GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Oct 2019, 11:42

### GMAT 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# $4800 becomes$6000 in 4 years at a certain rate of compound interest.

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 25 Dec 2018
Posts: 484
Location: India
Concentration: General Management, Finance
GMAT Date: 02-18-2019
GPA: 3.4
WE: Engineering (Consulting)
$4800 becomes$6000 in 4 years at a certain rate of compound interest.  [#permalink]

### Show Tags

06 Mar 2019, 01:19
1
4
00:00

Difficulty:

25% (medium)

Question Stats:

80% (02:25) correct 20% (01:55) wrong based on 20 sessions

### HideShow timer Statistics

$4800 becomes$6000 in 4 years at a certain rate of compound interest. What would have been the amount after 12 years?

A) $9500 B)$9375
C) $9250 D)$9775
E) $9250 Math Expert Joined: 02 Aug 2009 Posts: 7953 Re:$4800 becomes $6000 in 4 years at a certain rate of compound interest. [#permalink] ### Show Tags 07 Mar 2019, 07:42 1 1 mangamma wrote:$4800 becomes $6000 in 4 years at a certain rate of compound interest. What would have been the amount after 12 years? A)$9500
B) $9375 C)$9250
D) $9775 E)$9250

Calculation of r and then finding amount after 12 years will be calculation intensive.
A very innovative way to avoid cumbersome calculation would be ..
The formula for CI is Amount = P$$(1+\frac{r}{100})^t$$.... 6000 = 4800$$(1+\frac{r}{100})^4..........(1+\frac{r}{100})^4=\frac{6000}{4800}.$$..
we are looking for 4800$$(1+\frac{r}{100})^12$$=4800$$((1+\frac{r}{100})^4)^3$$=4800$$(\frac{6000}{4800})^3=4800*\frac{5*5*5}{4*4*4}=75*125=9375$$

B
_________________
Re: $4800 becomes$6000 in 4 years at a certain rate of compound interest.   [#permalink] 07 Mar 2019, 07:42
Display posts from previous: Sort by

# $4800 becomes$6000 in 4 years at a certain rate of compound interest.

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne