Please help...
During an experiment, some water was removed from each of 6 water tanks. If the standard deviation of the volumes of water in the tanks at the beginning of the experiment was 10 gallons, what was the standard deviation of the volumes of water in the tanks at the end of the experiment?
1) For each tank 30% of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment
2) The average (arithmetic mean) volume of water in the tanks at the end of the experiment was 63 gallons
GMAT prep test
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sd at the beginning of expt = 10
we nned to find sd at the end of the expt.
Stm1
30% water has been removed from each tank, so new sd now is 0.30(old sd) = 0.30*10 = 3...suff
stm2
Does not tell us anything about the variation...insuff
answer is A
we nned to find sd at the end of the expt.
Stm1
30% water has been removed from each tank, so new sd now is 0.30(old sd) = 0.30*10 = 3...suff
stm2
Does not tell us anything about the variation...insuff
answer is A
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I should have added this info before, but some general rules for std dev (sd) are:
1. sd (x+c) = sd(x) where c is a const
2. sd (xc) = c*sd(x)
To illustrate 1., 6 water tanks have different water levels with sd =10 units at the beginning of the expt. 50 units was added to each of the 6 tanks., new sd will still be 10
To illustrate 2., 6 water tanks have different water levels with sd =10 units at the beginning of the expt. 30% vol of water of each tank was removed, new sd will be 0.3*10 = 3 units
1. sd (x+c) = sd(x) where c is a const
2. sd (xc) = c*sd(x)
To illustrate 1., 6 water tanks have different water levels with sd =10 units at the beginning of the expt. 50 units was added to each of the 6 tanks., new sd will still be 10
To illustrate 2., 6 water tanks have different water levels with sd =10 units at the beginning of the expt. 30% vol of water of each tank was removed, new sd will be 0.3*10 = 3 units
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1.
each be a1. a2 a3 a4 a5 a6
new .7a1, .7a2 , .7a3 etc
old mean = x
new mean mean == .7 x
SD new = .70 SD old
Hence sufficient So IMO A
each be a1. a2 a3 a4 a5 a6
new .7a1, .7a2 , .7a3 etc
old mean = x
new mean mean == .7 x
SD new = .70 SD old
Hence sufficient So IMO A
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