anon1 wrote:
For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was $1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was $1,000, what was the amount of the deposit on June 21?
A. $1,000
B. $1,150
C. $1,200
D. $1,450
E. $1,600
The DAILY BALANCE is the amount in the account AT THE END OF THE DAY.
Since the daily balance for the first 10 days ($1000) is the same as the average daily balance for the entire month ($1000), we can ignore the first 10 days.
We need to determine the amount that must be deposited on June 21 so that the average daily balance for the LAST twenty days is $1000.
Sum of the daily balances for June 11-30 = (number of days)(daily balance) = 20*1000 = 20,000.
When $350 is withdrawn on June 11, the daily balance decreases to $650.
Sum of the daily balances for June 11-15 = (number of days)(daily balance) = 5*650 = 3250.
When another $150 is withdrawn on June 16, the daily balance decreases to $500.
Sum of the daily balances for June 16-20 = (number of days)(daily balance) = 5*150 = 750.
Thus:
Sum of the daily balances for June 21-30 = (sum for June 11-30) - (sum for June 11-15) - (sum for 16-20) = 20,000 - 3250 - 750 = 16000.
Daily balance for June 21-30 \(= \frac{sum-of-the-daily-balances}{number-of-days} = \frac{16,000}{10} = 1600\).
Since the daily balance on June 20 = 150, the amount deposited on June 21 = 1600-150 = 1450.
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