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125
F – Murderer
G – Victim
H – Judge
M – Police
C – Witness
W – Hangman
33
b
as i assume the time taken to complete the job is not to be
changed while increasing the employees as it does not
clearly states so……….
30 men wrk for 9 hrs
therefore for the work to be done if no of workers is
increases the no of hrs / day to work should decrease so
here s a inverse relation
solving further we get
30 = K /9 ==> K= 270
now
40 = 270/y ==> y= 6.75
therefore ans is d)none
Monday =y cheques
Tuesday =3y cheques
wednesday=4000 cheques
totaling to 16000 cheques
therefore y+3y+4000=16000 cheques
4y+4000=16000 cheques therefore 4y=16000-4000 cheques= 4y=12000cheques
therefore y=12000/4=3000 cheques
monday=3000 cheques
tuesday=3×3000=9000cheques
300%
Let x be fathers present age and y be son present age.
5 yrs ago, the age of father and son be x-5 & y-5.
Then,
x-5+y-5=40
x+y-10=40
x+y=50
y=50-x ———–> (1)
ratio between father and son in present age
x:y=4:1
x/y=4/1
x=4y
Apply eq (1) ,
x=4(50-x)
x=200-4x
x+4x=200
5x=200
x=200/5
x=40,,
=> The present age of father is 40.
430
5x5x6 = 150 cube
150 – 20( corner cube) = 130 cube
130 x 3 (side remaining) = 390
20 x 2 (side remaining) = 40
total side remaining = 390 + 40 = 430
Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.
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.