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Thread: A Couple of Hard Questions from a CAT Exam

  1. #1
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    A Couple of Hard Questions from a CAT Exam

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    Hello,

    I recently did a CAT exam. I missed a few questions on Math and, unfortunately, there is no explanation for the solution. I hoped the wizards here could please help. I would love to hear how you would solve any of these problems

    1. For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is

    A. b/n 2 and 10
    B. b/n 10 and 20
    C. b/n 20 and 30
    D. b/n 30 and 40
    E. Greater than 40

    2. If x <> 0, then square root (x^2) / x =

    A. -1
    B. 0
    C. 1
    D. x
    E. abs(x)/x

    I thought it was c, but no dice

    3. For which of the following functions is f(a+b) = f(a) + f(b) for all positive numbers a and b
    A. f(x) = x^2
    B. f(x) = x +1
    C. f(x) = sqr (x)
    D. f(x) = 2/x
    E. f(x) = -3x

    Thank you kindly

  2. #2
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    3.Ans (E)

    f(x) =-3x

    f(a)= -3a--(1)
    f(b)= -3b---(2)

    f(a+b)= -3 (a+b)

    (1) + (2)

    f(a) + f (b) = -3a + (-3b) = -3 (a+b)

    f (a+b) =f (a) + f (b)



    2. ans (E)

    (sqrt x^2)/ x = (+/- x)/x

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    What does b/n mean in question n&#176;1?

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    dissa,
    sqrt(x^2)= x
    not +x and -x

    sqrt(9) = 3

    if x^2 = 9
    then x can be +3 and -3..

    I also got C..

    jsloan,
    are you sure that C is wrong.. and what is b/n in a.. I do not see b in the question.

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    Quote Originally Posted by abhishek_mumbai View Post
    dissa,
    sqrt(x^2)= x
    not +x and -x

    sqrt(9) = 3

    if x^2 = 9
    then x can be +3 and -3..

    I also got C..

    jsloan,
    are you sure that C is wrong.. and what is b/n in a.. I do not see b in the question.
    @ Abhishek,

    x^2 = 9==> x = +/- sqrt (9) = +/- 3

    y^2 = (x^2)==> y = sqrt(x^2) =+/-x
    sqrt (x^2) = +/- x

    PS: keep in mind; when we take sqrt of anything there are two solutions(+ or -)!!
    Last edited by dissa; 05-27-2009 at 09:29 PM.

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    1. For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is

    A. b/n 2 and 10
    B. b/n 10 and 20
    C. b/n 20 and 30
    D. b/n 30 and 40
    E. Greater than 40

    I think in this question b/n means between.


    And the answer is E
    Last edited by Md. Minuddin; 05-29-2009 at 11:09 AM.

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    Quote Originally Posted by Md. Minuddin View Post
    1. For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is

    A. b/n 2 and 10
    B. b/n 10 and 20
    C. b/n 20 and 30
    D. b/n 30 and 40
    E. Greater than 40

    I think in this question b/n means between.


    And the answer is A. Because the smallest prime factor is 7

    h(100) +1=100* 98* 96* ....*2+1

    Now, 2*4*6*8*10+1=3840+1=3841 Smallest prime factor is 7
    2*4*6*8*10* 12+1=46080+1=46081 Smallest prime factor is 7

    And so on.................................

    Here unit digit will always be 1. So is not divisible 2 or, 5. So we should check by 3 and then by 7 and if it is not factor then option is 11, 13 etc. But here it is divisible by 7. So answer is A
    Minuddin and everyone else,

    Thanks for the responses. I understand 2 and 3, but question 1 is a mystery. The answer is E. The question is from the GMATPrep software.

    Is there a way to view an explanation to the solution? The software tells you the answer, but does not explain the answer.

    Hopefully someone can figure out why it is E

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    hmm.. b/n between? I didn't get it.

    My answer is here,
    h(n) = 2 x 4 x .....n
    h(100) = 2 x 4 x ...100
    h(100) + 1 = (2 x 4 x ...100 + 1) = 2^50 (1 x 2 x...50) +1

    if we divide this any number below 50;
    1 would be the remainder. So that the smallest prime factor would be greater than 50.

    Ans (E)
    Last edited by dissa; 05-29-2009 at 02:32 AM.

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    take simple eg,,, h(100) = 2 *4*8*10......100 so ,= 2^50(1.2.3......25)= 2^50*2*2^2*2^3*2^4( 1.3.5.6.7.9.10.11.12.13.14.15.17.18.19.20.21.22.23 .24.25) and so on h(100)+1 will have a smallest prime no greater than 50 hence E

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    Quote Originally Posted by dissa View Post
    hmm.. b/n between? I didn't get it.

    My answer is here,
    h(n) = 2 x 4 x .....n
    h(100) = 2 x 4 x ...100
    h(100) + 1 = (2 x 4 x ...100 + 1) = 2 (1 x 2 x...50) +1

    if we divide this any number below 50;
    1 would be the remainder. So that the smallest prime factor would be greater than 50.

    Ans (E)
    Just a small correction.
    It (red 2) will be 250
    your explanation is good

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