thanks
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….
the pool will be half full at the 19th day .so that it doubles on the 20th day and becomes full
When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.
Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.
So, in a race of 1000 meters, A beats C by (1000 – 787.5) = 212.5 meters to C.
So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.
6084
7431 is largest number and 1347 is the least number from digits 1,3,4,7
so 7431-1347=6084
A
pass English =80%
fail=100-80=20%
pass Math = 70%
fail in Math=100-70=30%
fail in both=10%
total fail students= fail in Eng+fail in Math-common
= 20+30-10=40%
if 40% fail then 60% will pass
let total students=x
hence
60% 0f (total students)=144
60/100 of (x)= 144
x=(144×100)/60
X=240
total students=240
P+C+M=80
P+C=70
Therefore M=10
Given , P+M=90
If M=10
P=80
Given:
Speed of man downstream= 30/3=10km/h
Speed of man in upstream=18/3=6km/h
Speed of man in still water=1/2 (speed of upstream+ speed of down stream)
= 1/2 (10+6)
=1/2*16
=8km/h
9, 12, 11, 14, 13, (…..), 15
16
19, 26, 33, 46, 59, 74, 91
Answer 33
Adding 7,9.13,15,17
Correct is 35
Creating a virtual image of the search topic