How much distance was traveled by first train of India? (a) 33 km (b) 36 km (c) 34 km (d) 46 km

How much distance was traveled by first train of India?
(a) 33 km (b) 36 km (c) 34 km (d) 46 km
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1 Answer

Answer :

(c) 34 km
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Related questions

How much distance was traveled by first Train of India? A. 30 km B. 34 km C. 40 km D. 44 km

Description : How much distance was traveled by first Train of India? A. 30 km B. 34 km C. 40 km D. 44 km

Last Answer : B [34 km] Explanation: India's first train ran from Mumbai to Thane in 1853. The first train on the Indian sub-continent ran from Bori Bunder in Mumbai to Thane at 3.30 pm on April 16, 1853.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is: A.40 m B.55 m C.65 m D.50 m E.None of these

Description : Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is: A.40 m B.55 m C.65 m D.50 m E.None of these

Last Answer : Answer- D (50) Explanation – Relative speed= total lengths/time (46-36) x 5/18 = [L + L ] / 36 10 x (5/18) x36 x (1/2)= L L=50 m

A locomotive of weight 100 tonnes is attached to train of wagons weighing 1000 tonnes. The locomotive pulls the wagons with a constant velocity of 36 km/hour. When the coupling between the train and wagons breaks, the wagons roll on 95 m in 10 sec. The distance between the locomotive and the wagons at this instant assuming that the forces acting on the train are unaltered would be a.0.36 m b.0.44 m c.0.77 m. d.0.5 m e.0.66 m

Description : A locomotive of weight 100 tonnes is attached to train of wagons weighing 1000 tonnes. The locomotive pulls the wagons with a constant velocity of 36 km/hour. When the coupling between the train and wagons breaks, the wagons roll on 95 m in ... would be a.0.36 m b.0.44 m c.0.77 m. d.0.5 m e.0.66 m

Last Answer : d. 0.5 m

A train accelerates from 36 km/h to 54 km/h in 10 sec. (i) Acceleration (ii) The distance travelled by car. -Science

Description : A train accelerates from 36 km/h to 54 km/h in 10 sec. (i) Acceleration (ii) The distance travelled by car. -Science

Last Answer : a. Acceleration is given by a=ΔvΔta=ΔvΔt Δv=54−36=18km/hr=18×10003600m/s=5m/sΔv=54−36=18km/hr=18×10003600m/s=5m/s So a=.5 m/s2 b. Distance is given by S=ut+12at2S=ut+12at2 Now u=36 km/hr =10m/s s=10×10+12×.5×102s=10×10+12×.5×102 So s=125m

25,?,45,58,73,90,109 a) 33 b) 34 c) 35 d) 38 e) 36

Description : 25,?,45,58,73,90,109 a) 33 b) 34 c) 35 d) 38 e) 36

Last Answer : The series is 3×5+10=25,4×6+10=34,5×7+10=45,6×8+10=58,7×9+10=73,8× 10+10=90,9×11+10=109 Hence missing term=34 Answer: b)

Find the root value of 36.1 / 102.4 (1) 61 / 34 (2) 19 / 31 (3) 19 / 32 (4) 19 / 33

Description : Find the root value of 36.1 / 102.4 (1) 61 / 34 (2) 19 / 31 (3) 19 / 32 (4) 19 / 33

Last Answer : (3) 19 / 32

The thiazolidinediones are mainly used as: A. Sole drug in type 1 diabetes mellitus B. Sole drug in type 2 diabetes mellitus C. Addon drug to a sulfonylurea and/or a biguanide in type 2 diabetes mellitus D. Addon drug to insulin in type 1 diabetes mellitus (p. 250) 18.33 1 8 . 3 3 C 18.34 1 8 . 3 4 B 18.35 1 8 . 3 5 D 18.36 1 8 . 3 6 C

Description : The thiazolidinediones are mainly used as: A. Sole drug in type 1 diabetes mellitus B. Sole drug in type 2 diabetes mellitus C. Addon drug to a sulfonylurea and/or a biguanide in type 2 diabetes mellitus D. Addon drug to insulin in type 1 ... 18.34 1 8 . 3 4 B 18.35 1 8 . 3 5 D 18.36 1 8 . 3 6 C

Last Answer : C. Addon drug to a sulfonylurea and/or a biguanide in type 2 diabetes mellitus

EAN of [Ni(CO)4] complex is? a) 36 b) 33 c) 35 d) 34

Description : EAN of [Ni(CO)4] complex is? a) 36 b) 33 c) 35 d) 34

Last Answer : a) 36

Ganesh is travelling on his motorbike and has calculated to reach golden temple at 8 p.m if he travels at 30 km/hr; he will reach there at 6 p.m if he travels at 45 km/hr. At what speed must he travel to reach the golden temple at 7 p.m? a) 36 b) 46 c) 32 d) 30

Description : Ganesh is travelling on his motorbike and has calculated to reach golden temple at 8 p.m if he travels at 30 km/hr; he will reach there at 6 p.m if he travels at 45 km/hr. At what speed must he travel to reach the golden temple at 7 p.m? a) 36 b) 46 c) 32 d) 30

Last Answer : A Let the distance travelled be x km x/30 – x/45 = 2 (3x – 2x) / 90 = 2 x/90 = 2 x = 180km Time taken to travel 180 km at 30 km/hr = 180 / 30 = 6 hrs So ganesh started 6 hrs before 8 pm That is 2 pm Required speed = distance / time =(180/5) =36 km/hr

kajal rides his scooter 14km at an average speed of 16 km/hr and again travels 16km at an average speed of 14 km/hr. What is her average speed for the entire trip approximately? a) 15.36 km/hr b) 17.46 km/hr c) 13.56 km/hr d) 14.86 km/hr

Description : kajal rides his scooter 14km at an average speed of 16 km/hr and again travels 16km at an average speed of 14 km/hr. What is her average speed for the entire trip approximately? a) 15.36 km/hr b) 17.46 km/hr c) 13.56 km/hr d) 14.86 km/hr

Last Answer : D Total distance travelled =14+16=30Km Time taken to travel 14 km at an average speed of 16 km/hr =14/16=7/8 Hr Time taken to travel 16 km at an average speed of 14 km/hr =16/14=8/7 hr Total time taken = ... /Total time taken =30/(7/8+8/7) =30/(113/56) =30*56/113=1680/113 =14.86 km/hr

1/2of a certain travel is covered at the rate of 30km/hr, one-third at the rate of 40 km/hr and the rest at 35km/hr. Find the average speed for the whole travel. A) 33 1/3 B) 33 3/5 C) 34 3/7 D) None of these

Description : 1/2of a certain travel is covered at the rate of 30km/hr, one-third at the rate of 40 km/hr and the rest at 35km/hr. Find the average speed for the whole travel. A) 33 1/3 B) 33 3/5 C) 34 3/7 D) None of these

Last Answer : B) let the total travel be X km.  Then X/2 km at the speed of 30 km/hr and X/3 km at 40 km/hr and the rest distance( X - X/2 - X/3) =1/6 X at the speed of 35km/hr. Total time taken during the travel ... 2*30 hrs+ X/3*40hrs+ X/6*35hrs = 5X/168 hrs Average speed =X/(5x/168) = 168/5 =33 3/5km hr

If diameter of a wheel is 1.26 m, what the distance covered in 500 revolutions? (a) 1.38 km (b) 4.64 km (c) 2.46 km (d) 1.98 km

Description : If diameter of a wheel is 1.26 m, what the distance covered in 500 revolutions? (a) 1.38 km (b) 4.64 km (c) 2.46 km (d) 1.98 km

Last Answer : (d) 1.98 km

Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km/hr) is: A.12 kmph B.24 kmph C.36 kmph D.48 kmph E.None of these

Description : Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km/hr) is: A.12 kmph B.24 kmph C.36 kmph D.48 kmph E.None of these

Last Answer : Answer – C (36 kmph) Explanation – Let the speed of each train be x m/sec. Then, relative speed of the two trains = 2x m/sec. So, 2x = (120 + 120)/12 ⇔ 2x = 20 ⇔ x = 10. ∴ Speed of each train = 10 m/sec = [10 x 18/5] km/hr = 36 km/hr

It takes 60 sec for a train running at 162 km/hr to cross a bridge. And it takes 36 sec for the same speed to cross a boy walking at the rate of 18km/hr in the same direction in which the train is running. What is the length of the train and length of the platform. A) 1440 &1260m B) 1220 &60m C) 1000 & 1460m D) 1260 &1450m

Description : It takes 60 sec for a train running at 162 km/hr to cross a bridge. And it takes 36 sec for the same speed to cross a boy walking at the rate of 18km/hr in the same direction in which the train is ... the train and length of the platform. A) 1440 &1260m B) 1220 &60m C) 1000 & 1460m D) 1260 &1450m

Last Answer : ANSWER: A  Explanation:  Relative speed of the train to man = (162 - 18)  = 144km/hr  Convert km/hr into m/s  144km/hr = (144 * 5 /18) m/s  = 40 m/s  When the train passes ... of bridge = 2700  length of bridge = 1260 m  the length of the bridge & train are 1260 and 1440m

The average speed of a bus in the upward journey is 50% more than that in the return journey. The bus halts for 2 hrs on reaching the destination. The total time taken for the complete to and from trip is 34 hrs covering a distance of 1600 km. the speed of the bus in the upward journey is. a) 63.24 b) 65.67 c) 63.49 d) 62.49

Description : The average speed of a bus in the upward journey is 50% more than that in the return journey. The bus halts for 2 hrs on reaching the destination. The total time taken for the complete to and from trip is 34 hrs ... speed of the bus in the upward journey is. a) 63.24 b) 65.67 c) 63.49 d) 62.49

Last Answer : D Let the speed in return journey be x km/hr Then speed in upward journey = 150 /100 x = (3/2 x)km/hr Average speed = (2*3x/2*x) / (3x/2 +x) =3x2 / (5x/2) =6x2 / 5x =6x/5 km/hr Therefore (1600 ... 192x = 8000 X = 41.66 Speed in upward journey = (3/2 x ) km/hr =(3/2*41.66) =62.49 km/hr 

For transmission of power over a distance of 200 km, the transmission voltage should be (a) 132 kV (b) 66 kV (c) 33 kV (d) 11 kV

Description : For transmission of power over a distance of 200 km, the transmission voltage should be (a) 132 kV (b) 66 kV (c) 33 kV (d) 11 kV

Last Answer : (a) 132 kV

Mr. Grumper grumbles about bad time-keeping trains like everybody else. On one particular morning he was justified, though. The train left on time for the one hour journey and it arrived 5 minutes late. However, Mr. Grumper's watch showed it to be 3 minutes early, so he adjusted his watch by putting it forward 3 minutes. His watch kept time during the day, and on the return journey in the evening the train started on time, according to his watch, and arrived on time, according to the station clock. If the train traveled 25 percent faster on the return journey than it did on the morning journey, was the station clock fast or slow, and by how much? -Riddles

Description : Mr. Grumper grumbles about bad time-keeping trains like everybody else. On one particular morning he was justified, though. The train left on time for the one hour journey and it arrived 5 minutes late. ... did on the morning journey, was the station clock fast or slow, and by how much? -Riddles

Last Answer : The station clock is 3 minutes fast. The morning journey took 65 minutes, and the evening journey therefore took 52 minutes, and the train arrived 57 minutes after it should have left, that is, 3 minutes early.

The average speed of a train is 1 3/7 times the average speed of a car. The car covers a distance of 616 km in 8 hours. How much distance will the train cover in 13 hours? a) 1340 km b) 1423 km c) 1430 km d) 1936 km e) None of these

Description : The average speed of a train is 1 3/7 times the average speed of a car. The car covers a distance of 616 km in 8 hours. How much distance will the train cover in 13 hours? a) 1340 km b) 1423 km c) 1430 km d) 1936 km e) None of these

Last Answer : speed of the car = 616/8 = 77km/hr speed of the train = 10/7 × 77 = 110 km/hr Distance covered by train = 110 × 13 = 1430 km. Answer: c)

A train covers the first half of the distance between two stations with a speed of 40 km/h and the other half with 60 km/h. Then its average speed is

Description : A train covers the first half of the distance between two stations with a speed of 40 km/h and the other half with 60 km/h. Then its average speed is

Last Answer : A train covers the first half of the distance between two stations with a speed of 40 km/h and the other half with ... 48 km/h C. 52 km/h D. 100 km/h

A train starts from Delhi and reaches Lucknow in 24.5 hours. If it travels the first half of journey at 30 kmph and second half at 40 kmph then what is the total distance it travelled? 1) 750 km 2) 800 km 3) 840 km 4) 900 km 5) 920 km

Description : A train starts from Delhi and reaches Lucknow in 24.5 hours. If it travels the first half of journey at 30 kmph and second half at 40 kmph then what is the total distance it travelled? 1) 750 km 2) 800 km 3) 840 km 4) 900 km 5) 920 km

Last Answer : 3) 840 km

The altitudinal distance of a geostationary satellite from the earth is about: (A) 26,000 km (B) 30,000 km (C) 36,000 km (D) 44,000 km

Description : The altitudinal distance of a geostationary satellite from the earth is about: (A) 26,000 km (B) 30,000 km (C) 36,000 km (D) 44,000 km

Last Answer : Answer: Option C

In a braking test, a vehicle travelling at 36 km ph was stopped at a braking distance of 8.0 m. The average value of the vehicle's skid resistance (friction coefficient) is (A) 0.64 (B) 6.25 (C) 0.16 (D) None of these

Description : In a braking test, a vehicle travelling at 36 km ph was stopped at a braking distance of 8.0 m. The average value of the vehicle's skid resistance (friction coefficient) is (A) 0.64 (B) 6.25 (C) 0.16 (D) None of these

Last Answer : Answer: Option C

The maximum speed of a train is 90 km/h. It takes 10 hours to cover a distance of 500 km. Find the ratio of its average speed to maximum speed? -Science

Description : The maximum speed of a train is 90 km/h. It takes 10 hours to cover a distance of 500 km. Find the ratio of its average speed to maximum speed? -Science

Last Answer : Average speed=total distancetime=50010=50Average speed=total distancetime=50010=50 km/hr Ratio of average speed to maximum speed= 50:90=5:9

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train. -Maths 10th

Description : A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train. -Maths 10th

Last Answer : Let the speed of train be x km/hr Distance to be travelled = 360km We know that, Time take by the train intially = If the speed was increased by 5km/hr Time taken by train = Difference in the time taken is 48 minutes, x2+5x -2250 = 0 (x+50)(x-45) Speed = 45 km/hr as speed cannot be negative

The distance between two cities A and B is 330km. A train starts from A at 8 (a)m. and travels towards B at 60 km/hr. Another train starts from B at 9 (a)m. and travels towards A at 75 km/hr. At what time do they meet? a) 10 am. b) 10 : 30 am. c) 11 am. d) 11 : 30 am. e) None of these

Description : The distance between two cities A and B is 330km. A train starts from A at 8 (a)m. and travels towards B at 60 km/hr. Another train starts from B at 9 (a)m. and travels towards A at 75 km/hr. At what time do they meet? a) 10 am. b) 10 : 30 am. c) 11 am. d) 11 : 30 am. e) None of these

Last Answer : Distance travelled by first train in one hour = 60 x 1 = 60 km Therefore, distance between two train at 9 a.m. = 330 – 60 = 270 km Now, Relative speed of two trains = 60 + 75 = 135 km/hr Time of meeting of two trains =270/135=2 hrs. Therefore, both the trains will meet at 9 + 2 = 11 A.M. Answer: c)

The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet? A.10:30 am B.10:45 am C.11 am D.11:25 am E.None of these

Description : The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet? A.10:30 am B.10:45 am C.11 am D.11:25 am E.None of these

Last Answer : Answer – C (11 am) Explanation – Suppose they meet x hrs after 8 a.m. Then, (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs] = 330 60x + 75(x – 1) = 330 x = 3 So, they meet at (8 + 3), i.e. 11 a.m

The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet? A.10:30 am B.10:45 am C.11 am D.11:25 am E.None of these

Description : The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet? A.10:30 am B.10:45 am C.11 am D.11:25 am E.None of these

Last Answer : Answer – C (11 am) Explanation – Suppose they meet x hrs after 8 a.m. Then, (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs] = 330 60x + 75(x – 1) = 330 x = 3 So, they meet at (8 + 3), i.e. 11 a.m

Anitha travels 400 km by bus at 70 km/hr, 600 km by train at 80km/hr, 250 km by bike at 60 km/hr and 50 kn by car at 40 km/hr. what is the average speed for the total distance? a) 68.45 b) 69.45 c) 67.45 d) 69.77

Description : Anitha travels 400 km by bus at 70 km/hr, 600 km by train at 80km/hr, 250 km by bike at 60 km/hr and 50 kn by car at 40 km/hr. what is the average speed for the total distance? a) 68.45 b) 69.45 c) 67.45 d) 69.77

Last Answer : D Total distance = (400+250+50+600)km =1300 km Total time taken = (400/70 + 600/80+250/60+50/40) hrs =(40/7 +60/8+25/6+5/4) =(24*40)+(21*60)+(28*25)+(42*5) / 168 =960+1260+700+210 / 168 =(3130 / 168) Average speed =(1300*168 / 3130) km/hr =218400/3130 =69.77 km/hr

Distance between two stations P & Q is 1556km. A passenger train covers the journey from P to Q at 168km per hr and return back to P with a uniform speed of 112km/hr. Find the Average speed of the train during the whole journey? A) 124.4 km/hr B) 130.4km/hr C) 134.4 km/hr D) 130.0 km/hr

Description : Distance between two stations P & Q is 1556km. A passenger train covers the journey from P to Q at 168km per hr and return back to P with a uniform speed of 112km/hr. Find the Average speed of the train during the whole journey? A) 124.4 km/hr B) 130.4km/hr C) 134.4 km/hr D) 130.0 km/hr

Last Answer : C) Given x= 168 , y = 112 Required average speed = (2xy / x + y) km/hr = 2 * 168 *112 / 168 + 112 = 37632 / 280 =134.4 km/hr

A train leaves Mumbai for chennai at 6 : 45 a.m. and goes at the rate of 100 km/h. Another train leaves Chennai for Mumbai at 4.30a.m. and goes at the rate of 120 km/h. If the distance between both is 1240 km, at what distance from Mumbai will the two trains meet? A) 210km B) 324km C) 115km D) 440km

Description : A train leaves Mumbai for chennai at 6 : 45 a.m. and goes at the rate of 100 km/h. Another train leaves Chennai for Mumbai at 4.30a.m. and goes at the rate of 120 km/h. If the distance between both is 1240 km, at what distance from Mumbai will the two trains meet? A) 210km B) 324km C) 115km D) 440km

Last Answer : ANSWER: D  Explanation:  Given speed of the first train = 100kmph  speed of the second train = 120kmph  The second train leaving Chennai starts its journey earlier and it travels = 120 *(6.45a.m. ... train covers 4.40 *100 = 440 km  Therefore the first train meet to the second train at 440km

The distance between two stations ooty and chennai is 615km. A train with speed of 75 km/h leaves ooty at 8:00 am towards chennai. Another train with speed of 105 km/h leaves Chennai at 9 : 00 am towards ooty . Then, at what time both trains meet? A) 6 B) 12 C) 18 D) 24

Description : The distance between two stations ooty and chennai is 615km. A train with speed of 75 km/h leaves ooty at 8:00 am towards chennai. Another train with speed of 105 km/h leaves Chennai at 9 : 00 am towards ooty . Then, at what time both trains meet? A) 6 B) 12 C) 18 D) 24

Last Answer : ANSWER:B  Explanation:  First train leaves at 8.00am and second leaves at 9.00am  So,first train that is , from ooty to Chennai has covered  75 km distance in 1 hr.  So , distance left by between the station = ... the trains will meet , 3 hr after second train left  i.e., 9.00 am +3 = 12.00 pm

A shatabadi express takes 240 seconds to cover a distance of length 700 meters at a speed of 270 km/hr. What will be the time taken by the train to cross a standing pole? A) 160sec B) 465sec C) 310sec D) 231sec

Description : A shatabadi express takes 240 seconds to cover a distance of length 700 meters at a speed of 270 km/hr. What will be the time taken by the train to cross a standing pole? A) 160sec B) 465sec C) 310sec D) 231sec

Last Answer : ANSWER:D Explanation:  Speed of the train = 270 km/hr. Converting the speed into m/sec  Speed = 270 km/hr = 270 x 5/18 m/sec = 75 m/sec  First, we have to find the length of the train  Let ... a standing Pole at 75 m/s =( 17300/75)seconds = 231 seconds. Hence the answer is 231 sec(approx.)

`IP_(1),IP_(2),IP_(3),IP_(4) and IP_(5)` for an element is 7.1,14.3,34.5,46.8 and 162.2 eV respectively then element is :-

Description : `IP_(1),IP_(2),IP_(3),IP_(4) and IP_(5)` for an element is 7.1,14.3,34.5,46.8 and 162.2 eV respectively then element is :-

Last Answer : `IP_(1),IP_(2),IP_(3),IP_(4) and IP_(5)` for an element is 7.1,14.3,34.5,46.8 and 162.2 eV respectively then element is :- A. Na B. Si C. F D. Ca

f the mass is of 10 Kg, find the natural frequency in Hz of the free longitudinal vibrations. The displacement is 0.01mm. a) 44.14 b) 49.85 c) 43.43 d) 46.34

Description : f the mass is of 10 Kg, find the natural frequency in Hz of the free longitudinal vibrations. The displacement is 0.01mm. a) 44.14 b) 49.85 c) 43.43 d) 46.34

Last Answer : b) 49.85

Which of the following couldnot be an Ethernet unicast destination? A) 43:7B:6C:DE:10:00 B) 44:AA:C1:23:45:32 C) 46:56:21:1A:DE:F4 D) 48:32:21:21:4D:34

Description : Which of the following couldnot be an Ethernet unicast destination? A) 43:7B:6C:DE:10:00 B) 44:AA:C1:23:45:32 C) 46:56:21:1A:DE:F4 D) 48:32:21:21:4D:34

Last Answer : 43:7B:6C:DE:10:0

Mr. Shrimant inherits 4325 gold coins and divides them among his three sons; Bharat, Parat and Marat; in a certain ratio. Out of the total coins received by each of them, Bharat sells 40 coins; Parat donates his 20 coins and Marat loses 30 coins. Now, the ratio of gold coins with them is 41:34:46, respectively. How many coins did Parat receive from his father? a) 1210 b) 1211 c) 1212 d) 1213 e) None of these

Description : Mr. Shrimant inherits 4325 gold coins and divides them among his three sons; Bharat, Parat and Marat; in a certain ratio. Out of the total coins received by each of them, Bharat sells 40 coins; Parat donates his ... did Parat receive from his father? a) 1210 b) 1211 c) 1212 d) 1213 e) None of these 

Last Answer : Answer: A 41x+40+34x+20+46x+30=4325 =>121x=4325-90 =>x=35 Required answer, number of coins received by parat =34x+20=34×35+20=1210

Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Description : Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Last Answer : Here's ur answer Hope it has helped u..

Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Description : Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Last Answer : Here's ur answer Hope it has helped u..

The numerical aperture of a coaxial cable with core and cladding indices given by 2.33 and 1.4 respectively is a) 3.73 b) 0.83 c) 3.46 d) 1.86

Description : The numerical aperture of a coaxial cable with core and cladding indices given by 2.33 and 1.4 respectively is a) 3.73 b) 0.83 c) 3.46 d) 1.86

Last Answer : d) 1.86

A vertical column of water will be supported to what height by standard atmospheric pressure.  a. 33.9 ft  b. 45 ft  c. 67 ft  d. 25.46 ft ho= Po/Yo

Description : A vertical column of water will be supported to what height by standard atmospheric pressure.  a. 33.9 ft  b. 45 ft  c. 67 ft  d. 25.46 ft ho= Po/Yo

Last Answer : 33.9 ft

A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days, 4/7 of the work completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day? a) 79 b) 80 c) 81 d) 82 e) 83

Description : A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days, 4/7 of the work completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day? a) 79 b) 80 c) 81 d) 82 e) 83

Last Answer : work finished by 33 days = 33 117 8 = 30888 man hours 4/7th of the job is finished. remaining work = 1-(4/7) = 3/7 if 4/7th of the work is 30888 hours, then 3/7th of the work = [30888 ... 9 number of men = 198 since, 117 men are already working, additional men required = 198 - 117 = 81 Answer: c)

V=100Sin (1000t+46 deg), I=2Sin (1000t+80 deg) what are the elements in the circuit a) R=30 ohm, L=30 mH b) R=30 ohm, C=33.3 micro farads c) R=40 ohm, L=30 mH d) R=40 ohm, L=33.3 micro farads

Description : V=100Sin (1000t+46 deg), I=2Sin (1000t+80 deg) what are the elements in the circuit a) R=30 ohm, L=30 mH b) R=30 ohm, C=33.3 micro farads c) R=40 ohm, L=30 mH d) R=40 ohm, L=33.3 micro farads

Last Answer : Ans: R=40 ohm, L=33.3 micro farads

The area under irrigation in India is about A. 39 %. B. 36 %. C. 64%. D. 46 %.

Description : The area under irrigation in India is about A. 39 %. B. 36 %. C. 64%. D. 46 %.

Last Answer : B. 36 %.

What is the distance traveled by the turtle during the seventh minute if it walked 0.3 m in the first minute then it tripled its speed each minute?

Description : What is the distance traveled by the turtle during the seventh minute if it walked 0.3 m in the first minute then it tripled its speed each minute?

Last Answer : Initial (average) speed = distance/time = 0.3 m / 1 min = 0.3m/minIf it triples its speed every minute:for the second minute it walks at 3 0.3m/minfor the third minute it walks at 3² 0.3 m/minfor the nth minute it ... time→ distance in 7th minute = 3⁶ 0.3 m/min 1 min = 3⁶ 0.3 m =218.7 m

The speed of a car increases by 2 kms after every one hour. If the distance travelled in the first one hour was 35 kms, what was the total distance traveled in 12 hours? A.456 kms B.482 kms C.552 kms D.556 kms E.None of these

Description : The speed of a car increases by 2 kms after every one hour. If the distance travelled in the first one hour was 35 kms, what was the total distance traveled in 12 hours? A.456 kms B.482 kms C.552 kms D.556 kms E.None of these

Last Answer : Answer – C (552 kms) Explanation – Total distance travelled in 12 hours = (35 + 37 + 39 + …… upto 12 terms) This is an A.P. with first term, a = 35, number of terms, n = 12, common difference, d =2. Required distance =12/2 (2 x 35 + (12 – 1) x 2) = 6(70 + 22) = 552 kms

5, 8, 12, 18, 24, 30, 36, 42, ____ a) 52 b) 46 c) 48 d) 54 e) 56

Description : 5, 8, 12, 18, 24, 30, 36, 42, ____ a) 52 b) 46 c) 48 d) 54 e) 56

Last Answer : Each term in the series is the addition of successive prime numbers. Like Prime numbers are2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and so on. So first term is addition of (2 + 3). Ie 5, then second term is (3 + 5) ie ... +17)ie. 30, then (17+19)ie.36, then (19+23)ie., 42, then (23+29)i.e. 52. Answer: a)

What percent of the total number of girls in the college do not know Hindi? (rounded off to nearest integer) (1) 38 (2) 46 (3) 48 (4) 36 (5) 43

Description : What percent of the total number of girls in the college do not know Hindi? (rounded off to nearest integer) (1) 38 (2) 46 (3) 48 (4) 36 (5) 43

Last Answer : (5) 43

Assume postal rates for 'light letters' are: $0.25 up to 10 grams; $0.35 up to 50 grams; $0.45 up to 75 grams; $0.55 up to 100 grams. Which test inputs (in grams) would be selected using boundary value analysis? A. 0, 9,19, 49, 50, 74, 75, 99,100 B. 10, 50, 75,100, 250, 1000 C. 0, 1,10,11, 50, 51, 75, 76,100,101 D. 25, 26, 35, 36, 45, 46, 55, 56

Description : Assume postal rates for 'light letters' are: $0.25 up to 10 grams; $0.35 up to 50 grams; $0.45 up to 75 grams; $0.55 up to 100 grams. Which test inputs (in grams) would be selected using boundary value analysis? A. 0, 9,19, ... C. 0, 1,10,11, 50, 51, 75, 76,100,101 D. 25, 26, 35, 36, 45, 46, 55, 56

Last Answer : C. 0, 1,10,11, 50, 51, 75, 76,100,101

24 tins of water fill a bunker when the capacity of each bunker is 27litres. How many tins will be needed to fill the same bunker, if the capacity of each tin is 18litres? A) 36 B) 26 C) 46 D) 56

Description : 24 tins of water fill a bunker when the capacity of each bunker is 27litres. How many tins will be needed to fill the same bunker, if the capacity of each tin is 18litres? A) 36 B) 26 C) 46 D) 56

Last Answer : A Capacity of the bunker = (24 * 27)litres =648litrs Capacity of each tin = 18litres Number of tins needed = 648/18 =36.