Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 14)
14.
A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:
Answer: Option
Explanation:
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle = (61 -x) km.
So, | x | + | (61 -x) | = 9 |
4 | 9 |
9x + 4(61 -x) = 9 x 36
5x = 80
x = 16 km.
Discussion:
37 comments Page 2 of 4.
Nikita said:
4 years ago
Here in this question, multiple of 4 is the distance traveled. Which is 16.
(3)
NOOWE ANNAH said:
2 years ago
Very interesting, Thanks, everyone.
(3)
Amit said:
6 years ago
By Using alligation method.
4:5=9.
So, 4 * 4 = 16 by walking,
5*9 = 45 by cycle.
4:5=9.
So, 4 * 4 = 16 by walking,
5*9 = 45 by cycle.
(2)
Sani ahire, roshan ahire said:
8 years ago
Total distance D = 61
4km/hr for 4 hr = 16
9km/hr for 5 hr = 45
Adding 45+16=61.
Hence answer =16.
4km/hr for 4 hr = 16
9km/hr for 5 hr = 45
Adding 45+16=61.
Hence answer =16.
(2)
Dhivagar said:
1 decade ago
Hi, its simple,
Total distance = 61 km.
Time taken = 9 hr.
Speed by foot = 4 km/hr.
Speed by bicycle = 9 km/hr.
So split the total hours into 2 parts like 4 hr + 5 hr = 9 hrs.
Multiply it with the speed by foot and bicycle like.
4 hr*4 km/hr = 16 km.
5 hr*9 km/hr = 45 km.
Add 16 + 45 = 61 km (total distance covered).
They asked distance covered by foot. So answer is 16. :).
Total distance = 61 km.
Time taken = 9 hr.
Speed by foot = 4 km/hr.
Speed by bicycle = 9 km/hr.
So split the total hours into 2 parts like 4 hr + 5 hr = 9 hrs.
Multiply it with the speed by foot and bicycle like.
4 hr*4 km/hr = 16 km.
5 hr*9 km/hr = 45 km.
Add 16 + 45 = 61 km (total distance covered).
They asked distance covered by foot. So answer is 16. :).
(1)
Aloki said:
1 year ago
@All.
Here's my explanation.
Basically, I deduced the given information into two equations:
df = 4tf -----> (1) [ Distance = Speed x Time (Speed mentioned)]
db = 9 x (9 - tb) --->(2) (Same as above)
Where,
df = distance on foot,
tf = time taken on foot,
db = distance on a bicycle,
tb = distance on bicycle
Now, db = 61 - df, as df + db = 61 (given),
Also, tb = 9 - tf, as tf + tb = 9 (given),
So now keeping values together we get,
df = 4tf-->from (1)
61 - df = -5tf + 81 -->from (2)
Solving these two equations, we get the value of tf which comes out to be 4 hours and hence putting this value in (1) we get, df = 16km!
Here's my explanation.
Basically, I deduced the given information into two equations:
df = 4tf -----> (1) [ Distance = Speed x Time (Speed mentioned)]
db = 9 x (9 - tb) --->(2) (Same as above)
Where,
df = distance on foot,
tf = time taken on foot,
db = distance on a bicycle,
tb = distance on bicycle
Now, db = 61 - df, as df + db = 61 (given),
Also, tb = 9 - tf, as tf + tb = 9 (given),
So now keeping values together we get,
df = 4tf-->from (1)
61 - df = -5tf + 81 -->from (2)
Solving these two equations, we get the value of tf which comes out to be 4 hours and hence putting this value in (1) we get, df = 16km!
(1)
Kawla said:
8 years ago
How can 5x=80? just explain it.
Sidharth.j said:
9 years ago
Please someone helps me, they have asked only for bare foot then why are we considering the bicycle also?
v = d/t:
we know the time for the full trip (i.e) 9 hours.
He walks bare foot at speed of 4kmph;
why can't we just multiply and find the answer?
d = v * t;
d = 9 * 4.
d = 36km.
v = d/t:
we know the time for the full trip (i.e) 9 hours.
He walks bare foot at speed of 4kmph;
why can't we just multiply and find the answer?
d = v * t;
d = 9 * 4.
d = 36km.
JPriya said:
8 years ago
Partly means is it not (x/2 )/4+(x/2)/9=9 if d is taken as the total distance?
Please tell me.
Please tell me.
Mohanapriyasavithri said:
8 years ago
This can be solved easily by the rule of alligation.
Take average speed as 61÷9 (speed formula is ditance÷time).
Take the dearer value cheaper value & apply alligation rule.
Take average speed as 61÷9 (speed formula is ditance÷time).
Take the dearer value cheaper value & apply alligation rule.
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