# Aptitude - Time and Distance - Discussion

Discussion Forum : Time and Distance - General Questions (Q.No. 12)

12.

Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. 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 P.M.?

Answer: Option

Explanation:

Let the distance travelled by *x* km.

Then, | x |
- | x |
= 2 |

10 | 15 |

3*x* - 2*x* = 60

*x* = 60 km.

Time taken to travel 60 km at 10 km/hr = | 60 | hrs | = 6 hrs. | |

10 |

So, Robert started 6 hours before 2 P.M. *i.e.,* at 8 A.M.

Required speed = | 60 | kmph. | = 12 kmph. | |

5 |

Discussion:

90 comments Page 1 of 9.
Vivek said:
4 months ago

Case1: speed = 10kmph.

Ce2: speed = 15kmph.

Given d/10-d/15 = 2,

d=60,

case1 time=60/10, time=6,

case2 time=60/15, time=4,

so, for 12noom time is 4, and for 2 pm is 6

And 1 pm is between them so it will be 5.

And 60/5=12.

Ce2: speed = 15kmph.

Given d/10-d/15 = 2,

d=60,

case1 time=60/10, time=6,

case2 time=60/15, time=4,

so, for 12noom time is 4, and for 2 pm is 6

And 1 pm is between them so it will be 5.

And 60/5=12.

(13)

Anushree Kulkarni said:
2 years ago

@Farzana.

Best and neat explanation, Thank you.

Best and neat explanation, Thank you.

(5)

Anuj said:
2 years ago

10 kmph (2 pm) = t h ----> eq(1)

x kmph (1 pm) = (t-1)h ----> eq(2)

15 kmph (2 pm) = (t-2)h ----> eq(3)

Since, distance is the same for all cases we can equate these equations to get the answer.

x kmph (1 pm) = (t-1)h ----> eq(2)

15 kmph (2 pm) = (t-2)h ----> eq(3)

Since, distance is the same for all cases we can equate these equations to get the answer.

(2)

Vicky said:
3 years ago

How 3 came? Please explain.

(4)

Akash said:
3 years ago

@All.

Use formula like (2 (a.b) /a+b) a is speed 10kmph and b is 15kmph.

Hope you get it.

Use formula like (2 (a.b) /a+b) a is speed 10kmph and b is 15kmph.

Hope you get it.

(24)

Prathamesh Pendal said:
3 years ago

1pm is in between 12 and 2 pm.

So, if we find average speed we can also get the answer.

Avg speed = (2*15*10)/(10+15) = 12.

So, if we find average speed we can also get the answer.

Avg speed = (2*15*10)/(10+15) = 12.

(41)

Havoc said:
3 years ago

We can also use average speed formula.

Because 12 am to 2 pm time 2 hr.

Thus 12 am to 1 am 1 hr.

So, we can find use avg speed also,

The mid point time of 12 am and 2 pm is 1 pm.

Because 12 am to 2 pm time 2 hr.

Thus 12 am to 1 am 1 hr.

So, we can find use avg speed also,

The mid point time of 12 am and 2 pm is 1 pm.

(2)

Farzana said:
4 years ago

Let the distance be x.

According to statement 1.

If he travels at 10 kmph and let's imagine the time taken to reach at 2 pm is 't' hrs

i.e, x/10=t ---> equation(1).

According to statement 2.

If he travels at 15 kmph and reaches at 12 noon, then he reaches 2 hrs earlier than the previous time (i.e, t-2 hrs)

then, x/15=t-2 ------> equation(2)

If you get two equations with same two variables, always subtract one equation from the other.

Here, the equations become,

x/10-x/15=t-(t-2) or x/15-x/10=(t-2)-t

either way you will get the same answer i.e, x=60 km.

Just Put this in equation(1) or equation(2) and you will get the original time.

equation(1), 60/10=t, t=6.

equation(2), 60/15=t-2, 4=t-2, t=6.

So if Robert wants to reach the destination at 1 pm, We have to calculate his speed, we got distance and time taken.

If he wants to reach at 1 pm which is 1 hr less than the normal time.

then equation becomes distance/speed=time-1 hr.

i.e, 60/speed=6-1, speed=60/5, speed=12 kmph.

Therefore, the answer is 12 kmph.

According to statement 1.

If he travels at 10 kmph and let's imagine the time taken to reach at 2 pm is 't' hrs

i.e, x/10=t ---> equation(1).

According to statement 2.

If he travels at 15 kmph and reaches at 12 noon, then he reaches 2 hrs earlier than the previous time (i.e, t-2 hrs)

then, x/15=t-2 ------> equation(2)

If you get two equations with same two variables, always subtract one equation from the other.

Here, the equations become,

x/10-x/15=t-(t-2) or x/15-x/10=(t-2)-t

either way you will get the same answer i.e, x=60 km.

Just Put this in equation(1) or equation(2) and you will get the original time.

equation(1), 60/10=t, t=6.

equation(2), 60/15=t-2, 4=t-2, t=6.

So if Robert wants to reach the destination at 1 pm, We have to calculate his speed, we got distance and time taken.

If he wants to reach at 1 pm which is 1 hr less than the normal time.

then equation becomes distance/speed=time-1 hr.

i.e, 60/speed=6-1, speed=60/5, speed=12 kmph.

Therefore, the answer is 12 kmph.

(31)

Ritik Florian said:
4 years ago

Simply, find the average speed u can get the answer. Since 1 pm comes between 12 noon and 2 pm.

The formula for finding AVG speed is= 2xy/X+y.

i.e. 2*10*15/10+15= 12.

The formula for finding AVG speed is= 2xy/X+y.

i.e. 2*10*15/10+15= 12.

(5)

Ishwar said:
4 years ago

Let calculate distance firs say d.

D/10-d/15=2 (he reach 12noon 2hr early).

By calculating we get d = 60km.

Let's speed is S to calculate S difference between travel at first speed rate and S rate is.

60/10-60/S=1 (because he reaches by 1 am, 1hr early).

By calculating this we get.

5S=60.

S = 12kmph.

D/10-d/15=2 (he reach 12noon 2hr early).

By calculating we get d = 60km.

Let's speed is S to calculate S difference between travel at first speed rate and S rate is.

60/10-60/S=1 (because he reaches by 1 am, 1hr early).

By calculating this we get.

5S=60.

S = 12kmph.

Post your comments here:

Quick links

Quantitative Aptitude

Verbal (English)

Reasoning

Programming

Interview

Placement Papers