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

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
by

1 Answer

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
by

Related questions

There are two points on a highway a,b. They are 70 km apart. An auto starts from A and another auto starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other they meet in 1 hour. Find how fast the two autos are -Maths 10th

Description : There are two points on a highway a,b. They are 70 km apart. An auto starts from A and another auto starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other they meet in 1 hour. Find how fast the two autos are -Maths 10th

Last Answer : When they travel in same direction, suppose they meet when B travels for x m, then A will have travelled 70+x m in the same time Since, Speed =TimeDistance Speed of auto at A =7x+70 And Speed of auto at B =7x ... So, Speed of auto at A =7x+70 =7280 =40km/hr Speed of auto at B =7x =7210 =30km/hr

A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. -Maths 10th

Description : A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. -Maths 10th

Last Answer : The given speed of the motor boat in the still water is equal to 18 km/ hr. The given distance travelled by motor boat is equal to 24 km. The time taken to travel 24 km downstream by motor boat = 1 hour. x=122=6 km/ hr.

A diver rowing at the rate of 5 km/h in still water takes double the time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream. -Maths 10th

Description : A diver rowing at the rate of 5 km/h in still water takes double the time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream. -Maths 10th

Last Answer : 5/3 km/h Step-by-step explanation: Speed in still water = 5km/h Define x: Let the speed of the steam be x Upstream speed = (5 - x) km/h Downstream speed = (5 + x) km/h Time taken going upstream: Time taken = ... 40x = 400 - 80x 120x = 200 x = 200 120 = 5/3 The speed of the stream is 5/3 km/h

A certain thing is thrown twice from a place with the gap of 45 minutes between the two shots. A girl approaching this point in a train heard the second shot 44 minutes after she heard the first shot. What is the speed of train (in kmph) if sound travels at 660 m/s? a) 62 km/hr b) 54 km/hr c) 48 km/hr d) 27 km/hr

Description : A certain thing is thrown twice from a place with the gap of 45 minutes between the two shots. A girl approaching this point in a train heard the second shot 44 minutes after she heard the first shot. What is the speed of train ... travels at 660 m/s? a) 62 km/hr b) 54 km/hr c) 48 km/hr d) 27 km/hr

Last Answer : B Actual time between the two shots being fired = 45 minutes. If a girl was stationary she would have heard the shots after 44 minutes. But since the train was moving towards the source, she heard the second shot after ... = (15 *18/5) km/hr =54 km/hr Hence the speed of the train is 54 km/hr

A train travels between two stations 15 km apart in 18 minutes. If the train accelerates for a part of journey uniformly followed by uniform retardation, the maximum speed attained by the train during the journey will by a.125 km/hr b.107 dynes c.80 km/hr d.100 km/hr e.60 km/hr

Description : A train travels between two stations 15 km apart in 18 minutes. If the train accelerates for a part of journey uniformly followed by uniform retardation, the maximum speed attained by the train during the journey will by a.125 km/hr b.107 dynes c.80 km/hr d.100 km/hr e.60 km/hr

Last Answer : d. 100 km/hr

A train travels between two stations 15 km, apart in 18 minutes. If the train accelerates for a part of journey uniformly followed by uniform relardation, the maximum speed attained by the train, during, the journey, will be a.60 km/hr b.80 km/hr c.125 km/hr d.100 km/hr e.107 dynes

Description : A train travels between two stations 15 km, apart in 18 minutes. If the train accelerates for a part of journey uniformly followed by uniform relardation, the maximum speed attained by the train, during, the journey, will be a.60 km/hr b.80 km/hr c.125 km/hr d.100 km/hr e.107 dynes

Last Answer : d. 100 km/hr

At �t� minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than minutes. Find ‘t’. -Maths 10th

Description : At �t� minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than minutes. Find ‘t’. -Maths 10th

Last Answer : answer:

If a train crosses a tree in 24 seconds while travelling at a speed of 90km/hr, then in how much time will the train cross a telegraph post of length 1200m at same speed? A) 26 sec B) 48 sec C) 72 sec D) 38 sec

Description :  If a train crosses a tree in 24 seconds while travelling at a speed of 90km/hr, then in how much time will the train cross a telegraph post of length 1200m at same speed? A) 26 sec B) 48 sec C) 72 sec D) 38 sec 

Last Answer : ANSWER:C Explanation: Speed of the train = 90 km/hr Converting the speed into m/sec 90 km/hr = (90* 5/18)m/s = 25m/s length of the train = distance travelled we know distance = speed * ...  Time = distance / speed  = ( 1800/25) sec  = 72 seconds  Hence the required answer is 72 seconds

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 boy sitting in a train which is travelling at 25 km/hr observes that a goods train travelling in opposite direction, takes 4.5 seconds to pass him. If the train is 140 m long, find its speed. A) 77km/hr B) 79km/hr C) 87km/hr D) 89km/hr

Description : A boy sitting in a train which is travelling at 25 km/hr observes that a goods train travelling in opposite direction, takes 4.5 seconds to pass him. If the train is 140 m long, find its speed. A) 77km/hr B) 79km/hr C) 87km/hr D) 89km/hr

Last Answer : ANSWER : C Explanation: Relative speed = distance / time = (140 / 4.5) m/s = 31.11 m/s Convert m/s into km/hr  31.11 m/s = (31.11 * 18 / 50 km/hr  = (560 / 5) km/hr  = ... train (by sitting) = (112 - 25) km/hr = 87 km/hr  Therefore the speed of goods train is 87 km/hr.

A train 125 m long passes a man, running at 5 kmph in the same direction in which the train is going, in 10 seconds. The speed of the train is: A.35 km/hr B.50 km/hr C.48 km/hr D.55 km/hr E.None of these

Description : A train 125 m long passes a man, running at 5 kmph in the same direction in which the train is going, in 10 seconds. The speed of the train is: A.35 km/hr B.50 km/hr C.48 km/hr D.55 km/hr E.None of these

Last Answer : Answer – B (50 km/hr) Explanation – Let speed be S Relative speed = total lengths/time [ in this case man length is neglible compare to train so neglected] (S – 5) x 5/18 = 125 /10 S- 5 = (125/10)x(18/5) S-5 = 45 S=50 km/hr

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

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.)

Thanu can travel 24 miles downstream in a certain river in 12hrs less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for his 48 mile round trip, the downstream 24 miles would then take only 2hr less than the upstream 24miles. What is the speed of the current in miles per hour? A) 2 / 3 miles/hr B) 8 / 3 miles/hr C) 7 / 5 miles/hr D) 3 / 5 miles/hr

Description : Thanu can travel 24 miles downstream in a certain river in 12hrs less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for his 48 mile round trip, the downstream 24 miles would ... 3 miles/hr B) 8 / 3 miles/hr C) 7 / 5 miles/hr D) 3 / 5 miles/hr 

Last Answer : ANSWER : B 

Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is: A.58 sec B.50 sec C.48 sec D.56 sec E.None of these

Description : Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is: A.58 sec B.50 sec C.48 sec D.56 sec E.None of these

Last Answer : Answer – C (48 sec) Explanation – Relative speed = all lengths/time [60+90] = [1.10 +0.9 ]/time [ PLUS when opposite direction] time=2/150 = 1/75 h 1 hour______3600 sec 1/75 hr______? ?= 3600/75=48 sec

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

Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (- a, – b) -Maths 10th

Description : Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (- a, – b) -Maths 10th

Last Answer : Distance between two points (x1 ,y1 ) and (x2 ,y2 ) is (x1 −x2 )2+(y1 −y2 )2 (i) Distance between(2,3) and (4,1) is =(2−4)2+(3−1)2 =(−2)2+(2)2 =4+4 =8 =22 (ii) Distance between (−5,7) and (−1,3) ... 42 (iii )Distance between (a,b) and(−a,−b) is =(a−(−a))2+(b−(−b))2 =(2a)2+(2b)2 =4a2+4b2 =2a2+b2

Find the distance between the points (-8/5, 2) and (2/5, 2). -Maths 10th

Description : Find the distance between the points (-8/5, 2) and (2/5, 2). -Maths 10th

Last Answer : answer:

A boat takes 38 hrs for travelling downstream from point p to point q and coming back to a point r midway between p and q. If the velocity of the stream is 8 km/hr and the speed of the boat in still water is 28 km/hr. What is the distance between p and q? A) 720 B) 640 C) 510 D) 450

Description : A boat takes 38 hrs for travelling downstream from point p to point q and coming back to a point r midway between p and q. If the velocity of the stream is 8 km/hr and the speed of the boat in still water is 28 km/hr. What is the distance between p and q? A) 720 B) 640 C) 510 D) 450

Last Answer : ANSWER: A Explanation: Speed downstream = (28 + 8)km/hr = 36 km/hr Speed upstream = (28 – 8)km/hr = 20 km/hr Let the distance between p and q be ‘x’km Then, x/36 + (x/2)/20 = 38 x/36 +x/40 = 38 19x = 13680 X = 720 km Therefore the distance between p and q is 720km

Anand takes 9 hours more than Babu to cover 150 km. Suppose the time taken by Anand is 30 minutes less than Babu he must double his speed. Then the speed of Anand will be: a) 7.64 km/hr. b) 7.43km/hr. c) 7.89 km/hr. d) 7.25 km/hr.

Description : Anand takes 9 hours more than Babu to cover 150 km. Suppose the time taken by Anand is 30 minutes less than Babu he must double his speed. Then the speed of Anand will be: a) 7.64 km/hr. b) 7.43km/hr. c) 7.89 km/hr. d) 7.25 km/hr.

Last Answer : C Let Anand 's speed be X km/hr. And let the time taken by Babu be Y. Since Anand takes 9+Y hours to cross 150km at X km/hr. i.e., 150 / X = 9+Y ---eqn1 And he takes Y - 1/2 hours to cross 150km at 2X km ... )/ 2x = 19/2 150/2X =19/2 150/X = 19 X = 7.89 Hence Anand 's speed is 7.89 km/hr.

The expenditure on fuels in running a train varies as the square of its speed measured in km/hr and it is Rs. 48 per hour at the speed of 16 km/hr. If

Description : The expenditure on fuels in running a train varies as the square of its speed measured in km/hr and it is Rs. 48 per hour at the speed of 16 km/hr. If

Last Answer : The expenditure on fuels in running a train varies as the square of its speed measured in km/ ... hour find the best speed of the train economically.

A girl covered a definite distance at same speed. Had she moved 6 km/hr faster, she would have taken 30 mins less. If she had moved 4 km/hr slower, she would have taken 30 mins more. The distance (in km) is. a) 50 b) 40 c) 70 d) 60

Description : A girl covered a definite distance at same speed. Had she moved 6 km/hr faster, she would have taken 30 mins less. If she had moved 4 km/hr slower, she would have taken 30 mins more. The distance (in km) is. a) 50 b) 40 c) 70 d) 60

Last Answer : D Let distance = x km Rate = y km/hr x/y - x / (y+6) = 30/60 (x(y+6)-xy) / y(y+6) = ½ (xy +6x - xy) / y(y+6) = ½ 12x = y(y+6) ------------>1 And x/ (y-4) - x/y = 30/60 (Xy - x(y-4 ... = (y+6) / (y-4) 3y - 12 = 2y +12 Y = 24km Put y value in 1 we get 12x = 24 (24+6) 12x = 24 *30 X=60km

In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Description : In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Last Answer : Let P has obtained x in mathematics and y in science. Then by the problem, x+y=28.......(1). If P would have got 3 marks more in mathematics, then P would have got (x+3) in mathematics ... P obtained 12 marks in mathematics and 16 in science. or, P obtained 9 marks in mathematics and 19 in science.

A bus can travel 25% faster than a jeep. Both start from point P at the same time and reach point Q, 225 kms away from P, at the same time. On the way, however, the bus lost about 37.5 minutes while stopping at the certain place. What is the speed of the jeep? a) 64 km/hr b) 72 km/hr c) 68 km/hr d) 56 km/hr

Description : A bus can travel 25% faster than a jeep. Both start from point P at the same time and reach point Q, 225 kms away from P, at the same time. On the way, however, the bus lost about 37.5 minutes while stopping at the ... is the speed of the jeep? a) 64 km/hr b) 72 km/hr c) 68 km/hr d) 56 km/hr

Last Answer : B Let speed of the jeep =x kmph Then, speed of the bus =(100+25)x/100 =125x/100 =5x/4 kmph Time taken by the jeep to travel from P to Q =225/x hours Time taken by the bus to travel from p to Q =225/(5x/4) + ... 225/x=900/5x+37.5/60 225/x=180/x+37.5/60 45/x =37.5/60 37.5x=45*60 X=72 km/hr

A person runs on the platform of 90 m at a speed of 5km/hr in the same direction of the passenger train. Find the time taken by the train to cross the running person if speed of the train is 35.5 km/hr? (Length of train = Length of platform) A) 5.32sec B) 7.5sec C) 12.6sec D) 10.62sec

Description : A person runs on the platform of 90 m at a speed of 5km/hr in the same direction of the passenger train. Find the time taken by the train to cross the running person if speed of the train is 35.5 km/hr? (Length of train = Length of platform) A) 5.32sec B) 7.5sec C) 12.6sec D) 10.62sec

Last Answer : ANSWER:D Explanation: speed of the train = 35.5 km/hr speed of the person = 5 km/hr As the train and the running person move in same direction, their speed values are subtracted to find the relative ...  =10.62 sec  Hence, the time taken by the train to cross the running person is 10.62 sec

Find the time taken by mailtrain 200m long running at a speed of 120 km/hr to cross another train of length 160m running at a speed of 96km/hr in the same direction. A) 34sec B) 52sec C) 45sec D) 54sec

Description : Find the time taken by mailtrain 200m long running at a speed of 120 km/hr to cross another train of length 160m running at a speed of 96km/hr in the same direction. A) 34sec B) 52sec C) 45sec D) 54sec

Last Answer : ANSWER: D  Explanation:  Total distance covered to cross each other = 200 + 160  = 360m  If the direction is given in the same direction then we subtract  Relative speed of two trains (same direction) = ... * 3 / 20  = 54sec  Time taken by the faster train to cross the slower train is 54sec.

The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. -Maths 10th

Description : The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. -Maths 10th

Last Answer : Step-by-step explanation: Let x be the cost of 1 bat Let y be the cost of 1 ball Cost of 3 bats = 3x Cost of 6 balls = 6y Cost of 3 balls = 3y The coach of a cricket team buys 3 ... 6 balls for ₹ 3900 She buys another bat and 3 more balls of the same kind for ₹1300. So, situation algebraically :

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? -Maths 10th

Description : 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? -Maths 10th

Last Answer : Given, Number of army contingent members=616 Number of army band members = 32 If the two groups have to march in the same column, we have to find out the highest common factor between the two groups. HCF(616, ... HCF (616, 32) = 8. Hence, the maximum number of columns in which they can march is 8.

A passenger train without stopping running at an average speed of 120 km/hr and with stoppages at an average speed of 80 km/hr. What is the total time taken by the train for stoppages on a route of length 480km? a) 2 hours b) 3 hours c) 4 hours d) 5 hours

Description : A passenger train without stopping running at an average speed of 120 km/hr and with stoppages at an average speed of 80 km/hr. What is the total time taken by the train for stoppages on a route of length 480km? a) 2 hours b) 3 hours c) 4 hours d) 5 hours

Last Answer : A Let r = running time of the train s= stoppage time of the train D= total distance travelled by train We have: D/r= 120 and ..(1) D/(r+s) =80 ..(2) Diviving (1)&(2), D/r *(r+s)/D = ... , D=480 kms 480/r= 120 r=4 put r value in (3),we get s/4 =1/2 S = 2 hours

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. -Maths 10th

Description : Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. -Maths 10th

Last Answer : Then , seven years ago, Aftab's age = x − 7 His daughter's age = y − 7 According to the question, x − 7 = 7 ( y − 7 ) x − 7 = 7 y − 4 9 x − 7 y = − 4 9 + 7 x − 7 y = − 4 2 ( ... these two equations x − 7 y = − 4 2 ⟹ x = 7 y − 4 2 x = 7 y − 4 2 4 2 3 5 4 9 y 1 2 1

The rajdhani express of 400m runs at a speed of 124 km/hr and a person runs on the platform at a speed of 40km/hr in the direction opposite to that of the train. Find the time taken by the train to cross the running person? A) 7 sec B) 8.78sec C) 7.78sec D) 8 sec

Description : The rajdhani express of 400m runs at a speed of 124 km/hr and a person runs on the platform at a speed of 40km/hr in the direction opposite to that of the train. Find the time taken by the train to cross the running person? A) 7 sec B) 8.78sec C) 7.78sec D) 8 sec

Last Answer : ANSWER : B Explanation:  Length of the train = 400m  Speed of train = 124 km/hr  Speed of person = 40 km/hr  Relative speed (speed of trai relative to man) = (124 + 40) km/hr  = 164 km/hr [As the ... train to cover 400m at a relative of 45.55 m/s  = 400 / 45.55  = 8.78 sec.

A bike starts with a speed of 60 km/hr at 8a.m. Due to the problem in engine it reduces its speed as 20 km/hr for every 1 hour. After 9 am, the time taken to covers 15 km is: a) 13 minutes and 20 seconds b) 15 minutes and 09 seconds c) 18 minutes and 15 seconds d) 22 minutes and 30 seconds

Description : A bike starts with a speed of 60 km/hr at 8a.m. Due to the problem in engine it reduces its speed as 20 km/hr for every 1 hour. After 9 am, the time taken to covers 15 km is: a) 13 minutes and 20 seconds b) 15 minutes and 09 seconds c) 18 minutes and 15 seconds d) 22 minutes and 30 seconds

Last Answer : 15 min

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

Anushiya covered a definite distance at some speed. If she had moved 5 kmph faster, she would have taken 55minutes less. If he had moved 4 kmph slower, she would have taken 55 minutes more. What is the distance in km? a) 110 Km b) 220 km c) 330 km d) 440 km

Description : Anushiya covered a definite distance at some speed. If she had moved 5 kmph faster, she would have taken 55minutes less. If he had moved 4 kmph slower, she would have taken 55 minutes more. What is the distance in km? a) 110 Km b) 220 km c) 330 km d) 440 km

Last Answer : C speed=2V1V2/(V1-V2) = 2*5*4/(5-4) =40 km/hr Distance= vt1 (1+v/v1) =40* 55/60(1+40/5) =40*11/12(1+8) =40*11/12*9 =330 km Therefore the distance is 330km

Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? -Maths 10th

Description : Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? -Maths 10th

Last Answer : answer:

Tharun has to cover a distance of 84 km in 40 minutes. If he covers one-half of the distance in one-fourth of the total time, to cover the remaining distance in the remaining time, what should be his speed in km/hr? a) 42 km/hr b) 64km/hr c) 84km/hr d)76km/hr

Description : Tharun has to cover a distance of 84 km in 40 minutes. If he covers one-half of the distance in one-fourth of the total time, to cover the remaining distance in the remaining time, what should be his speed in km/hr? a) 42 km/hr b) 64km/hr c) 84km/hr d)76km/hr

Last Answer : C Tharun needs to cover 84 km in 40minutes Given that he covers one-half of the distance in one-fourth of the total time ⇒ he covers half of 84km in one-fourth of 40 minutes ⇒ He covers 42 km in ¼ * ... Distance =42km Time =30minutes =1/2 hr Required Speed = Distance / Time =42/(1/2)=84 km/hr

Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x – 2x = (- 2) (3-x) (iii) (x – 2) (x + 1) = (x – 1) (x + 3) (iv) (x – 3) (2x + 1) = x (x + 5) (v) (2x – 1) (x – 3) = (x + 5) (x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x(x2 – 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

Description : Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x - 2x = (- 2) (3-x) (iii) (x - 2) (x + 1) = (x - 1) (x + 3) (iv) (x - 3) (2x + 1) = x (x + 5) (v) (2x - 1) (x - 3) = (x ... vi) x2 + 3x + 1 = (x - 2)2 (vii) (x + 2)3 = 2x(x2 - 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

Last Answer : this is the correct answer!

How many solutions does the pair of equations y = 0 and y = -5 have? -Maths 10th

Description : How many solutions does the pair of equations y = 0 and y = -5 have? -Maths 10th

Last Answer : y = 0 and y = -5 are Parallel lines, hence no solution.

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. -Maths 10th

Description : 2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. -Maths 10th

Last Answer : Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r

Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. -Maths 10th

Description : Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. -Maths 10th

Last Answer : Let a' be any positive integer and b = 6. ∴ By Euclid's division algorithm, we have a = bq + r, 0 ≤ r ≤ b a = 6q + r, 0 ≤ r ≤ b [ ∵ b = 6] where q ≥ 0 and r = 0,1, 2, 3, 4, ... numbers. Therefore, any odd integer can be expressed is of the form 6q + 1, or 6q + 3, or 6q + 5 where q is some integer

Is polynomial y4 + 4y2 + 5 have zeroes or not? -Maths 10th

Description : Is polynomial y4 + 4y2 + 5 have zeroes or not? -Maths 10th

Last Answer : Y4+4y2+5=y4+4y2+4 +1 =(y2+2)2+1 The first term being square,it will be equal to zero or greater than zero. Therefore (y2+2)2+1 will not be zero for any value of y. Therefore given polynomial have no zeroes.

Find a point on the y-axis equidistant from (-5, 2) and (9, -2). -Maths 10th

Description : Find a point on the y-axis equidistant from (-5, 2) and (9, -2). -Maths 10th

Last Answer : answer:

A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5 m, find the increase in area of the grassy lawn in which the calf can graze. -Maths 10th

Description : A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5 m, find the increase in area of the grassy lawn in which the calf can graze. -Maths 10th

Last Answer : Given, The initial length of the rope = 6 m Then the rope is said to be increased by 5.5m So, the increased length of the rope = (6 + 5.5) = 11.5 m We know that, the corner of the lawn is a quadrant of a circle. Thus ... – 62) = 1/4 x 22/7 (132.25 – 36) = 1/4 x 22/7 x 96.25 = 75.625 cm2

If the product of two zeroes of polynomial 2x3 + 3x2 – 5x – 6 is 3, then find its third zero. -Maths 10th

Description : If the product of two zeroes of polynomial 2x3 + 3x2 – 5x – 6 is 3, then find its third zero. -Maths 10th

Last Answer : The third zero is 1. Step-by-step explanation: Given the polynomial The product of two zeroes is 3 we have to find the third zero As product of two zeroes is 3 Let ab=3 Therefore, 3c=3 ⇒ c=1 hence, the third zero is 1.

A number say z is exactly the four times the sum of its digits and twice the product of the digits. Find the numbers -Maths 10th

Description : A number say z is exactly the four times the sum of its digits and twice the product of the digits. Find the numbers -Maths 10th

Last Answer : This is the Answer

When you add two numbers and the number obtained by reversing the order of its digits is 165. If the both numbers differ by three, find the number. -Maths 10th

Description : When you add two numbers and the number obtained by reversing the order of its digits is 165. If the both numbers differ by three, find the number. -Maths 10th

Last Answer : Let the No. Be xy ATGC , xy + yx = 165 10x + y + 10y+x = 165 11x + 11y = 165 Divide Both sides By 11 x + y = 15 __________(1) Also,x ... By 1 and 2 we get 2x = 18 x = 9 y = 6 so the number is So the number is 96 or 69 ______________________

At the same speed a boat travelling 50km upstream and 78km downstream in 16hrs. Also it can travel 70km upstream and 104km downstream, in 22hrs at the same speed of the stream is. A) 5km/hr B) 3km/hr C) 4km/hr D) 2km/hr

Description : At the same speed a boat travelling 50km upstream and 78km downstream in 16hrs. Also it can travel 70km upstream and 104km downstream, in 22hrs at the same speed of the stream is. A) 5km/hr B) 3km/hr C) 4km/hr D) 2km/hr

Last Answer : ANSWER : C

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically. -Maths 10th

Description : The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically. -Maths 10th

Last Answer : y 20 40 60 80Let the cost of 1 kg of apples be x and that of 1 kg be y So the algebraic representation can be as follows: 2x+y=160 4x+2y=300⇒2x+y=150 The situation can be represented ... that the lines do not intersect anywhere, i.e. they are parallel. Hence we can not arrive at a solution.

What are trigonometric ratios? -Maths 10th

Description : What are trigonometric ratios? -Maths 10th

Last Answer : The trigonometric ratios of the angle A in the right triangle ABC are defined as follows: sine of ∠A = side opposite to angle Ahypotenuseside opposite to angle Ahypotenuse = BCACBCAC cosine of ∠A = side ... of angle A = ACABACAB cotangent of ∠A = 1tangent of angle A1tangent of angle A = ABBCABBC

If ax + by = a2 – b2 and bx + ay = 0, find the value of (x + y). -Maths 10th

Description : If ax + by = a2 – b2 and bx + ay = 0, find the value of (x + y). -Maths 10th

Last Answer : Given, ax+by=a2−b2......(1) bx+ay=0.......(2). Now adding (1) and (2) we get, x(a+b)x+(a+b)y=a2−b2 or, (a+b)(x+y)=a2−b2 or, x+y=a−b.