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 from June 1 to June 10 (10 days) was $1,000 --> the sum of the balances 10*$1,000 = $10,000;
The daily balance from June 11 to June 15 (5 days) was $1,000-$350=$650 --> the sum of the balances 5*$650 = $3,250 ;
The daily balance from June 16 to June 20 (5 days) was $650-$500=$150 --> the sum of the balances 5*$150 = $750;
The daily balance from June 21 to June 30 (10 days) was $150 --> the sum of the balances 10*($150 + x).
We need the sum of the balances for 30 days to be $30,000 (since the average for 30 days was $1,000), thus we need
\(10,000+3,250+750+10*($150+x)=30,000\);
\(x=$1,450\).
Answer: D.
OR:For 5 days the total balance was less by 5*$350 = $1,750;
For another 5 days the total balance was less by 5*($350 + $500) = $4,250;
The final 10 days of June should compensate \($1,750 + $4,250 = $6,000\), thus the sum of the balances for the final 10 days must be \($10,000+$6,000=$16,000\):
\(($100-($350+$500)+x)*10=$16,000\);
\(x=$1,450\).
Answer: D.
Hope it's clear.