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Numerical Ability
Time Distance and Speed
An Engine length 1000 m moving at 10 m/s. A bird is flying from engine to end with x kmph and coming back at 2x. Take total time of bird traveling as 187.5 s. Find x and 2x.
Read Solution (Total 3)
-
- suppose birds speeds in mtrs/sec is s and 2s .
While flying from Engine to end, relative speed = (s+10) m/sec
from end to engine, flying speed = (2s-10) mtr/sec
so
1000/(s+10) + 1000/(2s-10) = 187.5 secs
solving it, we get
so s= 8.728 m/sec and 2s= 17.456 m/sec - 12 years agoHelpfull: Yes(36) No(10)
- total velocity is 1000/3x = 187.5s
x=0.5625m/s...
- 10 years agoHelpfull: Yes(1) No(4)
- Ans:
x
=
31.43
x=31.43 km/hr
2
x
=
62.86
2x=62.86 km/hr
Let bird is flying from engine to end with
y
y m/s and coming back at
2
y
2y m/s
Relative speed when bird is flying from engine
=
y
+
10
=y+10 m/s
Relative speed when bird is coming back to engine
=
2
y
−
10
=2y−10 m/s
Total time taken by the bird = 187.5 s. Therefore,
1000
y
+
10
+
1000
2
y
−
10
=
187.5
1000y+10+10002y−10=187.5
1000
(
2
y
−
10
)
+
1000
(
y
+
10
)
1000(2y−10)+1000(y+10)
=
187.5
(
y
+
10
)
(
2
y
−
10
)
=187.5(y+10)(2y−10)
1000
(
2
y
−
10
+
y
+
10
)
1000(2y−10+y+10)
=
187.5
(
2
y
2
−
10
y
+
20
y
−
100
)
=187.5(2y2−10y+20y−100)
1000
(
3
y
)
=
187.5
(
2
y
2
+
10
y
−
100
)
3000
y
=
375
y
2
+
1875
y
−
18750
375
y
2
−
1125
y
−
18750
=
0
y
2
−
3
y
−
50
=
0
y
=
3
±
√
9
+
200
2
=
3
±
√
209
2
=
3
±
14.46
2
=
8.73
1000(3y)=187.5(2y2+10y−100) 3000y=375y2+1875y−18750 375y2−1125y−18750=0 y2−3y−50=0 y=3±9+2002=3±2092 =3±14.462=8.73
i.e., bird is flying from engine to end with 8.73 m/s
=
8.73
×
18
5
=
31.43
=8.73×185=31.43 km/hr
=>
x
=
31.43
x=31.43 km/hr
2
x
=
31.43
×
2
=
62.86
2x=31.43×2=62.86 km/hr - 7 years agoHelpfull: Yes(0) No(2)
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