GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Nov 2019, 00:04 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Math Revolution DS Expert - Ask Me Anything about GMAT DS

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

[GMAT math practice question]

(function) $$f(x)$$ is a function. What is the value of $$f(2006)$$?

$$1) f(11)=11$$

$$2) f(x+3)=\frac{(f(x)-1)}{(f(x)+1)}$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

[GMAT math practice question]

(number properties) $$m$$ and $$n$$ are integers. What is the value $$(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}$$?

$$1) m = n + 1$$

$$2) m = 3$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(function) $$f(x)$$ is a function. What is the value of $$f(2006)$$?

$$1) f(11)=11$$

$$2) f(x+3)=\frac{(f(x)-1)}{(f(x)+1)}$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have many variables to determine a function and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

Since we have $$f(x+3) = \frac{(f(x)-1)}{(f(x)+1)}$$ and $$f(11)=11$$, we have $$f(14)=\frac{(f(11)-1)}{(f(11)+1)}=\frac{10}{12}=\frac{5}{6}$$ when we substitute $$11$$ for $$x.$$

We have $$f(17) = \frac{(f(14)-1)}{(f(14)+1)} = ((\frac{5}{6})-1)/(\frac{5}{6})+1) = (-(\frac{1}{6}))/(\frac{11}{6}) = \frac{-1}{11}$$ when we substitute $$14$$ for $$x.$$

We have $$f(20) = \frac{(f(17)-1)}{(f(17)+1)} = (-(\frac{1}{11})-1)/(-(\frac{1}{11})+1) = (-(\frac{12}{11}))/(\frac{10}{11}) = \frac{-12}{10} = \frac{-6}{5}$$, when we substitute $$17$$ for $$x.$$

We have $$f(23) = \frac{(f(20)-1)}{(f(20)+1)} = (-(\frac{6}{5})-1)/(-(\frac{6}{5})+1) = (-(\frac{11}{5}))/(-(\frac{1}{5}))=11$$, when we substitute $$20$$ for $$x.$$

Then we have the following patterns.
$$f(11) = f(23) = f(35) = … = f(12k-1) = 11$$

$$f(14) = f(26) = f(38) = … = f(12k+2) = \frac{5}{6}$$

$$f(17) = f(29) = f(41) = … = f(12k+5) = \frac{-1}{11}$$

$$f(20) = f(32) = f(44) = … = f(12k+8) = \frac{-6}{5}$$

So, $$f(2006) = f(12*167+2) = \frac{5}{6}.$$

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions when the answer is A, B, C, or D.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

[GMAT math practice question]

(statistics) $$f(x)$$ is a function. What is the value of $$f(x)$$?

$$1) f(x)+f(1-x)=7$$

$$2) x + f(\frac{x}{3})= \frac{f(x)}{2}$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(number properties) $$m$$ and $$n$$ are integers. What is the value $$(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}$$?

$$1) m = n + 1$$

$$2) m = 3$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have $$2$$ variables ($$m$$ and $$n$$) and $$0$$ equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since we have $$m = n + 1$$ and $$m = 3$$, we can substitute $$m = 3$$ into $$m = n + 1$$ to get $$3 = n + 1$$ and $$n = 2.$$

$$(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}$$

$$=(-1)^{3-2} +(-1)^{3+2} +(-1)^{3*2} +(-1)^{2*2}$$

$$=(-1)^1 +(-1)^5 +(-1)^6 +(-1)^4$$

$$=(-1) + (-1) + 1 + 1 = 0$$

Since both conditions together yield a unique solution, they are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Since $$m = n + 1$$, $$m$$ and $$n$$ are consecutive integers, they have different parities, which means that if $$m$$ is an odd integer, then $$n$$ is an even integer, and if $$m$$ is an even integer, then $$n$$ is an odd integer.

Case 1: $$m$$ is an odd integer and $$n$$ is an even integer.
Then, $$m+n$$ is an odd integer, $$m – n$$ is an odd integer, $$mn$$ is an even integer and $$2n$$ is an even integer.
$$(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}$$

$$=(-1)^{odd} +(-1)^{odd} +(-1)^{even} +(-1)^{even}$$

$$=(-1) + (-1) + 1 + 1 = 0$$

Case 2: $$m$$ is an even integer and $$n$$ is an odd integer.
Then, $$m+n$$ is an odd integer, $$m – n$$ is an odd integer, $$mn$$ is an even integer and $$2n$$ is an even integer.
$$(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}$$

$$=(-1)^{odd}+(-1)^{odd} +(-1)^{even} +(-1)^{even}$$

$$=(-1) + (-1) + 1 + 1 = 0$$

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)

Case 1: $$n$$ is an even integer.
Then, $$m+n$$ is an odd integer, $$m – n$$ is an odd integer, $$mn$$ is an even integer and $$2n$$ is an even integer, since $$m = 3.$$

$$(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}$$

$$=(-1)^{odd} +(-1)^{odd} +(-1)^{even} +(-1)^{even}$$

$$=(-1) + (-1) + 1 + 1 = 0$$

Case 2: $$n$$ is an odd integer.
Then, $$m+n$$ is an even integer, $$m – n$$ is an even integer, $$mn$$ is an odd integer and $$2n$$ is an even integer.

$$(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}$$

$$=(-1)^{even} +(-1)^{even} +(-1)^{odd} +(-1)^{even}$$

$$=1 + 1 + (-1) + 1 = 2$$

Since condition 2) does not yield a unique solution, it is not sufficient.

If the question has both C and A as its answer, then A is an answer rather than C by the definition of DS questions. Also, this question is a 50/51 level question and can be solved by using the Variable Approach and the relationship between Common Mistake Type 3 and 4 (A or B).

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

[GMAT math practice question]

(Number) What is a positive integer $$p$$?

1) $$p$$ is a prime number

2) $$p^2+2$$ is a prime number
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(statistics) $$f(x)$$ is a function. What is the value of $$f(x)$$?

$$1) f(x)+f(1-x)=7$$

$$2) x + f(\frac{x}{3})= \frac{f(x)}{2}$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have many variables to determine a function and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

We have $$0 + f(0) = (\frac{1}{2})f(0)$$ or $$f(0) = 2$$, when we substitute $$0$$ for $$x$$, since we have $$x+f(\frac{x}{3})=\frac{f(x)}{2}.$$

We have $$f(0) + f(1) = 7$$, when we substitute $$0$$ for $$x$$, since we have $$f(x)+f(1-x)=7.$$

When we substitute $$1$$ for $$x$$ in condition 2), we have $$1+f(\frac{1}{3}) = (\frac{1}{2})f(1)$$ or $$f(\frac{1}{3})=(\frac{1}{2})f(1)-1=\frac{7}{2}-1-\frac{5}{2}.$$

When we substitute $$\frac{1}{3}$$ for $$x$$ in condition 2), we have $$\frac{1}{3}+f(\frac{1}{9})=(\frac{1}{2})f(\frac{1}{3})$$ or $$f(\frac{1}{9})=(\frac{1}{2})f(\frac{1}{3})-\frac{1}{3} = (\frac{1}{2})(\frac{5}{2})-\frac{1}{3}=\frac{5}{4}-\frac{1}{3}=\frac{11}{12}.$$

Since both conditions together yield a unique solution, they are sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions when the answer is A, B, C, or D.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

[GMAT math practice question]

(Statics) The table shows the heights of students in a class. What is the value of $$y$$?

Attachment: 10.18ds.png [ 7.89 KiB | Viewed 173 times ]

1) The number of students with a height greater than 155 is 4 times that of students with a height less than 155.
2) The students with a height less than 160 are 40% of all the students in the class.
_________________
Intern  B
Joined: 24 Sep 2019
Posts: 12

### Show Tags

Hello,

1. If you have a DS question and it involves area and you solve the solution, but it has two answers: one positive and one negative. Since you have a negative answer, that can't be correct as area cant be negative. Therefore, you have one solution and it is SUFFICIENT to solve the question.

2. If you solve the question and you have a quadratic equation like 2x^2+7x+20. Since you can't solve for this using (X+@)(X+\$), it is not sufficient to solve the question? We don't use quadratic formula for GMAT, as this can get us decimals too.
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(Number) What is a positive integer $$p$$?

1) $$p$$ is a prime number

2) $$p^2+2$$ is a prime number

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have $$1$$ variable ($$p$$) and $$0$$ equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
Since we have an infinite number of prime numbers, we don’t have a unique value of $$p$$, and condition 1) is not sufficient.

Condition 2)
If $$p$$ has a remainder $$1$$ when it is divided by $$3$$ or $$p=3k+1$$ for some integer $$k$$, then $$p^2+2 = (3k+1)^2+2 = 9k^2+6k+1+2 = 3(3^k2+2k+1)$$ is a multiple and it is a prime number. We have $$3k^2+2k+1=1, 3k^2+2k=0, k(3k+2)=0$$ and $$k=0$$ or $$k=\frac{-2}{3}$$. However, $$k$$ is an integer so only $$k=0$$ works. Then $$p=3(0)+1 = 1$$. However, $$p = 1$$ is not a solution since $$1$$ is not a prime number.

If $$p$$ has a remainder $$2$$ when it is divided by $$3$$ or $$p=3k+2$$ for some integer $$k$$, then $$p^2+2 = (3k+2)^2+2 = 9k^2+12k+4+2 = 3(3k^2+4k+2)$$ is a multiple and it is a prime number. Since we have $$3k^2+4k+2=1, 3k^2+4k+1=0$$ or $$(3k+1)(k+1)=0$$ and we have $$k =-1$$ and $$k=\frac{-1}{3}$$. However, $$k$$ must be an integer so then $$p=3(-1)+2 = -1.$$ However, $$p = -1$$ is not a solution since $$-1$$ is negative.

Assume $$p$$ has a remainder $$0$$ when it is divided by $$3.$$
If $$p=3$$, then $$p^2+2=11$$ is a prime number.
If $$p=9$$, then $$p^2+2=83$$ is a prime number.
Since condition 2) does not yield a unique solution, it is not sufficient.

Conditions 1) & 2)
$$p$$ is a multiple of $$3$$ from condition 2), and $$p$$ is a prime number from condition 1). Then $$p = 3.$$
Since both conditions together yield a unique solution, it is sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations,” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(Statics) The table shows the heights of students in a class. What is the value of $$y$$?

Attachment:
10.18ds.png

1) The number of students with a height greater than 155 is 4 times that of students with a height less than 155.
2) The students with a height less than 160 are 40% of all the students in the class.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have $$2$$ variables ($$x$$ and $$y$$) and $$0$$ equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Condition 1) tells us that $$(x+5+3+y)=4(1+6)$$ or $$x+y+8=28$$ and it is equivalent to $$x+y=20.$$

Condition 2) tells us that $$(1+6+x)=(\frac{40}{100})(1+6+x+5+3+y), (7+x)=(\frac{2}{5})(15+x+y)$$ or $$5(x+7)=2(x+y+15).$$ Since $$x+y=20$$ from condition 1) we have $$5(x+7)=2(20+15)$$ or $$5(x+7) = 70.$$ Therefore, $$x+7=14,$$ and $$x=7$$. Substituting $$x=7$$ into $$x+y=20$$ gives us $$y=13.$$

Then we have $$x=7$$ and $$y=13$$.

Since both conditions together yield a unique solution, they are sufficient.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Condition 1) tells $$x+y=20.$$
Since condition 1) does not yield a unique solution, it is obviously not sufficient.

Condition 2)
Condition 2) tells $$5(x+7)=2(x+y+15)$$ or $$3x+5 = 2y.$$

Since condition 2) does not yield a unique solution, it is not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

[GMAT math practice question]

(number properties) $$p$$ and $$q$$ are integers. Is $$(p-1)(q-1)$$ an even number?

1) $$p+q$$ is an odd number

2) $$pq$$ is an even number
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

[GMAT math practice question]

(algebra) What is the value of $$x + \frac{1}{y}$$?

$$1) y + \frac{1}{z}= 1$$

$$2) z + \frac{1}{x} = 1$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(number properties) $$p$$ and $$q$$ are integers. Is $$(p-1)(q-1)$$ an even number?

1) $$p+q$$ is an odd number

2) $$pq$$ is an even number

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

The following reasoning shows that in the question, either $$p$$ or $$q$$ is an odd integer.
$$(p-1)(q-1)$$ is an even integer
=> $$p – 1$$ or $$q – 1$$ is an even integer
=> $$p$$ or $$q$$ is an odd integer

Therefore, either $$p$$ and $$q$$ is an odd number, and the other one is an even number, according to condition 1. So, condition 1) is sufficient.

Condition 2)

If $$p$$ is an odd number and $$q$$ is an even number, then $$p-1$$ is an even number, $$q-1$$ is an odd number, and $$(p-1)(q-1)$$ is an even number, which means the answer is ‘yes’.
If both $$p$$ and $$q$$ are even numbers, then $$(p-1)(q-1)$$ is an odd number, and the answer is ‘no’ since both $$p-1$$ and $$q-1$$ are odd numbers.

Since condition 2) does not yield a unique solution, it is not sufficient.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

[GMAT math practice question]

(geometry) The figure shows that $$OA = 20, OB = 30$$ and $$OC = x$$ and $$□OCDE$$ is a rectangle. What is the area of rectangle $$OCDE$$?

$$1) x = 10$$

$$2) OE = 15$$

Attachment: 10.23DS.png [ 10.14 KiB | Viewed 93 times ]

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(algebra) What is the value of $$x + \frac{1}{y}$$?

$$1) y + \frac{1}{z}= 1$$

$$2) z + \frac{1}{x} = 1$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since we have $$3$$ variables ($$x, y,$$ and $$z$$) and $$0$$ equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since $$y + \frac{1}{z} = 1$$, we have $$y = 1 – \frac{1}{z}, y = \frac{(z-1)}{z}$$ and $$\frac{1}{y} = \frac{z}{(z-1)}.$$

Since $$z + \frac{1}{x} = 1$$, we have $$\frac{1}{x} = 1 - z$$ or $$x = \frac{1}{(1-z)}.$$

Then $$x + \frac{1}{y} = \frac{1}{(1-z)} + \frac{z}{(z-1)} = \frac{1}{(1-z)} - \frac{z}{(1-z)} = \frac{(1-z)}{(1-z)} = 1.$$

Since both conditions together yield a unique solution, they are sufficient.

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

[GMAT math practice question]

(propotional) What is $$\frac{z^2}{xy} + \frac{x^2}{yz} + \frac{y^2}{zx}$$ ?

$$1) x:y = 2:3$$

$$2) x:z = 1:2$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(geometry) The figure shows that $$OA = 20, OB = 30$$ and $$OC = x$$ and $$□OCDE$$ is a rectangle. What is the area of rectangle $$OCDE$$?

$$1) x = 10$$

$$2) OE = 15$$

Attachment:
10.23DS.png

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since the triangle $$OAB$$ and the triangle $$CAD$$ are similar, we have $$OA:OB = 2:3$$ and $$CA:CD = 2:3$$. Then we have $$3CA = 2CD or CD = (\frac{3}{2})(20-x).$$

So the area of the rectangle $$OCDE$$ is $$x*(\frac{3}{2})(20-x)$$. Therefore, we have one variable in this question.

Since we have $$1$$ variable ($$x$$) and $$0$$ equations, D is the most likely answer. So, we should consider each condition separately first.

Condition 1) is sufficient, since it yields a unique solution.

Condition 2)
Since $$CD = OE = 15$$ from condition 2), and from the original condition we know $$CD = (\frac{3}{2})(20-x).$$

$$=>15 = (\frac{3}{2})(20-x)$$

$$=>10 = 20-x$$

$$=>x = 10$$

$$=>3CA = 2CD$$

$$=>3CA = 2(15)$$

$$=>3CA = 30$$

$$=>CA = 10$$

We have $$CA = 10$$ and $$x = 10.$$

So, condition 2) is also sufficient, because it is equivalent to condition 1).

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

[GMAT math practice question]

(equation) What are the values of $$x+y$$ and $$xy$$?

$$1) x + y + xy = -2$$

$$2) (\frac{1}{x}) + (\frac{1}{y}) = 1$$
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8163
GMAT 1: 760 Q51 V42 GPA: 3.82

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

(propotional) What is $$\frac{z^2}{xy} + \frac{x^2}{yz} + \frac{y^2}{zx}$$ ?

$$1) x:y = 2:3$$

$$2) x:z = 1:2$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have $$3$$ variables ($$x, y$$, and $$z$$) and $$0$$ equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since $$x:y = 2:3$$ and $$x:z = 1:2$$, we have $$x:y:z = 2:3:4.$$

Then we have $$x = 2k, y = 3k$$, and $$z = 4k$$ for some number $$k.$$

$$\frac{z^2}{xy} + \frac{x^2}{yz} + \frac{y^2}{zx}$$

$$= \frac{(4k)^2}{(2k)(3k)} + \frac{(2k)^2}{(3k)(4k)} + \frac{(3k)^2}{(4k)(2k)}$$

$$= \frac{16k^2}{6k^2} + \frac{4k^2}{12k^2} + \frac{9k^2}{8k^2}$$

$$= \frac{16}{6} + \frac{4}{12} + \frac{9}{8}$$

$$= \frac{64}{24} + \frac{8}{24} + \frac{27}{24} = \frac{99}{24} = \frac{33}{8}.$$

Since both conditions together yield a unique solution, they are sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
_________________ Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS   [#permalink] 27 Oct 2019, 19:49

Go to page   Previous    1  ...  23   24   25   26   27   28   29   30   31   32   33   34    Next  [ 679 posts ]

Display posts from previous: Sort by

# Math Revolution DS Expert - Ask Me Anything about GMAT DS  