Question Bank - 100 Algebra Questions From Previous CAT Papers (Solved)


  • Skadoooosh!!!


    Solution: We have to consider positive and negative numbers for all the cases.
    f1(x) is positive for positive numbers, and 0 for negative numbers. (0 for x = 0)
    f2(x) is 0 for positive numbers, and positive for negative numbers. (0 for x = 0)
    f3(x) is 0 for positive numbers, and negative for negative numbers. (0 for x = 0)
    f4(x) is negative for positive numbers, and 0 for negative numbers. (0 for x = 0)
    So, we see that, out of the 3 products in the question, f1(x) * f2(x) and f2(x) * f4(x) are always zero, for any x.

    Second sub question, f4(x) = f3(-x) = -f2(-x) = -f1(x).
    Hence, 1st option is false.
    –f3(-x) = f2(-x) = f1(x). Hence this is true.


  • Skadoooosh!!!


    Q17. (CAT 2003)
    If 1/3 log_3M + 3 log_3N = 1 + log_0.0008 (5), then:
    (1) M^9 = N/9
    (2) N^9 = 9/M
    (3) M^3 = 3/N
    (4) N^9 = 3/M


  • Skadoooosh!!!


    Solution: 1st let us find the value of log_0.0008 (5).
    Let log_0.0008 (5) = y.
    So 5 = (0.008)^y.
    5 = (8/1000)^y
    5 = (2/10)^3y
    5 = (1/5)^3y
    5 = (5)^(-3y)
    -3y = 1. y = -1/3.
    So, 1 + log_0.0008 (5) = 1 – 1/3 = 2/3.

    Now, the options need us to find a relation between M & N, and hence it should be true for all values of M & N.
    We have the equation :1/3 log_3 (M) + 3 log_3 (N) = 2/3.
    For simplicity, let us split and give, 1/3 log_3 (M) = 1/3 and 3 log_3 (N) = 1/3.
    That means log_3 (M) = 1, and M = 3.
    Also, it means, log_3 (N) = 1/9. N = 3^(1/9).
    N^9 = 3 and M = 3.
    So M * N^9 = 3 * 3 = 9.
    This is correct as per the 2nd option.


  • Skadoooosh!!!


    Q18. (CAT 2003)
    If both a and b belong to the set {1, 2, 3, 4}, then the number of equations of the form ax^2 + bx + 1 = 0 having real roots is
    a. 10
    b. 7
    c. 6
    d. 12


  • Skadoooosh!!!


    Solution: For real roots, discriminant should be greater than or equal to 0.
    D = b^2 – 4ac. Here, c = 1.
    So D = b^2 – 4a ≥ 0.
    If b = 1, a cannot be any number among 1, 2, 3, 4 as D < 0.
    If b = 2, a can be 1 from the set. For other numbers D < 0.
    If b = 3, a can be 1 or 2 from the set. For other numbers D < 0.
    If b = 4, a can be 1, 2, 3 or 4.
    totally there are 0 + 1 + 2 + 4 = 7 numbers possible for which D ≥ 0, and hence real roots.


  • Skadoooosh!!!


    Q19. (CAT 2003)
    If log_10 (x) – log_10 (√x) = 2 log_x (10), then a possible value of x is given by:
    a. 10
    b. 1/100
    c. 1/1 000
    d. None of these


  • Skadoooosh!!!


    Solution: If you are not sure of the basics of logarithms, follow this method :
    Let us suppose log10x = y.
    Then x = 10y.
    x^(1/y) = 10
    logx10 = 1/y

    If you are pretty thorough with logarithms, then you know that if log_10 (x) = y, then log_x (10) = 1/y. Also, we know that log_10 (√x) is same as log_10 (x)^1/2, which is same as 1/2 log_10 (x).

    So, we have log_10 (x) – log_10 (√x) = 2 log_x (10)
    y - y/2 = 2/y
    y/2 = 2/y
    y^2 = 4
    y = 2 or y = -2.
    If y = 2, then log_10 (x) = 2, and x = 10^2 = 100.
    If y = -2, then log_10 (x) = -2, and x = 10^(-2) = 1/100


  • Skadoooosh!!!


    Q20. (CAT 2003)
    What is the sum of ‘n’ terms in the series: log m + log(m^2/n) + log(m^3/n^2) + log (m^4/n^3) + …
    (1) log (n^(n-1)/ m^(n + 1))^n/2
    (2) log (m^m/ n^n)^n/2
    (3) log (m^(1 - n)/ n^(1 - m))^n/2
    (4) log (m^(n+1)/ n^(n - 1))^n/2


  • Skadoooosh!!!


    Solution: log m + log(m^2/n) + log(m^3/n^2) + log (m^4/n^3) + ….
    = log m + 2 log m – log n + 3 log m – 2 log n + 4 log m – 3 log n + …
    = log m ( 1 + 2 + …n ) – log n ( 1 + 2 + ….(n-1)) (This is an AP)
    = (n)(n+1)/2 * log m - (n-1)(n)/2 * log n
    = log [m^(n)(n+1)/2 - log n^((n-1)n/2)]
    =log (m^[(n+1)/2]/n^[(n-1)/2])^n
    = log (m^(n+1)/n^(n-1))^(n/2)


  • Skadoooosh!!!


    Q21. (CAT 2003)
    If three positive real numbers x, y, z satisfy y – x = z – y and xyz = 4, then what is the minimum possible value of y?
    a. 2^1/3
    b. 2^2/3
    c. 2^1/4
    d. 2^3/4


  • Skadoooosh!!!


    Solution: Since all the options of the answer are in the power of 2, let us take x = 2^a and z = 2^b.
    So, we get y = 2^[(a + b)/2] (As y is AM of x and z)
    We know xyz = 4.
    So, 2^a * 2^b * 2^[(a + b)/2] = 2^2.
    i.e. 2^[(3a +3b)/2] = 2^2.
    So, (3a + 3b)/2 = 2.
    3a + 3b = 4.
    a + b = 4/3.
    (a + b)/2 = 2/3.
    We know that y = 2(a + b)/2 , and (a + b)/2 is 2/3.


  • Skadoooosh!!!


    Q22. (CAT 2002)
    For three numbers x, y and z, x + y + z = 5, and xy + yz + xz = 3. What is the largest value which x can take?
    (1) 3 √13
    (2) √19
    (3) 13/3
    (4) √15


  • Skadoooosh!!!


    Solution: (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2xz
    25 = x^2 + y^2 + z^2 + 2(xy +yz + xz)
    25 = x^2 + y^2 + z^2 + 6
    x^2 + y^2 + z^2 = 19.
    For x to be maximum, y^2, z^2 should be minimum. Minimum value of a “square” integer is 0.
    Hence y^2, z^2 = 0. x^2 will be 19, and x will be √19.


  • Skadoooosh!!!


    Q23. (CAT 2002)
    If x^2 + 5y^2 + z^2 = 2y (2x + z), then which of the following statements are necessarily true?
    I. x = 2y II. x = 2z III. 2x = z
    (1) Only I
    (2) Only II
    (3) Only III
    (4) Only I and II


  • Skadoooosh!!!


    Solution: x^2 + 5y^2 + z^2 = 2y (2x + z)
    x^2 + 5y^2 + z^2 - 4xy – 2yz = 0. This can be re-written as:
    x^2 + 4y^2 – 4xy + y^2 + z^2 -2yz = 0
    (x - 2y)^2 + (y – z)^2 = 0
    Sum of two “square numbers” can be 0, only if both of them are 0.
    Hence x – 2y = 0 and y – z = 0.
    Hence x = 2y and y = z.
    x = 2y = 2z.


  • Skadoooosh!!!


    Q24. (CAT 2002)
    The number of roots of (a^2/x) + (b^2/(x-1)) = 1
    (1) 1
    (2) 2
    (3) 3
    (4) None of these


  • Skadoooosh!!!


    Solution: We essentially have a quadratic equation.
    a^2 (x-1) + b^2 x = x(x-1)
    x^2 – x(1 + a^2+ b^2) + a^2 = 0
    This is a quadratic equation, and discriminant D = b^2 – 4ax = (1 + a^2+ b^2)^2 – 4a^2
    D is not equal to 0, and hence they will not have equal roots.
    So, they have 2 roots as any quadratic equation.


  • Skadoooosh!!!


    Q25. (CAT 2002)
    Mayank, Mirza, Little and Jagbir bought a motorbike for 60 Dollar. Mayank contributed half of the total amount contributed by others, Mirza contributed one-third of total amount contributed by others, and Little contributed one-fourth of the total amount contributed by others. What was the money paid by Jagbir?
    (1) 12 Dollar
    (2) 13 Dollar
    (3) 18 Dollar
    (4) 20 Dollar


  • Skadoooosh!!!


    Solution: Let Mayank’s contribution be x. So, the remaining amount is 60 – x.
    Mayank’s contribution is half of other’s contribution.
    So, x = (60 – x)/2
    60 = 3x => x = 20.
    So Mayank’s contribution = 20.
    Similarly Mirza’s contribution y = (60 – y)/3 => y = 60/4 = 15
    And Little’s contribution z = (60-z)/4. => z = 60/ 5 =12
    Hence Jagbir’s contribution is 60 – 20 – 15 – 12 = 13.


  • Skadoooosh!!!


    Q26. (CAT 2002)
    If f(x) = log((1+x)/(1-x)), then f(x) + f(y) =
    (1) f(x + y)
    (2) f(1 + xy)
    (3) (x + y) f(1 + xy)
    (4) f ((x +y) / (1+xy))


 

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