Last visit was: 11 Sep 2024, 05:37 It is currently 11 Sep 2024, 05:37
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

Tags:
Show Tags
Hide Tags
Manager
Joined: 03 Jun 2019
Posts: 220
Own Kudos [?]: 11674 [11]
Given Kudos: 49
GRE Forum Moderator
Joined: 02 Nov 2016
Posts: 14018
Own Kudos [?]: 38708 [1]
Given Kudos: 5865
GPA: 3.62
General Discussion
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Posts: 3127
Own Kudos [?]: 2840 [2]
Given Kudos: 1511
Location: India
GPA: 4
WE:Analyst (Internet and New Media)
VP
Joined: 14 Aug 2019
Posts: 1326
Own Kudos [?]: 864 [0]
Given Kudos: 381
Location: Hong Kong
Concentration: Strategy, Marketing
GMAT 1: 650 Q49 V29
GPA: 3.81
Re: Consider the sets S, T and U, where S={31,14,64,22,43,66} U={y,31,14,6 [#permalink]
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;
Senior Manager
Joined: 15 Jul 2018
Posts: 348
Own Kudos [?]: 478 [0]
Given Kudos: 94
Re: Consider the sets S, T and U, where S={31,14,64,22,43,66} U={y,31,14,6 [#permalink]
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
VP
Joined: 29 Oct 2015
Posts: 1204
Own Kudos [?]: 541 [0]
Given Kudos: 735
GMAT 1: 570 Q42 V28
Re: Consider the sets S, T and U, where S={31,14,64,22,43,66} U={y,31,14,6 [#permalink]
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} ?
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11513
Own Kudos [?]: 35932 [0]
Given Kudos: 333
Consider the sets S, T and U, where S={31,14,64,22,43,66} U={y,31,14,6 [#permalink]
parkhydel wrote:
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.

­
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
Johnson Moderator
Joined: 19 Jan 2024
Posts: 388
Own Kudos [?]: 290 [1]
Given Kudos: 145
Schools: Johnson '27
Re: Consider the sets S, T and U, where S={31,14,64,22,43,66} U={y,31,14,6 [#permalink]
1
Kudos
sayan640 wrote:
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} ?

­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.
Re: Consider the sets S, T and U, where S={31,14,64,22,43,66} U={y,31,14,6 [#permalink]
Moderators:
Math Expert
95450 posts
Retired Moderator
997 posts
RC & DI Moderator
11513 posts