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Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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21 Sep 2018, 04:08
Question Stats:
61% (03:21) correct 39% (03:01) wrong based on 109 sessions
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eGMAT Question of the Week #15Set 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. \(\frac{3}{2}\)
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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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25 Sep 2018, 09:59
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




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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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21 Sep 2018, 04:42
EgmatQuantExpert wrote: eGMAT Question of the Week #15Set 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
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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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24 Sep 2018, 03:06
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



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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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25 Sep 2018, 19:17
honneeey wrote: EgmatQuantExpert wrote: eGMAT Question of the Week #15Set 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?



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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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25 Sep 2018, 23:45
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
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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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02 Oct 2018, 09:40
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?



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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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02 Oct 2018, 09:45
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 >>??



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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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02 Oct 2018, 10:46
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



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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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22 Oct 2018, 23:02
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..



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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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26 Oct 2018, 15:47
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?



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Re: Set S contains all the integers from 10 to 99. S1, a subset of S, cont
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27 Oct 2018, 06:56
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
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27 Oct 2018, 06:56






