55924.A man can reach a certain place in 40 hours. If he reduces his speed by 1/15th, he goes 5 km less in that time. Find the total distance covered by him.
60 km.
85 km.
75 km.
52 km.
Explanation:
Let Distance = D km, Speed = S km/h
Time = Distance/Speed
40 = Distance/Speed .... (1)
After reducing speed,
Distance = Distance – 5 km, Speed = S × 14/15 km/h
Time = Distance/Speed
40 = $\dfrac{(Distance – 5)}{Speed ×14/15}$ .... (2)
After solving equation 1 and 2
$\dfrac{Distance}{Speed} $ = $\dfrac{(Distance – 5)}{Speed ×14/15}$
14 Distance = 15 (Distance – 5)
14 Distance = 15 Distance – 75
Distance = 75 km.
Let Distance = D km, Speed = S km/h
Time = Distance/Speed
40 = Distance/Speed .... (1)
After reducing speed,
Distance = Distance – 5 km, Speed = S × 14/15 km/h
Time = Distance/Speed
40 = $\dfrac{(Distance – 5)}{Speed ×14/15}$ .... (2)
After solving equation 1 and 2
$\dfrac{Distance}{Speed} $ = $\dfrac{(Distance – 5)}{Speed ×14/15}$
14 Distance = 15 (Distance – 5)
14 Distance = 15 Distance – 75
Distance = 75 km.
55925.A person travelled 132 km by auto, 852 km by train and 248 km by bike. It took 21 hours in all. If the speed of train is 6 times the speed of auto and 1.5 times speed of bike, what is the speed of train?
78 $kmh^{-1}$
104 $kmh^{-1}$
96 $kmh^{-1}$
88 $kmh^{-1}$
Explanation:
Let the speed of auto be x kmph-1 . So, the speed of train will be 6x and that of bike will be
= $\dfrac{6x}{1.5}= 4x$
As per the given information,
Time taken by auto + Time taken by train + Time taken by bike = 21 hours
$\dfrac{132}{x}$+$\dfrac{852}{6x}$+$\dfrac{248}{4x}$= 21 or
$\dfrac{132}{x} $+$\dfrac{142}{x}$+$\dfrac{62}{x}$= 21
or, 21x = 132 + 142 + 62 = 336
∴ x = 336/21 = 16
∴ Speed of the train = 6x = 6 × 16 = 96$kmh^{-1}$
Let the speed of auto be x kmph-1 . So, the speed of train will be 6x and that of bike will be
= $\dfrac{6x}{1.5}= 4x$
As per the given information,
Time taken by auto + Time taken by train + Time taken by bike = 21 hours
$\dfrac{132}{x}$+$\dfrac{852}{6x}$+$\dfrac{248}{4x}$= 21 or
$\dfrac{132}{x} $+$\dfrac{142}{x}$+$\dfrac{62}{x}$= 21
or, 21x = 132 + 142 + 62 = 336
∴ x = 336/21 = 16
∴ Speed of the train = 6x = 6 × 16 = 96$kmh^{-1}$
55926.Cars take 15 hr to cover the distance of 150 km between P and Q. A bus starts from point P at 6.00 a.m. and another bus starts from point Q at 9.00 a.m. on the same day. When do the two cars meet?
3 PM
2.50 PM
1.30 PM
12 PM
Explanation:
Distance 150 km.
Speed of the cars =150/15=10km/hr
By 9.00 a.m., the car from P covered 30 km.
Remaining distance = 150 – 30 = 120 km
Relative speed of the cars = 20 km/hr.
Time taken to meet = 120/20=6 hr after Q starts
ie 3pm meeting time.
Distance 150 km.
Speed of the cars =150/15=10km/hr
By 9.00 a.m., the car from P covered 30 km.
Remaining distance = 150 – 30 = 120 km
Relative speed of the cars = 20 km/hr.
Time taken to meet = 120/20=6 hr after Q starts
ie 3pm meeting time.
55927.A bike rider starts at 60 km/hr and he increases his speed in every 2 hours by 3 km/hr. Then the maximum distance covered by him in 24 hours is:
1000km
918km
899 km
none of these
Explanation:
Speed of the rider = 60km/hr.
Distance covered in 1st 2 hours = 60 km.
He increased his speed in every 2 hours by 3 km/hr.
Distance covered in every 2 hours will be, 60, 63, 66,... upto 12 terms.(for 24 hours).
The above series is an A.P series;
Sum of first n terms = (n/2)(2a+(n-1)d)
Here, a = 60, d = 3 and n = 12.
Sum of first 12 terms = (12/2)(2(60)+(11)3) = 6(120 + 33) = 6(153) = 918.
Hence, he covers 918 km in 24 hours.
Speed of the rider = 60km/hr.
Distance covered in 1st 2 hours = 60 km.
He increased his speed in every 2 hours by 3 km/hr.
Distance covered in every 2 hours will be, 60, 63, 66,... upto 12 terms.(for 24 hours).
The above series is an A.P series;
Sum of first n terms = (n/2)(2a+(n-1)d)
Here, a = 60, d = 3 and n = 12.
Sum of first 12 terms = (12/2)(2(60)+(11)3) = 6(120 + 33) = 6(153) = 918.
Hence, he covers 918 km in 24 hours.
55928.A car starts at 10 am with a speed of 50 km/hr. Due to the problem in engine it reduces its speed as 10 km/hr for every 2 hours. After 11 am, the time taken to covers 10 km is:
12 minutes and 10 seconds
15 minutes and 09 seconds
13 minutes and 20 seconds
none of these
Explanation:
Initial speed of the car = 50km/hr
Due to engine problem, speed is reduced to 10km for every 2 hours(i.e., 5 km per hour).
Speed of the car at 11 am = (50 - 5) = 45km/hr
Time to cover 10 km at 45 km/hr = distance/speed = 10/45 hours. = 2/9 hours
= 2/9 x 60 minutes = 40/3 minutes = 13 minutes + 1/3 minutes
= 13 minutes + 1/3 x 60 seconds
= 13 minutes and 20 seconds.
Initial speed of the car = 50km/hr
Due to engine problem, speed is reduced to 10km for every 2 hours(i.e., 5 km per hour).
Speed of the car at 11 am = (50 - 5) = 45km/hr
Time to cover 10 km at 45 km/hr = distance/speed = 10/45 hours. = 2/9 hours
= 2/9 x 60 minutes = 40/3 minutes = 13 minutes + 1/3 minutes
= 13 minutes + 1/3 x 60 seconds
= 13 minutes and 20 seconds.
55929.P, Q & R participated in a race. P covers the same distance in 49 steps, as Q covers in 50 steps and R in 51 steps. P takes 10 steps in the same time as Q takes 9 steps and R takes 8 steps. Who is the winner of the race?
P
Q
R
Can’t be determined
Explanation:
A Distance=> P*49 = Q*50 = R*51
P= Distance/49, Q= Distance/50, R= Distance/51
Time => P*10 = Q*9 = R*8
P = Time/10, Q = Time/9, R= Time/8
P’s speed = (Distance/49)/(Time/10) = 10/49
Same like that,
The ratio of speeds of P, Q, R = (10/49) : (9/50) : (8/51)
Hence, P is the fastest.
A Distance=> P*49 = Q*50 = R*51
P= Distance/49, Q= Distance/50, R= Distance/51
Time => P*10 = Q*9 = R*8
P = Time/10, Q = Time/9, R= Time/8
P’s speed = (Distance/49)/(Time/10) = 10/49
Same like that,
The ratio of speeds of P, Q, R = (10/49) : (9/50) : (8/51)
Hence, P is the fastest.
55930.Towns A and B are 225 km apart. Two cars P and Q travel towards each other from towns A and B respectively and meet after 3 hours. If the speed of P be 1/2 of its original speed and Q be 2/3 of its original speed, they would have met after 5 hours. Find the speed of the faster car.
50 km/hr
40 km/hr
45 km/hr
30 km/hr
Explanation:
Let speeds be x km/hr and y km/hr
So 225/(x+y) = 3
And 225/(x/2 + 2y/3) = 5
Solve, x = 30, y = 45
Let speeds be x km/hr and y km/hr
So 225/(x+y) = 3
And 225/(x/2 + 2y/3) = 5
Solve, x = 30, y = 45
55931.Raghav drives his truck very fast at 360 kmph. Moving ahead for some hours he finds some problem in headlights of the truck. So he takes 20 seconds in changing in the bulb of the headlight by stopping the truck. Mean while he notices that 2nd truck which was 400 m back is now 200 m ahead of his truck. What is the speed of 2nd truck?
100 kmph
92 kmph
108 kmph
300 kmph
Explanation:
Speed of 2nd truck = [(400 + 200)/20] * [18/5] kmph
= 108 kmph
Speed of 2nd truck = [(400 + 200)/20] * [18/5] kmph
= 108 kmph
55932.A motor car does a journey in 16 hours, covering the first half at 30 Km/hr and the second half at 50 Km/hr. What is the distance covered?
480 km
540 km
500 km
600 km
Explanation:
Since different speeds are traveled for equal distance, the average speed can be found out
Average speed = $\dfrac{(2 × 30 × 50)}{(30 + 50)8}=300/8$ km/h
Distance covered = 300/8 × 16 = 600 km
Since different speeds are traveled for equal distance, the average speed can be found out
Average speed = $\dfrac{(2 × 30 × 50)}{(30 + 50)8}=300/8$ km/h
Distance covered = 300/8 × 16 = 600 km
55933.If a person divides a distance into 3 equal parts and travels the three parts with speeds of 50, 40 and 30 km/hr respectively, what is his average speed (in km/hr) for the whole journey?
37.5
60
38.3
40
Explanation:
Let the total distance be 3x kms.
∴ The distance travelled in each part is x kms.br>
∴ The time taken to cover the three parts is x/50 ,x/40 and x/30 respectively.br>
Now, average speed is total distance covered divided by the total time taken, so we get,br>
Average speed = Total distance/Total timebr>
Average speed = $\dfrac{3x}{(x/50 + x/40 +x/30)}$br>
=$\dfrac{3}{(1/50 + 1/40 +1/30)}$br>
$\dfrac{3}{(12+15+20)/600}$br>
=$\dfrac{1800}{47}$br>
=38.3 kmph
Let the total distance be 3x kms.
∴ The distance travelled in each part is x kms.br>
∴ The time taken to cover the three parts is x/50 ,x/40 and x/30 respectively.br>
Now, average speed is total distance covered divided by the total time taken, so we get,br>
Average speed = Total distance/Total timebr>
Average speed = $\dfrac{3x}{(x/50 + x/40 +x/30)}$br>
=$\dfrac{3}{(1/50 + 1/40 +1/30)}$br>
$\dfrac{3}{(12+15+20)/600}$br>
=$\dfrac{1800}{47}$br>
=38.3 kmph