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24%
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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
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A. x/6 - 6
B. x/6 - 5
C. x/7 - 5
D. x/5 - 7
E. x/5 - 6
_________________
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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
Joined: 25 Jul 2011
Last visit: 06 Oct 2025
Posts: 53
Location: India
Concentration: Strategy, Operations
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
GPA: 3.5
WE:Engineering (Energy)
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
Posts: 53
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
Bunuel
AVG of y Numbers is x, so SUM is yx..
Add 30 to it yx+30..
So new avg = \(\frac{yx+30}{1+y}=x-5.....yx+30=x-5+xy-5y.....35=x-5y......y=\frac{x-35}{5}=\frac{x}{5}-7\)
D
_________________
Bunuel
Sum =Average *Number
Sum of y numbers is xy
Sum of y+1= (x-5)*y+1
xy+30 =(x-5)*(y+1)
xy+30=xy-5y+x-5
30+5=x-5y
x=5(y+7);
y=x/5-7
So Ans:D
Bunuel
If the equation is hard to set up, a simple A*n = S table can help.
The key is that one occurrence of a numeric value added or subtracted to the set of numbers gets its own row; then add column "n" to get total number and column "S" to get total sum (averages CANNOT be added).
__A__|__n__|___S____|
__x__|__y__|___xy____|
_30__|__1__|___30___|
-------|_y+1_|_xy+30__|
The new average is (x - 5). That doesn't get written in the table anywhere. It's the RHS of the equation.
LHS = \(\frac{Total Sum}{Total \\
Number}\)
\(\frac{(xy + 30)}{(y+1)} = (x - 5)\)
\(xy + 30 = (x - 5)(y + 1)\)
\(xy + 30 = xy + x - 5y - 5\)
\(5y = x - 35\)
\(y = \frac{x}{5} - 7\)
ANSWER D
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