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


  • 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))


  • Skadoooosh!!!


    Solution: f(x) + f(y) = log((1+x)/(1-x)) + log((1+y)/(1-y))
    = log (1+x) – log (1-x) + log(1+y) – log(1-y)
    = log (1+x) + log(1+y) – log (1-x) – log(1-y)
    = log ((1+x)(1+y) / (1-x)(1-y))
    = log (1 + x + y + xy / 1 – x – y + xy)

    Now, look at the options :
    1st option :f(x + y) = log (1 + x + y / 1 – x – y)
    2nd option : f(1+xy) = log (1 + xy/ 1 – xy)
    3rd option : Not true as it involves power of (x + y)
    4th option: f (x + y/1+xy) = log (1 + (x + y/1+xy) / 1 – (x + y / 1 + xy)
    = log (1 + x + y + xy / 1 – x – y + xy)
    Hence 4th option is right.


  • Skadoooosh!!!


    Q27. (CAT 2002)
    Three travelers are sitting around a fire, and are about to eat a meal. One of them has five small loaves of bread, the second has three small loaves of bread. The third has no food, but has eight coins. He offers to pay for some bread. They agree to share the eight loaves equally among the three travelers, and the third traveler will pay eight coins for his share of the eight loaves. All loaves were the same size. The second traveler (who had three loaves) suggests that he be paid three coins and that the first traveler be paid five coins. The first traveler says that he should get more than five coins. How much the first traveler should get?
    a. 5
    b. 7
    c. 1
    d. None of these


 

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