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What is the remainder of x^2 + y^2, when it is divided by 4?

1) x and y are different prime numbers. 2) x – y = 2

=>

Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution.

Since we have 2 variables and 0 equation, C is most likely to be the answer.

Conditions 1) & 2)

Since x and y are different prime numbers and x – y = 2, both x and y are odd integers. x = 2a + 1 and y = 2b + 1 for some integers a and b. x^2 + y^2 = (2a+1)^2 + (2b+1)^2 = 4a^2 + 4a + 1 + 4b^2 + 4b + 1 = 4(a^2 + a + b^2 + b ) + 2. Thus, its remainder is 2, when it is divided by 4. Both condition 1) and 2) are sufficient. The answer is C.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore, C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously, there may be cases where the answer is A, B, D or E.

If x and y are integers, is x + y an even integer?

1) x is an odd integer. 2) x^2 + y^2 has a remainder of 2 when it is divided by 4.

=> 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.

Since the question includes 2 variables (x and y) and no equation, C is most likely to be the answer. Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Conditions 1) & 2) Since x^2 + y^2 has a remainder of 2 when it is divided by 4, x^2 + y^2 must be even. Since x is odd, x^2 is odd and so y^2 must also be odd. Therefore, y is odd, and x + y is even. The answer is ‘yes’.

Condition 1) Since we don’t know whether y is even or odd, this is not sufficient.

Condition 2) The condition tells us that x^2+y^2=4k+2=2(2k+1) is even. Since x^2+y^2=(x+y)^2-2xy, and 2xy is even, this implies that (x+y)^2 is also even. But this can only happen if x+y is even. So, the answer is ‘yes’. This condition is sufficient.

Therefore, the answer is B.

Normally, in problems which require 2 or more additional 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). Answer: B
_________________

Re: Overview of GMAT Math Question Types and Patterns on the GMAT [#permalink]

Show Tags

20 Nov 2017, 21:37

MathRevolution wrote:

[GMAT math practice question]

If x and y are integers, is x + y an even integer?

1) x is an odd integer. 2) x^2 + y^2 has a remainder of 2 when it is divided by 4.

=> 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.

Since the question includes 2 variables (x and y) and no equation, C is most likely to be the answer. Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Conditions 1) & 2) Since x^2 + y^2 has a remainder of 2 when it is divided by 4, x^2 + y^2 must be even. Since x is odd, x^2 is odd and so y^2 must also be odd. Therefore, y is odd, and x + y is even. The answer is ‘yes’.

Condition 1) Since we don’t know whether y is even or odd, this is not sufficient.

Condition 2) The condition tells us that x^2+y^2=4k+2=2(2k+1) is even. Since x^2+y^2=(x+y)^2-2xy, and 2xy is even, this implies that (x+y)^2 is also even. But this can only happen if x+y is even. So, the answer is ‘yes’. This condition is sufficient.

Therefore, the answer is B.

Normally, in problems which require 2 or more additional 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). Answer: B

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.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer.

Conditions 1) & 2) |x|=10|y| ⇔ x = ±10y

Case 1: If x=10y, then x/2x+y= 10y/20y+y= 10y/21y=10/21<1, and the answer is ‘yes’.

Case 2: If x= -10y, then x/2x+y = -10y/-20y+y = -10y/-19y =10/19<1, and the answer is ‘yes’.

Both conditions, applied together, are sufficient.

Since this is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4 (A) tells us that we need to also consider conditions 1) and 2) separately.

Condition 1)

|x|=10|y| ⇔ x = ±10y

Case 1: If x = 10y, then x/2x+y= 10y/20y+y= 10y/21y=10/21<1, and the answer is ‘yes’.

Case 2: If x = -10y, then x/2x+y= -10y/-20y+y= -10y/-19y=10/19<1 and the answer is ‘yes’.

Since the answer is ‘yes’ in both cases, this condition is sufficient.

Condition 2) If x = -2 and y = 1, then x/2x+y = -2/-4+1 = -2/-3 = 2/3 <1, and the answer is ‘yes’. If x=-2 and y=3, then x/2x+y = -2/-4+3 = -2/-1 =2>1, and the answer is ‘no’. This is NOT sufficient.

Note: Since this condition is so trivial, it is unlikely to be sufficient by Tip 4) of the VA method.

The answer is A.

Normally, in problems which require 2 or more additional 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).

If 0<2x+3y<50 and -50<3x+2y<0, then which of the following must be true?

I. x>0 II. y>0 III. x<y

A. I only B . II only C. III only D. I and III E. I, II, and III

=>

When we add the two inequalities 0<2x+3y<50 and -50<3x+2y<0, we obtain -50<5x+5y<50, or -20<-2x-2y< 20.

Statement I. Adding the two inequalities -50<3x+2y<0 and -20<-2x-2y< 20 yields -70<x<20. So x may not be greater than zero. Statement I may not be true.

Statement II. Adding the two inequalities 0<2x+3y<50 and -20<-2x-2y< 20 yields -20<y<70. So y may not be greater than zero. Statement II may not be true, either.

Statement III. Since 0<2x+3y<50 is equivalent to -50<-2x-3y<0 and -50<3x+2y<0, adding the two inequalities yields -100<x-y<0. This implies that x < y. Statement III must be true.

i, j, and k are non-negative integers such that i+j+k=3. If p, q, and r are three fixed, but different, prime numbers, how many different values of p^iq^jr^k are possible?

A. 8 B. 9 C. 10 D. 11 E. 12

=>

The number of possible values of p^iq^jr^k is equal to the number of solutions of the equation i + j + k = 3. The solution set of the equation i + j + k = 3 includes all permutations of (3,0,0), (2,1,0), and (1,1,1). The number of permutations of (3,0,0) is 3!/2! = 3. The number of permutations of (2,1,0) is 3! = 6. The number of permutations of (1,1,1) is 1.

Therefore, the number of solutions of the equation i+j+k=3 is 3 + 6 + 1 = 10.

Nov. 18th fell on a Thursday in 1999. On which day did Nov. 18th fall in 2005?

A. Tuesday B. Wednesday C. Thursday D. Friday E. Saturday

=>

If a year is divisible by 400, it is a leap year. If a year is divisible by 100, but not divisible by 400, it is not a leap year. If a year is divisible by 4, but not divisible by 100, it is a leap year.

The day on which a particular date falls will be shifted by two days from one year to the next if the next year is a leap year, and by one day from one year to the next if the next year is not a leap year. For example, Nov. 18th fell on a Thursday in 1999, and a Saturday in 2000 since 2000 was a leap year. It fell on a Saturday in 2000, and a Sunday in 2001 since 2001 was not a leap year.

As there were two leap years (2000 and 2004) between 1999 and 2005, Nov. 18th shifted by 8 days over the 6-year period. If a date is shifted by 7 days, it will fall on the same day of the week. So, the net effect was to shift Nov. 18th by one day. Therefore, Nov 18th fell on a Friday in 2005, and the answer is D.

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question

We can modify the original condition as follows:

n>1,000^2-999^2 ⇔ n > (1000+999)(1000-999) ⇔ n > 1999

Since we have 1 variable (n) and 0 equations, D is most likely to be the answer.

Condition 1)

n<1,001^2-1,000^2 ⇔ n < (1001+1000)(1001-1000) ⇔ n < 2001

Since the original condition and condition 1) combine to give 1999 < n < 2001, we must have n = 2000. We have a unique solution. Therefore, condition 1) is sufficient.

Condition 2)

n<502^2-500^2 ⇔ n < (502+500)(502-500) ⇔ n < 1002*2 ⇔ n < 2004

Since the original condition and condition 2) combine to give 1999 < n < 2004, n = 2000, 2001, 2002, or 2003. We don’t have a unique solution. So, condition 2) is not sufficient.

Therefore, the answer is A.

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.

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.

The first step of the VA method is to modify the original condition and the question, and then recheck the question.

Modifying the question yields x^2-y^2 < x-y? ⇔ x^2 - y^2 – ( x - y ) < 0? ⇔ ( x + y )( x – y ) – ( x - y ) < 0? ⇔ ( x + y – 1 )( x – y ) < 0?

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer, and so we should consider conditions 1) & 2) together first.

Conditions 1) & 2) Since x – y < 0, asking if ( x + y – 1 )( x – y ) < 0 is equivalent to asking if x + y – 1 > 0. For x = 1, y = 1, x + y – 1 > 0 and the answer is ‘yes’. For x = 1/4, y = 1/4, x + y – 1 < 0 and the answer is ‘no’.

Thus, both conditions together are not sufficient.

Therefore, the answer is E.

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.

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