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13 Nov 2017, 18:37
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[GMAT math practice question]
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.
Answer: C
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.
Answer: C
15 Nov 2017, 18:05
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[GMAT math practice question]
What of the followings is the range of X = 1 + (1/2^2) + (1/3^2) + ...... + (1/100^2),
A. 0 < X < 1
B. 1 < X < 2
C. 2 < X < 3
D. 3 < X < 4
E. 4 < X < 5
=>
1/A^2 < 1/{(A-1}A} = 1/(A-1) – 1/A
It is clear that 1 + (1/2^2) + (1/3^2) + ...... + (1/100^2) > 1.
1 + (1/2^2) + (1/3^2) + ...... + (1/100^2)
< 1 + 1/(1∙2) + 1/(2∙3) +1/(3∙4) + … + 1/(99∙100)
= 1 + ( 1/1 – 1/2 ) + ( 1/2 – 1/3 ) + … + (1/99 – 1/100)
= 1 + 1/1 – 1/100 = 2 – 1/100 < 2
Thus 1 < X < 2.
Therefore, B is the answer.
Answer: B
What of the followings is the range of X = 1 + (1/2^2) + (1/3^2) + ...... + (1/100^2),
A. 0 < X < 1
B. 1 < X < 2
C. 2 < X < 3
D. 3 < X < 4
E. 4 < X < 5
=>
1/A^2 < 1/{(A-1}A} = 1/(A-1) – 1/A
It is clear that 1 + (1/2^2) + (1/3^2) + ...... + (1/100^2) > 1.
1 + (1/2^2) + (1/3^2) + ...... + (1/100^2)
< 1 + 1/(1∙2) + 1/(2∙3) +1/(3∙4) + … + 1/(99∙100)
= 1 + ( 1/1 – 1/2 ) + ( 1/2 – 1/3 ) + … + (1/99 – 1/100)
= 1 + 1/1 – 1/100 = 2 – 1/100 < 2
Thus 1 < X < 2.
Therefore, B is the answer.
Answer: B
20 Nov 2017, 17:15
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[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
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
