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15 Sep 2004, 19:01
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What will be the first arithmetic mean between 1/4 and 9 3/4 if certain arithmetic means are inserted between these?
1) If 19 arithmetic means are inserted
2) If 3 arithmetic means are inserted
1) If 19 arithmetic means are inserted
2) If 3 arithmetic means are inserted
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16 Sep 2004, 00:46
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Is this a DS question? If so, ans is D. First AM can be found for both cases.
Per Ian, the general GMAT rule is that the two DS statements do not contradict with each other. But here the number AMs seem to be different.
Btw, in case you want to dig deep on finding the first AM, see below.
In general, number of terms n = (l-a)/d + 1
A simple formula for count, that equals difference between last and first terms divided by difference, + 1.
What will be the first arithmetic mean between 1/4 and 9 3/4 if certain arithmetic means are inserted between these?
l = last no = 39/4
a = first no = 1/4
d = diff = ?
1) If 19 arithmetic means are inserted
n = count = 19 + 2 = 21 (2 added for the first and last included).
d = (l-a)/(n-1) = (39/4-1/4)/(21-1) = 38/40 = 19/20
First AM = 1/4 + 19/20 = 24/20 = 6/5
2) If 3 arithmetic means are inserted
n = count = 3 + 2 = 5 (2 added for the first and last included).
d = (l-a)/(n-1) = (39/4-1/4)/(5-1) = 38/16 = 19/8
First AM = 1/4 + 19/8 = 21/8
Per Ian, the general GMAT rule is that the two DS statements do not contradict with each other. But here the number AMs seem to be different.
Btw, in case you want to dig deep on finding the first AM, see below.
In general, number of terms n = (l-a)/d + 1
A simple formula for count, that equals difference between last and first terms divided by difference, + 1.
What will be the first arithmetic mean between 1/4 and 9 3/4 if certain arithmetic means are inserted between these?
l = last no = 39/4
a = first no = 1/4
d = diff = ?
1) If 19 arithmetic means are inserted
n = count = 19 + 2 = 21 (2 added for the first and last included).
d = (l-a)/(n-1) = (39/4-1/4)/(21-1) = 38/40 = 19/20
First AM = 1/4 + 19/20 = 24/20 = 6/5
2) If 3 arithmetic means are inserted
n = count = 3 + 2 = 5 (2 added for the first and last included).
d = (l-a)/(n-1) = (39/4-1/4)/(5-1) = 38/16 = 19/8
First AM = 1/4 + 19/8 = 21/8
