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]]>Let’s look at three different question formats today to understand the difference between them:
Case 1: Must Be True
If |-x/3 + 1| < 2, which of the following must be true?
(A) x > 0
(B) x < 8
(C) x > -4
(D) 0 < x < 3
(E) None of the above
We have two linked inequalities here. One is |-x/3 + 1| < 2 and the other is the correct answer choice. We need to think about how the two are related.
We are given that |-x/3 + 1| < 2. So we know that x satisfies this inequality. That will give us the universe which is relevant to us. x will take one of those values only. So let’s solve this inequality. (We will not focus on how to solve the inequality in this post – it has already been discussed here. We will just quickly show the steps.)
|x/3 – 1| < 2
(1/3) * |x – 3| < 2
|x – 3| < 6
The distance of x from 3 is less than 6, so -3 < x < 9. Now we know that every value that x can take will lie within this range.
The question now becomes: what must be true for each of these values of x? Let’s assess each of our answer options with this question:
(A) x > 0
Will each of the values of x be positive? No – x could be a negative number greater than -3, such as -2.
(B) x < 8
Will each of the values of x be less than 8? No – x could be a number between 8 and 9, such as 8.5
(C) x > -4
Will each of the values of x be more than -4? Yes! x will take values ranging from -3 to 9, and each of the values within that range will be greater than -4. So this must be true.
(D) 0 < x < 3
Will each of these values be between 0 and 3. No – since x can take any of the values between -3 and 9, not all of these will be just between 0 and 3.
Therefore, the answer is C (we don’t even need to evaluate answer choice E since C is true).
Case 2: Could Be True
If −1 < x < 5, which is the following could be true?
(A) 2x > 10
(B) x > 17/3
(C) x^2 > 27
(D) 3x + x^2 < −2
(E) 2x – x^2 < 0
Again, we have two linked inequalities, but here the relation between them will be a bit different. One of the inequalities is −1 < x < 5 and the other will be the correct answer choice.
We are given that -1 < x < 5, so x lies between -1 and 5. We need an answer choice that “could be true”. This means only some of the values between -1 and 5 should satisfy the condition set by the correct answer choice – all of the values need not satisfy. Let’s evaluate our answer options:
(A) 2x > 10
x > 5
No values between -1 and 5 will be greater than 5, so this cannot be true.
(B) x > 17/3
x > 5.67
No values between -1 and 5 will be greater than 5.67, so this cannot be true.
(C) x^2 > 27
x^2 – 27 > 0
x > 3*√(3) or x < -3*√(3)
√(3) is about 1.73 so 3*1.73 = 5.19. No value of x will be greater than 5.19. Also, -3*1.73 will be -5.19 and no value of x will be less than that. So this cannot be true.
(Details on how to solve such inequalities are discussed here.)
(D) 3x + x^2 < −2
x^2 + 3x + 2 < 0
(x + 1)(x + 2) < 0
-2 < x < -1
No values of x will lie between -2 and -1, so this also cannot be true.
(E) 2x – x^2 < 0
x * (x – 2) > 0
x > 2 or x < 0
If -1 < x < 5, then x could lie between -1 and 0 (x < 0 is possible) or between 2 and 5 (x > 2 is possible). Therefore, the correct answer is E.
Case 3: Complete Range
Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?
(A) 0 < |x| < ½
(B) |x| > ½
(C) -½ < x < 0 or ½ < x
(D) x < -½ or 0 < x < ½
(E) x < -½ or x > 0
We have two linked inequalities, but the relation between them will be a bit different again. One of the inequalities is x^3 – 4x^5 < 0 and the other will be the correct answer choice.
We are given that x^3 – 4x^5 < 0. This inequality can be solved to:
x^3 ( 1 – 4x^2) < 0
x^3*(2x + 1)*(2x – 1) > 0
x > 1/2 or -1/2 < x < 0
This is our universe of the values of x. It is given that all values of x lie in this range.
Here, the question asks us the complete range of x. So we need to look for exactly this range. This is given in answer choice C, and therefore C is our answer.
We hope these practice problems will help you become able to distinguish between the three cases now.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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]]>The post Using Special Formats on GMAT Variable Problems appeared first on Veritas Prep Blog.
]]>Here are some examples:
An even number: 2a
Logic: It must be a multiple of 2.
An odd number: (2a + 1) or (2a – 1)
Logic: It will not be a multiple of 2. Instead, it will be 1 more (or we can say 1 less) than a multiple of 2.
Two consecutive integers: 2a, (2a + 1) or (2a – 1), 2a
Logic: One number will be even and the other will be the next odd number (or the other way around).
Four consecutive odd numbers: (2a – 3), (2a – 1), (2a + 1), (2a + 3)
In this case, the sum of the numbers comes out to be a clean 8a. This can be very useful in many cases.
Five consecutive even numbers: (2a – 4), (2a – 2), 2a, (2a + 2), (2a + 4)
In this case, the sum of the numbers comes out to be a clean 10a. This can also be very useful in many cases.
A prime number: (6a+1) / (6a – 1)
Every prime number greater than 3 is of the form (6a + 1) or (6a – 1). Note, however, that every number of this form is not prime.
Three consecutive numbers:
If we know one number is even and the other two are odd, we will have: (2a – 1), 2a, (2a + 1).
Logic: They add up to give 6a.
In a more generic case, we will have: 3a, (3a+1), (3a+2).
This gives us some important information. It tells us that one of the numbers will definitely be a multiple of 3 and the other two numbers will not be. Note that the numbers can be in a different order such as (3a + 1), (3a + 2) and (3a + 3). (3a + 3) can be written as 3b, so the three numbers will still have the same properties.
Basically, try to pick numbers in a way that will make it easy for you to manage them. Remember, three numbers do not need to be a, b and c – there could be, and in fact often are, several other hints which will give you the relations among the numbers.
Now, let’s see how picking the right format of these numbers can be helpful using a 700-level GMAT question:
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
(A) 12
(B) 17
(C) 25
(D) 33
(E) 50
Let’s have the four consecutive odd numbers be the following, where “a” is any integer: (2a – 3), (2a – 1), (2a + 1), (2a + 3)
The sum of these numbers is: (2a – 3) + (2a – 1) + (2a + 1) + (2a + 3) = 8a
Now let’s have the three consecutive even numbers be the following, where “b” is any integer: (2b – 2), 2b, (2b + 2)
The sum of these numbers is: (2b – 2) + 2b + (2b + 2) = 6b
Note here that instead of 2a, we used 2b. There is no reason that the even numbers would be right next to the odd numbers, hence we used different variables so that we don’t establish relations that don’t exist between these seven numbers.
We are given that the sum 8a is equal to the sum 6b.
8a = 6b, or a/b = 3/4, where a and b can be any integers. So “a” has to be a multiple of 3 and “b” has to be a multiple of 4.
With this in mind, possible solutions for a and b are:
a = 3, b = 4;
a = 6, b = 8;
a = 9, b = 12
etc.
We are also given that the middle term of the even numbers is greater than 101 and less than 200.
So 101 < 2b < 200, i.e. 50.5 < b < 100.
B must be an integer, hence, 51 ≤ b ≤ 99.
Also, b has to be a multiple of 4, so the values that b can take are 52, 56, 60, 64 … 96
The number of values b can take = (Last term – First term)/Common Difference + 1 = (96 – 52)/4 + 1 = 12
For each of these 12 values of b, there will be a corresponding value of a and, hence, we will get 12 such sequences. Therefore, the answer to our question is A.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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]]>The post Quarter Wit, Quarter Wisdom: Beware of Sneaky Answer Choices on the GMAT! appeared first on Veritas Prep Blog.
]]>Today, we will look at an example of this concept – if it seems to be too easy, it is a trap!
In the figure given above, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?
(A) 8√(2)
(B) 24√(3)
(C) 72√(2)
(D) 144√(2)
(E) 384
The first thing I notice about this question is that we have an equilateral triangle. So I am thinking, the area = s^2 * √(3)/4 and/or the altitude = s*√(3)/2.
The irrational number in play is √(3). There is only one answer choice with √(3) in it, so will this be the answer?
Now, it actually makes me uncomfortable that there is only one option with √(3). At first glance, it seems that the answer has been served to us on a silver plate. But the question format doesn’t seem very easy – it links two geometrical figures together. So I doubt very much that the correct answer would be that obvious.
The next step will be to think a bit harder:
The area of the triangle has √(3) in it, so the side would be a further square root of √(3). This means the actual irrational number would be the fourth root of 3, but we don’t have any answer choice that has the fourth root of 3 in it.
Let’s go deeper now and actually solve the question.
The area of the equilateral triangle = Side^2 * (√(3)/4) = 48
Side^2 = 48*4/√(3)
Side^2 = 4*4*4*3/√(3)
Side = 8*FourthRoot(3)
Now note that the side of the equilateral triangle is the same length as the sides of the squares, too. Hence, all sides of the three squares will be of length 8*FourthRoot(3).
All nine sides of the figure are the sides of squares. Hence:
The perimeter of the nine sided figure = 9*8*FourthRoot(3)
The perimeter of the nine sided figure =72*FourthRoot(3)
Now look at the answer choices. We have an option that is 72√(2). The other answer choices are either much smaller or much greater than that.
Think about it – the fourth root of 3 = √(√(3)) = √(1.732), which is actually very similar to √(2). Number properties will help you figure this out. Squares of smaller numbers (that are still greater than 1) are only a bit larger than the numbers themselves. For example:
(1.1)^2 = 1.21
(1.2)^2 = 1.44
(1.3)^2 = 1.69
(1.414)^2 = 2
Since 1.732 is close to 1.69, the √(1.732) will be close to the √(1.69), i.e. 1.3. Also, √(2) = 1.414. The two values are quite close, therefore, the perimeter is approximately 72√(2). This is the reason the question specifically requests the “approximate” perimeter.
We hope you see how the Testmaker could sneak in a tempting answer choice – beware the “easiest” option!
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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]]>The post The Holistic Approach to Absolute Values – Part V appeared first on Veritas Prep Blog.
]]>(Before you continue reading, be sure to check out Part I, Part II, Part III and Part IV of this lesson.)
Let’s look at the following GMAT question:
For how many integer values of x, is |x – 6| > |3x + 6|?
(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite
In this question, we are given the inequality |x – 6| > 3*|x + 2|
Using the same logic as we did in the previous two posts, we will word the inequality like this: the distance from 6 should be more than three times the distance from -2.
At x = -2, the distance from 6 is 8 and the distance from -2 is 0. This means the distance from 6 is more than three times the distance from -2.
At x = -1, the distance from 6 is 7 and the distance from -2 is 1. Three times the distance from -2 is 3. This means the distance from 6 is more than three times the distance from -2.
At some point on the right of -1, the distance from 6 will be equal to three times the distance from -2. The distance between -2 and 6 is 8. If we split this 8 into 4 equal parts to get to x = 0, the distance from 6 will be equal to three times the distance from -2.
Now for every point to the right of 0, the distance from 6 will be less than three times the distance from -2.
Let’s try to go to the left of -2 instead. Will there be a point to the left of -2 where the distance from 6 will be equal to three times the distance from -2? Say that point is “a” units away from -2. -2 must then be 2a units away from 6 to ensure that 6 is a total of 3a units away from that point.
The distance between -2 and 6 is 8 – this 8 needs to be equal to 2a, so “a” must be 4 units.
The point where the distance from 6 will be equal to three times the distance from -2 will be 4 units to the left of -2, i.e. at -6. So at points to the right of -6 (but left of 0), the distance from 6 will be more than three times the distance from -2.
Note that for all values to the left of -6, the distance from 6 will be less than three times the distance from -2.
Hence, our x will lie in the range from -6 to 0.
-6 < x < 0
With these parameters, we will have 5 integer solutions: -5, -4, -3, -2 and -1. Hence, our answer is C.
Let’s look at a second question:
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?
(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite
Now the true value of this method is visible, as we have three or more terms. The arduous algebra involved in this given inequality makes our logical approach much more attractive.
First note that we have the term |5 – x|. This is the same as |x – 5| because |x| = |-x|.
We will word the inequality like this: the distance from 5 + the distance from 8 should be greater than the distance from -7.
Let’s find the point where the sum of the distance from 5 and the distance from 8 is equal to distance from -7. Say that point is “a” units to the left of 5.
a + a + 3 = 12 – a
a = 3
So the point is 3 units to the left of 5, which means it is at 2. For all points to the left of 2, the sum of the distance from 5 and the distance from 8 will be greater than the distance from -7.
How about the points that are to the right of 8? Say there is a point “b” units away from 8 where the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.
3 + b + b = 15 + b
b = 12
So if we go 12 units to the right of 8, i.e. at x = 20, the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.
For all points to the right of 20, the sum of the distance from 5 and the distance from 8 is greater than the distance from -7, so there will be infinite points for which the sum of the distance from 5 and the distance from 8 is greater than the distance from -7. Therefore, our answer is E.
Using this concept, try to answer the following question on your own: For how many integer values of x, is |x – 6| – |3x + 6| > 0?
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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]]>The post The Holistic Approach to Absolute Values – Part IV appeared first on Veritas Prep Blog.
]]>Imagine that there are 3 friends with houses at points -1, 0 and 3 in a straight line. They decide to meet at the point x.
So, the total distance the friends will cover to meet at x will be |x – 3| + |x + 1| + |x|.
Now we can choose to minimize this total distance, bring it to some particular value or make it more or less than some particular value.
If we want to minimize the total distance, we just make the friends meet at the second guy’s house, i.e. at the point 0. The friend at 3 and the friend at -1 need to travel 4 units total to meet anyway, so there’s no point in making the guy at 0 travel any distance at all. So the minimum total distance would be 4, which would then be the minimum value of |x – 3| + |x + 1| + |x|. This minimum value is given by the expression at x = 0.
With this in mind, when we move to the right or to the left of x = 0, the total distance will increase and, hence, the value of the expression |x – 3| + |x + 1| + |x| will also increase.
Thereafter, it is easy to solve for |x – 3| + |x + 1| + |x| = 10 or |x – 3| + |x + 1| + |x| < 10, etc., as seen in our previous post.
Today, let’s look at how to solve a more advanced GMAT question using the same logic:
For how many integer values of x, is |x – 5| + |x + 1| + |x| + |x – 7| < 15?
(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite
In our parallel situation of friends and houses, we now have 4 friends with houses at points -1, 0, 5 and 7.
The friends at -1 and 7 are 8 units apart, so they will need to cover at least this total distance together to meet. It doesn’t matter where they meet between -1 and 7 (inclusive), they will need to cover exactly 8 units.
The friends at 0 and 5 will need to travel a minimum distance of 5 to meet. They can meet anywhere between 0 and 5 (inclusive) and the distance they will cover will still be 5.
So, all four friends can meet anywhere between 0 and 5 (inclusive) and the total distance covered will be 8 + 5 = 13. This would be the minimum total distance, and hence, the minimum value of the expression |x – 5| + |x + 1| + |x| + |x – 7|.
When we move to the left of 0 or to the right of 5, the total distance covered will be more than 13. At any point between -1 and 7, the total distance covered by the friends at -1 and 7 will be only 8. When we move 1 unit to the left of 0 and reach -1, the total distance covered by the friends at 0 and 5 will be 1 + 6 = 7. So to meet at -1, the total distance traveled by all friends together will be 8 + 7 = 15.
Similarly, when we move 1 unit to the right of 5 and reach 6, the total distance covered by the four friends will be again 8 + 7 = 15. So at points x = -1 and x = 6, the value of the expression will be 15. Between these two points (excluding the points themselves), the value of the expression will be less than 15.
So now we know -1 < x < 6. With these parameters, x can take 6 integer values: 0, 1, 2, 3, 4, 5. Therefore, the answer is D.
Note that when we had 3 points on the number line, the minimum total distance was found at the second point. Now when we have 4 points on the number line, the minimum total distance has been found to be in the range between second and third points.
Let’s look at another question:
For how many integer values of x, is |2x – 5| + |x + 1| + |x| < 10?
(A) 1
(B) 2
(C) 4
(D) 5
(E) Infinite
|2x – 5| + |x + 1| + |x| < 10
2*|x – 5/2| + |x + 1| + |x| < 10
In this sum, now the distance from 5/2 is added twice.
In our parallel situation, this is equivalent to two friends living at 5/2, one living at 0 and one living at -1. Now note that the expression may not take the minimum value of x = 0 because there are 2 people who will need to travel from 5/2.
We have four friends in all, so we can expect to get a range in which we will get the minimum value of the expression. The second and third people are at 0 and 5/2, respectively.
The total distance at x = 0 will be 1 + 2*(5/2) = 6. The total distance at x = 5/2 will be 7/2 + 5/2 = 6.
So if we move to the left of 0 or to the right of 5/2, the total distance will increase. If we move 1 unit to the right of 5/2 and reach 7/2, the total distance covered by the four friends will be 9/2 + 7/2 + 2 = 10. If we move 1 unit to the left of 0 and reach -1, the total distance covered by the four friends will be 0 + 1 + 2*(7/2) = 8. Now all four friends are at -1. To cover a distance of another 2, they should move another 0.5 units to the left of -1 to reach -1.5.
Now the total distance covered by the four friends will be 0.5 + 1.5 + 2*4 = 10, so the total distance when x lies between the points -1.5 and 3.5 (excluding the points themselves) will be less than 10.
Now we know -1.5 < x < 3.5. With these parameters, x can take 5 integer values: -1, 0, 1, 2 and 3. Therefore, the answer is D.
Now use these concepts to solve the following question: For how many integer values of x, is |3x – 3| + |2x + 8| < 15?
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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]]>Instead of looking at how to solve equations, like we did in our previous post, we will look at how to solve inequalities using the same concept.
A quick review:
Let’s move ahead now and see how we can use these concepts to solve inequalities:
For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?
(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite
In the previous post, we saw the a similar question, except it involved an equation rather than an inequality. For that problem, we found that the two points where the total distance is equal to 10 are -2.667 and 4:
What will be the total distance at any value of x between these two points?
Say, x = 0
|x – 3| + |x + 1| + |x|
= 3 + 1 + 0
= 4
Say, x = 3
|x – 3| + |x + 1| + |x|
0 + 4 + 3
= 7
In both cases, we see that the total distance covered is less than 10. Note that the minimum distance covered will be 4 at x = 0 (discussed in the previous post) so by moving to the right of 0 or to the left of 0 on the number line, we get to the points where the distance increases to 10. So for every point in between, the total distance will be less than 10 (the entire red region).
Hence, at integer points x = -2, -1, 0, 1, 2 and 3 (which are all between -2.667 and 4), the total distance will be less than 10. The total distance will be less than 10 for all non-integer points lying between -2.667 and 4 too, but the question only asks for the integer values, so that is all we need to focus on. (Of course, there are infinite non-integer points between any two distinct points on the number line.) Hence, the answer will be 6 points, or D.
Along the same lines, consider a slight variation of this question:
For how many integer values of x, is |x – 3| + |x + 1| + |x| > 10?
(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite
What will the answer be here? We hope you immediately jumped to answer choice E – for every integer value of x to the right of 4 or to the left of -2.667, the total distance will be more than 10 (the blue regions). So there will be infinite such integer points (all integers greater than 4 or less than -2.667). Thus, the answer is E.
We hope this logic is clear. We will look at some other variations of this concept next week!
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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]]>The post Data Sufficiency Questions: How to Know When Both Statements Together Are Not Sufficient appeared first on Veritas Prep Blog.
]]>Let’s try to answer these questions in today’s post using using one of our own Data Sufficiency questions.
A certain car rental agency rented 25 vehicles yesterday, each of which was either a compact car or a luxury car. How many compact cars did the agency rent yesterday?
(1) The daily rental rate for a luxury car was $15 higher than the rate for a compact car.
(2) The total rental rates for luxury cars was $105 higher than the total rental rates for compact cars yesterday
We know from the question stem that the total number of cars rented is 25. Now we must find how many compact cars were rented.
There are four variables to consider here:
Let’s examine the information given to us by the statements:
Statement 1: The daily rental rate for a luxury car was $15 higher than the rate for a compact car.
This statement gives us the difference in the daily rental rates of a luxury car vs. a compact car. Other than that, we still only know that a total of 25 cars were rented. We have no data points to calculate the number of compact cars rented, thus, this statement alone is not sufficient. Let’s look at Statement 2:
Statement 2: The total rental rates for luxury cars was $105 higher than the total rental rates for compact cars yesterday.
This statement gives us the difference in the total rental rates of luxury cars vs. compact cars (we do not know the daily rental rates). Again, we have no data points to calculate the number of compact cars rented, thus, this statement alone is also not sufficient.
Now, let’s try to tackle both statements together:
The daily rate for luxury cars is $15 higher than it is for compact cars, and the total rental rates for luxury cars is $105 higher than it is for compact cars. What constitutes this $105? It is the higher rental cost of each luxury car (the extra $15) plus adjustments for the rent of extra/fewer luxury cars hired. That is, if n compact cars were rented and n luxury cars were rented, the extra total rental will be 15n. But if more luxury cars were rented, 105 would account for the $15 higher rent of each luxury car and also for the rent of the extra luxury cars.
Event with this information, we still should not be able to find the number of compact cars rented. Let’s find 2 cases to ensure that answer to this question is indeed E – the first one is quite easy.
We start with what we know:
The total extra money collected by renting luxury cars is $105.
105/15 = 7
Say out of 25 cars, 7 are luxury cars and 18 are compact cars. If the rent of compact cars is $0 (theoretically), the rent of luxury cars is $15 and the extra rent charged will be $105 (7*15 = 105) – this is a valid case.
Now how do we get the second case? Think about it before you read on – it will help you realize why the second case is more of a challenge.
Let’s make a slight change to our current numbers to see if they still fit:
Say out of 25 cars, 8 are luxury cars and 17 are compact cars. If the rent of compact cars is $0 and the rent of luxury cars is $15, the extra rent charged should be $15*8 = $120, but notice, 9 morecompact cars were rented than luxury cars. In reality, the extra total rent collected is $105 – the $15 reduction is because of the 9 additional compact cars. Hence, the daily rental rate of each compact car would be $15/9 = $5/3.
This would mean that the daily rental rate of each luxury car is $5/3 + $15 = $50/3
The total rental cost of luxury cars in this case would be 8 * $50/3 = $400/3
The total rental cost of compact cars in this case would be 17 * $5/3 = $85/3
The difference between the two total rental costs is $400/3 – $85/3 = 315/3 = $105
Everything checks out, so we know that there is no unique answer to this question – for any number of compact cars you use, you will come up with the same answer. Thus, Statements 1 and 2 together are not sufficient.
The strategy we used to find this second case to test is that we tweaked the numbers we were given a little and then looked for a solution. Another strategy is to try plugging in some easy numbers. For example:
Instead of using such difficult numbers, we could have tried an easier split of the cars. Say out of 25 cars, 10 are luxury and 15 are compact. If the rent of compact cars is $0 and the rent of luxury cars is $15, the extra rent charged should be 10*$15 = $150 extra, but it is actually only $105 extra, a difference of $45, due to the 5 additional compact cars. The daily rental rent of 5 extra compact cars would be $45/5 = $9. Using these numbers in the calculations above, you will see that the difference between the rental costs is, again, $105. This is a valid case, too.
Hence, there are two strategies we saw in action today:
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Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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]]>In this example, it relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.
We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.
But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:
Is x < 0?
(1) x^3 + x^2 + x + 2 = 0
(2) x^2 – x – 2 < 0
In this problem, x can be any real number – we have no constraints on it. Now, is x negative?
Statement 1: x^3 + x^2 + x + 2 = 0
If we try to solve this equation as we are used to doing, look at what happens:
If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0
We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.
Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.
Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.
This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.
Statement 2: x^2 – x – 2 < 0
This, we can easily solve:
x^2 – 2x + x – 2 < 0
(x – 2)*(x + 1) < 0
We know how to solve this inequality using the method discussed here.
This this will give us -1 < x < 2.
Since x can be a non-integer value too, x can be negative, 0, or positive. This statement alone is not sufficient,and therefore, the answer is A.
To evaluate Statement 1, we didn’t need to solve the equation at all. We figured out everything we wanted to know by simply using some logic.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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]]>The post Quarter Wit, Quarter Wisdom: A GMAT Quant Question That Troubles Many! appeared first on Veritas Prep Blog.
]]>In the same way, it is possible that a question may appear to be testing very obscure concepts, while it is really solvable by using only basic ones.
This happens with one of our own practice questions – we have often heard students exclaim that this problem isn’t relevant to the GMAT since it “tests an obscure number property”. It is a question that troubles many people, so we decided to tackle it in today’s post.
We can easily solve this problem with just some algebraic manipulation, without needing to know any obscure properties! Let’s take a look:
† and ¥ represent non-zero digits, and (†¥)² – (¥†)² is a perfect square. What is that perfect square?
(A) 121
(B) 361
(C) 576
(D) 961
(E) 1,089
The symbols † and ¥ are confusing to work with, so the first thing we will do is replace them with the variables A and B.
The question then becomes: A and B represent non-zero digits, and (AB)² – (BA)² is a perfect square. What is that perfect square?
As I mentioned before, we have heard students complain that this question isn’t relevant to the GMAT because it “uses an obscure number property”. Now here’s the thing – most advanced number property questions CAN be solved in a jiffy using some obscure number property such as, “If you multiply a positive integer by its 22nd multiple, the product will be divisible by …” etc. However, those questions are not actually about recalling these so-called “properties” – they are about figuring out the properties using some generic technique, such as pattern recognition.
For this question, the complaint is often that is that the question tests the property, “(x + y)*(x – y) (where x and y are two digit mirror image positive integers) is a multiple of 11 and 9.” It doesn’t! Here is how we should solve this problem, instead:
Given the term (AB)^2, where A and B are digits, how will you square this while keeping the variables A and B?
Let’s convert (AB)^2 to (10A + B)^2, because A is simply the placeholder for the tens digit of the number. If you are not sure about this, consider the following:
58 = 50 + 8 = 10*5 + 8
27 = 20 + 7 = 10*2 + 7
…etc.
Along those same lines:
AB = 10A + B
BA = 10B + A
Going back to our original question:
(AB)^2 – (BA)^2
= (10A + B)^2 – (10B + A)^2
= (10A)^2 + B^2 + 2*10A*B – (10B)^2 – A^2 – 2*10B*A
= 99A^2 – 99B^2
= 9*11*(A^2 – B^2)
We know now that the expression is a multiple of 9 and 11. We would not have known this beforehand. Now we’ll just use the answer choices to figure out the solution. Only 1,089 is a multiple of both 9 and 11, so the answer must be E.
We hope you see that this question is not as hard as it seems. Don’t get bogged down by unknown symbols – just focus on the next logical step at each stage of the problem.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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]]>The post Quarter Wit, Quarter Wisdom: Try to Answer This GMAT Challenge Question! appeared first on Veritas Prep Blog.
]]>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:
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 discussed 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:
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!
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
The post Quarter Wit, Quarter Wisdom: Try to Answer This GMAT Challenge Question! appeared first on Veritas Prep Blog.
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