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Of the 31 questions you will come across in the quantitative reasoning section of the GMAT exam, at least 50% will be data-sufficiency questions. This means that there will be 15 to 16 data-sufficiency questions in any quantitative reasoning section of the GMAT exam. About half of these will be word problems, which means that you will have about eight data-sufficiency word problem questions. The questions will have the same five constant options correlating to which condition(s) is/are sufficient to provide enough data to provide a definitive answer to the question asked. The options are:

**A – **(1) ALONE is sufficient, but (2) alone is not sufficient

**B** – (2) ALONE is sufficient, but (1) alone is not sufficient

**C **– TOGETHER are sufficient, but NEITHER ALONE is sufficient

**D** – EACH ALONE is sufficient

**E** – NEITHER ALONE NOR TOGETHER is the statements sufficient

You will have two types of data sufficiency word problems – the yes/no questions and the value questions.

You will often find data-sufficiency word problems to be long and winding. It is not easy to capture all the information in any word problem just by reading through and minimizing the impact of long ** given** statements and conditions by translating English into math. To do this, you must deliberately evaluate what is known and what is needed algebraically from given statements. Use initials and shorthand to efficiently define variables for each unknown. For example, if you were talking about Frank’s sheep and George’s sheep, you can use

It is important to recognize that one evaluation can be significantly longer. Either an individual statement or the initial conditions can be significantly longer than the subsequent conditions in a data-sufficiency word problem. For instance, you may find that statement (1) is rather long, but statement (2) is short and goes very quickly because you did most of your work in statement (1).

Data Sufficiency word problems are simply a system of linear equations. Basically, for any linear equations, to solve for a single variable, you need one equation. To solve for two variables, you require two equations. In theory, to solve for as many as 74 variables, you will require 74 unique equations. This implies that in a data sufficiency word problem need, to solve for *x* unique variables, you will require *x* unique equations.

If you determine that the number of unique equations you have is equivalent to the number of unique variables you need to determine, then you don’t unnecessarily need to solve for the exact values of the variables. You know you can solve, but you don’t have to solve because the format of data sufficiency questions does not require you to.

You need to know the algebraic paths to specifically sought variables. For instance, if you are being asked to solve for \(xy\), you don’t have to solve for *x* and *y* separately. What you need is a straight path to determining \(xy\) combined rather than determining *x*, then *y*, then combining them.

It is also important to recognize that inequalities do not function identically to equations. Inequalities don’t have one specific solution, so just because you have two inequalities does not mean you need to solve for two variables. However, inequalities with all values greater than zero can be manipulated in the same way as equations. You can add, subtract, divide, and multiply, provided that everything in the inequality is greater than zero. This is because if you multiply or divide an inequality by a negative, you have to remember to flip the sign of your inequality, and that can cause some complications in data insufficiencies.

Brant, Carl, and Earl each are paid a different constant rate for each bag one of them makes at Duffel Industries. How much does Carl get paid for each bag he makes at Duffel Industries?

- Earl’s rate per bag is twice that of Carl and one-third that of Brant.
- Brant makes $200 more per bag than Earl makes.

Always articulate what you know first. (K = What we know; N = What we need)

- K: b = Brant’s rate; c = Carl’s rate; e = Earl’s rate. We want to determine c = Carl’s rate so can highlight it.
- N:
*x*equations to solve for*x*variables. Since we have three unique variables,*b*,*c*, and*e*, we will need three unique equations with no new variables.

From statement:

(1) \(e = 2c; e = ⅓b\). We have 3 variables and 2 equations, not three equations. This statement on its own is not sufficient. We need not manipulate the algebra any further. We are left with options B, C, and E.

(2) \(b = 200 + e\). We have 3 variables and one equation, not three. This statement alone is not sufficient to solve Carl’s rate and eliminate choice B. We are left with C and E.

We can now combine the two statements. Together we have three unique equations for three unique variables. Theoretically, this is sufficient to solve for *c,* and we don’t have to solve the algebraic equations fully. Select choice C ** together are sufficient, but neither alone is sufficient**.

Duffel Industries received an order for 10 bags consisting of two of its bag models, the Amani and the Guddi. If each Amani bag costs $25 and a Guddi bag costs $75, are there more Amani bags than Guddi bags in the order?

- The total revenue from the bag order was less than $550.
- The total revenue for the bag order was greater than $450.

We start by articulating what we know.

K:

*a*= number of Amani bags;*g*= number of Guddi bags; \(a + g = 10\). We can make*g*the subject of the equation, \(g = 10 – a\).- Cost a = $25; cost g = 75.
**Is a > g, Y/N?**Always write down the Y/N so that you don’t lose sight of the question you seek to answer. In this case, we need to determine the “No”.

N: inequality definitively relating *a* to *g*.

From statement:

(1). \(25a + 75g < 550\) / substituting \(10 – \text{a}\) for *g*, \(25a + 75(10 – a) < 550\).

\(25a + 750 – 75a < 550\)

add 50*a* and subtract 550 from both sides \(200 < 50a\)

divide both sides by 50 \(4 < a\)

Since *a* > 4, if *a*= 5, then the answer is “No,” there aren’t more Amani bags than Guddi bags in the order. If *a* = 6, then “Yes,” there are more Amani bags than Guddi bags in the order. Since we have both outcomes, this condition alone is not sufficient, and therefore we are left with options B, C, and E.

(2). \(450 < 25a + 75g\) / substituting \(10 – a\) for g, \(450 < 25a + 75(10 -a)\)

\(450 < 25a + 750 -75a\)

* *Add 50a and subtract 450 from each side \(50a < 300\)

divide both sides by 50 \(a < 6\)

Since \(a < 6\), then our maximum \(a = 5\). This means that our answer to the question, “Are there more Amani bags than Guddi bags in the order?” is always “No”.

Since we have determined that there aren’t more Amani bags than Guddi bags in the order, the correct answer is choice* B, (2) ALONE is sufficient, but (1) alone is not sufficient.*

** Always No** answers and

Algebraically articulate the given conditions by assigning variables to all unknown values provided while asking these three questions:

- What do I know? Note it with a “K”
- What am I looking for? Highlight the question with a box. If it is a Yes/No question, indicate Y/N so that you don’t lose sight of it.
- What do I need to know? or what I’m I looking for? Note this with “N”.

Carefully articulate and manipulate statement (1) algebraically.

- If it is sufficient, then choices A and D will remain viable.
- If not, then your options will remain B, C, and E.

Carefully articulate and manipulate statement (2) algebraically.

- If (1) and (2) are each sufficient, then you will choose option D
- If (1) is sufficient but (2) is not, your correct choice will be A
- If (2) is sufficient but (1) is not, then select B
- If either (1) or (2) is sufficient, then stop here. You do not need to continue evaluating them together if one of them is sufficient.
- If neither 1 nor 2 is sufficient on its own, then only choices C and E remain viable.

Carefully articulate and manipulate algebraically the two statements together.

- If together they are sufficient, select choice C.
- If they are not sufficient, select choice E

If Heloise bought only 25-cent and 20-cent stamps at the post office and bought at least one of each denomination, how many 25-cent stamps did she buy?

- Heloise bought a total of 10 stamps.
- Heloise spent a total of $4 on stamps.

K: a = # 25 cent stamps; b = # 20 cent stamps; \(a > 0\); \(b > 0\).

N: Two new equations with no new variables.

- \(a + b = 10\). Many values of
*a*can satisfy this condition. On its own, it is not sufficient for determining the value of*a*. Therefore, we are left with BC&E. - \(25a + 20b = 400\)

Simplify: \(5a + 4b = 80\). Looking at this, we can see that there are multiple possible values of *a* and *b*. For instance, *b* could be equal to 5 or 10. Therefore condition (2) by itself is not sufficient. We eliminate choice B, and we are left with C&E.

If we consider the two statements together, we have two equations with no new variables. Without doing the math either in a system or by substitution, we know that these two conditions are sufficient.

2. During a 5-day rodeo, the least number of tickets sold for a single day was 800. Was the average (arithmetic mean) number of tickets sold per day during the rodeo greater than 1,000?

- For the three days with the highest number of tickets sold, the average (arithmetic mean) of tickets sold per day was 1,200.
- For the three days with the lowest number of tickets sold, the average (arithmetic mean) of tickets sold per day was 900.

K: days = 5; least # of tickets a day = 800

Was Ave. tickets/day > 1,000?

N: Sum of tickets = s. So, \(\frac{s}{5} > 1,000\)

\(5(\frac{s}{5} > 1000)\)

\(s > 5000\). ** We need to establish that the sum of tickets sold s **>

(1)Ave. for top 3 days = 1200, sum = 3600.

Minimum # tickets sold:

$$\begin{align*}&= 3600 + 2(800)\\&= 3600+1600=5200\\&=5200>5000\end{align*}$$

(2) Ave. of least 3 days = 900, sum = 2700.

Minimum # tickets sold:

$$\begin{align*}&= 2700 + 2(800)\\&=2700 + 1600 = 4300\\&=4300 < 5000\end{align*}$$

We have established that the minimum number of tickets sold is 4300, but the maximum could be any number, probably much higher. With this condition alone, we cannot determine that s > 5000, therefore, is not sufficient. Our answer is choice* A, (1) ALONE is sufficient, but (2) alone is not sufficient.*

Your ability to solve GMAT data sufficiency word problems will come in handy if you are going to score highly in the GMAT quantitative reasoning section and, consequently, in your GMAT exam. Take your time, a little bit every day, to do as much practice as you can. We’ve got a lot of GMAT study materials on our website, which you can use to practice adequately. Your GMAT goals are very much within your reach!

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