Quant Test for IBPS PO 2018 Prelim Exam Set – 3

Directions(1-5): Find the relation between x and y and choose a correct option.

1. I. 3√x– 18/ √x = √x
   II. 2√y + (y−36)/ √y = –√y
   x>=y
   x=y or relation cannot be established.
   y>=x
   x>y
   y>x
   Option B
   3√x– 18/ √x = √x
   =>3x–18 = x
   =>2x = 18
   =>x = 9
   2√y +( y−36)/ √y = –√y
   =>2y+y–36 = –y
   =>4y = 36
   =>y = 9
   x=y

    

2. I. 6 y^2 + 1/2 = 7/2 y
   II. 12 x^2 + 2 = 10 x
   y>=x
   x>y
   x>=y
   y>x
   x=y or relation cannot be established.
   Option C
   6y^2 + 1/2 = 7/2 y
   12y^2 – 7 y + 1 = 0
   12y^2 – 44y – 3y + 1 = 0
   4y (3y – 1) – 1 (3y –1) = 0
   (3y – 1) (4y – 1) = 0
   y = 1/3, ¼
   12x^2 – 10x + 2 = 0
   6x^2 – 5x + 1 = 0
   3x (2x –1) – 1 (2x –1)= 0
   x = 1/3, ½
   x>=y

    

3. I. x ^4 – 1000 = 296
   II. y^2 + 330 = 346
   x=y or relation cannot be established.
   x>=y
   x>y
   y>x
   y>=x
   Option A
   x = + 6,-6
   y = + 6, -6
   No relation

    

4. I. 21x^2 – 26x + 5 = 0
   II. y = √0.1024
   y>x
   x>y
   y>=x
   x=y or relation cannot be established.
   x>=y
   Option D
   x = 0.23, 1
   y = + 0.32
   No relation

    

5. I. (6^ 3+9^2)/ 11 = x^3
   II. 15y^3= (36×18) + 12y^3
   x>=y
   y>x
   x=y or relation cannot be established.
   y>=x
   x>y
   Option B
   (6^3+9^2)/ 11 = x^3
   =>216+81 11 = x ^3
   =>297 11 = x ^3
   =>x ^3 = 27
   x = +3
   15y^3= (36×18) + 12y^3
   =>15y^3 –12y^3 = (36×18)
   =>3y^3 = (36×18)
   =>3y^3 = 648
   =>y ^3 = 216
   y = +6
   y>x

    

6. How many 3-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?
   700
   660
   720
   656
   685
   Option C
   The word ‘LOGARITHMS’ has 10 different letters.
   Required answer = 10P3
   =10×9×8=720
   

    

7. Compound Interest for two years at 20% per annum would fetch a sum of money Rs. 482, if the interest was payable half- yearly then it was payable annually. What is the sum?
   35000
   22800
   33000
   20000
   25200
   Option D
   Let sum=Rs.x
   C.I. when compounded half yearly = [x(1+10/100)^4−x] = 4641/10000
   C.I. when compounded annually = [x(20/100)^2−x] = 11/25
   4641/10000 x−11/25x = 482
   => x=20000

    

8. A and B can do a piece of work in 45 and 40 days respectively.They began work together but A leaves after X days and B finished the rest of the work in X+14 days. After how many days did A leave?
   11
   7
   9
   10
   5
   Option C
   Total work in one day = (1/45+1/40) = 17/360
   In X days, 17X/360
   Remaining work = (1-17x/360)
   Now, B can do 1 pt. work in (x+14)/(1-17x/360) days.
   (x+14)/(1-17x/360) = 40
   X = 9 days.

    

9. A train travelling at a speed of 75 mph enters a tunnel 3(1/2) miles long. The train is ¼ mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?
   5
   3
   7
   4
   2
   Option B
   Total distance covered = (7/2+1/4) miles
   = 154 miles
   Time taken = (154*75) hrs = 120 hrs.
   = (1/20*60) min.= 3 min

    

10. In a city of an adult population, 45% of men and 25% of women are married. If no man marries more than one woman, and vice versa, What is the percentage of total population of adults who are married ?
    20.25%.
    32.14%.
    28.15%.
    25.23%.
    30.12%.
    Option B
    Let no. of men = x & total population be 100.
    Total women = (100-x)
    Married man = 45x /100 —————(1)
    & married women = 25/100 (100-x) ——— (2)
    From (1) & (2) x = 500/14
    So, Married man = 225 /14
    Married woman = 225/14
    Total married population = men + women = 450/14
    Required % = (450/14)/100 *100
    = 450/14% = 32.14%.