# 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
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
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
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)
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
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
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
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
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
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
Option C
Solution:
Total balls in the bag = 18
Required Probability = (6C2 * 4C1)/18C3 = 60/816 = 5/68

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