Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 10)
10.
A car travelling with
of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.

Answer: Option
Explanation:
Time taken = 1 hr 40 min 48 sec = 1 hr 40 | 4 | min = 1 | 51 | hrs = | 126 | hrs. |
5 | 75 | 75 |
Let the actual speed be x km/hr.
Then, | 5 | x x | 126 | = 42 |
7 | 75 |
![]() |
![]() |
42 x 7 x 75 | ![]() |
= 35 km/hr. |
5 x 126 |
Discussion:
117 comments Page 1 of 12.
NANDHINI V said:
9 months ago
Let's take the actual speed as X,
So,
The travelling speed is 5x/7,
Distance d = 42km,
Time t=1hr 40min 48sec is written as 126/75,
Now we need to find the actual speed,
We know the formula,
T = D/S,
126/75 = 42/5x/7,
126/75 = 42 * 7/5x,
5x = 42 * 7 * 75/126,
x = 35km/hr.
So,
The travelling speed is 5x/7,
Distance d = 42km,
Time t=1hr 40min 48sec is written as 126/75,
Now we need to find the actual speed,
We know the formula,
T = D/S,
126/75 = 42/5x/7,
126/75 = 42 * 7/5x,
5x = 42 * 7 * 75/126,
x = 35km/hr.
(9)
Tenzin said:
1 year ago
@Dhivya.
How you get 204? Please explain.
How you get 204? Please explain.
(12)
Dhivya said:
2 years ago
T = 1hr40min48sec.
= 1hr40(48/60)min,
= 1hr40(4/5)min,
= 1hr(204/5*60)min,
= 1(51/75)hr,
= 126/75 hr,
Then, Actual speed = (5/7x)*(126/75) = 42.
X = (42*7*75/15*126) = 35km/hr.
= 1hr40(48/60)min,
= 1hr40(4/5)min,
= 1hr(204/5*60)min,
= 1(51/75)hr,
= 126/75 hr,
Then, Actual speed = (5/7x)*(126/75) = 42.
X = (42*7*75/15*126) = 35km/hr.
(34)
Vivek said:
2 years ago
Time = 1 + 40/60 + 48/60*60 = 1.68hr
Speed=x,
Given speed 5x/7,
5x/7 = 42/1.68.
x = 35 km/hr.
Speed=x,
Given speed 5x/7,
5x/7 = 42/1.68.
x = 35 km/hr.
(32)
Suraj N Raikar said:
2 years ago
Let 42kmph is the average speed,
And travelling at 5/7 of average speed.
7/5 x 42 = 35kmph.
And travelling at 5/7 of average speed.
7/5 x 42 = 35kmph.
(68)
Sherief said:
2 years ago
Simplest form;
S = D ÷ T formula;
S = 5s ÷ 7.
42km~convert in meter so, 4200.
1hr+40min+48sec.
= (60min × 60sec) + (40min × 60sec) + 48sec,
= 6048.
S = D÷T.
5s ÷ 7 = 4200 ÷ 6048.
= 4200×7 ÷ 6048 ÷ 5
= 8400 ÷ 864,
= 350 ÷ 36.
To change in m/s to km/hr so;
= (350÷36)×18÷5.
= 35km/hr answer.
S = D ÷ T formula;
S = 5s ÷ 7.
42km~convert in meter so, 4200.
1hr+40min+48sec.
= (60min × 60sec) + (40min × 60sec) + 48sec,
= 6048.
S = D÷T.
5s ÷ 7 = 4200 ÷ 6048.
= 4200×7 ÷ 6048 ÷ 5
= 8400 ÷ 864,
= 350 ÷ 36.
To change in m/s to km/hr so;
= (350÷36)×18÷5.
= 35km/hr answer.
(48)
Akshay said:
2 years ago
Short trick;
= 5/7×42;
= 30.
= 2xy/x+y
= 2 * 42 * 30/42 + 30
= 2520/72.
= 35km/hrs.
= 5/7×42;
= 30.
= 2xy/x+y
= 2 * 42 * 30/42 + 30
= 2520/72.
= 35km/hrs.
(106)
Jenny said:
2 years ago
As the ratio of the car's own speed is given;
5 is the new speed of the car and 7 is the old speed of the car.
As the distance is constant;
The ratio of time is reverse that is 7:5
Thus if 7 = 100.8 (hours into minutes).
1 = 14.4.
5 = 72.
Convert this into an hour, you will get 1.2.
42km distance divided by 1.2 times = 35kmph.
5 is the new speed of the car and 7 is the old speed of the car.
As the distance is constant;
The ratio of time is reverse that is 7:5
Thus if 7 = 100.8 (hours into minutes).
1 = 14.4.
5 = 72.
Convert this into an hour, you will get 1.2.
42km distance divided by 1.2 times = 35kmph.
(15)
Mohd younis said:
3 years ago
1hrs 40 min 48sec.
1 + 40/60 + 48/60 * 60 = 42/25 = 1.68 hours.
1 + 40/60 + 48/60 * 60 = 42/25 = 1.68 hours.
(32)
Kailash said:
3 years ago
@Nikita,
Please solve and explain the following step.
(204/5)/60 = 51/75.
Like if I do (204/5)/60 = 4.8/60 = 0.08.
Please solve and explain the following step.
(204/5)/60 = 51/75.
Like if I do (204/5)/60 = 4.8/60 = 0.08.
(7)
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