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03 Apr 2020, 06:04
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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:
65%
(hard)
Question Stats:
67% (02:52) correct
33%
(02:36)
wrong
based on 104
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An aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then what is the total number of tests taken by him ?
A. 50
B. 55
C. 60
D. 65
E. 70
Are You Up For the Challenge: 700 Level Questions
A. 50
B. 55
C. 60
D. 65
E. 70
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 08:08
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Bunuel
Let the number of tests = \(n\) and average of all tests = \(a\)
--> Sum of all scores = \(n*a\)
His average score increases by 1 if the first 10 tests are not considered and his average scores for the first 10 tests is 20
--> \((n - 10)(a + 1) + 200 = n*a\)
--> \(n*a + n - 10a - 10 + 200 = n*a\)
--> \(10a - n = 190\) ....... (1)
His average score decreases by 1 if the last 10 tests are not considered and his average scores for the last 10 tests is 30
--> \((n - 10)(a - 1) + 300 = n*a\)
--> \(n*a - n - 10a + 10 + 300 = n*a\)
--> \(10a + n = 310\) ....... (2)
(1) + (2),
--> \(10a - n + 10a + n = 190 + 310\)
--> \(20a = 500\)
--> \(a = 25\)
From (1),
\(10a - n = 190\)
--> \(10*25 - n = 190\)
--> \(n = 250 - 190\)
--> \(n = 60\)
Option C
04 May 2020, 10:26
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Bunuel
first 10 games average is 20, sum these = 200
last 10 games average is 30, sum these = 300
(n-10)(a+1)=na-200, na+n-10a-10=na-200, n-10a=-190
(n-10)(a-1)=na-300, na-n-10a+10=na-300, -n-10a=-310
-20a=-500, a=500/20=25
n-10(25)=-190, n=-190+250=60
ans (C)
