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# If in a certain sequence of consecutive multiples of 50, the median is

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Retired Moderator
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If in a certain sequence of consecutive multiples of 50, the median is [#permalink]

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25 Jun 2015, 12:01
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58% (01:29) correct 42% (01:16) wrong based on 196 sessions

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If in a certain sequence of consecutive multiples of 50, the median is 625, and the greatest term is 950, how many terms that are smaller than 625 are there in the sequence?

A. 6
B. 7
C. 8
D. 12
E. 13

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Re: If in a certain sequence of consecutive multiples of 50, the median is [#permalink]

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27 Jun 2015, 12:27
Since the median is 625 we know there must be a even number of integers because 50 is not a multiple of 625.

So the list around 625 must go. 600 (625) 650 700 750 800 850 900 950

Since we know there are 7 numbers greater than 625 then there must be 7 numbers less then 625.
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Re: If in a certain sequence of consecutive multiples of 50, the median is [#permalink]

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01 Sep 2015, 09:07
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Correct, if I am doing wrong here :

If the numbers in a set are evenly spaced, then median = mean .
Now, For any evenly spaced set, the mean is the average of 1st and the last term
Therefore , mean = median will give following equation , here let the first term be x and last term in the sequence is 950 ( given in the question )

so $$\frac{x+950}{2}$$ = 625

On solving above equation we get the value of x i.e first term as 300

Now 625 is not a multiple of 50 so , term which is less then 625 and multiple of 50 is 600.
To calculate the number of elements less then 625 is
=== > $$tn = a + (n-1)d$$

600 = 300 + (n-1)50
on solving n we get n = 7
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Re: If in a certain sequence of consecutive multiples of 50, the median is [#permalink]

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08 Sep 2017, 09:46
Quick Approach would be
since 625 is median and its not a multiple of 50, the value to the right of 625 is 650 and to the left is 600
So 950-650= 300. Then 600-x= 300 so x=300 where x is first value

Say we have a sequence 2,4,6,8,10,12 the median here is 7 ok now 12-8=4 also 6-2=4
now 50*6=300 and 50*12=600
So no of values between 6 to 12 is ( 12-6)+1= 7
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Re: If in a certain sequence of consecutive multiples of 50, the median is [#permalink]

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09 Sep 2017, 06:16
950/5=19 so
and 625 is the medain so it can not be divisible by 50 therefore it must be the average of the two numbers and there are even number of elements
x+x+50=1350
2x=1300
x=650
since there are even number of elements and last term is 50*19=90
we must have out first term as 50*2=100
so there are 7 numbers less than 625 leaving middle two numbers i.e 9th and 10th
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Re: If in a certain sequence of consecutive multiples of 50, the median is [#permalink]

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14 Feb 2018, 16:53
Hi All,

When dealing with large sequences of numbers, it often helps to find a logical 'starting point' so that you can 'map out' the overall sequence. It's important to note that the 'starting point' might not be the first term in the sequence....

Since we're dealing with CONSECUTIVE multiples of 50, and the MEDIAN is 625...we can 'map out' 2 of the terms in the sequence. Since 625 is NOT a multiple of 50, it must be the AVERAGE of the two 'middle terms'....so we need two consecutive multiples of 50 that average to 625.....those would be...

600 and 650

600 is below the median; 650 is above it. From here, we can map 'up' to 950 (try counting in 'sets of 2'), count those total terms and then map 'down' in the other direction (which would be the same number of terms). Since the question just asks for the number of terms, we don't have to physically write them all out.

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Re: If in a certain sequence of consecutive multiples of 50, the median is [#permalink]

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18 Mar 2018, 19:50
since 625 is not a multiple of 50, it must be true that the median is the average of 2 multiples of 50, namely 600 and 650

now, since the largest value is 950, there is a total of (950 - 650) = 300 values to the right of 650
and, we can see 300/50 = 6 numbers sum to 300

thus, since there are 6 values to the right of "(600 + 650)" there are 7 values less than 625 = B the answer

thanks
Re: If in a certain sequence of consecutive multiples of 50, the median is   [#permalink] 18 Mar 2018, 19:50
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