GMAT Sample Questions: Quantitative
At a bakery, all donuts are priced equally and all bagels are priced equally. What is the total price of 5 donuts and 3 bagels at the bakery?
(1) At the bakery, the total price of 10 donuts and 6 bagels is $12.90.
(2) At the bakery, the price of a donut is $0.15 less than the price of a bagel.
A. If statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked;
B. If statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked;
C. If BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient;
D. If EACH statement ALONE is sufficient to answer the question asked;
E. If statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
Here we need to evaluate how sufficient the statements (1) and (2) are to answer the question asked, and choose the correct alternative accordingly.
If we let d be the price of 1 Donut and b be the price of 1 Bagel, statements (1) and (2) can be re-expressed as:
1. 10d + 6b = 12.90
2. d = b - 0.15
Mathematically, the only way to get the values of d and b is to use the equations from both statements at the same time and solve a typical school-exercise of simultaneous equations. This will help us get the values of d and b. We would then be enabled to answer any question related to those variables.
However, we are faced with twi facts that should be examined carefully:
Fact 1) The only alternative above that considers using both statements being used together is (C) If BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient.
Fact 2) The question is actually not asking for the values of b and d; rather it requires the price of a combo: 5d + 3b.
Let’s look attentively to statement 1, in the question, on its own: 10d + 6b = 12.90.
If we divide both sides by 2, we will have 5d + 3b = 6.45
This means that the price for 5 donuts and 3 bagels is $6.45!! That’s the answer we were looking for!
Statement (1) alone allows us to answer the question proposed; statement (2) alone only informs us about the difference in prices between a donut and bagel is clearly not sufficient to answer the question asked.
A. If statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked
Problem solving: Arithmetic
If 893 × 78 = p, which of the following is equal to 893 × 79?
(A) p + 1
(B) p + 78
(C) p + 79
(D) p + 893
(E) p + 894
1. Common approach
The unprepared approach to this question would lead us to determine the value of p.
Indeed, we could obtain p after some strenuous calculations. p would be 893 x 78 which leads to the value of p: p = 69654.
Then we need to move on to another hard, time-consuming sum: 893 x 79, which gives us 70547. If that’s not enough work, we need now to relate both numbers, as all the answers (including the correct one!) require us to express the new number (70547) in terms of p.
Since 70547 – 69654 = 893, we conclude that 70547 = 69654 + 893, which is the same as saying that 70547 is p + 893. Alternative (D) is the correct one.
2. The GMAT approach
As a rule of thumb, we should know that no question on GMAT is either too easy or too hard. They are suited for the test constraints: calculators are not allowed and the average time available per questions is no more than 2 minutes! There must be another way!
Indeed, the luxury of calculators makes us forget that some basic algebraic operations could be perfectly used with numbers just as well.
When it comes to Algebra, almost everyone recalls the multiplication property a(b + c) = ab +ac. The good news is that we could replicate this simple model here! After all: 893 x 79 = 893 x (78 +1) = 893 x 78 + 893 x 1. Having in mind that that question told us that 893 x 78 = p, then we have that:
893 x 79 = p + 893 and alternative (D) is correct. No big numbers and the question is straight-forwardly addressed.
Problem solving: Geometry, Circles
In the figure shown, if the area of the shaded region is 3 times the area of the smaller circular region, then the circumference of the larger circle is how many times the circumference of the smaller circle?
Before we examine the different approaches to this question, we should have a clear idea of what is happening in this case:
There are two circles; a small one and a big one and the small one is completely contained within the big circle.
The shaded area represents the difference in the area between them:
Area of big circle = Area of small circle + shaded area
The shaded area, according to the question, is 3 times the area of the small circle:
Shaded area = 3 x area of small circle.
Substituting in the above equation:
Area of big circle = Area of small circle + 3 x area of small circle
Thus: area of big circle = 4 x area of small circle
So how many times is the circumference of the big circle greater than circumference of the small circle?
The academic approach
We need to recall the basic formulas that relate to circles:
Area of circle: A=πR^2
Circumference of circle: C=2πR
Where r is the radius of the circle.
In terms of the areas, we saw above that: area of big circle = 4 x area of small circle
We can now say that: πR_BIG^2=4× πR_SMALL^2
Cancelling p in both sides, we will have R_BIG^2=4× R_SMALL^2
Taking the square root on both sides: RBIG = 2 x RSMALL
Now, the circumference of big circle is 2πR_BIG
But if we replace RBIG with 2 x RSMALL, we will have:
Circumference of big circle is 2π2×R_SMALL.
However, π2×R_SMALL= 2πR_SMALL, which is the circumference of the small circle.
Therefore, the circumference of the big circle is 2 x the circumference of the small circle and the correct answer is C.
A simpler approach: similarity
The small and big circles have identical shape (quite obviously!). We could see the big circle as a ‘magnified version’ of the small circle. When this happens, we have ‘similarity’: all linear measures of both circles are proportional: radii, diameters, circumferences.
That means if the radius of the big circle is k times bigger than the radius of the small circle, then:
The diameter of the big circle will be k times bigger that the diameter of the small circle
The circumference of the big Circle will be k times bigger that the circumference of the small circle
What about the two areas? They will also be proportional, but to k2. Therefore:
The area of the big circle will be k2 times bigger that the area of the small circle.
Thus, in our case, k2 = 4 and k = 2. So, the circumference of the big circle is 2x the circumference of the small one. The correct answer is C.
Problem solving: Algebra, Plug-in numbers, Quant
If 1 < x < y < z, which of the following has the greatest value?
A. z(x + 1)
B. z(y + 1)
C. x(y + z)
D. y(x + z)
E. z(x + y)
Method 1: Algebraic Approach
As we are dealing with numbers greater than 1, there are 2 rules that must be noted:
The product of two numbers is greater than each one of them;
The larger the numbers, the larger the product
Thus if 1 < x < y < z, we can derive the following inequailities:
xy < xz < yz
x < y < z < xz
When we open up the multiplications on each alternative:
A. z(x + 1) = xz + z
B. z(y + 1) = yz + z
C. x(y + z) = xy + xz
D. y(x + z) = xy + yz
E. z(x + y) = xz + yz
We have the following conclusions:
(B) is greater than (A) as yz > xz
Comparing (C), (D) and (E), (E) will be the greatest, as xy < xz < yz
(E) is greater than (B), as xz > z
Method 2: Plug-in numbers
Recalling the fact that the GMAT doesn’t regard the workings towards the correct answer, we could experiment with numbers instead of letters, since the alternatives have to be true for any values of x, y and z, as long as the condition 1 < x < y < z is respected:
Let’s try with x = 2, y = 3 and z = 4 and check which alternative gives us the largest number:
A. z(x + 1) = 4(2 + 1) = 4 × 3 = 12
B. z(y + 1) = 4(3 + 1) = 4 × 4 = 16
C. x(y + z) = 2(3 + 4) = 2 × 7 = 14
D. y(x + z) = 3(2 + 4) = 3 × 6 = 18
E. z(x + y) = 4(2 + 3) = 4 × 5 = 20
And we can quickly spot (E) as the correct alternative!
Problem solving: Quant, Quantitative, Estimation
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