshaded circle. ie 615.44-153.86=461.58 Do the answer is 22 of this question area of OA-area of OB =πR^2-πr^2R=14cm r=7 oa=14Ob=7circle shape = oa-ob=14-7=7 area of big circle = 22÷7 × 14 ×14 =22 ×28=616 cm.sqarea of small circle = 22÷7 ×7 ×7=22×7 cm.sq.=154area of shaded part = area of big circle - area of small circle=616 - 154=462 cm.sq.= पहले छोटे वृत्त का चेत्रफल निकाल के फिर बड़े वृत्त का और उसके बाद जोड़ दे फॉर्मूला ,"πr2 πका मान 22/7 हो गा |
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| 48. |
fcos x .cos 7xdr村丽而动fTU |
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| 49. |
4. If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, thfind(i) radius of its base(ii) its volume. (Use Tt 3.14) |
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| 50. |
1.Find the area of a rectangle whose length and breadth are 25 om and 16 om |
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Answer» area of the rectangle= l×b = 25cm×16cm =. 400 cm s.q |
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