SOLUTION: Once a month, Benfort put some money into the bamboo jar. Each month, he put 50 centavos more into the jar than the month before. On its 12th yr, he counted his money for his gradu

Algebra ->  Sequences-and-series -> SOLUTION: Once a month, Benfort put some money into the bamboo jar. Each month, he put 50 centavos more into the jar than the month before. On its 12th yr, he counted his money for his gradu      Log On


   



Question 1199553: Once a month, Benfort put some money into the bamboo jar. Each month, he put 50 centavos more into the jar than the month before. On its 12th yr, he counted his money for his graduation; and he had Php 5, 436. How much money did he put in the jar in the last month?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Once a month for 12 years means 12*12 = 144 deposits.

The final amount was 5436; the average deposit was 5436/144 = 37.75.

The average deposit is the average of the first and last deposits:
(first + last)/2 = 37.75
first + last = 2*37.75 = 75.5 [1]

The last deposit is the first deposit, plus the common difference of 50c = 0.5p 119 times:
last = first + 119(0.5) = first + 59.5 [2]

From [1] and [2]...
first + last = first + first + 59.5 = 75.5
2*first = 16
first = 8 [3]

From [2] and [3], last = 8 + 59.5 = 67.5

ANSWER: He put Php 67.5 in the jar the last month

CHECK: 144%28%288%2B67.5%29%2F2%29=144%2A37.75=5436


Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
Once a month, Benfort put some money into the bamboo jar. Each month, he put 50 centavos more
into the jar than the month before.  highlight%28At_the_end_of%29 highlight%28cross%28On%29%29  its 12th yr,
he counted his money for his graduation; and he had Php 5, 436. How much money did he put in the jar in the last month?
~~~~~~~~~~~~~~~~~


            There is an error in calculations by @greenestamps,
            So I came to fix this deficiency.
            The place where I changed the @greenestamps' numbers, is marked in red in my post.


Once a month for 12 years means 12*12 = 144 deposits.

The final amount was 5436; the average deposit was 5436/144 = 37.75.

The average deposit is the average of the first and last deposits:
(first + last)/2 = 37.75
first + last = 2*37.75 = 75.5 [1]

The last deposit is the first deposit, plus the common difference of 50c = 0.5p   highlight%28cross%28119%29%29   highlight%28highlight%28143%29%29   times:
last = first + 143(0.5) = first + 71.5 [2]     <<<--- starting from this point, the numbers are different from that of @greenestamps

From [1] and [2]...
first + last = first + first + 71.5 = 75.5
2*first = 4
first = 2 [3]

From [2] and [3], last = 2 + 71.5 = 73.5

ANSWER: He put Php 73.5 in the jar the last month

CHECK: 144%28%282%2B73.5%29%2F2%29=144%2A37.75=5436


//////////////


Notice that I edited you post to make the meaning absolutely clear,
as it  SHOULD  BE  in any  Math problem.