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Mixed Quantitative Aptitude Questions (Memory based SBI PO Mains)

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Below are Memory Based Questions SBI PO Mains 2017 Exam in Quantitative Aptitude Section. 

  1. MLKJ is a trapezoid. ABCD and PQRS are two rhombus. Diagonals of ABCD are 6 cm and 8 cm. One of the angle of PQRS is 120 degree and diagonal bisecting that angle measures 15 cm. Side of ABCD = ML, Side of PQRS = JK. FInd ON (median of trapezoid). (single question on this)

    A) 10 cm
    B) 12 cm
    C) 8 cm
    D) 15 cm
    E) 9 cm
    View Answer
      Option A
    Solution:
    Rhombus ABCD:
    Side of ABCD = (√(62 + 82)/2 = 10/2 = 5 cm
    Rhombus PQRS:

    Angle PQO is 60. Because diagonals of rhombus bisect each other.
    Also diagonals bisect each other at 90 degree. SO angle POQ is 90
    Hence angle OPQ is 30 degree. Also SQ = 15, so OQ = 7.5
    In triangle OPQ. With OP as base
    sine 30 = perpendicular/Hypotenuse(or side of PQRS)
    1/2 = (15/2)/Hypotenuse(or side of PQRS)
    Solve, Side of PQRS  = 15 cm [sine 30 = 1/2]
    Now  ML = 5 cm, JK = 15 cm
    So Median ON = (5+15)/2 = 10 cm

Directions (2-4): There are 3 bags containing 3 colored balls – Red, Green and Yellow.
Bag 1 contains:
15 red balls. Y yellow balls and G green balls. Probability of drawing one yellow ball is 2/9. The ratio of number of green and yellow balls is 9 : 4

Bag 2 contains:
Number of green balls is 2/3rd of G. Total number of balls in bag is 5/6th of balls in Bag A. Number of yellow balls is 3 greater than number of red balls

Bag 3 contains:
Number of red balls is 1/3rd of total number of red balls in bags A and B. Number of yellow balls is 20% more than number of yellow balls in bad B. Probability of drawing one green ball is 7/16. (3 questions asked as below)

  1. One ball from each bag is drawn. Find the probability that these are yellow balls.
    A) 5/16
    B) 3/16
    C) 7/36
    D) 1/36
    E) 9/16
    View Answer
      Option D
    Solution:
    First find balls in each bag. Start with bag A:
    R = 15, Y = 4x, G = 9x. Prob of Y = 2/9
    So 4x/(15+9x+4x) = 2/9
    Solve, x = 3
    So in bag A: R = 15, Y = 12, G = 27. Total = 54
    Bag B:
    G = 2/3 * 27 = 18
    Total = 5/6 * 54 = 45
    Now 18 + Y + Y – 3 = 45
    So, yellow balls = 15, and then red = 12
    So in bag B: R = 12, Y = 15, G = 18. Total = 45
    Bag C:
    R = 1/3 (15+12) = 9
    Yellow = 120/100 * 15 = 18
    Probability of green ball = 7/16
    So z/(9+18+z) = 7/16 [Let z green balls]
    Solve, z = 21
    So in bag C: R = 9, Y = 18, G = 21. Total = 48
    Now probability of yellow ball from each bag = 12/54 * 15/45 * 18/48 = 1/36
  2. 5/6th of red balls, 4/9th of green balls from bag B are placed in bad D. What is the probability that 2 out of 3 balls from bag D are red?
    A) 15/37
    B) 15/34
    C) 17/33
    D) 11/34
    E) 19/34
    View Answer
      Option B
    Solution:
    In bag D:
    Red balls = 5/6 * 12 = 10
    Green balls = 4/9 * 18 = 8
    So required probability = 10C2 * 8C1/18C3 = 15/34
  3. P yellow balls are transferred from bag A to C. One ball is picked at random from bag C and probability of yellow ball is 2/5. Find P.
    A) 3
    B) 1
    C) 2
    D) 4
    E) Other than those given in options.
    View Answer
      Option C
    Solution:
    In bag C:
    Probability of yellow ball = (18+P)/(48+P) = 2/5
    Solve, P = 2 

Directions (5-6): Distance between stations P and Q is 516 km. Train A starts from station P at 9:45 AM with speed ‘X’ km/hr. Train B starts in opposite direction from station Q half an hour later with speed ‘1.5X’ km/hr. They meet at 12:55 PM. (2 questions asked as below)

  1. If both trains were moving in same direction, find the time at which they will meet.
    A) 50 minutes
    B) 60 minutes
    C) 108 minutes
    D) 40 minutes
    E) 72 minutes
    View Answer
      Option B
    Solution:
    Speed of A = x km/hr
    So distance it traveled in 1/2 hr = x/2 km
    Now remaining distance = (516 – x/2) km
    Time taken to meet = 12:55 – 10:15 = 8/3 hours
    So 8/3 = (516 – x/2)/(x+1.5x) [speed is added when in opposite direction]
    Solve, x = 72 km/hr
    So speed of A = 72 km/hr, B = 108 km/hr
    Now when moving in same direction:
    In half hour, A covered = 1/2 * 72 = 36 km
    and relative speed = 108-72 = 36 km/hr
    So time = 36/36 = 1 hour
  2. What is the relative speed of B with respect to A in m/sec?
    A) 50
    B) 60
    C) 36
    D) 48
    E) 72
    View Answer
      Option A
    Solution:
    Relative speed =  108+72 = 180 km/hr = 180 * 5/18 = 50 m/sec  

Directions (7-10): Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly. (5 questions asked, 1 was based on mensuration, 1 on geometry, rest 3 are:)

  1. There are 65 cards (numbered 1, 2, 3, ….65) in a box. Two cards are picked at random.
    Quantity I: Probability that both cards show 2 digit numbers which increase by 36 when its digits are reversed.
    Quantity II: Probability that both cards show a number divisible by 8 but not 16.
    A) Quantity I > Quantity II
    B) Quantity I < Quantity II
    C) Quantity I > Quantity II
    D) Quantity I > Quantity II
    E) Quantity I = Quantity II or relationship cannot be determined
    View Answer
      Option A
    Solution:
    15+36 = 51, 26+36 = 62, 37+36 = 73, 48+36 = 82, 59+36 = 95
    So there are 5 such numbers (15,26,37,48,59)
    So required prob = 5/65 * 4/64 = 20/65*64
    II: Such numbers are = 8, 24, 40, 56
    So required prob = 4/65 * 3/64 = 12/65*64
  2. There are 5 couples who want to sit at different positions.
    Quantity I: Number of ways in which men and women sit at alternate positions
    Quantity II: Number of ways in which all men sit together and all women sit together.
    A) Quantity I > Quantity II
    B) Quantity I < Quantity II
    C) Quantity I > Quantity II
    D) Quantity I > Quantity II
    E) Quantity I = Quantity II or relationship cannot be determined
    View Answer
      Option A
    Solution:
    I: ways for men = 5!, now 6 seats for 5 women, so for choosing 5 seats 6C5, and then arrangement of these 5 women is 5!
    So total number of ways = 5! * 6C5 * 5! = 5! * 5! * 6
    II: Ways for all men together = 5!, for all women together is 5!, now arrangement of men and women as group = 2!
    So total number of ways = 5! * 5! * 2! = 5! * 5! * 2
    So I > II
  3. Two parallel lines are given. s is an acute angle.

    Quantity I: Angle a
    Quantity II: 25 degree
    A) Quantity I > Quantity II
    B) Quantity I < Quantity II
    C) Quantity I > Quantity II
    D) Quantity I > Quantity II
    E) Quantity I = Quantity II or relationship cannot be determined
    View Answer
      Option B
    Solution:

    s is acute means <90 degree, so x > 90
    In triangle
    a + (a+40)+ x = 180
    2a + 40 = 140 – x
    As x > 90
    So 2a is < 50
    So a < 25
    So I < II

 

We will post more questions soon. Keep Checking the Quant Section 

 

 

46 thoughts on “Mixed Quantitative Aptitude Questions (Memory based SBI PO Mains)”

  5. Shilpa Thakur

    R = 15, Y = 9x, G = 4x. Prob of Y = 2/9
    So 4x/(15+9x+4x) = 2/9
    Solve, x = 3???????????????? is yellow 9x nhi hoga kya??????????????// please es ko explain kru

  6. Shilpa Thakur

    ???????????????????????????
    First find balls in each bag. Start with bag A:
    R = 15, Y = 9x, G = 4x. Prob of Y = 2/9
    So 4x/(15+9x+4x) = 2/9????????????????????????????????????
    Solve, x = 3
    So in bag A: R = 15, Y = 27, G = 12. Total = 54
    Bag B:
    G = 2/3 * 27 = 18
    Total = 5/6 * 54 = 45
    Now 18 + Y + Y – 3 = 45
    So, yellow balls = 15, and then red = 12
    So in bag B: R = 12, Y = 15, G = 18. Total = 45
    Bag C:
    R = 1/3 (15+12) = 9
    Yellow = 120/100 * 15 = 18
    Probability of green ball = 7/16
    So z/(9+18+z) = 7/16 [Let z green balls]
    Solve, z = 21
    So in bag C: R = 9, Y = 18, G = 21. Total = 48
    Now probability of yellow ball from each bag = 27/54 * 15/45 * 18/48 = 1/16

        1. Al! Z Well

          Helloo ma’am..can we olso get memory based questions of reasoning section..
          Your dis post is really vry helpful..!! Thnkx a ton

  7. Diya

    Rhombus PQRS:

    sine 30 = (15/2)/Hypotenuse(or side of PQRS)

    Solve, Side of PQRS = 15 cm [sine 30 = 1/2]
    //…….
    sin = perpendicular/hypotenuse………is it correct?
    yaha par 15/2 kaise aaya?
    PQRS ka ek angle 120 degree hai and 15cm diagonal bisect it then angle 60 degree nahi hoga??means sin 60

        1. Sawan Sharma

          component needs right angle and for the right angle we need to draw a perpendicular on PQ from point S ….after that we got “PQ= 15cos60 (length from perpendicular to point Q) and x (length from point P to perpendicular) … now we need to calculate x .. so we have another right angle tringle ( we have length of perpendicular and we need to determine base “x” and we have angle SPQ =60 ) so tan60 =15sin60/x …now we can determine the vallue of x and that is 15cos60 also that is why PQ=15cos60+15cos60. hope you will get it .. 🙂

      2. Diya

        in PQRS
        cos 60 = base/hypotenus = x/15
        1/2 = x/15
        x = 15/2 = 7.5
        after this what I can do?

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