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16 Dec 2018, 07:38
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B
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Difficulty:
35%
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Question Stats:
75% (02:00) correct
25%
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wrong
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The table above shows present rates and proposed rates for electricity for residential customers. For which of the monthly kilowatt-hours shown would the proposed rate be the greatest percent increase over the present rate?
(A) 500
(B) 1,000
(C) 1,500
(D) 2,000
(E) Each of the percent increases is the same.
Attachment:
kwatt.JPG [ 27.18 KiB | Viewed 15818 times ]
11 Feb 2019, 02:14
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HKD1710
Recall that smaller the denominator, larger the number.
You need to compare the percentage increases:
500 -> 2/24 = 1/12
1000 -> 4/41 = 1/10.2
1500 -> 5/57 = 1/11.4
2000 -> 6/73 = 1/12
The smallest denominator belongs to 1000 units and hence the percentage increase is the highest.
Answer (B)
General Discussion
16 Dec 2018, 11:50
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Solution
Given:
- • The table shows present rates and proposed rates for electricity for residential customers.
To find:
- • For which monthly kilowatt-hours shown, the proposed rate shown the greatest percent increase over the present rate.
Approach and Working:
- • For 500 kilowatt-hours:
- o Percentage change= \(\frac{{26-24}}{24} *100 = \frac{2}{24}*100= \frac{1}{12} *100 =\frac{100}{12}=8.33%\)
- • For 1,000 kilowatt-hours:
- o Percentage change= \(\frac{{45-41}}{41} *100 = \frac{4}{41}*100= \frac{1}{10.25} *100 =\frac{100}{10.25}=9.756%\)
- • For 1,500 kilowatt-hours:
- o Percentage change= \(\frac{{62-57}}{57} *100 = \frac{5}{57}*100= \frac{1}{11.4}*100 =\frac{100}{11.4}=8.77%\)
- • For 2,000 kilowatt-hours:
- o Percentage change= \(\frac{{79-73}}{73} *100 = \frac{6}{73}*100= \frac{1}{12.166} *100 =\frac{100}{12.166}=8.219%\)
Hence, for 1000 kilowatt-hours, the percentage change is greatest.
Hence, the correct answer is option B.
Answer: B
