GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
For which two uses of electricity was the ratio of the amoun
[#permalink]
09 May 2015, 10:22
09 May 2015, 10:22
1
Expert Reply
1
Bookmarks
For which two uses of electricity was the ratio of the amounts of electricity used most nearly 3 to 1?
(A) Water heater and lights/small appliances
(B) Large appliances and lights/small appliances
(C) Air conditioner and water heater
(D) Air conditioner and lights/small appliances
(E) Air conditioner and large appliances
Show: :: OA
E
The electricity used by the water heater was measured separately and its cost per kilowatt-hour was one-half the cost per kilowatt-hour of the rest of the electricity used. The cost of the electricity used by the water heater was most nearly what fraction of the total cost of all the electricity used?
(A) 1/11
(B) 1/9
(C) 1/8
(D) 1/5
(E) It cannot be determined from the information given.
Show: :: OA
B
In November the Smythe household used the same total amount of electricity as in July, but the water heater used 33 percent of this total amount. By approximately what percent did the amount of electricity used by the water heater increase from July to November?
(A) 13%
(B) 33%
(C) 50%
(D) 65%
(E) 130%
Show: :: OA
D
Show: ::
Attachment:
Quant2.jpg [ 35.23 KiB | Viewed 8836 times ]
Re: DI Set #9- Electricity usage in a House
[#permalink]
16 Jun 2016, 16:19
16 Jun 2016, 16:19
Expert Reply
1. Air Condition/ Large Appliance = \(\frac{53}{18}\) ~= 3:1
Answer: E
2. Now let per unit Cost be $1 for normal and %0.5 for water heating.
So Water heating Cost: 20% of 800 * 0.5 = $80.
Rest Cost: 80% of 800 * 1 = $640
Total Cost $80 + $640 = $720
So cost of the electricity used by the water heater/ total cost of all the electricity used = \(\frac{$80}{$720}\) = \(\frac{1}{9}\)
Answer: B
3. In July : 20% of 800 and
In November : 33% of 800.
Increase Percentage = \(\frac{((.33 - .20) * 800) * 100}{(.20 * 800)}\) = (13 * 5) = 65%.
Or (33 -20)/20 *100 = 65%
Answer: D
Answer: E
2. Now let per unit Cost be $1 for normal and %0.5 for water heating.
So Water heating Cost: 20% of 800 * 0.5 = $80.
Rest Cost: 80% of 800 * 1 = $640
Total Cost $80 + $640 = $720
So cost of the electricity used by the water heater/ total cost of all the electricity used = \(\frac{$80}{$720}\) = \(\frac{1}{9}\)
Answer: B
3. In July : 20% of 800 and
In November : 33% of 800.
Increase Percentage = \(\frac{((.33 - .20) * 800) * 100}{(.20 * 800)}\) = (13 * 5) = 65%.
Or (33 -20)/20 *100 = 65%
Answer: D
