The total price of 5 pounds of regular coffee and 3 pounds of decaffei

We can create the equation:

5r + 3d = 21.5

We need to determine 5r.

Statement One Alone:

If the price of the 5 pounds of regular coffee had been reduced 10 percent and the price of the 3 pounds of decaffeinated coffee had been reduced 20 percent, the total price would have been $18.45.

We can create the equation:

5(0.9r) + 3(0.8d) = 18.45

4.5r + 2.4d = 18.45

45r + 24d = 184.5

Multiplying the equation in the stem analysis by 8, we have 40r + 24d = 172. Subtracting this from the equation above, we have:

5r = 12.5

Therefore, we see that the value of 5r is 12.5. Statement one alone is sufficient to answer the question.

Statement Two Alone:

The price of the 5 pounds of regular coffee was $3.50 more than the price of the 3 pounds of decaffeinated coffee.

We can create the equation:

5r = 3d + 3.50

3d = 5r - 3.50

Substituting 5r - 3.50 for 3d in the equation in the stem analysis, we have:

5r + 5r - 3.50 = 21.5

10r = 25

5r = 12.5

We see that the value of 5r is 12.5. Statement two alone is sufficient to answer the question.

Answer: D
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We're given that 5*regular + 3*decaf = 21.5 and asked to calculate 5*regular
So we need to know 3*decaf or to have another equation with both variables.
We'll look for this information, a Logical approach.
(1) gives another equation with both variables -- sufficient! (it gives 5*regular*0.9 + 3*decaf*0.8 = 18.45)
(2) this also gives another equation with both variables. Sufficient.

(D) is our answer.

regular = x and decaffianated = y
given 5x+3y=21.5
find 5x
#1
5x*.9+.8*3y=18.45
sufficient solve
#2
5x=3.5+3y
sufficient
IMO D

Let \(r\) and \(d\) be the regular prices per pound of regular coffee and decaffeinated coffee, respectively. We know that \(5r+3d=21.5\).
The original question: \(5r=?\)
The rephrased question: \(r=?\)

1) We know that \(0.9\cdot 5r+0.8\cdot 3d=18.45\). The system of equations from the original information and from 1) has a unique solution. Thus, we could get a unique value to answer the rephrased question. \(\implies\) Sufficient

2) We know that \(5r=3d+3.5\). The system of equations from the original information and from 2) has a unique solution. Thus, we could get a unique value to answer the rephrased question. \(\implies\) Sufficient

Answer: D

A system of 2-variable linear equations does not guarantee a unique solution. However, by looking at the coefficients of the variables in the two equations, we can usually quickly evaluate whether there is a unique solution, an infinite number of solution, or no solution at all.

Hi All,

We're told that the total price of 5 pounds of regular coffee and 3 pounds of decaffeinated coffee was $21.50. We're asked for the price of the 5 pounds of regular coffee. The information in the first sentence can be used to create a 2 variable equation, which should get us thinking about 'System math' (re: 2 variables and 2 unique equations):

5(R) + 3(D) = $21.50

R represents the price of a pound of regular coffee while D represents the price of a pound of decaffeinated coffee. If we enough information to create a second, unique equation using those 2 variables, then we can stop working - that information would be enough for us to get to the correct answer and solve for the prices/pound of the two coffees.

(1) If the price of the 5 pounds of regular coffee had been reduced 10 percent and the price of the 3 pounds of decaffeinated coffee had been reduced 20 percent, the total price would have been $18.45.

The information in Fact 1 can be used to create another equation:

5(.9R) + 3(.8D) = $18.45

While the two equations might look a bit 'ugly', it's still a System of equations, so we CAN solve for the two variables. Thankfully, we don't actually have to do that work to know that we COULD, so we would know the exact value of 5R.
Fact 1 is SUFFICIENT

(2) The price of the 5 pounds of regular coffee was $3.50 more than the price of the 3 pounds of decaffeinated coffee.

With the information in Fact 2, we can create an equation relating the values of 5R and 3D:

5R = 3D + $3.50

Again, we end up with a System of equations, so we CAN solve for the two variables. Thankfully, we don't actually have to do that work to know that we COULD, so we would know the exact value of 5R.
Fact 2 is SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

We're given that 5*regular + 3*decaf = 21.5 and asked to calculate 5*regular
So we need to know 3*decaf or to have another equation with both variables.
We'll look for this information, a Logical approach.
(1) gives another equation with both variables -- sufficient! (it gives 5*regular*0.9 + 3*decaf*0.8 = 18.45)
(2) this also gives another equation with both variables. Sufficient.

(D) is our answer.

Posted from my mobile device

Hi, ZoltanBP
could you please elaborate a little more on "A system of 2-variable linear equations does not guarantee a unique solution"? and how then should we tackle this ques? should we solve each statement to be sure that a sol is possible? are there any questions where such a trick has been utilized?

If none of the coefficients or the constants are zero, then a system of 2-variable linear equations

\(a_1x+b_1y=c_1\)
\(a_2x+b_2y=c_2\)

must have one of the following cases.

Case 1: Unique solution if

\(\cfrac{a_1}{a_2}\neq\cfrac{b_1}{b_2}\)

Case 2: No solution if

\(\cfrac{a_1}{a_2}=\cfrac{b_1}{b_2}\neq\cfrac{c_1}{c_2}\)

Case 3: Infinite number of solutions if

\(\cfrac{a_1}{a_2}=\cfrac{b_1}{b_2}=\cfrac{c_1}{c_2}\)

In DS problems, we can often evaluate statements very quickly by using this techinque.

Hi Kritisood,

You might find it helpful to draw a picture to visualize all of this. Start off with an xy-coordinate plane and draw 2 lines. Since those two lines 'go on forever', they will almost certainly cross one another UNLESS the lines are PARALLEL or the lines are IDENTICAL. Parallel lines have the 'same slope', so for example...

Y = 2X + 5
Y = 2X - 1

There is NO solution for this pair of equations. While you might be tested on this concept in a graphing question, you won't see it in a standard "story problem" on the GMAT such as the one in which this prompt is based.

IDENTICAL lines might 'look' different, but when you simplify the equations, if they are the SAME equation, then there will be an INFINITE number of solutions. For example:

Y = 2X + 5
2Y = 4X + 10

The second equation is just the first equation with everything "doubled."

2Y = 4X + 10...... divide everything by 2 and you end up with...

Y = 2X + 5

Every solution to the first equation is also a solution to the second equation because that second equation is NOT a unique equation. This issue can sometimes appear in rare DS prompts on the GMAT (including in 'story problems').

GMAT assassins aren't born, they're made,
Rich

Answer: Option D

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