GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Aug 2019, 00:06 ### 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
Your Progress

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. ### Request Expert Reply # Set S contains all the integers from 10 to 99. S1, a subset of S, cont

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3018
Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

11 00:00

Difficulty:   85% (hard)

Question Stats: 61% (03:20) correct 39% (03:01) wrong based on 110 sessions

### HideShow timer Statistics

Question of the Week #15

Set S contains all the integers from 10 to 99. $$S_1$$, a subset of S, contains all the numbers of S, in which both the digits are even. $$S_2$$, also a subset of S, contains all the numbers of S, in which both the digits are odd. What is the ratio of sum of all elements in $$S_1$$ to sum of all elements in $$S_2$$?

A. $$\frac{108}{275}$$

B. $$\frac{216}{275}$$

C. $$\frac{2}{3}$$

D. $$\frac{275}{216}$$

E. $$\frac{3}{2}$$

_________________
##### Most Helpful Community Reply
Intern  B
Joined: 03 Sep 2018
Posts: 7
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

5
4
Analysis (20 seconds): Looks like I need to find the size of each set, use the sum of a series formula to find their respective sums and simplify the fraction. I also notice that the numerators of the answer choices are distinct so I'll be focussing on reducing the numerator until I see a match.

Strategy: Find n for both sets, Calculate sum of each series, Reduce the fraction focussing on the numerator

Find n (40 seconds)
S1 -> n = 4 * 5 = 20, first element = 20, last = 88
S2 -> n = 5 * 5 = 25, first element = 11, last = 99

Calculate sums (45 seconds)
$$Sum = \frac{n(a1 + an)}{2}$$
$$S1 = \frac{20(20 + 88)}{2} = 10 * 108$$
$$S2 = \frac{25(11 + 99)}{2} = \frac{25 * 110}{2} = 25 * 11 * 5$$

Reduce & Eliminate (30 seconds)
$$\frac{10 * 108}{25 * 11 * 5} = \frac{2 * 108}{25 * 11} = \frac{216}{...}$$

Answer = B
Total Time: 2:15
##### General Discussion
Manager  G
Joined: 23 Aug 2016
Posts: 105
Location: India
Concentration: Finance, Strategy
GMAT 1: 660 Q49 V31 GPA: 2.84
WE: Other (Energy and Utilities)
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

1
EgmatQuantExpert wrote:
Question of the Week #15

Set S contains all the integers from 10 to 99. $$S_1$$, a subset of S, contains all the numbers of S, in which both the digits are even. $$S_2$$, also a subset of S, contains all the numbers of S, in which both the digits are odd. What is the ratio of sum of all elements in $$S_1$$ to sum of all elements in $$S_2$$?

A. $$\frac{108}{275}$$

B. $$\frac{216}{275}$$

C. $$\frac{2}{3}$$

D. $$\frac{275}{216}$$

E. $$\frac{3}{2}$$

Quite a lengthy One!

Set S1= { 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62,64,66,68,80,82,84,86,88}= Sum(1080)
Set S2={ 11, 13,15,17,31,33,35,37,39,51,53,55,57,59,71,73,75,77,79,91,93,95,97,99}= Sum(1375)
Sum of S1/Sum of S2= 1080/1375= 216/275.

Answer B
_________________
Thanks and Regards,

Honneeey.

In former years,Used to run for "Likes", nowadays, craving for "Kudos". :D
Intern  B
Joined: 08 Apr 2018
Posts: 16
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

Answer would be 'B'.

Within {0 ... 9}, the even digits would be {0, 2, 4, 6 and 8} and similarly the odd digits would be {1, 3, 5, 7, 9}

Here, since we are referring to 2 sets, S1 where both the digits are even, hence all of the numbers which are going to be included will contain both the digits from {0, 2, 4, 6, 8}.

... and for S2, where it contains all the numbers of S, in which both the digits are odd, hence all of the numbers which are going to be included will contain both the digits from {1, 3, 5, 7, 9}.

S1 would contain the following sets - {20, 22, 24, 26, 28} , {40, 42, 44, 46, 48} , {60, 62 , 64, 66, 68} and {80, 82, 84, 86, 88}

Similarly, S2 would be containing the following - {11, 13, 15, 17, 19} , {31, 33, 35, 37, 39} , {51, 53, 55, 57, 59} , {71, 73, 75, 77, 79} and {91, 93, 95, 97, 99}

All of the individual sets within S1 and S2 is having the same common difference as 2 and the number of terms as 5. Utilizing the formula for a sequence in arithmetic progression, we would be able to determine the sum of the indivual series and add those up to determine the final sum.

S1 = 5/2 [ 40 + (4*2) ] + 5/2 [ 80 + (4*2) ] + 5/2 [ 120 + (4*2) ] + 5/2 [ 160 + (4*2) ]
= 120 + 220 + 320 + 420
= 1080

S2 = 5/2 [ 22 + (4*2) ] + 5/2 [ 62 + (4*2) ] + 5/2 [ 102 + (4*2) ] + 5/2 [ 142 + (4*2) ] + 5/2 [ 182 + (4*2) ]
= 75 + 175 + 275 + 375 + 475
= 1375

Now, we are being asked to determine the ratio of the sum of all elements in S1 to the sum of all elements in S2.

S1 / S2 = 1080/1375 = 216 / 275
Intern  B
Joined: 07 Jan 2012
Posts: 14
Location: Canada
Concentration: International Business, Entrepreneurship
GMAT Date: 04-30-2012
WE: Information Technology (Computer Software)
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

honneeey wrote:
EgmatQuantExpert wrote:
Question of the Week #15

Set S contains all the integers from 10 to 99. $$S_1$$, a subset of S, contains all the numbers of S, in which both the digits are even. $$S_2$$, also a subset of S, contains all the numbers of S, in which both the digits are odd. What is the ratio of sum of all elements in $$S_1$$ to sum of all elements in $$S_2$$?

A. $$\frac{108}{275}$$

B. $$\frac{216}{275}$$

C. $$\frac{2}{3}$$

D. $$\frac{275}{216}$$

E. $$\frac{3}{2}$$

Quite a lengthy One!

Set S1= { 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62,64,66,68,80,82,84,86,88}= Sum(1080)
Set S2={ 11, 13,15,17,31,33,35,37,39,51,53,55,57,59,71,73,75,77,79,91,93,95,97,99}= Sum(1375)
Sum of S1/Sum of S2= 1080/1375= 216/275.

Answer B

Quick question - can't see 19 am I missing something?
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3018
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

Solution

Given:
• Set S contains all the integers from 10 to 99, both inclusive
o $$S_1$$, contains all the numbers of set S, in which both the digits are even
o $$S_2$$, contains all the numbers of set S, in which both the digits are odd

To find:
• $$\frac{Sum of all elements in S_1}{Sum of all elements in S_2}$$

Approach and Working:
$$S_1$$ = {20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88}

• The sum of first set of five elements in $$S_1$$ = 20 + 22 + 24 + 26 + 28
o (2*10) + (2*10 + 2) + (2*10 + 4) + (2*10 + 6) + (2*10 + 8) = 2*10*5 + (2 + 4 + 6 + 8) = 120

• The sum of next set of five elements in $$S_1$$= 40 + 42 + 44 + 46 + 48
o Now, if we compare the elements in the first and second set of 5 numbers each, we can see that each element in the second set is 20 more than the corresponding element in the first set.
o Thus, we can write the sum of the five elements in the second set = the sum of the five elements in the first set + 20*5 = 120 + 100 = 220

• Similarly, the sum of next set of five elements = sum of the five elements in the second set + 20 * 5 = 220 + 100 = 320
• And, the sum of last set of five elements = 320 + 100 = 420

Thus, sum of all elements in $$S_1$$ = 120 + 220 + 320 + 420 = 1080

$$S_2$$ = {11, 13, 15, 17, 19, 31, 33, 35, 37, 39, 51, 53, 55, 57, 59, 71, 73, 75, 77, 79, 91, 93, 95, 97, 99}

• Sum of first five elements in $$S_2$$ = 11 + 13 + 15 + 17 + 19
o (10 + 1) + (10 + 3) + (10 + 5) + (10 + 7) + (10 + 9) = 10*5 + (1 + 3 + 5 + 7 + 9) = 75

• The sum of next five elements in $$S_2$$= 31 + 33 + 35 + 37 + 39
o Which can be written as (11+ 20) + (13 + 20) + (15 + 20) + (17 + 20) + (19 + 20)
o (11 + 13 + 15 + 17 + 19) + 20*5 = 75 + 100 = 175

• Similarly, the sum of next five elements = 175 + 100 = 275
• And, the sum of next five elements = 275 + 100 = 375
• And, the sum of last five elements = 375 + 100 = 475

Thus, sum of all elements in $$S_2$$ = 75 + 175 + 275 + 375 + 475 = 75*5 + 1000 = 1375

Therefore, $$\frac{S_1}{S_2} = \frac{1080}{1375} = \frac{216}{275}$$

Hence, the correct answer is option B.

Answer: B

_________________
Manager  B
Joined: 28 Jun 2018
Posts: 73
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

jameslewis could you explain a little clearly how you derived N? Isn't N suppose to be the intervals between the sequence? In this case how would you find that?
Manager  B
Joined: 28 Jun 2018
Posts: 73
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

jameslewis sorry, N is number of terms, but doesn't the AS formula only apply when the increase or decrease is by the same amount? How did we apply it here then? More importantly, how did it work??? chetan2u >>??
Manager  G
Joined: 14 Jun 2018
Posts: 222
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

3
1
I think this might be the quickest way

Even :
20 , 22 , 24, 26 , 28.
40 , 42 , 44, 46 ,48
60...
80...
The mean of each group is 24 , 44 , 64 , 84
The mean of the entire set is (44+64) / 2 = 54
Therefore , the sum of the set is 54*4*5 = 1080 (4= no of group ; 5 = no of terms in each group)

Do the same for odd
11 , 13 , 15 , 17 , 19
31..
51..
71..
91..

The mean of the set will be 55
The sum of the entire set is 55*5*5 = 1375 (Here total number of group is 5 and each group has 5 terms)

Required ratio = 1080/1375 = 216/275
Intern  B
Joined: 18 Jul 2018
Posts: 18
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

Hi JAMES,

I see you have used the AP formula to do the calculation but neither is S1 nor S2 in arithmetic progression from what I understand. S1- 20,22,24,26,28,40,42....can you please explain.Thankyou..
Intern  B
Joined: 17 Dec 2017
Posts: 17
Location: United States
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

hibobotamuss wrote:
jameslewis sorry, N is number of terms, but doesn't the AS formula only apply when the increase or decrease is by the same amount? How did we apply it here then? More importantly, how did it work??? chetan2u >>??

chetan2u I am also stumped on why we were able to apply this formula here. It is my understanding that N(Last + First)/2 only works with sets consisting of evenly spaced integers. Could you please help explain it's application here?
Intern  B
Joined: 21 Aug 2018
Posts: 3
Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont  [#permalink]

### Show Tags

This is how i solved (hopefully it's right)
S1:
20, 22, 24, 26, 28 (even spaced set => mean = median = 24) => Sum = 24*5
4....
6...
8...
S2:
11, 13, 15, 17, 19, same as above => Sum = 15*5
3...
5...
7...
9....

Just dont do any calculation yet.
S1/S2 = (24*5 + 44*5 + 64*5 + 84*5) /(15*5 + 35*5 + 55*5 + 75*5 + 95*5)

Cancel 5 and quickly you can identify the last digit of the numerator (6) and the denominator (5) => C

Double check by actually do the calculation if you want Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont   [#permalink] 27 Oct 2018, 06:56
Display posts from previous: Sort by

# Set S contains all the integers from 10 to 99. S1, a subset of S, cont

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

#### MBA Resources  