we know,
area=b*h
b gets increased by 20% i.e (b+0.20b)
h gets decreased by 20% i.e (h-0.20h)
rewriting the equation(area=b*h),
area=(b+0.20b)*(h-0.20h)
area=b(1+0.20)*h(1-0.20)
area=b(1.20)*h(0.80)
area=b*h*(1.20)*(0.80)
area=b*h*(0.96)
i.e new area=0.96 times the original area
if 100% was the original area,it has decreased to 96%
so,100%-96%= 4%
10C3 = 120
Time taken to fill repaired cistern = 12 mins
In 1 min, pipe will fill = 1/12
Time taken to fill cistern with leak = 12 + 18 = 30 mins
In 1 min, it will fill = 1/30
Consider the leak as an outlet pipe, so
1/30 = 1/12 – [ in 1 min how much the outlet pipe leaks ]
= 1/12 – 1/30
= 5/60 – 2/60 = 3/60 = 1/20
Therefore, it will take 20 mins for the leak to empty the cistern
NCPQJG
– 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
Diagonal of square=1.414* a, where a is side of square.
ans=1.414*20
ans=28.28
( a ) 10 metres
a = 2
d = 5 – 2 = 3
Then 12th term is :
This in arithmetic progression
L = a + (12 – 1) d
L = 2 + 11 * 3
L = 2 + 33 = 35
Ramji sir
1056/(2*pi*r)=1056/(2*3.14*14)=1059/88 =12 revolution
X-340 = X x 4 x 8/ 100
100X – 34000 = 32X
68X = 34000
X = 34000/68
X= 500.
5A=3B,
*3 SO,
15A=9B
5
108
the answer is At 9:48 PM
At 1:00 pm the difference between A & B = 8 km
after 2:00 pm ………………. = 11 km (as B’s speed
is 1 and A’s 4 km, then eqv speed=(4-1)=3 km)
After 3:00………………….. = 13 km (as B’s speed 2 km)
After 4:00………………….. = 14 km
after 5:00………………….. = 14 km (A’s speed= B’s
speed)
after 6:00………………….. = 13 km
after 7:00………………….. = 11 km
after 8:00………………….. = 8 km
after 9:00………………….. = 4 km
and now the eqv speed is= (9-4) =5 km/hr;
and the renaming distance is 4 km;
then, time=(60*4)/5=48 min;
then the meeting time is=9:00+48 min=9:48 pm;