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

Question

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.

B.

C.

D.

E.

A.  \frac{108}{275}

B.  \frac{216}{275}

C.  \frac{2}{3}

D.  \frac{275}{216}

E.  \frac{3}{2}

e-GMAT         Question of the Week #15

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pandeyashwin · Accepted Answer

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

Emma263 · Answer

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

honneeey · Answer

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

sayankh · Answer

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   + 5/2   + 5/2   + 5/2  
 = 120 + 220 + 320 + 420
 = 1080
 
S2 = 5/2   + 5/2   + 5/2   + 5/2   + 5/2  
 = 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

jameslewis · Answer

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

EgmatQuantExpert · Answer

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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exc4libur · Answer

Assuming that set S contains all the integers from 10 to 99, inclusive.

S_1:(20,22,24,26,28…40…60…80…) 
 Tens: 20(5)+40(5)… = 5(20+40+60+80) = 5(200) = 1000 
 Units: (0+2+4+6+8)(4)=20(4)=80 
 Total:1000+8=1080

S_2:(11,13,15,17,19…31…51…71…91…) 
 Tens: 10(5)+30(5)… = 5(10+30+50+70+90) = 5(250) = 1250 
 Units: (1+3+5+7+9)(5)=25(5)=125 
 Total:1250+125=1375

\frac{S_1}{S_2}=\frac{1080}{1375}=\frac{5(200)+4(4*5)}{5(250)+5(25)}=\frac{5(200+16)}{5(250+25)}=\frac{216}{275}

Ans (B)

GMATGuruNY · Answer

For any EVENLY SPACED SET:
sum = (count)(median)

EVEN CASES: 
Options with an even tens digits and a units digit of 0:
20, 40, 60, 80 
Sum of the evenly spaced set above = (count)(median) = 4*50 = 200
In each subsequent case, the tens digits will remain the same, while the four units digits will each increase by 2:
22...82
24...84
26...86
28...88
As a result, each subsequent case will increase the sum by 8, yielding the following list:
200, 208, 216, 218, 220
Sum of the evenly spaced set above = (count)(median) = 5*216

ODD CASES: 
Options with an odd tens digit and a units digit of 1:
11, 31, 51, 71, 91
Sum of the evenly spaced set above = (count)(median) = 5*51 = 255
In each subsequent case, the tens digits will remain the same, while the five units digits will each increase by 2:
13...93
15...95
17...97
19...99
As a result, each subsequent case will increase the sum by 10, yielding the following list:
255, 265, 275, 285, 295
Sum of the evenly spaced set above = (count)(median) = 5*275

Resulting ratio:
 \frac{even-sum}{odd-sum} = \frac{5*216}{5*275} = \frac{216}{275}

B

perseverance2021 · Answer

@

Hi  jameslewis  ,  Bunuel  , VeritasKarishma  , and  chetan2u  - Can you guys help me understand why the sum of S1 is calculated as  S1 = \frac{20(20 + 88)}{2} = 10 * 108  , and the same applies for S2 while S1 and S2 do not contain equally spaced sets ?

Any help is greatly appreciated! Thanks!

chetan2u · Answer

It is not that you will get your answer in every question of this kind.

Here, it does because there are two items that are equally spaced from Center of the set. That is, if something is x less than mean or Center, then there is an item x more than the mean.

For example: 
(20+88)/2=54
1) 20 and 88 are 34 away from 54.
2) 22 and 86 are 32 away from 54.
3) 24&84, 26&82 and 28&80 are 30, 28 and 26 respectively away from 54.
4) Next set of number is 40&68, 42&66 and so on, which are 14, 12, away from 54.

KarishmaB · Answer

First check this out: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... eviations/ Now think: What will be the mean of 20, 22, 86, 88? It will be 54. Why? Because 20 is 34 less than 54 and 88 is 34 more than 54. Also, 22 is 32 less than 54 and 86 is 32 more than 54. So excess = shortfall. Hence mean = 54. (20 + 88)/2 = 54 After this 20 * 54 because we have 20 terms. Hope it all makes sense now.

rubyrehal · Answer

Why do we consider 20, 40, 60 and 80 in S1? The question specifies that both digit must be even but 0 is not even nor odd

chetan2u · Answer

Properties of 0: It is even. It is neither positive nor negative. It is a multiple of every number.

Bambi2021 · Answer

S1 will contain every even in the 4 subsets:  ,  ,  ,

S2 will contain every odd in the 5 subsets:  ,  ,  ,  ,

The ratio should be something close to 4/5.

Only choice B is a good fit.

Crytiocanalyst · Answer

S1:(20,22,24,26,28…40…60…80…)S1:(20,22,24,26,28…40…60…80…) 
Tens:20(5)+40(5)…=5(20+40+60+80)=5(200)=1000Tens:20(5)+40(5)…=5(20+40+60+80)=5(200)=1000
Units:(0+2+4+6+8)(4)=20(4)=80Units:(0+2+4+6+8)(4)=20(4)=80
Total:1000+8=1080Total:1000+8=1080

S2:(11,13,15,17,19…31…51…71…91…)S2:(11,13,15,17,19…31…51…71…91…)
Tens:10(5)+30(5)…=5(10+30+50+70+90)=5(250)=1250Tens:10(5)+30(5)…=5(10+30+50+70+90)=5(250)=1250
Units:(1+3+5+7+9)(5)=25(5)=125Units:(1+3+5+7+9)(5)=25(5)=125
Total:1250+125=1375Total:1250+125=1375

S1S2=10801375=5(200)+4(4∗5)5(250)+5(25)=5(200+16)5(250+25)=216275S1S2=10801375=5(200)+4(4∗5)5(250)+5(25)=5(200+16)5(250+25)=216275
Hence IMO B

punit2020 · Answer

For set S1, total such numbers will be 20. Each of the 4 possible number for tens place will repeat five times and each of the possible number at units digit will repeat 5 times. SO unit digit sum comes to 20*4 =80 and tens digit sum comes to 20*5=100, carrying forward 8 total comes to 1080). Based on similiar concept for S2 total comes to 1375. THere you have it!

Rohanx9 · Answer

Sum of evens:
- Number =  4*5 = 20 
- Average =  \frac{(10 + 98)}{2}= 54

Therefore sum of even = 20*54 (don't solve yet, potentially can be eliminated for simplification in the final fraction to save time)

Sum of odds:
- Number =  5*5 = 25 
- Average =  \frac{(11 + 99)}{2} = 55 
Therefore sum of odd =  25*55

Fraction =  \frac{(20*54)}{(25*55)} = \frac{4}{5}*\frac{54}{55}= \frac{216}{275}

impeditrem · Answer

S1 Summary (Even Digits)  
 
   Numbers:  20 total.  
   Tens Digits:  2, 4, 6, 8 (each appears 5 times).  
   Units Digits:  0, 2, 4, 6, 8 (each appears 4 times).  
   Sum Calculation:  =   +   = 1080
   
  S2 Summary (Odd Digits)  
 
   Numbers:  25 total.  
   Tens Digits:  1, 3, 5, 7, 9 (each appears 5 times).  
   Units Digits:  1, 3, 5, 7, 9 (each appears 5 times).  
   Sum Calculation:  =   +   = 1375
   
  Final Ratio  
 
   1080 / 1375  =  216 / 275 .

To solve this the  GMAT Way , we use  Set Distribution logic  and the  Average-Sum relationship . Trying to list and add all numbers in $S_1$ and $S_2$ would be too slow, so we look at the structure of the digits. 
 1. Analyze Subset $S_1$ (Both Digits Even) 
 
   Tens Digit:  Can be $\{2, 4, 6, 8\}$ (Note: 0 cannot be the tens digit in a 2-digit number). There are  4 options .  
   Units Digit:  Can be $\{0, 2, 4, 6, 8\}$. There are  5 options .  
   Total numbers in $S_1$:  $4 	imes 5 = \mathbf{20}$.
   
  To find the sum of $S_1$:  
 Each tens digit $\{2, 4, 6, 8\}$ appears 5 times (once for each unit). 
 Each unit digit $\{0, 2, 4, 6, 8\}$ appears 4 times (once for each ten). 
 $$	ext{Sum } S_1 =   +  $$ 
 $$	ext{Sum } S_1 =   +   = 1000 + 80 = \mathbf{1080}$$ 
 
 2. Analyze Subset $S_2$ (Both Digits Odd) 
 
   Tens Digit:  Can be $\{1, 3, 5, 7, 9\}$. There are  5 options .  
   Units Digit:  Can be $\{1, 3, 5, 7, 9\}$. There are  5 options .  
   Total numbers in $S_2$:  $5 	imes 5 = \mathbf{25}$.
   
  To find the sum of $S_2$:  
 Each tens digit $\{1, 3, 5, 7, 9\}$ appears 5 times. 
 Each unit digit $\{1, 3, 5, 7, 9\}$ appears 5 times. 
 $$	ext{Sum } S_2 =   +  $$ 
 $$	ext{Sum } S_2 =   +   = 1250 + 125 = \mathbf{1375}$$ 
 
 3. Calculate the Ratio 
 The question asks for the ratio of the sum of $S_1$ to the sum of $S_2$: 
 $$	ext{Ratio} = \frac{1080}{1375}$$ 
 Divide both by 5: 
 $$	ext{Ratio} = \mathbf{\frac{216}{275}}$$ 
  Correct Option: B. 216/275

zachueber · Answer

So I did this problem by counting up the number of integers that were in S1 and in S2 then dividing S1/S2 which got me 4/5. That was 80% which was really close to the 78.54% from Answer B, so I chose Answer B. Is there anything wrong with this method? Did I get the correct answer on accident or is this a valid approach for these types of problems?

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

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GMAT Problem Solving (PS) Questions

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

Prep Toolkit

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

More than Quant/Verbal Abilities: Speed vs. Accuracy

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From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards