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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