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13 Sep 2017, 03:32
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D
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Difficulty:
25%
(medium)
Question Stats:
76% (02:02) correct
24%
(02:09)
wrong
based on 187
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The average (arithmetic mean) of y numbers is x. If 30 is added to the set of numbers, then the average will be x - 5. What is the value of y in terms of x ?
A. x/6 - 6
B. x/6 - 5
C. x/7 - 5
D. x/5 - 7
E. x/5 - 6
A. x/6 - 6
B. x/6 - 5
C. x/7 - 5
D. x/5 - 7
E. x/5 - 6
13 Sep 2017, 03:56
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Let sum of y numbers by S(y)
\(So Average = \frac{S(y)}{y}= x => S(y) = x*y\)
Now 30 is added to the list.
=> Sum = S(y)+30
No. of terms = y+1
\(So average = \frac{S(y) +30}{y+1}= x-5\)
=> S(y)+30 = (x-5)(y+1)
=> xy+30 = xy+x-5y-5
=> x-5y=35
=>5y=x-35
=>\(y= \frac{x}{5} -7\)
Answer: D
\(So Average = \frac{S(y)}{y}= x => S(y) = x*y\)
Now 30 is added to the list.
=> Sum = S(y)+30
No. of terms = y+1
\(So average = \frac{S(y) +30}{y+1}= x-5\)
=> S(y)+30 = (x-5)(y+1)
=> xy+30 = xy+x-5y-5
=> x-5y=35
=>5y=x-35
=>\(y= \frac{x}{5} -7\)
Answer: D
devarshi9283
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GMAT 1: 740 Q49 V41
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Schools: NUS '20 (D) Jones '20 (A$) Mays '20 (A$) Jindal '20 (A$) Cox '20 (A$) Kenan-Flagler '20 (WL)
GMAT 1: 740 Q49 V41
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13 Sep 2017, 03:57
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Bunuel
Since mean of "y" numbers is "x"...sum of these no. = mean (or average)* number of nos=x*y
again..if a no 30 is added..sum will be (xy+30) and no of numbers will be y+1;
reapplying same formula... (xy+30)=(x-5)(y+1) as new mean is (x-5) as per question stem;
so...solving we get (xy+30)=xy+x-5y-5
35=x-5y
5y=x-35
or y=x/5-7 hence (D)
hit kudos if this was helpful
