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

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

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

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

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

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)

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:
\(\frac{even-sum}{odd-sum} = \frac{5*216}{5*275} = \frac{216}{275}\)

@

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!

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.

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.

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

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

S1 will contain every even in the 4 subsets: [20-28], [40-48], [60-68], [80-88]

S2 will contain every odd in the 5 subsets: [11-19], [31-39], [51-59], [71-79], [91-99]

The ratio should be something close to 4/5.

Only choice B is a good fit.

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

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!