44035.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:
2 : 3
3 : 2
3 : 4
4 : 3
Explanation:
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then,$\dfrac{120}{x}+\dfrac{480}{y}=8 \Rightarrow \dfrac{1}{x}+\dfrac{4}{y}=\dfrac{1}{15}$...(i)
And, $\dfrac{200}{x}+\dfrac{400}{y}=\dfrac{25}{3} $$\Rightarrow \dfrac{1}{x}+\dfrac{2}{y}=\dfrac{1}{24}$...(ii)
Solving (i) and (ii), we get: x = 60 and y = 80.
$\therefore$Ratio of speeds = 60 : 80 = 3 : 4.
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then,$\dfrac{120}{x}+\dfrac{480}{y}=8 \Rightarrow \dfrac{1}{x}+\dfrac{4}{y}=\dfrac{1}{15}$...(i)
And, $\dfrac{200}{x}+\dfrac{400}{y}=\dfrac{25}{3} $$\Rightarrow \dfrac{1}{x}+\dfrac{2}{y}=\dfrac{1}{24}$...(ii)
Solving (i) and (ii), we get: x = 60 and y = 80.
$\therefore$Ratio of speeds = 60 : 80 = 3 : 4.
44038.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.
220 km
224 km
230 km
234 km
Explanation:
$\dfrac{(1/2)x}{21}+\dfrac{(1/2)x}{24}=10$
$\Rightarrow \dfrac{x}{21}+\dfrac{x}{24}=20$
$\Rightarrow$ 15x = 168 x 20
$\Rightarrow x=\dfrac{168 \times 20}{15}$
=224 km.
$\dfrac{(1/2)x}{21}+\dfrac{(1/2)x}{24}=10$
$\Rightarrow \dfrac{x}{21}+\dfrac{x}{24}=20$
$\Rightarrow$ 15x = 168 x 20
$\Rightarrow x=\dfrac{168 \times 20}{15}$
=224 km.
44041.A car travelling with $\dfrac{5}{7}$ of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.
17 $ \dfrac{6}{7}$ km/hr
25 km/hr
30 km/hr
35 km/hr
Explanation:
Time taken = 1 hr 40 min 48 sec = 1 hr 40$\dfrac{4}{5}$min=1$\dfrac{51}{75}$hrs=$\dfrac{126}{75}$hrs
Let the actual speed be x km/hr.
Then, $\dfrac{5}{7}x \times \dfrac{126}{75}$=42
$\Rightarrow \left( \dfrac{42\times7\times75}{5\times126}\right)$
=35km/hr.
Time taken = 1 hr 40 min 48 sec = 1 hr 40$\dfrac{4}{5}$min=1$\dfrac{51}{75}$hrs=$\dfrac{126}{75}$hrs
Let the actual speed be x km/hr.
Then, $\dfrac{5}{7}x \times \dfrac{126}{75}$=42
$\Rightarrow \left( \dfrac{42\times7\times75}{5\times126}\right)$
=35km/hr.
44042.An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in $ 1\dfrac{2}{3} $ hours, it must travel at a speed of:
300 kmph
360 kmph
600 kmph
720 kmph
Explanation:
Distance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write $ 1\dfrac{2}{3} $ hours as 5/3 hours]
Required speed =$\left(1200 \times \dfrac{3}{5}\right) $km/hr
= 720 km/hr.
Distance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write $ 1\dfrac{2}{3} $ hours as 5/3 hours]
Required speed =$\left(1200 \times \dfrac{3}{5}\right) $km/hr
= 720 km/hr.
44043.A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
35
36$\dfrac{2}{3}$
37$\dfrac{2}{3}$
40
Explanation:
Let distance = x km and usual rate = y kmph.
Then,$\dfrac{x}{y}-\dfrac{x}{y+3}=\dfrac{40}{60} \Rightarrow$ 2y(y + 3) = 9x ....(i)
And, $\dfrac{x}{y-2}-\dfrac{x}{y}=\dfrac{40}{60} \Rightarrow$ y(y - 2) = 3x ....(ii)
On dividing (i) by (ii), we get: x = 40.
Let distance = x km and usual rate = y kmph.
Then,$\dfrac{x}{y}-\dfrac{x}{y+3}=\dfrac{40}{60} \Rightarrow$ 2y(y + 3) = 9x ....(i)
And, $\dfrac{x}{y-2}-\dfrac{x}{y}=\dfrac{40}{60} \Rightarrow$ y(y - 2) = 3x ....(ii)
On dividing (i) by (ii), we get: x = 40.
44045.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:
1 hour
2 hours
3 hours
4 hours
Explanation:
Let the duration of the flight be x hours.
Then,$\dfrac{600}{x}-\dfrac{600}{x+(1/2)}=200$
$\Rightarrow \dfrac{600}{x}-\dfrac{1200}{2x+1}=200$
$\Rightarrow$ x(2x + 1) = 3
2x2 + x - 3 = 0
(2x + 3)(x - 1) = 0
x = 1 hr. [neglecting the -ve value of x]
Let the duration of the flight be x hours.
Then,$\dfrac{600}{x}-\dfrac{600}{x+(1/2)}=200$
$\Rightarrow \dfrac{600}{x}-\dfrac{1200}{2x+1}=200$
$\Rightarrow$ x(2x + 1) = 3
2x2 + x - 3 = 0
(2x + 3)(x - 1) = 0
x = 1 hr. [neglecting the -ve value of x]
44232. The speed of a bus increases by 2 kmph after every one hour. If the distance traveled in the first one hour was 35 km, what was the total distance traveled in 12 hours?
550 km
552 km
402 km
452 km
Explanation:
Distance traveled in 1st hour =35 km
Speed of the bus increases by 2 kmph after every one hour.
Hence,distance travelled in 2nd hour =37 km
distance travelled in 3rd hour =39 km
and so on
Total distance travelled=[35+37+39+⋯(12 terms)]
=12/2[2×35+(12−1)2]
=6(70+22)
=6×92
=552
Distance traveled in 1st hour =35 km
Speed of the bus increases by 2 kmph after every one hour.
Hence,distance travelled in 2nd hour =37 km
distance travelled in 3rd hour =39 km
and so on
Total distance travelled=[35+37+39+⋯(12 terms)]
=12/2[2×35+(12−1)2]
=6(70+22)
=6×92
=552
44233.Arun is travelling on his cycle and has calculated to reach point A at 2 pm if he travels at 10 kmph. He will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 pm?
8kmph
10 kmph
12 kmph
14 kmph
Explanation:
Let the distance be x km
Travelling at 10 kmph, Arun will reach point A at 2 pm.
Travelling at 15 kmph, Arun will reach point A 12 noon.
Therefore, time taken when travelling at 10 km - time taken when travelling at 15 km =2hours
⇒x/10−x/15=2
⇒3x−2x=2×30
⇒x=60
Time needed if travelled at 10 kmph =60/10=6 hours
Therefore, to reach at 1 pm, his travelling time must be (6−1)=5 hours.
Hence, required speed =60/5=12 kmph
Let the distance be x km
Travelling at 10 kmph, Arun will reach point A at 2 pm.
Travelling at 15 kmph, Arun will reach point A 12 noon.
Therefore, time taken when travelling at 10 km - time taken when travelling at 15 km =2hours
⇒x/10−x/15=2
⇒3x−2x=2×30
⇒x=60
Time needed if travelled at 10 kmph =60/10=6 hours
Therefore, to reach at 1 pm, his travelling time must be (6−1)=5 hours.
Hence, required speed =60/5=12 kmph
44235.A train traveled at an average speed of 100 km/hr, stopping for 3 minutes after every 75 km. How long did it take to reach its destination 600 km from the starting point?
6 hrs 21 min
7 hrs 21 min
6 hrs 2 min
6 hrs 10 min
Explanation:
Time needed to travel 600 km =600/100=6 hour
Now we need to find out the number of stops in the 600 km journey. Given that the train stops after every 75 km.
600/75=8
It means, the train stops 7 times before 600 km and 1 time just after 600 km.
Hence we need to take only 7 stops into consideration for the 600 km journey.
Hence, total stopping time in the 600 km journey
=7×3=21 minutes
Total time needed to reach the destination
=6 hours +21 minutes
=6 hours 21 minutes
Time needed to travel 600 km =600/100=6 hour
Now we need to find out the number of stops in the 600 km journey. Given that the train stops after every 75 km.
600/75=8
It means, the train stops 7 times before 600 km and 1 time just after 600 km.
Hence we need to take only 7 stops into consideration for the 600 km journey.
Hence, total stopping time in the 600 km journey
=7×3=21 minutes
Total time needed to reach the destination
=6 hours +21 minutes
=6 hours 21 minutes
44237.The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time will they meet?
10.30 a.m.
10 a.m.
12 a.m.
11 a.m.
Explanation:
Assume that they meet x hours after 8 a.m.
Then, train 1, starting from A, travels x hours till the trains meet.
Distance travelled by train 1 in x hours =60x km
Train 2, starting from B, travels (x−1) hours till the trains meet.
Distance travelled by train 2 in (x−1) hours =75(x−1) km
Total distance travelled
= Distance travelled by train 1 + Distance travelled by train 2
⇒330=60x+75(x−1)
⇒12x+15(x−1)=66
⇒12x+15x−15=66
⇒27x=66+15=81
⇒3x=9
⇒x=3
Hence, the trains meet 3 hours after 8 a.m., i.e. at 11 a.m.
Assume that they meet x hours after 8 a.m.
Then, train 1, starting from A, travels x hours till the trains meet.
Distance travelled by train 1 in x hours =60x km
Train 2, starting from B, travels (x−1) hours till the trains meet.
Distance travelled by train 2 in (x−1) hours =75(x−1) km
Total distance travelled
= Distance travelled by train 1 + Distance travelled by train 2
⇒330=60x+75(x−1)
⇒12x+15(x−1)=66
⇒12x+15x−15=66
⇒27x=66+15=81
⇒3x=9
⇒x=3
Hence, the trains meet 3 hours after 8 a.m., i.e. at 11 a.m.