if initially the price was Re 1.the new price nw bcoms to b
0.7..
so ,if the price z 0.7 ,the increment required is 0.3
so % in crement= 0.3/0.7*100=42.87…
NTFS — New Technology File System
Fat — file allocation table
NTFS having a quota, commpress system with securites base as
administrator giving, multipal user, groups, to set permission
ans is C.
let H be present age of husaband.W be present age of wife.
H+W=91.
now let diff between their ages be x.that is H-W=x.
now when husband is W yaers old, his wife must be W-x years
old.and it is given that H=2(W-x). so 2W-H=2x. and H-
W=x.eliminating x we get 4W=3H. but H+W=91, so solving thse
two H=52 W=39.
Ans: 60 kph
Suppose Person meets the train everyday at 3 PM at Station A.
His speed is 12kph.
So normally he reaches 5 km before the meeting point (pt B) at (5/12 hr = 25 min before) 2:35PM.
But if he is late by 30 min, then he will reach that point (pt B) by 3:05 PM.
Train is traveling at its normal speed so it covers the distance of 5 Km in 5 min starting from Station A and reaches the meeting point (pt B) at 3:05 PM.
So speed of the train is 5KM/5min = 60 kph.
b
5c1 / 9c1
3/7
No, the tank will be filled after 36 minutes
24/16= 1,5
24×1,5= 36
let square be x (squares are 4 sides)
i.e., X+X+X+X=4X
4X+3=1460
4X=1460-3
4X=1457
X=364.25
Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125
speed = distence / time
60km/hr=distance / 9 sec
km / hr to m / s==> 60*1000 / 1* 3600 = 16.667
16.667*9 = distance
150 m is the answer
Speed = (45×518)m/sec=252m/sec
Total distance covered = (360+140) m = 500 m.
∴Required time= 500×225sec = 40 sec.
One day work = 1 / 20
One mans one day work = 1 / ( 20 * 75)
Now:
No. Of workers = 50
One day work = 50 * 1 / ( 20 * 75)
The total no. of days required to complete the work = (75 * 20) / 50 =
30