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….
810
x = 5/3y
x – 4 = y + 4
x – 8 = y
x = 5/3 * (x-8)
5/3x – 40/3 = x
2/3x = 40/3
2x = 40
x = 20
y = 20 – 8
y = 12
x + 4 = 24
y – 4 = 8
Ratio is 24:8 , so 3:1
let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240
volume of cylinder=volume of plane
pi*r*R*H=L*B*H
16PI*H=176
SO h= length=176/16*pi
ANser is 44pi
1.5
190, 166, 145, 128, 112, 100, 91
128 is incorrect. There is a difference of multiples of 3 in the number pattern, which starts at 24 and reduces by 3 each time. Therefore the correct pattern would read 190, 166, 145, 127, 112, 100, 91.
this would leave the pattern with differences in the pattern 24, 21, 18, 15, 12, 9. The next number in the pattern would be 6 lower than 91. 91 – 6 = 85. Then 85 -3 – 82, followed by 82 – 0 = 82. After this, if -3 were taken the pattern would start to rise so 82 – – 3 = 85.
7 min clock:|——-7——-|
4 min clock:|—–4—-|—-4—–|—–4—-|—–4—-|
you got 9 min: |—————9————–|
original cost is 100^2 / (100^2 – p^2) Rs
2:1