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A total of $20,000 was invested in two certificates of depos

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A total of $20,000 was invested in two certificates of depos  [#permalink]

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New post 05 Apr 2011, 15:37
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Question Stats:

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A total of $20,000 was invested in two certificates of deposit at simple annual interest rates of 6 percent and 8 percent, respectively. If the total interest on the two certificates was $1,440 at the end of one year, what fractional part of the $20.000 was invested at the higher rate?

(A) 3/8
(B) 2/5
(C) 1/2
(D) 3/5
(E) 3/4
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Re: Rate problem  [#permalink]

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New post 05 Apr 2011, 18:26
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Acer86 wrote:
A total of $20,000 was invested in two certificates of deposit at simple annual interest rates of 6 percent and 8 percent, respectively. If the total interest on the two certificates was $1,440 at the end of one year, what fractional part of the $20.000 was invested at the higher rate?
(A) 3/8
(B) 2/5
(C) 1/2
(D) 3/5
(E) 3/4


You can use weighted average in this question. One amount is deposited at 6% and another at 8%. The overall interest rate is \((\frac{1440}{20,000})*100 = 7.2%\)
\(\frac{w1}{w2}= \frac{(8 - 7.2)}{(7.2 - 6)} = \frac{2}{3}\)
So 2/5 of the 20,000 was invested at 6% and 3/5 was invested at 8%.

For explanation of the formula used above, see: http://www.veritasprep.com/blog/2011/03 ... -averages/
and
http://www.veritasprep.com/blog/2011/04 ... ge-brutes/
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Re: Rate problem  [#permalink]

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New post 05 Apr 2011, 15:48
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1
D
1440/20000*100%=7.2
7.2-6=1.2
8-6=2
1.2/2=3/5 D

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Re: Rate problem  [#permalink]

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New post 05 Apr 2011, 15:48
D
1440/20000*100%=7.2
7.2-6=1.2
8-6=2
1.2/2=3/5 D

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Re: Rate problem  [#permalink]

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New post 05 Apr 2011, 15:52
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Acer86 wrote:
A total of $20,000 was invested in two certificates of deposit at simple annual interest rates of 6 percent and 8 percent, respectively. If the total interest on the two certificates was $1,440 at the end of one year, what fractional part of the $20.000 was invested at the higher rate?
(A) 3/8
(B) 2/5
(C) 1/2
(D) 3/5
(E) 3/4


Let "$x" be invested with the higher rate of 8%
Thus, "20000-x" must have been invested with lower rate of 6%

Simple Interest = Principal* Rate* Time

For the Amount invested at 8%
Principal = x
Rate = 0.08
Time = 1
Simple Interest = x* 0.08* 1= 0.08x

For the Amount invested at 6%
Principal = 20000-x
Rate = 0.06
Time = 1
Simple Interest = (20000-x)* 0.06* 1= 1200-0.06x

Total Interest = $1440

0.08x+1200-0.06x=1440
0.02x=240
x = 240/0.02=12000

Fraction of x per total amount = 12000/20000 = 12/20 = 3/5

Ans: "D"
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Re: Rate problem  [#permalink]

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New post 05 Apr 2011, 18:32
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VeritasPrepKarishma wrote:


Known as Alligation technique...
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Re: Rate problem  [#permalink]

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New post 05 Apr 2011, 19:05
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x * 6/100 * 1 + (20000 - x) * 8/100 * 1 = 1440

6x - 8x = 144000 - 160000

=> -2x = -16000

=> x = 8000

So 12000/20000 = 3/5

Answer - D
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Re: Rate problem  [#permalink]

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New post 05 Apr 2011, 19:07
VeritasPrepKarishma wrote:
Acer86 wrote:
A total of $20,000 was invested in two certificates of deposit at simple annual interest rates of 6 percent and 8 percent, respectively. If the total interest on the two certificates was $1,440 at the end of one year, what fractional part of the $20.000 was invested at the higher rate?
(A) 3/8
(B) 2/5
(C) 1/2
(D) 3/5
(E) 3/4


You can use weighted average in this question. One amount is deposited at 6% and another at 8%. The overall interest rate is \((\frac{1440}{20,000})*100 = 7.2%\)
\(\frac{w1}{w2}= \frac{(8 - 7.2)}{(7.2 - 6)} = \frac{2}{3}\)
So 2/5 of the 20,000 was invested at 6% and 3/5 was invested at 8%.

For explanation of the formula used above, see: http://www.veritasprep.com/blog/2011/03 ... -averages/
and
http://www.veritasprep.com/blog/2011/04 ... ge-brutes/

i didnt know this technique... very good and efficient...
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Re: Rate problem  [#permalink]

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New post 05 Apr 2011, 19:32
Acer86 wrote:
A total of $20,000 was invested in two certificates of deposit at simple annual interest rates of 6 percent and 8 percent, respectively. If the total interest on the two certificates was $1,440 at the end of one year, what fractional part of the $20.000 was invested at the higher rate?
(A) 3/8
(B) 2/5
(C) 1/2
(D) 3/5
(E) 3/4


My Solution:
Interest Earned = Principal * Rate * Time

x + y = 20,000
.06x + .08y = 1,440

Combine equations:
.06(20,000 - y) + .08y = 1,440
1200 - .06y + .08y = 1440
.02y = 240
2y = 24,000
y = 12,000


12,000 / 20,000 = 6/10 = 3/5
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Re: Rate problem  [#permalink]

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New post 17 Apr 2011, 17:51
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8x/100 + (20000 - x) (6/100) = 1440
=> x = 12000
Which is a fraction of 12000/20000
I.e 3/5

Answer is D.

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Re: A total of $20,000 was invested in two certificates of depos  [#permalink]

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New post 01 Dec 2014, 08:10
Acer86 wrote:
A total of $20,000 was invested in two certificates of deposit at simple annual interest rates of 6 percent and 8 percent, respectively. If the total interest on the two certificates was $1,440 at the end of one year, what fractional part of the $20.000 was invested at the higher rate?

(A) 3/8
(B) 2/5
(C) 1/2
(D) 3/5
(E) 3/4



Backsolving as option

1) take C to get that 1/2*20000=10000*8/100=800 and 1/2*20000=10000*6/100=600, so 800+600=1400 less than given. Eliminate A,B,C

2) go D to take 3/5*20000=12000*8/100=960 and 8000*6/100=480, so 960+480=1440. It is correct


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Re: A total of $20,000 was invested in two certificates of depos  [#permalink]

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New post 02 Dec 2014, 00:29
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...................... 6% .................... 8% ............ Total

Principle ........ 20000-x ................ x .............. 20000

Interest ......... \((20000-x)\frac{6}{100}\) ..... \(\frac{8x}{100}\) ....... 1440

\(\frac{(20000-x)6}{100} + \frac{8x}{100} = 1440\)

120000 - 6x + 8x = 144000

x = 72000-60000 = 12000

Fraction \(= \frac{12000}{24000} = \frac{3}{5}\)

Answer = D
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Re: A total of $20,000 was invested in two certificates of depos  [#permalink]

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Re: A total of $20,000 was invested in two certificates of depos   [#permalink] 25 Jul 2017, 05:32
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