Reasoning Questions (Inequality) for LIC AAO/IBPS SO Exams

    Reasoning Questions (Inequality) for LIC AAO/IBPS SO Exams
    Reasoning Questions (Inequality) for LIC AAO/IBPS SO Exams Set-35:
    Dear Readers, Important Practice Reasoning Questions for Upcoming AAO/SO Exams was given here with Explanations. Aspirants those who are preparing for the examination can use this.

    Directions (Q.1-5): In each of these questions two equations are given. You have to solve these equations and given answer.

    1).I.7P + 3Q = 30                          II.6P – 2Q = 12
    a)   If P> Q
    b)   If P< Q
    c)   If P = Q
    d)   If P> Q
    e)   If P ≤ Q or relationship can’t be established

    2).I.P2+ 2P – 8 = 0                                   II.Q2– 5Q + 6 = 0
    a)   If P> Q
    b)   If P< Q
    c)   If P= Q
    d)   If P> Q
    e)   If P ≤ Q or relationship can’t be established

    3). I.21P2– 17P + 2 = 0                 II.56Q2– 15Q + 1 = 0
    a)   If P> Q
    b)   If P< Q
    c)   If P = Q
    d)   If P> Q
    e)   If P ≤ Q or relationship can’t be established

    4).I.2P2– 11P + 15 = 0                 II.2Q2+ 3Q – 9 = 0
    a)   If P> Q
    b)   If P< Q
    c)   If P = Q
    d)   If P> Q
    e)   If P ≤ Q or relationship can’t be established

    5).I.2P2– 15P + 28 = 0                 II.Q2= 9
    a)   If P> Q
    b)   If P< Q
    c)   If P = Q
    d)   If P> Q
    e)   If P ≤ Q or relationship can’t be established

    Directions (Q.6-10): For the two given equations I and II give answer

    6).I.3p + 4q = 5                             II.p + q = 2
    a)   If p is greater than q
    b)   If p is smaller than q
    c)   If p is equal to q
    d)   If p is either equal to or greater than q
    e)   If p is either equal to or smaller than q

    7).I.2q2– 3q + 1 = 0                       II.2p2– 5p + 3 = 0
    a)   If p is greater than q
    b)   If p is smaller than q
    c)   If p is equal to q
    d)   If p is either equal to or greater than q
    e)   If p is either equal to or smaller than q

    8).I.4p2– 5p + 1 = 0                       II.80q2– 18q + 1 = 0
    a)   If p is greater than q
    b)   If p is smaller than q
    c)   If p is equal to q
    d)   If p is either equal to or greater than q
    e)   If p is either equal to or smaller than

    9).I.4p + 8q = 3                             II.12p +16q = 7
    a)   If p is greater than q
    b)   If p is smaller than q
    c)   If p is equal to q
    d)   If p is either equal to or greater than q
    e)   If p is either equal to or smaller than q

    10).I.q2– 9q + 14 = 0                    II.p2– 23p + 112 = 0
    a)   If p is greater than q
    b)   If p is smaller than q
    c)   If p is equal to q
    d)   If p is either equal to or greater than q
    e)   If p is either equal to or smaller than q

    Answers:                         
    1). c)   2). e)   3). d)   4). a)   5). a)   6). a)   7). d)   8). a)   9). c)   10). d)

    Solution:

    1.Answer: c)

    2. I.P2+ 2P – 8 = 0
    =P2+ 4P – 2P – 8 = 0
    =(P + 4) ( P – 2) = 0
    P = 2, -4
    II.Q2– 5Q + 6 = 0
    =(Q – 3) (Q – 2) = 0
    Q = 2, -3
    Hence, relationship can’t be established.
    Answer: e)

    3. I.21P2– 17P + 2 = 0
    =>21P2– 14P – 3P + 2 =0
    => (7P – 1) (3P – 2) = 0
    P = (1/7), (2/3)
    II.56Q2– 15Q + 1 = 0
    =>(7Q – 1) (8Q – 1) = 0
    Q = (1/7), (1/8)
    Answer: d)

    4. P = (5/2), 3;
    Q = (3/2), -3
    Answer: a)

    5.P = 4, (7/2);
    Q = ± 3
    Answer: a)

    6. I.3p + 4q = 5
    or, 6 – 6q + 4q = 5
    or, 2q =1 or, q = (1/2)
    II.p + 2q = 2
    Or, 3p + 6q = 6
    Or, 3p = 6 – 6q
    Also, 3p = 6 – 6q = 6 – 6 × (1/2) = 6 – 3 = 3
    Or, 3p = 3
    Or, p = 1
    Hence, p> q
    Answer: a)

    7. I.2q2– 3q + 1 = 0
    Or, q =1, (1/2)
    II.2P2– 3q + 1 = 0
    Or, p = 1, (3/2)
    Hence,  p ≥ q.
    Answer: d)

    8. I.4p2– 5p + 1 = 0
    Or, p = 1, (1/4)
    II.80q2– 18q + 1 = 0
    Or, 80q2– 10q – 8q + 1 = 0
    Or, (8q – 1) (10q – 1) = 0
    Or, q = (1/8), (1/10)
    Hence, p> q
    Answer: a)

    9. I.4p + 8q = 3
    Or, 12p + 24q = 9
    Or, 12p = 9 – 24q
    II.12p +16q = 7
    Or, 9 – 24q + 16q = 7
    Or, 8q = 2
    Or, q = (2/8) = (1/4)
    Also, 12p = 9 – 24 × (1/4) = 9 – 6 = 3
    Or, p = (1/4)
    Hence, p = q
    Answer: c)

    10. I.q2– 9q + 14 = 0
    Or, q2– 7q – 2q + 14 = 0
    Or, (q – 7) (q – 2) = 0
    Or, q = 2 or, 7
    II.p2– 23p + 112 = 0
    Or, p2– 16p – 7p + 112 = 0
    Or, (p – 16) (p – 7) =0
    Or, p = 7 or, 16
    Hence,  p ≥ q.
    Answer: d)

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