Consider the sets S, T and U, where S={31,14,64,22,43,66} U={y,31,14,6

Question

Consider the sets S, T and U, where

S = {31, 14, 64, 22, 43, 66},

T = {x, 31, 14, 64, 22, 43, 66}, and

U = {y, 31, 14, 64, 22, 43, 66}

The mean of T is 5 less than the mean of S. The median of U is 4 less than the median of S.

Select the value of x and the value of y consistent with the statements given. Make two selections, one in each column.

Sajjad1994 · Accepted Answer

Official Explanation

RO1

To find the mean of T, sum the values in the set and divide by the number of values:  \frac{x+31+14+64+22+43+66}{7} = \frac{x+240}{7} . Similarly, the mean of S is  \frac{31+14+64+22+43+66}{6} = \frac{240}{6} = 40.  Since the mean of T is 5 less than the mean of S,  \frac{x+240}{7} = 40 – 5 = 35.  Multiplying each side of the equation by 7, x + 240 = 245. Thus, x= 245 − 240 = 5.

The correct answer is 5.

RO2

To find the median of S, first arrange the values in order: 14, 22, 31, 43, 64, 66. Because there is an even number of values, the median is half the sum of the two middle values:  \frac{31+43}{2} = 37 . The median of U is 4 less than the median of S, or 33. U contains an odd number of values, so its median must be one of the values. Since 33 is not among the known values in U, it must be the value of y.

The correct answer is 33.

yashikaaggarwal · Answer

S={31,14,64,22,43,66} = (14,22,31,43,64,66)
Mean = 14+22+31+43+64+66/6
Mean = 240/6
Mean = 40
Median = average of middle term of even no. set
Median = 31+43/2
Median = 37

T={x,31,14,64,22,43,66} and => (X,14,22,31,43,64,66)
Mean = 14+22+31+43+64+66+x/7
Mean = 240+X/7
Set T Mean is 5 less than Mean of Set S
Mean S -5 = Mean T
40 -5 = 240+x/7
35*7 = 240+x
245-240 = x
x = 5

U={y,31,14,64,22,43,66} =>(14,22,31,Y,43,64,66)
The median of U is 4 less than the median of S.
Median S = 37
Median U = 37-4 = 33 = Y

(X,Y) = (5,33)

mSKR · Answer

Mean of S –Mean of T= 5
(other values )/6- ( x + other values )/7= 5
7o-6x-6o= 210
O= 31+14+64+22+43+66= 240
240-6x = 210
x= 5

2. median of U is less than median of S (14,22,31,43,64,66)= 31+43)/2= 37
Median of u = 33 ; 14,22,31,y,43, 64,66
>> y= 33;

Apt0810 · Answer

Men of set S = Men of set T + 5
Men of set S= 14+22+31+43+64+66/6 = 240/6=40

Mean of set T =14+22+31+43+64+66+x/7 = 240+x/7

So here 40 = 240+x/7 +5
35*7 = 240+x
245=240+x
x=5

Now Median of U = Median of S - 4
Median of S =37
So,median of U = 37-4=33=y

Posted from my mobile device

sayan640 · Answer

KarishmaB   MartyMurray  Is there any quick and innovative way to find the mean of so many numbers in set S = 
S = {31, 14, 64, 22, 43, 66} ?

chetan2u · Answer

You should observe that all the values in S are also there in other sets, so you just require to calculate mean once.

sayan640 , you have a calculator to ease your job. But if you wanted to do the calculations by yourself, S = {31, 14, 64, 22, 43, 66}: the units digit (4,4,2) and (1,3,6) will give you 10 each, so total 20. Ten's digit = 3+1+6+2+4+6 = 22.....Total = 220+20 = 240 and mean = 240/6 = 40

Now, average of T is 5 less.  x would be 40 if the average remained same.   But 5 less means each of the 7 elements in T are to be reduced by 5,  and this reduction is catered by x, so x= 40-5*7 = 5.
OR
Average of T =  \frac{x+ 31+14+64+ 22+ 43+ 66}{7} = 35......\frac{x+240}{7}=35......x=245-240=5

Next, let us check on Median
Median of S = average of middle two values as there are even number of elements =  \frac{31+43}{2 } = 37 
So Median of U = 37-4 = 33.  As U has odd number of elements, the median should be an element ,  but there is no 34 in the given values.  So, y=33

playthegame · Answer

You could try to use the assumed mean method for S, meaning assume the mean is 35 for that set and then find deviation of each number from the assumed mean, -21, -13, -4, +8, +29, +31 which comes out to +30 divided by 6 gives you 5 so the mean is 40. Now, waht can you do to neutralize this 5 point increase for set T, add - 30 in or in other words add 5 to the set so the deviation from the assumed mean is now 0 to get to the mean at 35 which is 5 less than the mean for S.

Hope this helps, you can always also use the calculator as well or test options....

Thanks and good luck.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Consider the sets S, T and U, where S={31,14,64,22,43,66} U={y,31,14,6

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Two-part Analysis (TPA) Questions

Consider the sets S, T and U, where S={31,14,64,22,43,66} U={y,31,14,6

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards