Quant Test for IBPS Clerk 2018 Prelim Exam Set – 10

Directions(1-5): What will come in place of question mark “?” in the following questions.

1. ? = (84.97 *102.06)/(17.03)^2.01 + 11.9*4.9
   112
   100
   98
   90
   85
   Option D
   ? = (84.97 *102.06)/(17.03)^2.01 + 11.9*4.9
   => ? = (85*102)/(17*17)+12*5
   => ? = 30+60
   => ? = 90

    

2. ?^2 = 17*23 + 131 – 9^2
   21
   32
   30
   25
   20
   Option A
   ?^2 = 17*23 + 131 – 9^2
   => ?^2 = 391 + 131 – 81
   => ?^2 = 441
   => ? = 21

    

3. [18*(?^3 /5)]/3^2 + 18 = 68
   3
   7
   5
   8
   4
   Option C
   [18*(?^3 /5)]/3^2 + 18 = 68
   => 18*(?^3/5)/9 = 50
   => ?^3/5= 25 =>?^3 = 125
   => ? = 5

    

4. ? = 47% of (14^2 + 4)+35% of (14*5*4)
   192
   180
   177
   185
   199
   Option A
   ? = 47% of (14^2 + 4)+35% of (14*5*4)
   => ? = 47% of (196+4)+ 35% of 280
   => ? = 0.47*200+ 0.35*280
   => ? = 94+98
   =>? = 192

    

5. ? + 14 = (40% of 215 / (1369)^1/2)*3
   333
   300
   355
   398
   390
   Option C
   ? + 14 = (40% of 215 / (1369)^1/2)*3
   => ? + 14 = (86+37)*3
   => ? = 369 – 14
   => ? = 355

    

6. If the total number of biscuit packets of cuboidal shape each of the dimension 2.5 cm*2cm *1cm is stored in a cuboidal carton of dimensions 25 cm * 7 cm *4 cm , then find the total cost of all the biscuit packet stored in the carton if each biscuit packet costs Rs. 10.
   Rs.1500
   Rs.1800
   Rs.1400
   Rs.1200
   Rs.1000
   Option C
   Volume of the cuboidal carton = 25*7*4 = 700 cm^3
   Volume of each biscuit packet = 2.5 *2*1 = 5 cm^3
   Required number of biscuit packet = 700/5 = 140
   Total cost of all the biscuit packet stored in the carton = 140*10 = Rs.1400

    

7. If a two-digit number when increased by 20% then we get the reverse of the number. If the sum of the original number and its reverse is 99, then find the square of the original number.
   2500
   2025
   2121
   2030
   2000
   Option B
   Let the original number be x.
   Then, the reverse of the number = 1.2x x+1.2x = 99
   => x = 45
   So, x^2 = 45^2 = 2025

    

8. The average age of 11 players of the Indian cricket team is decreased by 1 year when Ishant Sharma aged 31 years and Suresh Raina aged 28 years are replaced by two new players (Shikhar Dhawan and Bhubaneshwar Kumar). Find the average age of those new players.
   15 years
   24 years
   27 years
   18 years
   20 years
   Option B
   Decrease in the sum of the ages of all the 11 players on replacedment = 11*1 = 11 years
   Sum of the ages of the two new players = 31+28- 11 = 48 years
   Average age of new players = 48/2 = 24 years

    

9. X deposited Rs. 800000 at 6% simple interest for 5 years whereas Y deposited Rs. 400000 at 4% compound interest for 2 years. What was the ratio of the accumulated amounts of X and Y ?
   125:57
   122:53
   121:50
   125:52
   132:51
   Option D
   Amount for X = 800000*(1+0.06*5)= Rs. 1040000
   Amount for Y = 400000*(1+0.04)^2 = Rs. 432640
   Required ratio = 1040000:432640 = 125:52

    

10. A player gets 2 chances to roll a die in a game . He wins if the sum of the numbers obtained on the faces of the die in the two attempts is divisible by 6. What is the probability that the player wins the game?
    1/3
    1/5
    1/7
    1/6
    1/8
    Option D
    Favourable cases
    = (1,5),(5,1),(4,2),(2,4),(3,3),(6,6)
    Required Probability = 6/36 = 1/6