BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

% problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 124
Joined: Mon May 19, 2008 1:12 pm
Thanked: 3 times

% problem

by Nycgrl » Tue Jul 01, 2008 3:24 pm
MONTHLY KILOWATT-HOURS
500 1,000 1,500 2,000
Present $24.00 $41.00 $57.00 $73.00
Proposed $26.00 $45.00 $62.00 $79.00

12. 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.
Top
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 25
Joined: Tue Jul 01, 2008 11:54 am
Location: San Diego

by somail » Tue Jul 01, 2008 3:51 pm
The most important formula for this question is ((change)/original)*100 = %change

Using the first value 500KW

500X24.00= $12,000
500x26.00= $13,000

13k-12k = $1,000 ($1,000/$12,000)*100= 8.33% change.

Use the same process for the remaining values.
Top
Master | Next Rank: 500 Posts
Posts: 124
Joined: Mon May 19, 2008 1:12 pm
Thanked: 3 times

by Nycgrl » Tue Jul 01, 2008 4:07 pm
Is there ant shor method to do it
Top
Master | Next Rank: 500 Posts
Posts: 144
Joined: Thu Apr 17, 2008 3:33 pm
Thanked: 8 times

by Sunny22uk » Tue Jul 01, 2008 4:49 pm
As Somail said, the formula is ((change)/original)*100 = %change
Instead of multiplying MONTHLY KILOWATT-HOURS,We can rewrite the solution as:-
500(26-25)/ (500*24) x 100 i.e. 1/24 * 100
As seen in this equation, there is no need to multiply MONTHLY KILOWATT-HOURS, the rest of the equations will be easy to solve.

Hope this helps.
Cheers
Top
Post new topic Post Reply
• Page 1 of 1