suppose
pipe:
A -30 hours A’s effeciency (60/30) =2
60( lcm of 30 and 20)
B- 20 hours B’s effeciency (60/20)=3
time taken by both to fill = 60/5 =12 as given in question (effeciencies of both a+b =2+3=5)
time taken by faster pipe i.e b = 60/3 =20
let x be lenghth of candles.
thicker one—- in 1 hr 1/6th will go.
thinner one— in 1 hr 1/4th will go.
let he lit for n hrs.
so in n hrs thicker one goes n/6.
and thinner one goes n/4.
so remainini length are x-n/6 and x-n/4.so x-n/6=2(x-n/4)
implie n=3x. if x=1 n=3hrs.
d
temp at 3.oo pm is = 22.5
temp at 6.oo pm is = 30
thus, percentage rise= 7.5/30 *100
=25%
Let the number of males be given the name M.
Let the number of females be given the name F.
If 15 females are absent, then M will be twice that of
present females.
This means that M = 2 * (F – 15)
M = 2 * F – 30.
or 2 * F – M = 30.
Now if in addition to the 15 females being absent, we also
have 45 males being absent,
then this gives the equation,
(F – 15) = 5 * (M – 45)
which simplifies to
F – 15 = 5 * M – 225
5 * M – F = 210
Pulling the equations together, we get
5 * M – F = 210
-M + 2 * F = 30
Multiply the first equation by 2, and keep the second
equation as is.
10 * M – 2 * F = 420
– M + 2 * F = 30
Add the equations.
9 * M = 450
M = 50
Verify answer.
Calculate F
from – M + 2 * F = 30
-50 + 2 * F = 30
2 * F = 30 + 50
F = 40.
If 15 females are absent, then number of males will be twice
that of females.
40 – 15 = 25.
50 = 2 * 25. Confirmed.
If also 45 males were absent, then female strength would be
5 times that of males.
Female strength is 25 due to the 15 females being absent.
50 – 45 = 5.
25 = 5 * 5. Confirmed.
65 KM
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
12
x,y sides of rectangle
area=xy
increased by 100% =double length
hence
new area=4xy
increased area=4xy-xy=3xy
percentage of area will be increased
by 300%
41, 43, 47, 53, 61, 71, 73, 81
81
East
Statements :
All poles are guns. Some boats are not poles.
Conclusions :
I. All guns are boats.
II. Some boats are not guns.
C
20 %
Since, there are 10 points on the circle and to draw a chord we need to connect any two points on the circle to make it a straight line, which implies that the number of chords = No of lines connecting any two points out of the 10 points
= 10C2 = 10*9/2 = 45 chords.