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01 Jul 2020, 04:38
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
60% (01:47) correct
40%
(01:41)
wrong
based on 394
sessions
History
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The circulation for magazine X in 1981 was 5 times the average (arithmetic mean) yearly circulation for magazine X for the years 1982-1990. What is the ratio of the circulation in 1981 to the total circulation during 1981-1990 for magazine X ?
(A) 1/5
(B) 1/4
(C) 5/14
(D) 5/9
(E) 5/6
(A) 1/5
(B) 1/4
(C) 5/14
(D) 5/9
(E) 5/6
yashikaaggarwal
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01 Jul 2020, 05:13
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The circulation for magazine X in 1981 was 5 times the average (arithmetic mean) yearly circulation for magazine X for the years 1982-1990.
Let the circulation for magazine x in 1981 = A
the average circulation for magazine x in 1982-1990 = B
B = total of circulation 1982-1990/no. of years
B = total of circulation 1982-1990/9
9B = total of circulation 1982-1990
also 5B = A => B = A/5
the ratio of the circulation in 1981 to the total circulation during 1981-1990 for magazine X
=> A/(A+9B)
=> A/ A+9A/5
=> 5A/14A
=> 5/14
answer should be C
Let the circulation for magazine x in 1981 = A
the average circulation for magazine x in 1982-1990 = B
B = total of circulation 1982-1990/no. of years
B = total of circulation 1982-1990/9
9B = total of circulation 1982-1990
also 5B = A => B = A/5
the ratio of the circulation in 1981 to the total circulation during 1981-1990 for magazine X
=> A/(A+9B)
=> A/ A+9A/5
=> 5A/14A
=> 5/14
answer should be C
01 Jul 2020, 05:17
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Explanation:
let average for 1982-1990 = 500
so for 1981 = 500x5 =2500
So total magazine = 500x9 + 2500 = 7000
Required Ratio: 2500/7000
= 5/14
IMO-C
let average for 1982-1990 = 500
so for 1981 = 500x5 =2500
So total magazine = 500x9 + 2500 = 7000
Required Ratio: 2500/7000
= 5/14
IMO-C
