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Re: Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
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is there a logical way of doing this?
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Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
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yuvrawj wrote:
is there a logical way of doing this?


Sort of.

Since the interest earned for A is double that of B, but the interest rate for A (5%) is a little less than double that of B (3%), this means A needs a slightly bigger piece of the $33,000 in order for its interest earned to be double.

From here, you can take a number that is slightly more than half of $33,000 to test. $18,000 for A and $15,000 are clean numbers and good places to start. From here, you can quickly see that $18,000*0.05 would yield $900 and $15,000*0.03 would yield $450—perfectly double.

$900 + $450 = $1,350
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Re: Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
yuvrawj wrote:
is there a logical way of doing this?


Since the interest earned by Account A was twice the interest earned by Account B and their interest rates were 5% and 3% respectively, it is obvious that Alfred had to invest more than 50% in Account A, otherwise Account A would earn 5%/3% = 1.66 or 66% more than Account B.
Now we could approximate how many more dollars Alfredo was supposed to deposit to Account A:
2 (times more earned)/1.66 (difference in interest rates)=1.2 or 20% more money went to Account A.
Then you could start playing with numbers: more than 50% of total 33k (let's assume 18k) went to Account A, then 15k went to Account B.
18k more than 15k by exactly 20% (3k/15k).
All is left is just to calculate interest earned by each account:
18k*0.05=900 and 15k*0.03=450 that gives us $1350 in total.
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Re: Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
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Let, a & b the principal of the respective of accounts,

Given ,
5a= 2*3b
Then principal ratio, a : b = 6 :5
By weighted average concept, effective % = 6*5+5*3/ 6+5 = 45/11 %
Total interest= 45/11% of 33000= 1350 (Ans)
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Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
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yuvrawj wrote:
is there a logical way of doing this?

­Some of these other solutions involve guessing the numbers 18,000 and 15,000 for A and B, which is an interesting method, but I think a lot of people will have trouble getting to these numbers quickly.

In this case, although it's longer than most questions, I think the algebra is your best bet, and it does benefit you to get very quick with your translations and algebra.

However, for many other mixture questions, I do encourage the Teeter Totter method. (It doesn't work well on this one, because we don't have the weights for the Principals; rather, we have a ratio of the amount of interest.)

Here's a playlist with 4 basic examples going over the process for this Teeter Totter method: https://www.youtube.com/playlist?list=PL2exXfCUscn8Hvafet5-IPH1eNNLSjQBP
 ­
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Re: Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
Bunuel this is a GMAT Prep Mock Question. Appeared in Mock 2
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Re: Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
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Freddy12 wrote:
Bunuel this is a GMAT Prep Mock Question. Appeared in Mock 2

­Thank you. The tag is already there.
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Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
gmatophobia Can alligation diagram be used in any way to solve this problem ?
gmatophobia wrote:
Bunuel wrote:
Alfredo invested a total of $33,000 in 2 accounts, A and B, with annual interest rates of 5% and 3%, respectively. For the first year, the interest earned by Account A was twice the interest earned by Account B. What was the total interest earned by the 2 accounts for the first year?

A. $1,290
B. $1,320
C. $1,350
D. $1,430
E. $2,475


  • The amount deposited in account A = \(a\)
  • The amount deposited in account B = \(b\)

\(a + b = 33000\)

  • Interest accumulated in account A = \(a * \frac{5}{100}\)
  • Interest accumulated in account B = \(b * \frac{3}{100}\)

Given

\(a * \frac{5}{100} = 2*b * \frac{3}{100}\)

\(5a = 6b\)

\(a + b = 33000\)

Multiplying by \(5\) on both the sides

\(5a + 5b = 5*33000\)

\(6b + 5b = 5*33000\)

\(11b = 5*33000\)

\(b = 15000\)

Interest accumulated on account B = \(150000 * \frac{3}{100}\) \(= 450\)
Interest accumulated on account A = \(2*450\) \(= 900\)

Total Interest = \(450 + 900 = 1350\)

Option C


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Re: Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
Bunuel wrote:
Freddy12 wrote:
Bunuel this is a GMAT Prep Mock Question. Appeared in Mock 2

­Thank you. The tag is already there.

­Where can I find this tag with the mock questions?
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Re: Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
We can directly use Simple interest formula and skip some steps to arrive at below ratio:
Pa/Pb= 2*3/5= 6/5

Therefore, Pa= (6/6+5)* 33k= 6*3k= 18k
Similarly, Pb= 5/11*33k= 15k

SIa= 18k*5*1/100
SIb= 15k*3*1/100

Combined SI= (90k+45k)/100= 1350.
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Re: Alfredo invested a total of $33,000 in 2 accounts, A and B, with [#permalink]
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