Math tricks for GMAT

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  • 7/29/2019 Math tricks for GMAT

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    If , then But if then

    ) --> if both unknowns are positive or both unknowns are negative then (if both are

    positive cross multiply to get and if both are negative cross multiply and flip the sigh twice to

    get again) and the answer will be NO but if given inequality also holds true and inthis case the answer will be YES (if is any negative number and is any positive number

    then ). Not sufficient.

    ---- multiplying by xy

    --> and have the same sign, next: : if both and are positive (for

    example 3 and 2 respectively) then and the answer will be NO but if both and are

    negative (for example -3 and -2 respectively) then and the answer will be NO. Not sufficient.

    Is x^2 + y^2 > 100?

    (1) 2xy < 100 --> clearly insufficient: if then the answer will be NO but

    if and then the answer will be YES.

    (2) (x + y)^2 > 200 --> . Now, as (square of any number is more

    than or equal to zero) then so we can safely

    substitute with (as is at least as big as then the inequality will still

    hold true) --> --> --> . Sufficient.

    If r and s are negative, is less than 1 ?

    Since r and s are negative then either or

    (1) . Divided both parts by : . We can not have the second case because

    if then its reciprocal and it can not equal to , hence .

    Sufficient.

    (2) is 2 less than --> . The same info as above. Sufficient.

    Is ?

    (1) --> . Clearly insufficient, for example: if and then the

    answer is NO, but if and then the answer is YES. Two different answers, hence not

    sufficient.

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    (2) --> --> so either or . Also insufficient:

    if and , then answer is NO, buy if and , then the answer is YES. Two

    different answers, hence not sufficient.

    (1)+(2) As from (1) , then from (2) must be true that --> so -->

    substitute in (1) --> --> , as , then , so (or which is

    the same ) will always be more than , thus the answer to the question "Is " is NO.

    Sufficient.

    To elaborate more as , the only chance for to hold true (or which is the same

    for to hold true) would be if is fraction ( ). For example

    if and then . But the fact that rules out this option

    In the fraction x/y, where x and y are positive integers, what is the value of y?

    (1) The least common denominator of x/y and 1/3 is 6 --> LCM of and 3 is is 6 --

    > or (the least common denominator of and is 6 and the least common denominator

    of and is also 6). Not sufficient.

    (2) --> no info about . Not sufficient.

    (1)+(2) still can be 2 or 6. Not sufficient.

    x = y+1 =>

    If is ?

    First of all as ( equals to absolute value of some number) then , as absolute value is

    always non-negative.

    (1) ( ) --> so and . But thus we can not

    say for sure that , because if then (so in case

    , is not more than it equals to ). Not sufficient.

    (2) --> --> . Sufficient. as we already

    know: and therefore, , which is a

    prime number. as a,b,c are positive, b-a must be 1 and b+a must be 23. As a result, we can calculate

    that a is 11 and b is 12, so c is 9.

    5x+4 is divisible by 6 5x+4=6,12,18,24, 30 ...

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    anaik100 wrote:

    If x is positive is x divisible by 2 ?

    (1) x^3+x is divisible by 4

    (2) 5x+4 is divisible by 6

    I don't think that this is a GMAT question. In it's current form it's way beyond the GMAT scope.

    First of all note that we are not told that is an integer.

    Given: . Question: is even (integer)?

    (1)

    If x is an integer then either:A. is a multiple of 4 and is odd;

    OR

    B. is odd and is a multiple of 4. But this scenario is not possible. Why? Because in

    case ( is odd), , which is not a multiple of 4, (it's even never

    multiple of 4.)

    So if is an integer it must be multiple of 4.

    But also can be a non-integer, for example equation has one real root ( ),

    which is not an integer OR equation also has one real root ( ), which is not an

    integer.

    Not sufficient.

    (2)

    If is an integer, then must be even as --> --

    > .

    But can also be a fraction, for example , or - basically can be a fraction of a form ,

    where is an integer.

    Not sufficient.

    (1)+(2) Now if is an integer then it's even as we concluded.

    But can x be reduced fraction of a form ? We'll have --

    > --> as and are integers, must be multiple of 5 (at least) -

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    -> hence , so can not be reduced fraction --> and if is an integer, then we

    know it must be even.

    From (3,4), on a perpendicular line (with slope 3/2), to find a point the same distance from (3,4) as

    (0,6) is, we decrease x by 2 and decrease y by 3, thus getting vertex (1,1), or we increase x by 2 and

    increase y by 3, thus getting vertex (5,7).

    If (0,6) and (6,2) are endpoints, (3,4) is the midpoint.

    From (0,6) to (3,4), we go right 3 and down 2; that is, we increase x by 3 and decrease y by 2: the

    slope is -2/3. Consider the perpendicular diagonal- its slope is the negative reciprocal, i.e. 3/2. From

    (3,4), on a perpendicular line, to find a point the same distance from (3,4) as (0,6) is, we can decrease

    x by 2 and decrease y by 3, or we can increase x by 2 and increase y by 3. The endpoints of the other

    diagonal are (1,1) and (5,7).