Time and distance. Numerical Reasoning. Test 03

If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is
Let the actual distance travelled be x km.
Then x/10=(x+20)/14
--> 14x = 10x + 200
--> 4x = 200
--> x = 50 km.
A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
0.5x/21 + 0.5x/24 = 10
--> x/21 + x/24 = 20
--> 15x = 168 x 20
--> x = (168 x 20)/15 = 224 km.
A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:
Total time taken = 160/64 + 160/80 = 9/2 hours
--> Average speed = 320 x 2/9 = 71.11 km/hr.
In covering a distance of 30 km, A takes 2 hours more than B. If Abhay doubles his speed, then he would take 1 hour less than B. A's speed is:
Let A's speed be x km/hr.
Then, 30/x -30/2x = 3
--> 6x = 30
--> x = 5 km/hr.
It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, 120/x + 480/y = 8 --> 1/x + 4/y = 1/15 ------ (i)
Also, 200/x + 400/y = 25/3 --> 1/x + 2/y = 1/24 ----(ii)
Solving (i) and (ii), we get: x = 60 and y = 80.
--> Ratio of speeds = 60 : 80 = 3 : 4.
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is
Let distance = x km and usual rate = y kmph.
Then, x/y - x/(y+3) = 40/60 --> 2y (y+3) = 9x ----- (i)
Also, x/(y-2) - x/y = 40/60 --> y(y-2) = 3x -------- (ii)
On dividing (i) by (ii), we get: x = 40.
A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
Let speed of the car be x kmph.
Then, speed of the train=(150/100)*x=(3/2)*x kmph
(75/x)-(75/(3/2x))=125/(10*60)
(75/x)-(50/x)=5/24
x = 25x24/5 = 120kmph.
In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is
Let the duration of the flight be x hours.
Then, 600/x - 600/0.5x = 200
--> 600/x - 1200/(2x+1) = 200
--> x(2x + 1) = 3
--> 2x^2 + x - 3 = 0
--> (2x + 3)(x - 1) = 0
--> x = 1 hr.
The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 kms in 4 hours, then the speed of the first train is:
Let the speed of two trains be 7x and 8x km/hr.
Then, 8x = 400/4 = 100
Let the speed of two trains be 7x and 8x km/hr.
--> x = 100/8 = 12.5
--> Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.
A car travelling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.
Time taken = 1 hr 40 min 48 sec = 1 hr 40 4/5 min. = 126/75 hrs.
Let the actual speed be x km/hr.
Then, (5/7)x X 126/75 = 42
--> x = 35 km/hr.

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