# Aptitude - Time and Distance - Discussion

Discussion Forum : Time and Distance - General Questions (Q.No. 11)

11.

In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:

Answer: Option

Explanation:

Let Abhay's speed be *x* km/hr.

Then, | 30 | - | 30 | = 3 |

x |
2x |

6*x* = 30

*x* = 5 km/hr.

Discussion:

200 comments Page 1 of 20.
JaSHaN SaGaR said:
10 months ago

Take the ratio of the speed of Abhay:

1:2

Then the inverse time we get Time ratio,

So:

2:1.

The Gap is 1 unit and this one is given as the sum of hours which is 3 hours.

So 2=6 hours and 1= 3 hours.

So we get the ans by applying a formula that is S= D/T

S= 30/6, which is equal to 5.

1:2

Then the inverse time we get Time ratio,

So:

2:1.

The Gap is 1 unit and this one is given as the sum of hours which is 3 hours.

So 2=6 hours and 1= 3 hours.

So we get the ans by applying a formula that is S= D/T

S= 30/6, which is equal to 5.

(32)

Urvesh Radadiya said:
11 months ago

@All.

My explanation is very simple;

T = D/S.

Suppose the speed of Abhay is x.

First case:

It takes 2 hours more than Samir.

T + 2 = 30/x ---> (1)

Second case:

when speed is double then 2x

it takes 1 hour less so

T-1=30/2x ----> (2)

Compare eq 1 &2;

30/x-2=T.

30/2x+1=T.

30/x-30/2x = 3.

Solving the above eq we get x=5km/h=> Abhay's speed.

My explanation is very simple;

T = D/S.

Suppose the speed of Abhay is x.

First case:

It takes 2 hours more than Samir.

T + 2 = 30/x ---> (1)

Second case:

when speed is double then 2x

it takes 1 hour less so

T-1=30/2x ----> (2)

Compare eq 1 &2;

30/x-2=T.

30/2x+1=T.

30/x-30/2x = 3.

Solving the above eq we get x=5km/h=> Abhay's speed.

(137)

Akshay said:
1 year ago

We know that,

1st case ---> (1)

If Sameer takes 2 hours,

Then, Abhay takes 4 hours.

So,

Sameer takes 2 * 60 = 120 seconds.

&Abhay takes 4 * 60 = 240 seconds.

2nd case ----> (2)

Abhay takes 1hour less than Sameer,

So,

2-1 = 1 * 60 = 60seconds

So,

Abhay's speed is

Case(1)&(2)

240 + 60 = 300Seconds;

Then 300/ 60 = 5.

1st case ---> (1)

If Sameer takes 2 hours,

Then, Abhay takes 4 hours.

So,

Sameer takes 2 * 60 = 120 seconds.

&Abhay takes 4 * 60 = 240 seconds.

2nd case ----> (2)

Abhay takes 1hour less than Sameer,

So,

2-1 = 1 * 60 = 60seconds

So,

Abhay's speed is

Case(1)&(2)

240 + 60 = 300Seconds;

Then 300/ 60 = 5.

(14)

Abhay the guy walking said:
1 year ago

Let's assume that Sameer's speed is S km/h and Abhay's speed is A km/h.

We know that Sameer takes time T to cover 30 km, so:

T = 30/S

We also know that Abhay takes 2 hours longer than Sameer, so:

T + 2 = 30/A

If Abhay doubles his speed, he will cover 30 km in half the time, which is:

T/2

We also know that this time is 1 hour less than Sameer's time, so:

T/2 = T - 1

Solving for T in terms of S:

30/S/2 = 30/S - 1

15/S = 30/S - 1

Multiplying both sides by S(S-15):

15(S-15) = 30(S-15) - S(S-15)

15S - 225 = 30S - 450 - S^2 + 15S

S^2 - 60S + 225 = 0

(S-15)^2 = 0

S = 15 km/h

Now we can use the equation T + 2 = 30/A to solve for Abhay's speed:

T + 2 = 30/A

30/15 + 2 = 30/A

4 = 30/A

A = 7.5 km/h

Therefore, Abhay's speed is 7.5 km/h.

We know that Sameer takes time T to cover 30 km, so:

T = 30/S

We also know that Abhay takes 2 hours longer than Sameer, so:

T + 2 = 30/A

If Abhay doubles his speed, he will cover 30 km in half the time, which is:

T/2

We also know that this time is 1 hour less than Sameer's time, so:

T/2 = T - 1

Solving for T in terms of S:

30/S/2 = 30/S - 1

15/S = 30/S - 1

Multiplying both sides by S(S-15):

15(S-15) = 30(S-15) - S(S-15)

15S - 225 = 30S - 450 - S^2 + 15S

S^2 - 60S + 225 = 0

(S-15)^2 = 0

S = 15 km/h

Now we can use the equation T + 2 = 30/A to solve for Abhay's speed:

T + 2 = 30/A

30/15 + 2 = 30/A

4 = 30/A

A = 7.5 km/h

Therefore, Abhay's speed is 7.5 km/h.

(2)

Krishnan said:
1 year ago

Here's a ChatGpt answer method to solve the problem:

Let's denote Abhay's speed as "a" and Sameer's speed as "s".

We know that distance = speed × time.

According to the problem, Abhay takes 2 hours more than Sameer to cover 30 km. So, we can write:

30 = s × (t + 2) (Equation 1)

where t is Sameer's time taken to cover the distance.

If Abhay doubles his speed, he would take 1 hour less than Sameer to cover 30 km. So, we can write:

30 = 2a × (t - 1) (Equation 2)

where t - 1 is Abhay's time taken to cover the distance at double the speed.

Now, we can solve these two equations for a and s:

From Equation 1, we get:

t = (30/s) - 2

Substituting this value of t in Equation 2, we get:

30 = 2a × ((30/s) - 3).

Simplifying this equation, we get:

a = (45s)/(4s - 60).

Now, substituting this value of a in Equation 1, we get:

30 = s × ((30/s) + 2).

Simplifying this equation, we get:

s^2 = 450.

Therefore, s = 15 km/h.

Substituting this value of s in the expression for a, we get:

a = (45 × 15)/(4 × 15 - 60) = 45 km/h

Therefore, Abhay's speed is 45 km/h.

Let's denote Abhay's speed as "a" and Sameer's speed as "s".

We know that distance = speed × time.

According to the problem, Abhay takes 2 hours more than Sameer to cover 30 km. So, we can write:

30 = s × (t + 2) (Equation 1)

where t is Sameer's time taken to cover the distance.

If Abhay doubles his speed, he would take 1 hour less than Sameer to cover 30 km. So, we can write:

30 = 2a × (t - 1) (Equation 2)

where t - 1 is Abhay's time taken to cover the distance at double the speed.

Now, we can solve these two equations for a and s:

From Equation 1, we get:

t = (30/s) - 2

Substituting this value of t in Equation 2, we get:

30 = 2a × ((30/s) - 3).

Simplifying this equation, we get:

a = (45s)/(4s - 60).

Now, substituting this value of a in Equation 1, we get:

30 = s × ((30/s) + 2).

Simplifying this equation, we get:

s^2 = 450.

Therefore, s = 15 km/h.

Substituting this value of s in the expression for a, we get:

a = (45 × 15)/(4 × 15 - 60) = 45 km/h

Therefore, Abhay's speed is 45 km/h.

(2)

Niraj katwal said:
1 year ago

Let the speed of Sameer be x.

Then, Ajay's speed = x+2.

When Ajay doubles his speed,

2(x+2) = x -1

or,2x+4 = x-1

Since speed can never be negative.

x=5.

Then, Ajay's speed = x+2.

When Ajay doubles his speed,

2(x+2) = x -1

or,2x+4 = x-1

Since speed can never be negative.

x=5.

(62)

Gramophone said:
1 year ago

t+2 = 30/s ----> 1

t-1 = 30/2s ---- > 2

Solving this equation, we get;

s = 5.

t-1 = 30/2s ---- > 2

Solving this equation, we get;

s = 5.

(23)

Jabir said:
2 years ago

@All.

Here, is my explanation;

When Abhay moves at normal speed;

Let the time taken by Sameer, time = X.

Then, Abhay needs 2 more hours time than Sameer. ie...time=X+2.

Consider the speed of Abhay, speed =Y

Now, Abhay moves at double speed, so he reaches 1 hour before Sameer.

So speed becomes speed=2Y and time =X-1

Now you can make the equation for distance

Distance is already given d=30

Case 1 (during the normal speed of Abhay)

Distance =speed*time

ie...30=Y*(X+2)----> equation 1

Similarly,

Case 2(during the double speed of Abhay)

30=2Y*(X-1)--->equation 2

Now solve equations 1 and 2 with your logic;

You, will get the answer 5.

Here, is my explanation;

When Abhay moves at normal speed;

Let the time taken by Sameer, time = X.

Then, Abhay needs 2 more hours time than Sameer. ie...time=X+2.

Consider the speed of Abhay, speed =Y

Now, Abhay moves at double speed, so he reaches 1 hour before Sameer.

So speed becomes speed=2Y and time =X-1

Now you can make the equation for distance

Distance is already given d=30

Case 1 (during the normal speed of Abhay)

Distance =speed*time

ie...30=Y*(X+2)----> equation 1

Similarly,

Case 2(during the double speed of Abhay)

30=2Y*(X-1)--->equation 2

Now solve equations 1 and 2 with your logic;

You, will get the answer 5.

(22)

Sarang said:
2 years ago

d = (s1*s2*time diff)/( s1-s2).

30 = x * 2x * 3/x * 2x.

x = 5.

30 = x * 2x * 3/x * 2x.

x = 5.

(7)

Anomie said:
2 years ago

Good explanation. Thanks all.

(4)

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