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Today, we will give you a GMAT challenge question. 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  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:
on June 21 – so on June 21, the account balance goes up to 150 + .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 + ) + (150 + ) + … (for 10 days, from June 21 to June 30)] / 30 = 1000
One might then end up doing this calculation to find the value of :
![\frac{[(1000 \times 10) + (650 \times 5) + (150 \times 5) + ((150 + x) \times 10)]}{30} = 1000](https://anaprep.com/wp-content/ql-cache/quicklatex.com-30e9071d3d442872bb21580eb7f81adb_l3.png "Rendered by QuickLaTeX.com")
= €1,450
The answer is (D).
But this calculation is rather tedious and time consuming. Can’t we use the deviation method we discussed in our weighted averages module, 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:
in excess.Through the deviation method, we can see the shortfall = the excess:
= 1,450
Answer (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!
