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02 Jun 2010, 22:05
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Difficulty:
95%
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Question Stats:
43% (02:52) correct
57%
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When Mrs. T's students answer the bonus question correctly, she awards a bonus. If the base score is between 10 and 99, the bonus is equal to 2 times the tens digit in the base score. The last test Mrs. T scored was between 10 and 99, and the student answered the bonus question correctly. Was the bonus given greater than 17% of the base score?
(1) The base score of the test was between 50 and 90.
(2) Mrs. T added 16 bonus points to the last test she graded.
(1) The base score of the test was between 50 and 90.
(2) Mrs. T added 16 bonus points to the last test she graded.
03 Jun 2010, 06:45
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shekar123
Given:
Score is a two digit number - \(S=10x+y\) (as the score was between 10 and 99);
Bonus - \(B=2x\).
Question: is \(2x>\frac{17}{100}(10x+y)\)? --> is \(30x>17y\)?
(1) The base score of the test was between 50 and 90 --> \(5\leq{x}\leq{9}\) --> the range of \(30x\) is \(150\leq{30x}\leq{270}\) and the range of \(17y\) is \(0\leq{17y}\leq{153}\) (as \(y\) can be from 0 to 9). Hence \(30x\) may or may not be more than \(17y\). Not sufficient.
(2) Mrs. T added 16 bonus points to the last test she graded --> \(B=2x=16\) --> \(x=8\) --> \(30x=240\). Max value of \(17y=153\) (as max value of \(y\) is 9), hence \(30x\) is more than \(17y\). Sufficient.
Answer: B.
Hope it's clear.
16 Aug 2013, 08:17
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shekar123
Alternative solution :
(1)The base score of the test was between 50 and 90.
if base score 51 then Bonus score :10
17% of 51 = 8.67 so Bonus score is more than 17% of base Score.
if base score is 59 then also Bonus score :10
17% of 59 = 10.03 here bonus is not greater then 17% of base score
Hence two cases insufficient
(2)Mrs. T added 16 bonus points to the last test she graded.
if Bonus is 16 then Base score must be between 80 and 89
17% of 80=13.60 , So Bonus is more than 17% of base score
17% of 89 = 15.13, Here also Bonus is more than 17 % of base score
if Base score is between 80-89 bonus score 16, will always be more than 17% of base core
Hence B is sufficient.
