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.
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 |
 x = |
 |
42 x 7 x 75 |  |
= 35 km/hr. |
| 5 x 126 |
Discussion:
117 comments Page 2 of 12.
Some people said:
3 years ago
How come 126? Please explain.
(11)
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)
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)
Vijay said:
3 years ago
Thanks @Siddharth.
(4)
Varshini said:
5 years ago
1hr 40min 40sec = (60x60) + (40x60) + 48.
3600 + 2400 + 48.
6048.
Now converting into the min and then to hr.
6048/60 = 100.8 min.
100.8/60 = 1.68 hr
3600 + 2400 + 48.
6048.
Now converting into the min and then to hr.
6048/60 = 100.8 min.
100.8/60 = 1.68 hr
(2)
Poojith said:
4 years ago
1hr 40min and 48sec == 1.68hours.
5x/7 = 42/1.68,
5x/7 = 25.
5(25)/7 = 35km/hr.
5x/7 = 42/1.68,
5x/7 = 25.
5(25)/7 = 35km/hr.
(2)
Siddharth said:
1 decade ago
1 hr 40min 48sec= 1 + 40/60 + 48/(60*60) hrs
=1+ 2/3 + 4/300
=(300+200+4)/300
=504/300
=126/75
=1+ 2/3 + 4/300
=(300+200+4)/300
=504/300
=126/75
(1)
Sunil pawar said:
7 years ago
@All.
In simple method;
1 hour +40min +48 sec, we have to convert it to an hour.
1+(40/60)+(48/60*60),
1+2/3+(1/75),
=(75+2*25+1)/75,
=126/75.
Now speed * time = distance,
5x/7 * 126/75 = 42 , x is the actual speed of car,
Solving this will give answer x = 35.
In simple method;
1 hour +40min +48 sec, we have to convert it to an hour.
1+(40/60)+(48/60*60),
1+2/3+(1/75),
=(75+2*25+1)/75,
=126/75.
Now speed * time = distance,
5x/7 * 126/75 = 42 , x is the actual speed of car,
Solving this will give answer x = 35.
(1)
Goutham said:
6 years ago
Thanks @Nikita.
(1)
Vrushali said:
6 years ago
By using formula S = D/T.
S=(5/7)x is the actual speed.
(5/7)x = 42/{1+(40÷60)+(48÷3600)} by using DST
(5/7)x = 42/{1+(2÷3)+(1÷75)} by dividing terms we got;
(5/7)x = 42/{1+[(150+3)÷225]},
(5/7)x = 42/{[225+153]÷225},
(5/7)x = (42 * 225)÷(225+153),
(5/7)x = (9420)÷(378),
x = (9420÷378)*(7/5).
By cancelling terms we got;
X = 1890 ÷ 54.
By dividing we got;
X = 35kmph.
S=(5/7)x is the actual speed.
(5/7)x = 42/{1+(40÷60)+(48÷3600)} by using DST
(5/7)x = 42/{1+(2÷3)+(1÷75)} by dividing terms we got;
(5/7)x = 42/{1+[(150+3)÷225]},
(5/7)x = 42/{[225+153]÷225},
(5/7)x = (42 * 225)÷(225+153),
(5/7)x = (9420)÷(378),
x = (9420÷378)*(7/5).
By cancelling terms we got;
X = 1890 ÷ 54.
By dividing we got;
X = 35kmph.
(1)
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