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


  • Skadoooosh!!!


    Solution: For an equation with power 3, the co-efficient of x is the sum of the individual product of the roots.
    Let (n-1), n, (n+1) be the roots of the equation. So, n(n-1) + n(n+1) + (n-1)(n+1) = b
    n^2 – n + n^2 + n + n^2 – 1 = b
    3n^2 – 1 = b.
    We know that n^2 is positive, the minimum value of n^2 will be 0 when n = 0. So, the minimum value of b is -1 when n = 0.


  • Skadoooosh!!!


    Q14. (CAT 2005)
    For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive?
    x^2 – y^2 = 0
    (x – k)^2 + y^2 = 1
    (1) 2
    (2) 0
    (3) √2
    (4) -√2


  • Skadoooosh!!!


    Solution: From 1st equation x^2 = y^2.
    Substituting for y^2 in the 2nd equation, we get, (x – k)^2 + x^2 = 1
    x^2 + k^2 – 2kx + x^2 = 1
    2x^2 - 2kx + (k^2 – 1) = 0
    For unique solution, Discriminant D should be 0. (D = b^2 – 4ac)
    D = 4k^2 – 8 (k^2 – 1) = 8 – 4k^2
    So, for unique solution, 8 – 4k^2 is 0
    4k^2 = 8 => k^2 = 2 => k = √2 or -√2
    We require the positive solution, substituting, we get positive solution for k = √2


  • Skadoooosh!!!


    Q15. (CAT 2005)
    If x ≥ y and y > 1, then the value of the expression logx(x/y) + logy(y/x) can never be
    (1) -1
    (2) -0.5
    (3) 0
    (4) 1


  • Skadoooosh!!!


    Solution: Let us take everything to the same base, so that it is easier.
    log_x(x/y) = (log x – log y)/ log x
    log_y(y/x) = (log y – log x)/ log y
    Adding these 2, we have (log x – log y)/ log x + (log y – log x)/ log y,
    1 – (log_y/log x) + 1 – (log x/ log y)
    2 - (logy/log x) - (log x/ log y)
    2 – (log_x y + log_y x) (As, log y/log x = log_x y and log x/ log y = logy x)
    Now, we know y > 1 and x ≥ y. So, log_x y < 1, and logy x > 1 (You can try out by taking examples as x = 100 and y = 10 or similar).
    So, we have the expression 2 - (log x y + log y x) as 2 – (a number greater than 1).
    So, our result is lesser than 1 always.


  • Skadoooosh!!!


    Q16. (CAT 2004)
    f1 (x) = x when 0 ≤ x ≤ 1
    = 1 when x ≥ 1
    = 0 otherwise
    f2 (x) = f1(–x) for all x
    f3 (x) = –f2(x) for all x
    f4 (x) = f3(–x) for all x

    How many of the following products are necessarily zero for every x:
    f1(x) f2 (x)
    f2 (x) f3 (x)
    f2(x) f4 (x)
    a. 0
    b. 1
    c. 2
    d. 3

    Which of the following is necessarily true?
    a. f4(x) = f1(x) for all x
    b. f1(x) = –f3(–x) for all x
    c. f2(–x) = f4(x) for all x
    d. f1(x) + f3(x) = 0 for all x


  • 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


 

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