Mixed Quantitative Aptitude Questions Set 36

Quantitative Aptitude Questions for IBPS RRB/PO/Clerk, SBI PO, NIACL, NICL, RBI Grade B/Assistant, BOI, Bank of Baroda and other competitive exams

  1. Ria covers (2/3)rd of a certain distance in 2 hours 30 minutes at the rate of x kmph. He covers the remaining distance at the rate of (x+2)kmph in 50 minutes . What is the total distance?
    A) 10km
    B) 15km
    C) 20km
    D) 35km
    E) 40km
    View Answer
      Option B
    Solution:
    Let the total distance be 3y km .
    Now,
    Speed * Time = Distance
    x * (5/2) =2y
    => 5x = 4y——– (1)
    Again ,
    (x + 2)*(50/60) = y
    => (x +2)*5 =6y———(2)
    On dividing (2) by (1)
    [(x+2)*5]/5x = 6y/4y
    => (x+2)/x = 3/2
    =>x =4
    5*4 =4y => y =5
    Therefore ,Total dist. = 3y = 3 * 5 = 15 km
     
  2. The population of a city in the year 2013 was 93771. During 2011-12 , the rate of growth of population was 8% and that in 2012-13 it was 15%. What was the population of the city in the year 2011?
    A) 75400
    B) 75400
    C) 80000
    D) 75500
    E) 65270
    View Answer
      Option D
    Solution:
    Population of town in 2011 = P1
    Therefore ,
    P =P1(1 + (R1/100))(1 + (R2/100))
    => 93771 = P1(1 + (8/100))(1 + (15/100))
    => 93771 = P1 *(108/100)*(115/100)
    =>P1 = (93771 * 100 *100)/(108*115)
    =>P1 = 75500
  3. Anchal’s age 8 years ago is equal to the sum of the present ages of her son and her daughter . 5 years hence, the ratio between daughter’s age and her son’s age will be 7:6 resp. Anchal’s husband is 7 years elder than her. Her husband’s present age is thrice the present age of her son. What is their daughter’s present age ?
    A) 22years
    B) 20years
    C) 23years
    D) 18years
    E) 14years
    View Answer
      Option C
    Solution:
    After 5 years from today ,
    Daughters age = 7x years
    son’s age = 6x years
    Therefore ,
    Daughter’s present age  = (7x – 5) years
    son’d present age = (6x – 5)years
    Anchal’s present age =( 7x – 5 + 6x – 5 + 8 )years = (13x – 2)years
    Therefore ,
    13x + 5 = 3(6x -5)
    => x = 4
    Then , Daughter’s present age= 7x – 5 =23years
  4. A batsman played three matches in a tournament. The respective ratio between the scores of 1st and 2nd  matches was 5:4 and between the scores of 2nd and 3rd matches was  2:1. The difference between the 1st and 3rd matches was 48 runs. What was the batsman’s average score in all the three matches ?
    A) 58(1/3)
    B) 58(2/3)
    C) 51(1/4)
    D) 53(1/7)
    E) 55(1/2)
    View Answer
      Option B
    Solution:
    Match 1 : Match 2= 5:4
    Match 2 :Match 3= 2:1 = 4:2
    Now,
    5x – 2x = 48
    => 3x = 48
    =>x = 16
    Total runs scored in three matches = 5x + 4x + 2x
    = 11x = 11 *16 = 176
    Required average = 176/3 = 58(2/3)
     
  5. Village A has population of 6800, which is decreasing at the rate of 120 per year. Village B has a population of 4200, which is increasing at the rate of 80 per year. In how many years will the population  of the two villages be equal?
    A) 13years
    B) 11years
    C) 17years
    D) 9years
    E) 7years
    View Answer
      Option A
    Solution:
    After 13 years , population of village
    A = 6800 – 120*13 = 5240
    After 13 years , population of village
    B = 4200 + 80 * 13 = 5240
  6. A train overtakes two persons who are walking in the same direction in which the train is going at the rate 2kmph and 4 kmph, and passes them completely in 9 and 10 seconds respectively. The  length of the train(in metres) is :
    A) 70m
    B) 40m
    C) 30m
    D) 50m
    E) 60m
    View Answer
      Option D
    Solution:
    2kmph = (2*5)/18 = (5/9) m/sec
    and 4kmph = (4*5)/18 = 10/9 m/sec
    Let length of the train be x and its speed be y metre/sec.
    Then,
    x/(y –(5/9)) = 9
    => 9y – x = 5 ————(1)
    and x/(y – (10/9)) = 10
    =>90y – 9y = 100 ——-(2)
    On solving eq. (1) and (2)
    x = 50m
    Length of the train = 50m
  7. A cicular park is developed inside a square plot, which is exactly fits in the plot and the diameter of the park is equal to the side of the square plot which is 28 metres. What is the area of the space left out in the square plot after developing the park?
    A) 168 sq. m
    B) 150 sq. m
    C) 220 sq. m
    D) 250 sq. m
    E) 177 sq. m
    View Answer
      Option A
    Solution:
    Radius of the cicular park = 14m
    Its area = (22/7)*14*14 = 616 sq. m
    Area of the square plot = 28 * 28 = 784 sq. m
    Area of the shaded region = 784 – 616 = 168 sq. m


  8. The currencies in countries A and B are denoted by a and b resp. The exchange rate in 1990 was 1a, 0.6b the price level in 2006 in A and  B are 150 and 400 resp. with 1990 as a base of 100. Find the exchange rate in 2006 based solely on the purchasing power pairly consideration is 1a.
    A) 1.10a
    B) 0.205b
    C) 0.005a
    D) 0.015b
    E) 0.225b
    View Answer
      Option E
    Solution:
    1a/150 = 0.6b/400
    => 1a = (0.6/400)*150b
    =>1a = 0.225b
  9. There is a committee of 12 persons in which there are 9 women and 8 men. In how many ways this can be done if atleast 5 women have to be included in the committee ?
    A) 5280
    B) 6500
    C) 6062
    D) 5890
    E) 6600
    View Answer
      Option C
    Solution:
    Total no. of ways forming the committee = 9C5 * 8C7 + 9C6 * 8C6 + 9C7 * 8C5  + 9C8 * 8C4 + 9C9 * 8C3 = 126*8 + 84*28 + 36*56 + 9*70 + 1*56 = 6062
  10. In a bag there are 6 red balls, 4 green balls and 8 yellow balls. Three balls are drawn at random from the  bag . What is the probability that 2 balls will be red and 1 ball will be green?
    A) 7/74
    B) 3/68
    C) 5/68
    D) 4/71
    E) 2/74
    View Answer
      Option C
    Solution:
    Total balls in the bag = 18
    Required Probability = (6C2 * 4C1)/18C3 = 60/816 = 5/68

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4 Thoughts to “Mixed Quantitative Aptitude Questions Set 36”

  1. Prince

    8 ka math …koi thoda explain kar do ..plz

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