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
ANS = 24.
16 = 2^4
24 = 2^3 x 3
H.C.F of 16 and 24 = 8
L.C.M will be 48
3 hours ago.
Thin candle melts 3/4 in 3 hours leaving 1/4
Where as in the same time thick candle melts 3/6 leaving 3/6 which is 1/2. Now thick candle is exactly twice than the thin candle.
Or via modeling:
We need to find time at which the length of the thin candle is half the thick candle. Let x be the time. Thin candle melts at 1/4 an hour and thick candle melts at 1/6 an hour. In x hours they melt at x/4 and x/6 respectively. What’s left will be 1 – x/4 and 1 – x/6. We need to find x at which :
2 * (1 – (x/4)) = 1 – (x/6)
This equation results in x = 3
The number is greater than the number obtained in reversing the digits and the ten’s digit is greater than the unit’s digit
Let ten’s and unit’s digit be 2x and x respectively
Then, (10×2x+x)−(10x+2x)=36
9x=36
x=4
Required difference=(2x+x)−(2x−x)=2x=8
6 inch
i think it is a and b
6 cuts
80
10%
Yellow Red Green
Blue Blue Green
Yellow Green Yellow
d=5
g=1
a=4, b=6,c=2,d=5,e=8,f=3,g=1,h=9,i=7
Second
#1: N = 1, f(N) = 1
#2: N = 199981, f(N) = 199981
#3: N = 199982, f(N) = 199982
#4: N = 199983, f(N) = 199983
#5: N = 199984, f(N) = 199984
#6: N = 199985, f(N) = 199985
#7: N = 199986, f(N) = 199986
#8: N = 199987, f(N) = 199987
#9: N = 199988, f(N) = 199988
#10: N = 199989, f(N) = 199989
#11: N = 199990, f(N) = 199990
#12: N = 200000, f(N) = 200000
#13: N = 200001, f(N) = 200001
#14: N = 1599981, f(N) = 1599981
#15: N = 1599982, f(N) = 1599982
CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
other..
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
on each.
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….
Speed of Stream = 1/2 (Downstream Speed – Upstream Speed)
=1/2(40-14)
=1/2(26)
=13km/h
method 2:
for down stream case :speed=Boat speed + Stream speed
for up stream case :speed=Boat speed – Stream speed
therefore, Bs + Ss =40
Bs – Ss =14
(-) (-) (-)
———————
2Ss=26
Ss=26/2 = 13Km/h
Stream speed= 13Km/hr