Average = (1+2+3+4+5+6+7+8+9)/3
= 45/3 = 15
Every side will contain sum of 15.
1st side contains 8 and 7.
2nd side contains 9 and 6.
3rd side contains 1,2,3,4 and 5.
75/8 days
let x be the avg age of each person in the family ten years
back…
so total age=6x
after 10 years i.e,now the total age of the family =6x+60
let grand father age as y
then the average of the family=(6x+60-y)/5
(bcoz grand father has expired and the new born baby has no
age at present)
according to the given question (6x+60-y)/5=x
bcoz x is the avg age bfore 10 years.
while calc the equation according to the presnt age we have
to add 10 to the result….
so the answer is 70
(c) 25%
20*55*65=y*65*75
y=(20*55*65)/(65*75)
y=14.667 =15
y=15
% reduction=(20-15)/20=25%
125
8:20
– 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
value remains same
MONKEY=YEKNOM then each letter is moved backward =XDJMNL so for TIGER=REGIT the each of the reversed letters are moved backward=QDFHS
The number of ways of selecting a group of eight is
5 men and 3 women=5C5×6C3 =20
4 men and 4 women=5C4×6C4 =75
3 men and 5 women=5C3×6C5=60
2 men and 6 women=5C2×6C6=10
Thus the total possible cases is 20+75+60+10=165.
That will depend heavily on when in the journey the 200 men join.
1 : 2
put 1 red marble in one jar and all the rest (99) in the
other.
This gives you 50% from the first jar (if they pick that
jar they will get red 100% of the time). For the other jar
the chances are 49/99 or 49.494949%. Divide that by 2 and
its 24.7474%. Total odds are 50% plus 24.7474% = 74.7474%
A can copy 50 papers in 10 hrs
that means 5 papers in 1 hour
A & B can copy 70 papers in 10 hrs
that means both of them copy 7 papers in 1hour
we know A can copy 5 papers
there fore B can copy 2papers in 1 hour
so B can copu 26 papers in x hours
2*x=26
x=26/2=13 hours
(N * 1.1) * 0.9 = 7920
N=8000
just will draw the line first and then draw three concentric circles
no