Mixed Quantitative Aptitude Questions Set 78

Hello Aspirants

State Bank of India (SBI) is going to conduct examination for its recruitment for the post of Probationary Officers (SBI PO 2018) for a total of 2000 vacancies.

Click here to know the details of the Examination

The examination will be held in three phases i.e. Preliminary Examination, Main Examination and Group Exercise & Interview. The Preliminary Exam is scheduled on 1st, 7th & 8th of July 2018. Details of the exam are as under:

Practice the questions so as to familiarize yourself with the pattern of questions to be asked in the exam. 

1. A bucket contains a mixture of two liquids A and B in the proportion 4 : 1. If 45 litres of the mixture is replaced by 45 litres of liquid B, then the ratio of the two liquids becomes 2 : 5. How much of the liquid B was there in the bucket?
   60 l
   40 l
   50 l
   30 l
   70 l
   Option C
   Let the capacity of the bucket be 5x.
   [5x – 45]*4/5 / [5x – 45]*1/5 = ½
   => 5(4x – 36) = 2(x + 36)
   => x = 14 litre
   Quantity of the liquid B in the bucket
   = (5 * 14 – 45)*1/5 + 45
   = 50 litre

    

2. 25 men and 15 women can complete a piece of work in 12 days. All of them start working together and after working for 8 days the women stopped working. 25 men completed the remaining work in 6 days. How many days will it take for completing the entire job if only 15 women are put on the job?
   40 days
   36 days
   25 days
   45 days
   15 days
   Option B
   Work done in 8 days = 8 * 1/12 = 2/3
   Remaining work = 1/3
   Work done by 25 men per day = 1/3*1/6 = 1/18
   Work done by 15 women per day = 1/12 – 1/18 = 1/36
   Number of days taken by 15 women to complete the work = 36 days

    

3. A sum of Rs. 11000 is equally divided and invested at two different rates of interest. The difference between the interests got after 3 year is Rs. 300. What is the difference between the rates of interest?
   1.75%
   1.15%
   1.11%
   1.21%
   1.81%
   Option E
   Let the 2 different rates be x and y respectively. Difference between the interests is Rs 300
   = [5500 × 𝑥 × 3] /100 – [5500 × 𝑦 × 3] /100 = 300
   => 16500x-16500y = 300×100
   => 16500(x – y) = 30000
   => x-y = 1.81%

    

4. Two stations A and B are 287 km apart. Two trains started their journey from the two stations towards each other at 10 PM. After three and half hours they crossed each other. If the speed of the faster train is 6 km/hr more than the speed of the slower train, then find the speed of the faster train.
   25 km/hr
   44 km/hr
   52 km/hr
   38 km/hr.
   22 km/hr
   Option D
   Relative speed of both trains = 287/ 3.5 = 82 km/hr.
   If they were moving at the same speed, means each travelling at 82/2 = 41 km/hr
   But difference in speed of trains is 6 km/hr.
   So add 6/2 = 3 to 41 for speed of faster train and subtract 3 from 41 for speed of slower train.
   So, faster train’s speed = 41+3 = 44 km/hr, and of slower = 41-3 = 38 km/hr

    

5. A man and his wife appear in an examination for two vacancies for the same post. The probability of husband′s selection is (1/6) and the probability of wife’s selection is (1/4). What is the probability that only one of them is selected?
   1/3
   1/5
   1/9
   1/7
   1/2
   Option A
   P(Man selected) = 1/6 ;
   P(Man not selected) = 1 – 1/6 = 5/6
   P(Wife selected)= 1/4 ;
   P(Wife not selected) = 3/4
   Required Probability = (Man selected & Wife notselected) OR (Wife selected & Man not selected)
   = (1/6 *3/4 ) + ( 1/4 *5/6 ) = 1/3

    

6. The digit in the units place of a two-digit number is equal to the digit in the tens place of one-third of that number and the digit in the tens place of the original number is 2 less than the digit in the units place of one-third of the number. If the sum of the digits of the original number is 6, then what is the number?
   71
   51
   41
   61
   31
   Option B
   Let the original number be (10a + b) and one-third of that number be (10x + y).
   Given, b = x …(i) a = y – 2
   => y = (a + 2) …(ii) a + b = 6 …(iii) Also, 10a + b = 3 * (10x + y)
   => 10a + b = 3 *{10b + (a + 2)} [Replacing x and y from equations i and ii] => 10a + b = 30b + 3a + 6
   => 7a – 29b = 6 …(iv)
   => 36a = 180 [(29*(iii) + iv] => a = 5
   => b = 1 [substituting the value of a in (iii)] Original number is 51

    

7. The diameter of a smaller circle and a larger circle are the side of a square and the diagonal of the square respectively. If the side of the square is 4cm, what is the ratio of the areas of the smaller circle and the larger circle?
   2:7
   1:5
   7:9
   1:2
   4:5
   Option D
   The radius of the smaller circle (r) = side of the square Hence, d = 4; r = 2
   The radius of the larger circle (R) = diagonal of the square
   Hence, R = √2 × 2
   Area of smaller circle: Area of larger circle
   = (2) ^2: (√2 × 2) ^2
   = 4: 8 = 1:2

    

8. A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. Find the length of train ?
   40 m
   10 m
   50 m
   30 m
   20 m
   Option C
   Let the length of train be x km & its speed be y km/hr Then, x/(y-2) = 9/(60*60) and x/(y-4) = 10/(60*60)
   => 9y – 3600x = 18 and 10y – 3600x = 40
   => y-400x = 2 and y – 360x = 4
   Therefore , 40x = 2 or x = (2/40 *1000)m
   = 50 metres

    

9. An item was bought at Rs. X and sold it at Rs.Y by earning a profit of 20%. Had the C.P been 15% less and the S.P been Rs.90 less, a profit of 20% would have been earned. What is the value of Y?
   700
   600
   500
   400
   800
   Option B
   C.P = x
   S.P = y
   Profit = 20%
   So, y = 1.2x
   New C.P = .85x
   New S.P = .85x ×120 /100 = 1.02x
   = 1.2x – 1.02x = 0.18x
   0.18x = 90
   1.2x = 90×120/18 = 600
   S.P = Rs. 600

    

10. A man can type 20 lines in 10 minutes but he leaves 8% margin in each line. In how much time will he type 23 pages with 40 lines on each page leaving 25% more margin than before ?
    8 hr. 00 min.
    6 hr. 30 min.
    4 hr. 20 min.
    7 hr. 30 min.
    5 hr. 45 min.
    Option
    1 page has 23 lines
    => (100 – 8)*20/10
    = {100 – 125/100*8}*23*40/x
    => x = 450 min. (7 hr. 30 min.)