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# A man invests a sum of money in a bank in the beginning of a year and

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A man invests a sum of money in a bank in the beginning of a year and  [#permalink]

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10 Jul 2017, 02:33
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A man invests a sum of money in a bank in the beginning of a year and another sum of money, after an integer number of months, in another bank, both under simple interest. After how many months from the beginning of the year did he invest the second sum of money?

(1) The sums of money invested in the two banks are in the ratio 1 : 6, respectively. The ratio of rates of interest applicable in the two banks are in the ratio 3 : 4, respectively.
(2) The total interest accumulated after the year end was 4 percent of the total investment made in the year.

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Re: A man invests a sum of money in a bank in the beginning of a year and  [#permalink]

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10 Jul 2017, 03:31
(1) The sums of money invested in the two banks are in the ratio 1 : 6, respectively.
The ratio of rates of interest applicable in the two banks are in the ratio 3 : 4, respectively.

If the sum invested in first bank is x, the sum invested in second bank is 6x
The ratio of rate of interests in the two banks are 3:4
But we cannot clearly tell from this, how many months later was the sum invested in the second bank.(Insufficient)

(2) The total interest accumulated after the year end was 4 percent of the total investment made in the year.
Just knowing the interest, that was accumulated over the end of the year from both the investments made, we can't
clearly tell how many months later the sum was invested in the second bank.(Insufficient)

However, on combining the information from both the statements, we can clearly tell how
many months from the beginning of the year was the second sum invested. Sufficient(Option C)
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Re: A man invests a sum of money in a bank in the beginning of a year and  [#permalink]

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10 Jul 2017, 10:45
Combining the information given in both the statements,

Assume sum invested for 12 months [s1 = 100]
Now, sum invested for 2 months(10 months later) [s2 = 600]
Hence, total investment = 700

Let the interests which are in the ratio 3:4 be as follows
R1 = 12
R2 = 16

Formula used :
Simple Interest = $$\frac{P*N*R}{100}$$

SI(First sum) = $$\frac{100 * 12 * 1}{100} = 12$$(for 12 months)
SI(Second sum) = $$\frac{600 * 16 * 2}{1200} = 16$$(for 2 months)

SI(Total) =$$\frac{700 * 4 * 1}{100} = 28$$(total sum for 12 months)

Hope that helps!
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Re: A man invests a sum of money in a bank in the beginning of a year and  [#permalink]

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25 Oct 2017, 00:23
pushpitkc wrote:
Combining the information given in both the statements,

Assume sum invested for 12 months [s1 = 100]
Now, sum invested for 2 months(10 months later) [s2 = 600]
Hence, total investment = 700

Let the interests which are in the ratio 3:4 be as follows
R1 = 12
R2 = 16

Formula used :
Simple Interest = $$\frac{P*N*R}{100}$$

SI(First sum) = $$\frac{100 * 12 * 1}{100} = 12$$(for 12 months)
SI(Second sum) = $$\frac{600 * 16 * 2}{1200} = 16$$(for 2 months)

SI(Total) =$$\frac{700 * 4 * 1}{100} = 28$$(total sum for 12 months)

Hope that helps!

Hi pushpitkc
in the SI(First sum), shouldn't replace 1 with 12, which show the months?

and could you please explain in with algebra instead of the number?
Thank you
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A man invests a sum of money in a bank in the beginning of a year and  [#permalink]

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25 Oct 2017, 04:31
1
soodia wrote:
pushpitkc wrote:
Combining the information given in both the statements,

Assume sum invested for 12 months [s1 = 100]
Now, sum invested for 2 months(10 months later) [s2 = 600]
Hence, total investment = 700

Let the interests which are in the ratio 3:4 be as follows
R1 = 12
R2 = 16

Formula used :
Simple Interest = $$\frac{P*N*R}{100}$$

SI(First sum) = $$\frac{100 * 12 * 1}{100} = 12$$(for 12 months)
SI(Second sum) = $$\frac{600 * 16 * 2}{1200} = 16$$(for 2 months)

SI(Total) =$$\frac{700 * 4 * 1}{100} = 28$$(total sum for 12 months)

Hope that helps!

Hi pushpitkc
in the SI(First sum), shouldn't replace 1 with 12, which show the months?

and could you please explain in with algebra instead of the number?
Thank you

Here N must be the number of years the amount was invested

While doing the calculation for the second year
(when they have asked to find the simple interest for 2 months)
N will be 2 months, divided by 12 months of the year of 1/6th of the year.

Hope that helps!
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Re: A man invests a sum of money in a bank in the beginning of a year and  [#permalink]

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26 Oct 2017, 12:40
combining both 1 and 2.
Say $100 invested in Bank1 and$600 in Bank 2
Interest rates:
3(Bank1):4(Bank2)

Stmt2 says that final cumulative rate was 4% => interest rates would be multiple of 3&4 (simple 3% and 4% for a year won't give you 4% aggregate)
Now, total interest for the selected amount ($100+$600 or \$700) would be: 4% of 700 or 28

To summarize:
Total interest: 28
Bank1 interest 3x (x>1)
Bank 2 intrest 4x (x>1) for n number of months n <12

From above, i am not sure how can you find n which is the question.

E for me.
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Re: A man invests a sum of money in a bank in the beginning of a year and  [#permalink]

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27 Oct 2017, 04:26
Bunuel wrote:
A man invests a sum of money in a bank in the beginning of a year and another sum of money, after an integer number of months, in another bank, both under simple interest. After how many months from the beginning of the year did he invest the second sum of money?

(1) The sums of money invested in the two banks are in the ratio 1 : 6, respectively. The ratio of rates of interest applicable in the two banks are in the ratio 3 : 4, respectively.
(2) The total interest accumulated after the year end was 4 percent of the total investment made in the year.

(1) and (2) not sufficient alone

on combining

say Bank 1:
Amount invested is "1x",
Int rate is "3y" per annum and
time is 12 months
SI from Bank 1 = (3xy/100)

Bank 2:
Amount invested is "6x",
Int rate is "4y" per annum and
time is "n" months
SI from Bank 2 = (24nxy/1200)

total interest accumulated = ((24nxy)/1200) + ((3xy)/100)

Also given that
total amount invested = 7x
SI = 28x/100 (4% of total amount invested)

now
((24nxy)/1200) + ((3xy)/100) = ((28x)/100)
2nxy + 3xy = 28x
2ny + 3y = 28

since n is an integer. it can take any value between 1 and 12
n = 1 then y = 28/5
n = 2 then y = 4
n = 3 then y = 28/9
....
i think question might require integer rate of interests

Please correct me if i am wrong
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Re: A man invests a sum of money in a bank in the beginning of a year and  [#permalink]

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27 Oct 2017, 08:34
How can we take 2 months
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Joined: 02 Apr 2014
Posts: 471
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A man invests a sum of money in a bank in the beginning of a year and  [#permalink]

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29 Oct 2017, 10:29
pushpitkc wrote:
Combining the information given in both the statements,

Assume sum invested for 12 months [s1 = 100]
Now, sum invested for 2 months(10 months later) [s2 = 600]
Hence, total investment = 700

Let the interests which are in the ratio 3:4 be as follows
R1 = 12
R2 = 16

Formula used :
Simple Interest = $$\frac{P*N*R}{100}$$

SI(First sum) = $$\frac{100 * 12 * 1}{100} = 12$$(for 12 months)
SI(Second sum) = $$\frac{600 * 16 * 2}{1200} = 16$$(for 2 months)

SI(Total) =$$\frac{700 * 4 * 1}{100} = 28$$(total sum for 12 months)

Hope that helps!

Hi i have one question,
The above explanation depends on R1 and R2.

Let a - initial amount, 6a second investment
Let r - rate of interest, so second rate of interest (4/3)r
Let n be number of months for second investment interest calculated.

ar/100 + (6a (4/3)r * n)/1200 = (4/100) * 7a

Simplifying gives r = 84 /(3 + 2n)

So for 0<n<12, we can different value of r .
So answer must be E right?

if question had given r must be integer or something like that, then we can think particular of r and n.

A man invests a sum of money in a bank in the beginning of a year and &nbs [#permalink] 29 Oct 2017, 10:29
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