问题1.点A和B是70公里。分开在高速公路上。一辆车从A出发,另一辆车从B出发。如果他们朝同一个方向行驶,他们会在7小时内见面,但是如果彼此朝对方行驶,则他们会在一小时内见面。找到两辆车的速度。
解决方案:
Let’s assume that the car starting from point A as X and its speed as x km/hr.
and, the car starting from point B as Y and its speed as y km/hr.
There are two cases given in the question :
Case 1 : Car X and Y are moving in the same direction.
Case 2 : Car X and Y are moving in the opposite direction.
Let’s assume that the meeting point in case 1 as P and in case 2 as Q.
Case 1 : The distance travelled by car X = AP and,
the distance travelled by car Y = BP
As the time taken for both the cars to meet is 7 hours,
The distance travelled by car X in 7 hours = 7x km (As we know that distance = speed x time)
therefore AP = 7x
Similarly,
The distance travelled by car Y in 7 hours = 7y km
therefore BP = 7Y
As the cars are moving in the same direction (i.e. away from each other), we can write it as
AP – BP = AB
therefore, 7x – 7y = 70
x – y = 10 ——————(i)
Case 2 : In this case as it’s clearly seen that,
The distance travelled by car X = AQ and,
The distance travelled by car Y = BQ
As the time taken for both the cars to meet is 1 hour,
The distance travelled by car x in 1 hour = 1x km
therefore AQ = 1x
Similarly,
The distance travelled by car y in 1 hour = 1y km
therefore BQ = 1y
Now, since the cars are moving in the opposite direction (i.e. towards each other), we can write
AQ + BQ = AB
Hence x + y = 70 —————–(ii)
Hence, by solving eq. (i) and (ii), we get the required solution
From (i), we have x = 10 + y ——————-(iii)
Substituting this value of x in (ii) and we got,
(10 + y) + y = 70
y = 30
Now, put y = 30 in (iii), and we get
x = 40
Hence, Speed of car X = 40km/hr and Speed of car Y = 30 km/hr.
问题2.一个水手在40分钟内向下游行驶8公里,并在1小时内返回。确定水手在静止水中的速度和水流的速度。
解决方案:
Let’s assume that the speed of the sailor in still water as x km/hr and,
The speed of the current as y km/hr
As we know that,
Speed of the sailor in upstream = (x – y) km/hr
Speed of the sailor in downstream = (x + y) km/hr
therefore, time taken to cover 8 km upstream = 8/ (x – y) hr (As we know that distance = speed x time)
And, time taken to cover 8 km downstream = 8/ (x + y hr
It’s given that time taken to cover 8 km downstream in 40 minutes or, 40/ 60 hour or 2/3 hr.
8/ (x + y) = 2/3
8 × 3 = 2(x + y)
24 = 2x + 2y
x + y = 12 ——————(i)
Similarly, time taken to cover 8 km upstream in 1hour can be written as,
8/ (x – y) = 1
8 = 1(x – y)
x – y = 8 ——————–(ii)
Hence, by solving eq. (i) and (ii) we get the required solution
On adding (i) and (ii) we get,
2x = 20
x = 10
Now, putting the value of x in (i), we find y
10 + y = 12
y = 2
Hence, the speed of sailor is 10km/hr and the speed of the current is 2km/hr.
问题3.船在10小时内向上游行驶30公里,向下游行驶44公里。在13小时内,它可以上行40公里,下行55公里。确定溪流的速度以及在静止水中乘船的速度。
解决方案:
Let’s assume that the speed of the boat in still water as x km/hr and,
The speed of the stream as y km/hr
As we know that,
Speed of the boat in upstream = (x – y) km/hr and
Speed of the boat in downstream = (x + y) km/hr
therefore,
Time taken to cover 30 km upstream = 30/ (x − y) hr (As we know that distance = speed x time)
Time taken to cover 44 km downstream = 44/ (x + y) hr (As we know that distance = speed x time)
It’s given that the total time of journey is 10 hours. So, this can expressed as
30/ (x – y) + 44/ (x + y) = 10 —————-(i)
Similarly,
Time taken to cover 40 km upstream = 40/ (x – y) hr (As we know that distance = speed x time)
Time taken to cover 55 km downstream = 55/ (x + y) hr (As we know that distance = speed x time)
And for this case the total time of the journey is given as 13 hours.
Hence, we can write
40/(x – y) + 55/ (x + y) = 13 —————-(ii)
Hence, by solving (i) and (ii) we get the required solution
Taking, 1/ (x – y) = u and 1/ (x + y) = v in equations (i) and (ii) we have
30u + 44v – 10 = 0 —————–(iii)
40u + 55v – 13 = 0 ——————-(iv)
Solving these equations by cross multiplication we get,
u/(44x-13-55x-10) = -v/(30x-13-40x-10) = 1/(30×55 – 40×44)
u = 2/10
v = 1/11
Now,
1/ (x – y) = 2/10
1 x 10 = 2(x – y)
10 = 2x – 2y
x – y = 5 —————–(v)
And,
1/ (x + y) = 1/11
x + y = 11 —————(vi)
Again, solving (v) and (vi)
Adding (v) and (vi), we get
2x = 16
x = 8
Using x in (v), we find y
8 – y = 5
y = 3
Hence, the speed of the boat in still water is 8 km/hr and the speed of the stream is 3 km/hr.
问题4.一条船在6小时内向上游24公里,向下游28公里。它在6.5小时内向上游行驶30公里,向下游行驶21公里。查找在静止水中的船速以及溪流的速度。
解决方案:
Let’s assume that the speed of the boat in still water as x km/hr and,
The speed of the stream as y km/hr
As we know that,
Speed of the boat in upstream = (x – y) km/hr and
Speed of the boat in downstream = (x + y) km/hr
therefore,, time taken to cover 28 km downstream = 28/ (x+y) hr (As we know that distance = speed x time)
Time taken to cover 24 km upstream =24/ (x – y) hr (As we know that distance = speed x time)
It’s given that the total time of journey is 6 hours. So, this can be expressed as
24/ (x – y) + 28/ (x + y) = 6 —————-(i)
Similarly,
Time taken to cover 30 km upstream = 30/ (x − y) (As we know that distance = speed x time)
Time taken to cover 21km downstream =21/ (x + y) (As we know that distance = speed x time)
And for this case the total time of the journey is given as 6.5 i.e 13/2 hours.
Hence, we can write it as
30/ (x – y) + 21/ (x + y) = 13/2 —————-(ii)
Hence, by solving (i) and (ii) we get the required solution
Taking, 1/ (x – y) = u and 1/ (x + y) = v in equations (i) and (ii) and we have,
24u + 28v – 6 = 0 ————–(iii)
30u + 21v – 13/2 = 0 ————–(iv)
Solving these equations by cross multiplication we get,
u/(28x-6.5-21x-6) = -v/(24x-6.5-30x-6) = 1/(24×21 – 30×28)
u = 1/6 and v = 1/14
Now,
u = 1/ (x − y) = 1/ 6
x – y = 6 …. (v)
v = 1/ (x + y) = 1/ 14
x + y = 14 —————(vi)
On Solving (v) and (vi)
Adding (v) and (vi), we get
2x = 20
x = 10
Using x = 10 in (v), we find y
10 + y = 14
y = 4
Hence, Speed of the stream = 4km/hr and Speed of boat = 10km/hr.
问题5:一个人以一定的速度走一定的距离。如果他以每小时1/2公里的速度走,他的时间就会减少1小时。但是,如果他以每小时1公里的速度慢行,那么他还要多花3个小时。找到该人所覆盖的距离及其最初的行走速度。
解决方案:
Let’s assume that the actual speed of the man be x km/hr and y be the actual time taken by him in hours.
As we know that,
Distance covered = speed × distance
Distance = x × y = xy —————-(i)
Case I :
If the speed of the man increase by 1/2 km/hr, the journey time will reduce by 1 hour.
it can be written in the form of equation,
When speed is (x + 1/2) km/hr, time of journey = y – 1 hours
Now,
Distance covered = (x + 1/2) x (y – 1) km
Since the distance is the same i.e xy we can equate it, therefore
xy = (x + 1/2) x (y – 1)
And finally we get,
-2x + y – 1 = 0 ——————–(ii)
Case II : If the speed reduces by 1 km/hr then the time of journey increases by 3 hours.
When speed is (x-1) km/hr, time of journey is (y+3) hours
Since, the distance covered = xy
xy = (x-1)(y+3)
xy = xy – 1y + 3x – 3
xy = xy + 3x – 1y – 3
3x – y – 3 = 0 —————–(iii)
Add eq. (ii) and (iii), and we get
x – 4 = 0
x = 4
Now, y can be obtained by using x = 4 in (ii)
-2(4) + y – 1 = 0
y = 1 + 8 = 9
Hence, putting the value of x and y in equation (i), we find the distance
Distance covered = xy
= 4 × 9
= 36 km
Hence, the distance is 36 km and the speed of walking is 4 km/hr.
问题6:一个人在静止水中以5 km / h的速度划船,上游40 km和下游40 km所花费的时间是原来的三倍。查找流的速度。
解决方案:
Let’s assume x to be the speed of the stream.
As we know that,
Speed of boat in downstream = (5 + x) and,
Speed of boat in upstream = (5 – x)
It is given that,
The distance in one way is 40km. and,
Time taken during upstream = 3 × time taken during the downstream
Expressing it by equations, we have
40/ (5 – x) = 3 x 40/ (5 + x) (As we know that distance = speed x time)
By cross multiplication, we get
(5+x) = 3(5-x)
5 + x = 3(5 – x)
x + 3x = 15 – 5
x = 10/4 = 2.5
Hence, the speed of the stream is 2.5 km/hr.
问题7.拉梅什部分乘火车和汽车乘车到他家760公里。如果乘火车旅行160公里,然后乘汽车旅行,则需要8个小时。如果乘火车旅行240公里,然后乘汽车旅行,他要多花12分钟。分别找到火车和汽车的速度。
解决方案:
Let’s assume that the speed of the train be x km/hr and,
The speed of the car = y km/hr
From the question, it’s understood that there are two cases
Case 1 : When Ramesh travels 160 Km by train and the rest by car.
Case 2 : When Ramesh travels 240Km by train and the rest by car.
Case 1 : Time taken by Ramesh to travel 160 km by train = 160/x hrs (As we know that distance = speed x time)
Time taken by Ramesh to travel the remaining (760 – 160) km i.e., 600 km by car = 600/y hrs
therefore, the total time taken by Ramesh to cover 760Km = 160/x hrs + 600/y hrs
given that,
Total time taken for this journey = 8 hours
therefore, 160/x + 600/y = 8
20/x + 75/y = 1 —————–(i)
Case 2 : Time taken by Ramesh to travel 240 km by train = 240/x hrs,
Time taken by Ramesh to travel (760 – 240) = 520km by car = 520/y hrs
given that Ramesh will take a total of is 8 hours and 12 minutes to finish.
therefore, 240/x + 520/y = 8hrs 12mins = 8 + (12/60) = 41/5 hr
240/x + 520/y = 41/5
6/x + 13/y = 41/200 —————(ii)
Solving (i) and (ii), we get,
Let’s take 1/x = u and 1/y = v,
therefore, (i) and (ii) becomes,
20u + 75v = 1 —————(iii)
6u + 13v = 41/200 —————-(iv)
On multiplying (iii) by 3 and (iv) by 10, we get
60u + 225v = 3
60u + 130v = 41/20
Subtracting the above two equations, we get
(225 – 130)v = 3 – 41/20
95v = 19/ 20
v = 19/ (20 x 95) = 1/100
y = 1/v = 100
Using v = 1/100 in (iii) to find v,
20u + 75(1/100) = 1
20u = 1 – 75/100
20u = 25/100 = 1/4
u = 1/80
x = 1/u = 80
Therefore, the speed of the train is 80km/hr and the speed of car is100km/hr.
问题8.一名男子部分乘火车,部分乘汽车旅行600公里。如果他乘火车行驶400公里,其余乘车行驶,则需要6个小时30分钟。但是,如果他乘火车旅行200公里,其余乘汽车旅行,则需要花费半小时以上的时间。找到火车的速度和汽车的速度。
解决方案:
Let’s assume that the speed of the train be x km/hr and,
The speed of the car = y km/hr
From the question, it’s understood that there are two cases
Case 1 : When the man travels 400 km by train and the rest by car.
Case 2 : When Ramesh travels 200 km by train and the rest by car.
Case 1 : Time taken by the man to travel 400km by train = 400/x hrs (As we know that distance = speed x time)
Time taken by the man to travel (600 – 400) = 200km by car = 200/y hrs
Time taken by a man to cover 600km = 400/x hrs + 200/y hrs
Total time taken for this journey = 6 hours + 30 mins = 6 + 1/2 = 13/2
therefore,
400/x + 200/y = 13/2
400/x + 200/y = 13/2
400/x + 200/y = 13/2
200 (2/x + 1/y) = 13/2
2/x + 1/y = 13/400 —————-(i)
Case 2 : Time taken by the man to travel 200 km by train = 200/x hrs. (As we know that distance = speed x time)
Time taken by the man to travel (600 – 200) = 400km by car = 200/y hrs
For the part, the total time of the journey is given as 6hours 30 mins + 30 mins that is 7hrs,
200/x + 400/y = 7
200 (1/x + 2/y) = 7
1/x + 2/y = 7/200 ————-(ii)
Taking 1/x = u, and 1/y = v,
So, the equations (i) and (ii) becomes,
2u + v = 13/400 —————(iii)
u + 2v = 7/200 ————-(iv)
Solving (iii) and (iv), we get
3v = 14/200 – 13/400
3v = 1/400 x (28 – 13)
3v = 15/400
v = 1/80
y = 1/v = 80
Now, using v in (iii) we get u,
2u + (1/80) = 13/400
2u = 13/400 – 1/80
2u = 8/400
u = 1/100
x = 1/u = 100
Hence, the speed of the train is 100km/hr and the speed of the car is 80km/hr.
问题9. A和B的位置在高速公路上相距80公里。一辆汽车同时从A出发,从B出发。如果它们沿相同方向移动,则它们将在8个小时内相遇;如果它们沿相反方向移动,则它们将在1小时20分钟内相遇。查找汽车的速度。
解决方案:
Let’s consider the car starting from point A as X and its speed as x km/hr and,
the car starting from point B as Y and its speed as y km/hr.
From the question, it’s understood that there are two cases
Case 1 : Car X and Y are moving in the same direction
Case 2 : Car X and Y are moving in the opposite direction
Let’s assume that the meeting point in case 1 as P and in case 2 as Q.
Case 1 : The distance travelled by car X = AP and,
the distance travelled by car Y = BP.
As the time taken for both the cars to meet is 8 hours,
The distance travelled by car X in 7 hours = 8x km (As we know that distance = speed x time)
AP = 8x
Similarly, The distance travelled by car Y in 8 hours = 8y km
therefore, BP = 8Y
As the cars are moving in the same direction (i.e. away from each other), we can write it as
AP – BP = AB
therefore, 8x – 8y = 80
x – y = 10 ———————(i)
Case 2 : The distance travelled by car X = AQ and,
The distance travelled by car Y = BQ
As the time taken for both the cars to meet is 1 hour and 20 min, ⇒1 + (20/60) = 4/3 hr
The distance travelled by car x in 4/3 hour = 4x/3 km
AQ = 4x/3
Similarly, The distance travelled by car y in 4/3 hour = 4y/3 km
BQ = 4y/3
Now, since the cars are moving in the opposite direction (i.e. towards each other), we can write
AQ + BQ = AB
4x/3 + 4y/3 = 80
4x + 4y = 240
x + y = 60 ——————(ii)
Hence, by solving (i) and (ii), we get
From (i), we have x = 10 + y ——————-(iii)
Substituting this value of x in (ii).
(10 + y) + y = 60
2y = 50
y = 25
Now, using y = 30 in (iii), we get
x = 35
Hence, Speed of car X = 35 km/hr and Speed of car Y = 25 km/hr.
问题10.一条船在8小时内向上游12公里,向下游40公里。它可以同时上行16公里和下行32公里。求出在静水中的船速和溪流的速度。
解决方案:
Let’s assume that the speed of the boat in still water be 1 km/hr and the speed of the stream be r km/hr then
Speed upstream = (x – y) km/hr,
Sped down stream= (x + y) km/hr.
Now, Time taken to cover 12 km upstream = 12/(x-y) hrs,
Time taken to cover 40 km downstream = 40/(x+y) hrs.
But, total time of journey is 8 hours
12/(x-y) + 40/(x+y) = 8 —————–(i)
Time taken to cover 16 km upstream = 16/(x-y) hrs,
Time taken to cover 32 km downstream = 32/(x+y) hrs
In this case total time of journey is given to 8 hrs,
16/(x-y) + 32/(x+y) = 8 ————–(ii)
12u + 40 v = 8
16u + 32 v = 8
12u + 40v – 8 = 0 ————–(iii)
16u + 32v – 8 = 0 —————(iv)
Solving these equations by cross multiplication, we get
u = 1/4 and v = 1/8
Now,
u = x/(x-y)
1/(x-y) = 1/4
4 = x – y —————–(v)
and,
v = 1/(x+y)
1/(x+y) = 1/8
x + y = 8 —————(vi)
By solving equation (v) and (vi) we get,
x = 6
x + y = 8
6 + y = 8
y = 8 – 6
y = 2
Therefore, The speed of boat in still water is 6 km/hr. and the speed of the stream is 2 km/hr.
问题11:Roohi乘火车和公共汽车部分地到达她家300公里。如果乘火车旅行60公里,然后乘公共汽车旅行,则需要4个小时。如果她乘火车旅行100公里,其余乘公共汽车旅行,则需要花费10分钟以上的时间。分别找到火车和公共汽车的速度。
解决方案:
Let’s assume that the speed of the train be x km/hr that of the bus be y km/hr, we have the following cases,
Case 1 : When Roohi travels 300km by train and rest by bus,
Time taken by Roohi to travel 60km by train = 60/x hrs
Time taken by Roohi to travel (300 – 60) = 240 km by bus = 240/y hrs
Total time taken by Roohi to cover 300 km = 60/x + 240/y
It is given that total time taken in 4 hours therefore,
60/x + 240/y = 4
1/x + 4/y = 1/15 ————-(i)
Case 2 : When Roohi travels 100km by train and the rest by bus,
Time taken by Roohi to travel 100km by train = 100/x,
Time taken by Roohi to travel (300 – 100) = 200 km by bus = 200/y,
In this case total time of the journey is 4 hours 10 minutes
100/x + 200/y = 4 hrs 10 min
100/x + 200/y = 25/6
1/x + 2/y = 1/24 ————–(ii)
1u + 4v = 1/15 —————(iii)
1u + 2v = 1/24 —————–(iv)
Subtracting equation (iv) from (iii) we get,
v = 1/80
1u = (20-15)/300 = 1/60
u = 1/60 therefore,
x = 60
And,
v = 1/80
1/y = 1/80
Hence, The speed of the train is 60 km/hr. and the speed of the bus is 80 km/hr.
问题12. Ritu可以在2小时内向下游划20公里,在2小时内向上游划4公里。找到她在静水中划船的速度和水流的速度。
解决方案:
Let’s assume that the speed of rowing be ‘X’ km/hr and the speed of the current be ‘Y’ km/hr,
Speed upstream = (x – y) km/hr.
Speed downstream = (x + y) km/hr.
Now, Time taken to cover 20 km downstream = 20/(x+y) hrs
Time taken to cover 4 km upstream = 4/(x-y) hrs
But, time taken to cover 20 km downstream in 2 hours,
20/(x+y) = 2
20 = 2x + 2y —————(i)
Time Taken to cover 20 km downstream in 2 hours
20/(x+y) = 2
20 = 2(x + y)
20 = 2x + 2y ——————-(ii)
By solving (i) and (ii) we get,
2y = 8
y = 4
Hence, Speed of rowing in still water is 6 km/hr. and speed of current is 4 km/hr.
问题13.母船可以在7个小时内向上游行驶30公里,向下游行驶28公里。它可以向上游行驶21公里,并在5个小时内返回。求出在静水中的船速和溪流的速度。
解决方案:
Let’s assume that the speed of A and B be X km/hr and Y km/hr respectively. Then,
Time taken by A to cover 30km = 30/x hrs,
and, time taken by B to cover 30km = 30/y hrs.
given that,
30/x – 30/y = 3
(10/x – 10/y) = 1 —————(i)
If A doubles his pace, then speed of A is 2x km/hr.
Time taken by A to cover 30 km = 30/2x hrs,
Time taken by B to cover 30 km = 30/y hrs,
According to given condition we have,
30/y – 30/2x = 3/2
10u – 10v = 1
10u – 10v = 0 —————–(iii)
-10u + 20v = 1
-10u + 20v – 1 = 0 ——————(iv)
Adding equations (iii) and (iv), we get,
v = 1/5
Putting v = 1/5 in equation (iii), we get,
10u – 10v – 1 = 0
10u – 3 = 0
u = 3/10
1/x = 3/10
x = 10/3
and, v = 1/5
1/y = 1/5
y = 5
Hence, The boat A’s speed is 10/3 km/hr. and boat B’s speed is 5 km/hr.
问题14.阿卜杜勒乘火车旅行300公里,乘出租车旅行200公里,他花了5个小时30分钟。但是,如果他乘火车旅行260公里,乘出租车旅行240公里,则需要花费6分钟多的时间。找到火车和出租车的速度。
解决方案:
Let’s assume that the speed of the train be x km/hour that of the taxi be y km/hr, we have the following cases
Case I : When Abdul travels 300 Km by train and the 200 Km by taxi
Time taken by Abdul to travel 300 Km by train = 300/x hrs,
Time taken by Abdul to travel 200 Km by taxi = 200/y,
Total time taken by Abdul to cover 500 Km = 300/x + 200/y
It is given that total time taken in 5 hours 30 minutes
300/x + 200/y = 5 hours 30 minutes
100(3/x + 2/y) = 5 30⁄60 (3/x + 2/y) = 5 1⁄2
100(3/x + 2/y) = 1/2 x 1/100
3/x + 2/y = 11/200 —————-(i)
Case II : When Abdul travels 260 km by train and the 240 km by taxi
Time taken by abdul to travel 260 km by train = 260/x hrs
Time taken by Abdul to travel 240 km by taxi = 240/y hrs
In this case, total time of the journey is 5 hours 36 minutes.
260/x + 240/y = 5 hours 36 minutes
260/x + 240/y = 5 36⁄60
260/x + 240/y = 5 6⁄10
260/x + 240/y = 5 3⁄5
20(13/x + 12/y) = 28/5
(13/x) = 28/5 × 1/20
13/x + 12/y = 7/25 —————–(ii)
Putting 1/x = u and 1/y = v, the equations (i) and (ii) reduces to
3u + 2v = 11/200 —————(iii)
13u + 12v = 7/25 —————-(iv)
Multiplying eq. (iii) by 6, the above equation becomes
18u + 12v = 33/100 —————-(v)
Subtracting equation (iv) from (v), we get
5u = 33/100 – 7/25
5u = 33/100 – 28/100
5u = 5/100
u = 1/100
Putting u = 1/100 in equation (iii), we get
3u + 2v = 11/200
3 x 1/100 + 2v = 11/200
3/100 + 2v = 11/200
2v = 5/200
v = 1/2
v = 1/80
Now,
u = 1/100
x = 100
And,
v = 1/80
1/y = 1/80
y = 80
Hence, the speed of the train is 100 km/hr and the speed of the taxi is 80 km/hr.
问题15:一列火车以均匀的速度驶过一定的距离。如果火车可以快10公里/小时,那么它将比计划的时间少2小时,并且火车慢10公里/小时。这将比计划的时间多花费3个小时。找到火车覆盖的距离。
解决方案:
Let’s assume that the actual speed of the train be x km/hr and the actual time is taken by y hours. Then,
Distance = Speed x Time
Distance covered = (xy) km ——————-(i)
If the speed is increased by 10 Km I hr, then the time of journey is reduced by 2 hours.
When speed is (x + 10) km I hr, time of journey is (y – 2) hours.
Therefore, Distance covered = (x + 10) (y – 2)
xy = (x + 10)(y – 2)
xy = xy + 10y – 2x – 20
-2x + 10y – 20 = 0
-2x + 3y – 12 = 0 ——————-(ii)
When the speed is reduced by 10 Km/hr, then the time of journey is increased by 3 hours.
When speed is (x -10) Km/h, time of journey is (y + 3) hours.
Hence, Distance covered = (x – 10) (y + 3)
xy = (x – 10)(y + 3)
0 = – 10y + 3x – 30
3x – 10y – 30 = 0 … (iii)
Thus, we obtain the following system of equations: = x + 5y – 10 = 0
x + 5y – 10 = 0
3x – 10y – 30 = 0
By using cross-multiplication, we get
x/(5x-30-(-10)x-10) = -y/(-1 x -30)-(3x-10) = 1/(-1 x -10)-(3 x 5)
x/-250 = -y/60 = 1/-5
x = 50
y = 12
Putting the values of x and y in equation (i), we get
Distance = xy km
= 50 × 12 = 600 km
Hence, the length of the journey is 600 km.
问题16. A和B地点在高速公路上相距100公里。一辆车同时从A出发,另一辆车从B出发。如果汽车以相同的方向以不同的速度行驶,它们将在5个小时内相遇。如果他们彼此相遇,他们会在1小时内见面。两辆车的速度是多少?
解决方案:
Let’s assume that x and y be two cars starting from points A and B.
Let’s the speed of the car X be x km/hr and that of the car Y be y km/hr.
Case 1 : When two cars move in the same directions: Suppose two cars meet at point Q then,
Distance travelled by car X = AQ,
Distance travelled by car Y = BQ,
Given that two cars meet in 5 hours.
Distance travelled by car X in 5 hours = 5x km AQ = 5x and
Distance travelled by car Y in 5 hours = 5y km BQ = 5y
therefore, AQ – BQ = AB
5x – 5y = 100
Both sides divided by 5, and we get x – y = 20 —————–(i)
Case 2 : When two cars move in opposite direction
Suppose two cars meet at point P then,
Distance travelled by X car X = AP and
Distance travelled by Y car Y = BP
In this case it is given that two cars meet in 1 hour, therefore
Distance traveled by car y in hours = lx km and,
Distance traveled by car y in 1 hours = ly km
therefore AP + BP = AB
1x + 1y = 100
x + y = 100 ————–(ii)
By solving eq. (i) and (ii) we get x = 60,
Substituting x = 60 in equation, we get,
x + y = 100
60 + y = 100
Y = 40
Hence, The speed of car X is 60 km/hr. and speed of car Y = 40 km/hr.