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kiran120680
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Set X consists of the first 15 positive multiple of 8 that are not a 16 May 2019, 00:47
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C
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73% (03:08) correct 27% (02:58) wrong based on 510 sessions

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Set X consists of the first 15 positive multiple of 8 that are not a multiple of 16 and set Y consists of the first 15 positive multiples of 8 that are also a multiple of 16. If x and y denote the average (arithmetic mean) of all the numbers in set X and set Y respectively, what is the value of x+y?


A. 120
B. 124
C. 248
D. 480
E. 496
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C
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Set X consists of the first 15 positive multiple of 8 that are not a 19 May 2019, 19:40
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kiran120680
Set X consists of the first 15 positive multiple of 8 that are not a multiple of 16 and set Y consists of the first 15 positive multiples of 8 that are also a multiple of 16. If x and y denote the average (arithmetic mean) of all the numbers in set X and set Y respectively, what is the value of x+y?


A. 120
B. 124
C. 248
D. 480
E. 496
Show more


We see that X = {8(1), 8(3), 8(5), …, 8(29)} and Y = {8(2), 8(4), 8(6), …, 8(30)}. Since each set is evenly spaced, the mean is the average of the smallest and the largest elements in the set. Therefore, we have:

x = [8(1) + 8(29)]/2 = 8(30)/2 = 4(30) = 120

and

y = [8(2) + 8(30)]/2 = 8(32)/2 = 4(32) = 128

So x + y = 120 + 128 = 248.

Answer: C
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Set X consists of the first 15 positive multiple of 8 that are not a 16 May 2019, 00:57
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Average of first 15 positive multiple of 8 that are not a multiple of 16, x= (8*2 -1)*8=120
Average of first 15 positive multiples of 8 that are also a multiple of 16, y= 8*16=128
x+y=120+128=248
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Set X consists of the first 15 positive multiple of 8 that are not a 16 May 2019, 22:09
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nick1816
Average of first 15 positive multiple of 8 that are not a multiple of 16, x= (8*2 -1)*8=120
Average of first 15 positive multiples of 8 that are also a multiple of 16, y= 8*16=128
x+y=120+128=248
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Please explain in detail.
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Set X consists of the first 15 positive multiple of 8 that are not a 16 May 2019, 22:26
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You can write odd multiples of 8 as (2n-1)8 and even multiples of 8 as 2n*8

Now if you have a Arithmetic series of 15 numbers then 8th number will always the mean. You can also prove it.
If we consider series of odd multiples of 8
Mean= (First number of series + last number of series)/2= (1*8 + 29*8)/2=(1+29)*8/2=15*8= (2*8-1)*8 (value of n is 8)



Vallabh02
nick1816
Average of first 15 positive multiple of 8 that are not a multiple of 16, x= (8*2 -1)*8=120
Average of first 15 positive multiples of 8 that are also a multiple of 16, y= 8*16=128
x+y=120+128=248

Please explain in detail.
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Set X consists of the first 15 positive multiple of 8 that are not a 19 May 2019, 06:42
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nick1816 : How is it represented in your formula that it is the first 15 terms?
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Set X consists of the first 15 positive multiple of 8 that are not a 15 Oct 2024, 05:07
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an = 8 + 16(n-1)

bn = 16 + 16(n-1)

this the sum of the first 15 elements of an is
sum(an) = (16 + 16(15-1))/2 * 15

sum(bn) = (32 + 16(n-1))/2 * 15

thus
x = avg(an) = (16 + 16(15-1))/2 = (8 + 8*14)
y = avg(bn) = (32+ 16(15-1))/2 = (16 + 8*14)

x+y = (8 + 8*14) + (16 + 8*14) = 8 + 112 + 16 + 112 = 248


now that i write it, i recognize that i did not need to calculate the sum itself :( i just needed the average of both sequences.
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Set X consists of the first 15 positive multiple of 8 that are not a 15 Jul 2025, 09:32
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kiran120680
Set X consists of the first 15 positive multiple of 8 that are not a multiple of 16 and set Y consists of the first 15 positive multiples of 8 that are also a multiple of 16. If x and y denote the average (arithmetic mean) of all the numbers in set X and set Y respectively, what is the value of x+y?


A. 120
B. 124
C. 248
D. 480
E. 496
Show more
In AP, average of the sequence = middle term of the sequence.
Therefore, we just need to find the 8th term of both the sequences.
Using the formula, Tn = a+(n-1)d

s1 = 8 + (7)8
s2 = 16 + 7(16)

s1 + s2 = 8 [15+16] = 8[31]
=248

Option C
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