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 1^st and 2^nd  matches was 5:4 and between the scores of 2^nd and 3^rd matches was  2:1. The difference between the 1^st and 3^rd 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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