Saket and Rakesh are brothers whose shops are located at a distance of 10km and 12km respectively from their home. Both of them start at the same time from home every day and reach their respective shops exactly at 8 am. When they decided to swap shops for a day, Saket started 4 minutes earlier than Rakesh. If both of them reached on time, what is the speed of Rakesh?
Since, ratio between distance covered by Saket and Rakesh = 10 : 12 = 5 : 6 ∴ Ratio between Speed of Saket and Rakesh = 5 : 6 Let speed of Saket and Rakesh be 5 x and 6 x respectively. ∴ 12/5x - 10/6x = 4/60⟹ (72-50 )/30x = 4/60 ⟹ 22/30x = 4/60⟹ 60 × 22 = 120 x ∴x = (60 × 22)/120 ∴ x = 11 ∴ Speed of Rakesh = 66 km/hr
Two guns were fired from the same place at an interval of 20 minutes but a person in a train approaching the place hears the sound 16 minutes 30 seconds after the first. Find the speed of the train (in kmph) supposing that the sound travels 330 meters per second.
The distance travelled by the train in 16min 30 seconds will be equal to that travelled by sound (20 min – 16 min 30 sec) = 3 min 30 seconds The train travelled 330 × 210 meters in 16 1/2 minutes. ∴ Speed of the train per hour = (330 × 210 × 2 × 60)/(33 ×1000) = 252 km/hr ALTERNATE METHOD: Let speed of train = x m/s The distance covered by bullet with a speed of sound 330m/s in 20 minutes is equal to distance covered by train & bullet together in 16 min 30 sec So let speed of train = x m/s So 330 × 20 = (x+330) × 16 × 1/2 6600 × 2/33 = x + 330 So x + 330 = 400 Or x = 70 m/s = 70 × 18/5 kmph = 252kmph
Saket and Rakesh are brothers whose shops are located at a distance of 10km and 12km respectively from their home. Both of them start at the same time from home every day and reach their respective shops exactly at 8 am. When they decided to swap shops for a day, Saket started 4 minutes earlier than Rakesh. If both of them reached on time, what is the speed of Rakesh?
Since, ratio between distance covered by Saket and Rakesh = 10 : 12 = 5 : 6 ∴ Ratio between Speed of Saket and Rakesh = 5 : 6 Let speed of Saket and Rakesh be 5 x and 6 x respectively. ∴ 12/5x - 10/6x = 4/60⟹ (72-50 )/30x = 4/60 ⟹ 22/30x = 4/60⟹ 60 × 22 = 120 x ∴x = (60 × 22)/120 ∴ x = 11 ∴ Speed of Rakesh = 66 km/hr
If a man covered a certain distance D km at some speed S kmph, if he had moved 3km per hour faster, he would have taken 40 minutes less, If he had moved 2 km per hour slower, he would have been 40 minutes slower then which of the following we can find from the given statements.
1. His usual speed S.
2. Time taken by man to cover (D+20)km at a speed of (S+20) kmph.
3. Time taken by a man to cover a bridge at a speed of (S-20) kmph.
4. Distance travelled by him to cross a train of length 100m at twice of his usual speed.
D= ST = (S+3)(T-2/3) =(S-2)(T+2/3) By solving , D = 40km , S= 12kmph & T= 40/12 = 10/3 hrs So we can find 1), 2) & 4) but we cannot find 3) because length of bridge is not given.
A goes out to shop on a rainy day. He travels at a speed of n km/hr when it’s raining and at a speed of (n – 1) km/hr when it stops raining. If his average speed for the journey is 3.2 km/hr. Find the fraction of the distance which he covered while it was raining. Given, that n is an integer.
Train A started from station P towards station Q. At the same time, train B started from station R towards station Q. All the three stations are in a straight line such that station Q is between station P and station R. Station Q is equidistant from station P and station R. Distance between station P and station Q is 220 km. If the speed of train A and train B is ______ km/hr and ______ km/hr respectively, the distance between both the trains after five hours is 290 km.
Which of the following satisfies the two blanks given in the questions?
I. 14 km/hr, 16 km/hr
II. 20 km/hr, 10 km/hr
III. 12.5 km/hr, 17.5 km/hr
From I: Distance between station P and station R = 220 x 2 = 440 km Distance travelled by train A in 5 hours = 14 x 5 = 70 Km Distance travelled by train B in 5 hours = 16 x 5 = 80 Km Required distance = 440 – 70 – 80 = 290 km This is satisfies the given condition. From II: Distance between station P and station R = 220 x 2 = 440 km Distance travelled by train A in 5 hours = 20 x 5 = 100 km Distance travelled by train B in 5 hours = 10 x 5 = 50 km Required distance = 440 – 100 – 50 = 290 km This is satisfies the given condition. From III: Distance between station P and station R = 220 x 2 = 440 km Distance travelled by train A in 5 hours = 12.5 x 5 = 62.5 km Distance travelled by train B in 5 hours = 17.5 x 5 = 87.5 km Required distance = 440 – 62.5 – 87.5 = 290 km This is satisfies the given condition.
Find the distance travelled (in km) by the truck to reach point B from point A, if a car cover 45% the same distance in 2 hours 42 minutes and a truck covers 60% of the distance in 6 hours. The sum of speed of the car and truck is 160 km/hr.
Three persons named P, Q, R started their journey at 9.30 AM, 11.30 AM, 2 PM respectively from A to B. The speed of Q is √625 m/sec. Q reached the destination at 7.30 pm on the same day. Q is fastest & R is not the slowest.
The speed of P is 20% less than the speed of Q, then in what time P will reach the destination?
i. 6 hours more than Q
ii. P will reach B at 7.30 pm
iii. The total reaching time of P and Q is 18 hrs
Speed of Q = √625 m/sec = 25 m/sec = 25 x 18/5 = 90 km/hr Time taken to reach point B by Q = 8 hrs (11.30 am – 7:30 pm) Distance between A and B = 90 x 8 = 720 km Speed of P = 0.8 x 90 = 72 km/hr Now, From i: Time taken by P to cover the distance = 720/72 = 10 hrs Difference of time is 2 hrs, so i is not true. From ii: P started journey at 9:30 AM, so he will reach by 7:30 pm (10 hrs) So , ii is true. From iii: Total time = 8 + 10 = 18 hrs iii is true.
The speed of a car is 40% less than the speed of a bike. The time taken by a bus to cover (d-100) km distance in 9 hours. The sum of the speed of the bus and car is equal to the speed of the bike. If the time taken by car to cover (d+80) km distance is 2 hours more than the time taken by bike to cover (d+260) km distance, then which of the following statements is/are true?
(i) The value of ‘d’ is the multiple of 16.
(ii) The speed of the bike is 180 km/h.
(iii) The time taken by the car to cover (d+80) km is 8 hours.
The speed of a car is 40% less than the speed of a bike.
Let’s assume the speed of a bike is ‘15y’.
speed of a car = 15y of (100-40)%
= 15y of 60%
= 9y
The sum of the speed of the bus and car is equal to the speed of the bike.
speed of the bus = 15y-9y = 6y
The time taken by a bus to cover (d-100) km distance in 9 hours.
(d-100)/6y = 9
(d-100) = 54y
d = (54y+100) Eq.(i)
If the time taken by car to cover (d+80) km distance is 2 hours more than the time taken by bike to cover (d+260) km distance.
(d+80)/9y = 2+ (d+260)/15y
Put the value of ‘d’ from Eq.(i) in the above equation.
(54y+100+80)/9y = 2+ (54y+100+260)/15y
(54y+180)/9y = 2+ (54y+360)/15y
(54y+180)/9y - (54y+360)/15y = 2
After solving the above equation, 0.4y+20 = 24
0.4y = 24-20
0.4y = 4
y = 10
Put the value of ‘y’ in Eq.(i).
d = (54x10+100) = 540+100 = 640
(i) The value of ‘d’ is the multiple of 16.
The above given statement is true.
(ii) The speed of the bike is 180 km/h.
Speed of bike = 15y = 15x10 = 150 km/h
The above given statement is not true.
(iii) The time taken by the car to cover (d+80) km is 8 hours.
time taken by the car to cover (d+80) km = (d+80)/9y
= (640+80)/(9x10)
= 720/90
= 8
The above given statement is true.
The speed of a car is 60% more than the speed of a bus. The train can cover (d-250) km distance in (t-3.5) hours. The car covers (d+140) km distance in (t-2) hours. The speed of train is 2.4 times the difference between the speed of bus and car. The distance travelled by bus is 1500 km in which it will take (t+4) hours, then which of the following statements is/are true?
(i) The value of ‘t’ is an odd number.
(ii) The value of ‘d’ is a three digit number.
(iii) The ratio between the distance travelled by bus and train is (6t+2) : (3t+3) respectively.
Let’s assume the speed of car is ‘8a’. The speed of a car is 60% more than the speed of a bus. 8a = (100+60)% of speed of a bus 160% of speed of a bus = 8a 1.6 x speed of a bus = 8a speed of a bus = 5a The speed of train is 2.4 times the difference between the speed of bus and car. speed of train = 2.4x(8a-5a) = 2.4x3a = 7.2a The train can cover (d-250) km distance in (t-3.5) hours. 7.2a = (d-250)/(t-3.5) a = (d-250)/7.2(t-3.5) Eq.(1) The car covers (d+140) km distance in (t-2) hours. 8a = (d+140)/(t-2) a = (d+140)/8(t-2) Eq.(2) The distance travelled by bus is 1500 km in which it will take (t+4) hours. 5a = 1500/(t+4) a = 300/(t+4) Eq.(3) So Eq.(1) = Eq.(2) (d-250)/7.2(t-3.5) = (d+140)/8(t-2) (d-250)/0.9(t-3.5) = (d+140)/(t-2) (d-250)/(0.9t-3.15) = (d+140)/(t-2) d = (376t−941)/(0.1t+1.15) Eq.(4) So Eq.(1) = Eq.(3) (d-250)/7.2(t-3.5) = 300/(t+4) (d-250)/(t-3.5) = 2160/(t+4) d = (2410t−6560)/(t+4) Eq.(5) So Eq.(4) = Eq.(5) (376t−941)/(0.1t+1.15) = (2410t−6560)/(t+4) After solving the above equation, t = 8, 3.5 But if t = 3.5, then the time taken by train to cover distance will be zero which is not possible. So t = 8. Put the value of ‘t’ in Eq.(4). d = (376x8−941)/(0.1x8+1.15) = (3008−941)/(0.8+1.15) = 2067/1.95 = 1060 (i) The value of ‘t’ is an odd number. The above given statement is not true. Because the value of ‘t’ cannot be an odd number. (ii) The value of ‘d’ is a three digit number. The above given statement is not true. Because the value of ‘d’ is not a three digit number. (iii) The ratio between the distance travelled by bus and train is (6t+2) : (3t+3) respectively. distance travelled by bus and train = (6t+2) : (3t+3) = 5ax(t+4) : 7.2ax(t-3.5) Put the value of ‘t’ in the above equation. (6x8+2) : (3x8+3) = 5x(8+4) : 7.2x(8-3.5) (48+2) : (24+3) = 5x12 : 7.2x4.5 50 : 27 = 50 : 27 The above given statement is true. Because the above given expression is satisfied.
