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17 Jul 2021, 22:15
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MathRevolution
Solution: Let us apply the IVY approach to solve the question. As we are dealing with percentage, then let the original price of a car be $100.
Price after price reduced by 30%: (100 - 30)% of $100
=> \((\frac{70}{100})*$100 = $70\)
Price after the new price increased by 30%: (100 + 30)% of $70
=> \((\frac{130}{100})$70 = $91\)
As the base price is $100, the final price would be 91% of the base price.
Since Percent change = \(\frac{(After - Before)}{Before} * 100\)(%) and $100 (Before) and $91 (After),
We get Percent change = \(\frac{($91 - $100)}{$100} * 100\)(%) =\( \frac{(-9)}{100} * 100\)(%) = -9%
Therefore, A is the correct answer.
Answer A
17 Jul 2021, 22:24
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Que: An instructor gave the same test to three groups: A, B, and C. The average (arithmetic mean) scores for the three groups were 40, 50, and 60, respectively. The ratio of the numbers of candidates in the A, B, and C groups were 1 : 3: 5, respectively. What was the average score for the three groups combined?
(A) 21.11
(B) 36.66
(C) 50
(D) 54.44
(E) 80
(A) 21.11
(B) 36.66
(C) 50
(D) 54.44
(E) 80
19 Jul 2021, 02:16
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MathRevolution
Solution: Let the number of candidates in groups A, B, and C be 1k, 3k, and 5k, respectively, where ‘k's is a constant of proportionality.
The average (arithmetic mean) scores for the three groups were 40, 50, and 60, respectively. The combined average for the three groups is
=> \(\frac{[ (40 * k) + (50 * 3k) + (60 * 5k)] }{ (k + 3k + 5k)}\)
=> \(\frac{[(40k + 150k + 300k)]}{ (9k)}\)
=> \(\frac{490k }{ 9k} = 54.44\)
Therefore, D is the correct answer.
Answer D
