The challenge of reviewing this question is not that the question is hard to understand – it is that you will need to solve this official question within a minute using minimum calculations.
Let’s take a look at the question stem:
| Date of Transaction |
Type of Transaction |
| June 11 |
Withdrawal of $350 |
| June 16 |
Withdrawal of $500 |
| June 21 |
Deposit of x dollars |
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
Think about how you might answer this question:
The average of daily balances = (Balance at the end of June 1 + Balance at the end of June 2 + … + Balance at the end of June 30) / 30 = 1000
Now we have been given the only three transactions that took place:
- A withdrawal of $350 on June 11 – so on June 11, the account balance goes down to $650.
- A withdrawal of $500 on June 16 – so on June 16, the account balance goes down to $150.
- A deposit of $x on June 21 – So on June 21, the account balance goes up to 150 + x.
Now we can plug in these numbers to say the average of daily balances = [1000 + 1000 + …(for 10 days, from June 1 to June 10) + 650 + 650 + … (for 5 days, from June 11 to June 15) + 150 + … (for 5 days, from June 16 to June 20) + (150 + x) + (150 + x) + … (for 10 days, from June 21 to June 30)] / 30 = 1000
One might then end up doing this calculation to find the value of x:
[(1000 * 10) + (650 * 5) + (150 * 5) + ((150 + x) * 10)] / 30 = 1000
x = $1,450
The answer is D.
But this calculation is rather tedious and time consuming. Can’t we use the deviation method we have for averages and weighted averages, instead? After all, we are dealing with large values here! How?
Note that we are talking about the average of certain data values. Also, we know the deviations from those data values:
- The amount from June 11 to June 30 is 350 less.
- The amount from June 16 to June 30 is another 500 less.
- The amount from June 21 to June 30 is x in excess.
Through the deviation method, we can see the shortfall = the excess:
350 * 20 + 500 * 15 = x * 10
x = 1,450 (D)
This simplifies our calculation dramatically! Though saving only one minute on a question like this may not seem like a very big deal, saving a minute on every question by using a more efficient method could be the difference between a good Quant score and a great Quant score!
