– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds
a. (x+y)'=x'.y' b. (x'+y')'=x.y
c. (x'.y')'=x+y d. (x'+y')'=x'.y'
C
Nice
65292
Statements :
Some books are pens. No pen is pencil.
Conclusions :
I. Some books are pencils.
II. No book is pencil.
E
1/32(x)L=21h
10L=1h
X=21x10x32
=6720 Liters
Thus:
6720L=Yh
10L=1h
Yh=6720/10
Y=672h
x- sum of first 4 values
y – 5th value
x/4 = 45
x=45*4=180
(x+y)/5 = 46
(180+y)/5=46
180+y=230
y=50
EALIME
4 days
The rational number that does not have a reciprocal 0 because reciprocal of 0 is undefined.
38 years
Let Rajan’s present age be x years. Then, his age at the time of marriage = (x – 8) years.
x = 65(x−8)
⇒5x=6x−48
⇒x=48 years
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = (x – 18) = 30 years
∴ Rajan’s sister’s present age = (30 + 8) years = 38 years
60%
May 21
.
Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.
610 × 717 × 1127
= (2 × 3)10 × 717 × 1127
= 210 × 310 × 717 × 1127
Number of prime factors in the given expression
= (10 + 10 + 17 + 27)
= 64