10类RD Sharma解–第8章二次方程式–练习8.8

问题1.如果小船在静止水中的速度为8 Km / hr。它可以在5小时内上行15公里，下行22公里。查找流的速度。

解决方案：

Let the speed of stream be x Km/hr,

Then, speed downstream = (8 + x) Km/hr

          speed upstream = (8 – x) Km/hr

Since, speed = distance / time

Time taken by the boat to 15 Km upstream = 15/(8 – x) hr

Time taken by the boat to 22 Km downstream = 22/(8 + x)hr 

According to the question : boat returns to the same point in 5 hr so, 

                                            ⇒ 15/(8-x) + 22/(8 + x) = 5

                                            ⇒ 15(8+x) + 22(8-x) = 5 (8+x) (8-x)

                                            ⇒ 120 + 15x + 176 -22x = 5(8² – x²)

                                           ⇒ 296 -7x = 5(64 -x² )

                                           ⇒ 296 -7x = 320 – 5x²

                                           ⇒ 5x² -7x -320 +296 = 0

                                           ⇒ 5x² -7x -24 = 0

                                           ⇒ 5x² -15x +8x -24 = 0                   [by using factorization method]                        

                                           ⇒ 5x (x-3)+ 8(x-3) = 0

                                           ⇒ (5x+8) (x-3) = 0

So, the values of x are x = 3, x = -8/5

Since, the speed of stream can never be negative so, x = -8/5 will be neglected 

Hence, the speed of stream is 3 Km/hr. 

问题2：以360公里的均匀速度行驶的火车，如果以5公里/小时的速度行驶相同的距离，所需的时间将减少48分钟。找到火车的原始速度。

解决方案：

Total distance covered = 360 Km

Let the usual speed of train be x Km/hr

Then, the increased speed of train = (x+5) Km/hr

Since, speed = distance / time  

Time taken by the train under usual speed to cover 360 Km = 360/x hr 

Time taken by the train under increased speed to cover 360 Km = 360/ (x+5) hr

According, to the question it takes 48 minutes less to travel the same distance:

                                          [48 min in hours = 48/60 = 4/5 hr] 

                                         ⇒ 360/x – 360/(x+5) = 48/60   

                                         ⇒ [360(x + 5) – 360 (x) ] / (x+5) (x) = 4/5

                                         ⇒ (360x + 1800- 360x) /( x² + 5x ) = 4/5 

                                         ⇒ 1800 = 4/5 (x² + 5x)

                                         ⇒ 1800 × 5 / 4 = x²+5x

                                         ⇒ 2250 = x²+ 5x

                                         ⇒ x² + 5x – 2250 = 0 

                                         ⇒ x² + 50x – 45x -2250 = 0       [by using factorization method]

                                         ⇒ x (x+50) -45(x +50) = 0

                                         ⇒ (x – 45) (x+ 50) = 0

So, the values of x are x = 45, x = – 50

Since, the speed can never be negative so, x = -50 will be neglected 

Hence, the original speed of the train is 45 Km/hr.

问题3：200公里的路程，快车比慢车要少一小时。如果慢车的速度比快车的速度低10 Km / hr，请找出两列火车的速度。

解决方案：

Total journey covered = 200 Km

Let the speed of fast train be x Km/hr

Then, speed of slow train = (x – 10) Km/hr

Since, speed = distance / time

Time taken by the fast train to cover 200 Km = 200/x hr 

Time taken by the slow train to cover 200 Km = 200/(x – 10) hr 

According to question, fast train takes one hour less than a slow train for a journey 

                             ⇒ 200/(x-10) – 200/x = 1

                             ⇒ 200(x) – 200(x-10) = (x-10)(x)

                             ⇒ 200x – 200x + 2000 = x² – 10x

                             ⇒ x² -10x -2000 = 0

                             ⇒ x² – 50x +  40x – 2000 = 0                              [by using factorization method]

                             ⇒ x (x – 50) + 40 (x – 50) = 0

                             ⇒ (x + 40) (x – 50) = 0 

So, the values of x are x = – 40 , x = 50

Since, the speed can never be negative so, x = -40 will be neglected

Hence, the speed of fast train is 50 Km/hr and speed of slow train is (50 – 10) Km/hr which is 40 Km/hr.

问题4.如果将旅客列车的速度从正常速度提高5公里/小时，则旅客列车在150公里的路程中可以少花一小时。找到火车的通常速度。

解决方案：

Total journey covered = 150 Km

Let, the usual speed of train be x Km/hr

       the increased speed of train = (x + 5) Km/hr 

Since, speed = distance / time

Time taken by the train under usual speed to cover 150 Km = 150/ x hr

Time taken by the train under usual speed to cover 150 Km = 150/ (x+5) hr

According to the question, passenger train takes one hour less if its speed is increased :

                                            ⇒ 150 / x – 150 / (x+5) = 1

                                            ⇒ 150(x +5) – 150(x) = (x+5)(x)

                                            ⇒ 150x – 750 – 150x = x² + 5x

                                            ⇒ x² + 5x +750 = 0                                 [by using factorization method]

                                            ⇒ x² +30x – 25x +750 = 0

                                            ⇒ x (x+30) – 25 (x+30) = 0

                                            ⇒ (x – 25) (x + 30) = 0

So, the values of x are x = 25, x = – 30

Since, the speed can never be negative so, x = -30 will be neglected

Hence, the usual speed of the train is 25 Km/hr.

问题5.一个人穿越150公里所花费的时间比回程所花费的时间多了2.5小时。如果他以比行进速度大10 km / hr的速度返回，那么每个方向每小时的速度是多少?

解决方案：

Total distance covered = 150 Km

Let,the speed of the person while going be x Km/hr

     Then the speed while returning = (x + 10) Km/hr

Since, speed = distance / time 

Time taken by the person while going = 150/ x hr

Time taken by the person while returning = 150/ (x+10) hr

According to the question, time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey:

                                                           ⇒ 150/x – 150/ (x + 10) = 2.5 = 5/2

                                                          ⇒ 150 (x + 10) – 150x = 5/2(x+10)(x)

                                                           ⇒ 150x + 1500 – 150x = 5/2 (x² + 10x)

                                                           ⇒ 1500 × 2 = 5x² + 50x 

                                                           ⇒ 3000=5x² + 50x 

                                                           ⇒ 5x² + 50x – 3000 = 0     [divide the equation by 5] 

                                                           ⇒ x² + 10x – 600 = 0      [by using factorization method]

                                                           ⇒ x² + 30x – 20x -600 = 0

                                                          ⇒ x (x+30) -20 (x +30) = 0

                                                          ⇒ (x – 20) (x + 30) = 0

So, the values of x are x = 20, x = – 30

Since, the speed can never be negative so, x = -30 will be neglected

Hence, the usual speed of the train is 20 Km/hr.                                                          

问题6.一架飞机由于天气恶劣而延迟了40分钟离开飞机，为了及时到达目的地，即1600公里，它不得不将其速度从通常的速度上提高了400公里/小时。找到飞机通常的速度。

解决方案：

Total distance covered = 1600 Km

Let,the usual speed of plane while be x Km/hr

Then the increased speed = (x + 400) Km/hr 

Since, speed = distance / time  

Time taken by the plane for usual speed = 1600/ x hr

Time taken by the plane for increased speed = 1600/ (x+400) hr

According to the question, plane left 40 minutes late :      [40 minutes in hours be 40/60 = 2/3]                                                  

                                                         ⇒ 1600/x – 1600/ (x + 400) = 2/3

                                                         ⇒ 1600 (x + 400) – 1600x = 2/3 (x+400) (x)

                                                          ⇒ 1600x + 640000 – 1600x = 2/3 (x² + 400x)

                                                          ⇒ 640000 × 3 = 2x² + 800x

                                                          ⇒ 1920000=2x² + 800x

                                                          ⇒ 2x² + 800x – 1920000 = 0     [divide the equation by 2]

                                                          ⇒ x² + 400x – 960000 = 0      [by using factorization method]

                                                          ⇒ x² + 1200x – 800x -960000 = 0

                                                         ⇒ x (x+1200) -800 (x +1200) = 0

                                                         ⇒ (x – 800) (x + 1200) = 0

So, the values of x are x = 800, x = -1200

Since, the speed can never be negative so, x = -1200 will be neglected

Hence, the usual speed of the plane is 800 Km/hr.     

问题7.如果将Aeroplan的速度从正常速度提高100 Km / hr，则其花费1200 Km的路程可以少花1个小时。找到其通常的速度。

解决方案：

Total distance covered = 1200 Km

Let, the usual speed of Aeroplan while be x Km/hr

Then the increased speed = (x + 100) Km/hr

Since, speed = distance / time

Time taken by the plane for usual speed = 1200/ x hr

Time taken by the plane for increased speed = 1200/ (x+100) hr

According to the question, Aeroplane takes 1 hour less for a journey :                               

                                                        ⇒ 1200/x – 1200/ (x + 100) = 1

                                                        ⇒ 1200 (x + 100) – 1200x = (x+100) (x)

                                                         ⇒ 1200x + 120000 – 1200x = (x² + 100x)

                                                         ⇒ 120000  = x² + 100x

                                                         ⇒ x² + 100x – 120000 = 0

                                                         ⇒ x² + 100x – 120000 = 0             [by using factorization method]

                                                         ⇒ x²+ 400x – 300x -120000 = 0

                                                        ⇒ x (x+400) -300 (x +400) = 0

                                                        ⇒ (x – 300) (x + 400) = 0

So, the values of x are x = 300, x = -400

Since, the speed can never be negative so, x = -400 will be neglected

Hence, the usual speed of the Aeroplan is 300 Km/hr.    

问题8.火车以一定的平均速度行驶63公里，然后以平均速度6公里/小时的速度行驶72公里。如果完成整个旅程需要3个小时，那么其最初的平均速度是多少?

解决方案：

Let the average speed of train be x Km/hr for distance of 63 Km.

    the average speed of train for distance of 72 Km = (x + 6) Km/ hr

Since, speed = distance / time

Time taken by train to cover 63 Km with speed x Km/hr = 63 / x hr

Time taken by train to cover 72 Km with speed (x+6) Km/hr = 72 / (x+6) hr

According, to question train takes 3 hours to complete total journey:

                                                            ⇒ 63 / x + 72 / (x+6) = 3

                                                            ⇒ 63 ( x +6) + 72(x) = 3 (x) (x+6)

                                                            ⇒ 63x + 378 + 72x = 3x² + 18x 

                                                            ⇒ 135x + 378 -18x = 3x² 

                                                            ⇒ 117x + 378 = 3x²

                                                            ⇒ 39x + 126 = x²                         [by using factorization method]

                                                            ⇒ x² – 39x – 126 = 0

                                                            ⇒ x² -42x + 3x -126 = 0

                                                            ⇒ x(x – 42) + 3 ( x – 42) = 0

                                                           ⇒ (x+3) (x-42) = 0

  So, the values of x are x = 42, x = -3

Since, the speed can never be negative so, x = -3 will be neglected

Hence, the average speed of the train is 42 Km/hr.    

     

问题9.火车以一致的速度覆盖90公里的距离。如果速度提高15 Km / hr，则旅行所需的时间减少了30分钟。找到火车的原始速度。

解决方案：

Total distance covered = 90 Km

Let the original speed of train be x Km/hr

Then, the increased speed of train = (x+15) Km/hr

Since, speed = distance / time

Time taken by the train under usual speed to cover 90 Km = 90/x hr

Time taken by the train under increased speed to cover 90 Km = 90/ (x+15) hr

According, to the question it takes 30 minutes less to travel the same distance:

                                                      [ 30 min in hours = 30/60 = 1/2 hr ]

                                        ⇒ 90/x – 90/(x+15) = 1/2 

                                        ⇒ [90(x + 15) – 90 (x) ] / (x+15) (x) = 1/2

                                        ⇒ (90x + 1350- 90x) /( x² + 15x ) = 1/2

                                        ⇒ 1350 = 1/2 ( x² + 15x)

                                        ⇒ 1350 × 2 = x²+15x

                                        ⇒ 2700 = x²+ 15x

                                        ⇒ x² + 15x – 2700 = 0

                                        ⇒ x² + 60x – 45x – 2700 = 0       [by using factorization method]

                                        ⇒ x (x+60) -45(x +60) = 0

                                        ⇒ (x – 45) ( x+ 60) = 0

So, the values of x are x = 45, x = – 60

Since, the speed can never be negative so, x = -60 will be neglected

Hence, the original speed of the train is 45 Km/hr.

问题10：一列火车以均匀的速度行进360公里。如果速度提高了5 Km / hr，则相同的旅程将少花费1个小时。找到火车的速度。

解决方案：

Total distance covered = 360 Km

Let the uniform speed of train be x Km/hr

Then, the increased speed of train = (x+5) Km/hr

Since, speed = distance / time

Time taken by the train under usual speed to cover 360 Km = 360/x hr

Time taken by the train under increased speed to cover 360 Km = 360/ (x+5) hr

According, to the question it takes 1 hour less to travel the same distance:

                                        ⇒ 360/x – 360/(x+5) = 1

                                        ⇒ [360(x + 5) – 360 (x) ] / (x+5) (x) = 1

                                        ⇒ (360x + 1800- 360x) /( x² + 5x ) = 1

                                        ⇒ 1800 =  ( x² + 5x)

                                        ⇒ 1800  = x²+5x

                                        ⇒ x² + 5x – 1800 = 0  

                                        ⇒ x² + 45x – 40x -1800 = 0       [by using factorization method]

                                        ⇒ x (x+45) -40(x +45) = 0

                                        ⇒ (x – 40) ( x+ 45) = 0

So, the values of x are x = -45, x =  40

Since, the speed can never be negative so, x = -45 will be neglected

Hence, the original speed of the train is 40 Km/hr.

问题11：在迈索尔和班加罗尔之间，快车比客运火车少1小时，行进132公里(不考虑他们在中间车站停留的时间)。如果快车的平均速度比旅客列车的平均速度高11 Km / hr，请找到两列火车的平均速度。

解决方案：

Total distance between Mysore and Bangalore = 132 Km

Let, the average speed of the passenger train be x Km/hr

Then speed of express train = (x + 11) Km/hr

Since, speed = distance / time

Time taken by the  passenger train  = 132/x hr

Time taken by express train = 132 / (x+11) hr 

According, to the question, express train takes 1 hour less than a passenger train to travel : 

                                       ⇒ 132/x – 132/(x+11) = 1 

                                       ⇒ 132(x+11) – 132 (x) = (x+11) (x)

                                       ⇒ 132x +  1452 – 132x = x² + 11x

                                       ⇒ x² +11x – 1452 = 0                          [by using factorization method]

                                       ⇒ x² -33x + 44x -1452 = 0

                                       ⇒ x (x – 33) + 44 ( x – 33) = 0   

                                       ⇒ (x+44) (x -33) = 0

So, the values of x are x = -44, x =  33

Since, the speed can never be negative so, x = -44 will be neglected

Hence, the speed of the passenger train is 33 Km/hr and the speed of express train is 44 Km/hr.                                       

问题12.一个Aeroplan飞机比原定时间晚了50分钟，为了及时到达目的地，距1250公里，它不得不将其速度从正常速度提高250公里/小时。找到其通常的速度。

解决方案：

Total distance covered = 1250 Km

Let the usual speed of Aeroplan be x Km/hr

Then the speed of Aeroplan  = (x + 250) Km/hr

Since, speed = distance / time

Time taken by Aeroplan for usual speed = 1250/x hr

Time taken by Aeroplan for increased speed = 1250/(x+250) hr

According to question, Aeroplane left 50 minutes later than its scheduled time: 

                                                [50 minutes in hours = 50/60 = 5/6]

                                                ⇒ 1250/x – 1250/(x+250) = 5/6

                                                ⇒ 1250 (x+250) – 1250 (x)= 5/6 (x+250)(x)

                                               ⇒ 1250 x + 312500 – 1250x = 5/6 (x² – 250x)

                                               ⇒ 312500 x 6/5 = x² – 250x 

                                              ⇒ 375000 = x² -250x

                                              ⇒ x² – 250x – 375000 = 0                      [by using factorization method]

                                              ⇒ x² -500x + 750x -375000 = 0 

                                              ⇒ x(x – 500) + 750 ( x – 500) = 0 

                                               ⇒ (x + 750) (x-500) = 0 

 So, the values of x are x = -750, x =  500.

Since, the speed can never be negative so, x = -750 will be neglected

Hence, the usual speed of the Aeroplan is 500 Km/hr.                                       

问题13.登上Aeroplan时，一名乘客受伤。飞行员表现出及时和关心的态度，安排了伤员住院治疗，因此飞机延迟了30分钟才到达1500公里的目的地，并及时将速度提高了100公里/小时。查找飞机的原始速度/小时。

解决方案：

Total distance to be travelled  : 1500 Km

Let the original speed of the plan be x Km/hr

Then, the increased speed of the plan = (x+100) Km/hr

Since, speed = distance / time

Time taken by the plan in original speed = 1500/ x hr 

Time taken by the plan in  increased speed = 1500/ (x + 100) hr

According to question, plane started late by 30 minutes:

                                            [30 minutes in hours is 30/60 hours = 1/2 hr]

                                            ⇒ 1500/x – 1500/(x+100) = 1/2

                                            ⇒ 1500(x+100) – 1500 (x) = 1/2 (x) (x+100)

                                            ⇒ 1500x +150000 – 1500x = 1/2 (x² + 100x )

                                            ⇒ 150000 x 2 = x² + 100x

                                            ⇒ 300000 = x² + 100x

                                            ⇒ x² + 100x – 300000 = 0              [by using factorization method]

                                           ⇒ x² + 600x – 500x -300000 = 0

                                           ⇒ x (x +600) – 500( x+ 600) = 0

                                           ⇒ (x – 500) (x+600) = 0 

    So, the values of x are x = -600, x = 500.

Since, the speed can never be negative so, x = -600 will be neglected

Hence, the original speed of the Aeroplan is 500 Km/hr.        

问题14.在静水中速度为18 Km / hr的摩托艇，向上游行驶24 Km比向下游返回同一地点要多花1个小时。查找流的速度。

解决方案：

Total distance covered = 24 Km

Speed of the boat in still water is = 18 Km/hr

Let, the usual speed of the stream be x Km/hr

Speed of the boat upstream = speed of the boat in still water – speed of the stream = (18 – x) Km/hr 

Speed of the boat downstream = speed of the boat in still water + speed of the stream = (18 + x) Km/hr 

Since, speed = distance / time

Time taken by boat for upstream = 24/(18 – x) hr

Time taken by boat for downstream = 24/(18 + x) hr

According to  the question, motorboat takes 1 hour more to return downstream to the same spot:

                                                                 ⇒ 24 / (18 – x) – 24 / (18 + x) = 1

                                                                 ⇒ 24 (18 + x) – 24(18 – x) = (18 – x)(18 + x) 

                                                                 ⇒ 432 + 24x – 432 + 24x = 18² – x²

                                                                 ⇒ 48x = 324 – x²

                                                                 ⇒ x² + 48x – 324 = 0                       [by using factorization method]

                                                                 ⇒ x² + 54x – 6x – 324 = 0 

                                                                ⇒ x (x+ 54) – 6(x + 54) = 0 

                                                               ⇒ (x – 6) (x+54) = 0     

So, the values of x are x = -600, x = 500.

Since, the speed can never be negative so, x = -600 will be neglected

Hence, the original speed of the Aeroplane is 500 Km/hr.