IBPS PO 2017 Practice Test – Quant Day 18

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Welcome to Online Reasoning Section with the explanation in AffairsCloud.com. Here we are creating Best question samples for IBPS PO 2017. We have included Some questions that are repeatedly asked in bank exams !!!

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Q(1-5) Study the Line graph and Table given below to answer the following questions.

1. In which of the following village given has maximum no of children?
   A. Village A
   B. Village B
   C. Village C
   D. Village D
   E. Village E
   
   Answer & Explanation

   D. 4%
   Solution:
   By observation % of Children are more in Village E: 20*429 = 8580
   
   
2. The No of men from Village A forms approximately what percent of total No of men from all villages together?
   A. 15.75%
   B. 16.25%
   C. 17.35%
   D. 18.15%
   E. None
   
   Answer & Explanation

   D. 4%
   Solution:
   Village A: 48*423 = 20304
   Village B: 42*436 = 18312
   Village C: 45*427 = 19215
   Village D: 44*433 = 19052
   Village E: 47*429 = 20163
   Village F: 46*434 = 19964
   Total Men = 117010
   % = 20304/117010=17.35
   
   
3. What is the average no of Women’s in all villages together?
   A. 14702
   B. 15262
   C. 15862
   D. 16082
   E. None
   
   Answer & Explanation

   D. 4%
   Solution:
   423*(33+42+37+39+33+37)+ 42*13+37*4+39*10+33*6+37*11
   95172/6 = 15862
   
   
4. In village F what is the difference between No of Men and No of Children?
   A. 11246
   B. 12586
   C. 13256
   D. 14886
   E. None
   
   Answer & Explanation

   D. 4%
   Solution:
   434*(46-17) = 12586
   
   
5. Difference between Men and Women in which village of the following villages is maximum?
   A. Village A
   B. Village C
   C. Village D
   D. Village E
   E. Village F
   
   Answer & Explanation

   D. 4%
   Solution:
   Village A: 423*15
   Village B: 436*0
   Village C: 427*8
   Village D: 433*5
   Village E: 429*14
   Village F: 434*9
   By observation answer could be either A or E so calculating them
   A = 6345 E= 6006
   
   

Q(6-10) In the following questions two equations numbered I & II are given. You have to solve both equations and mark the option.

6. I. 63X² – 260X + 253 = 0
   II. 49Y² – 196Y + 187 = 0
   A. X > Y
   B. X <Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relationship cannot be established
   
   Answer & Explanation

   E. X = Y or relationship cannot be established
   Solution:
   X = 23/9 , 11/7
   Y = 11/7 , 17/7
   
   
7. I. 441X² – 945X + 506 = 0
   II. 484Y² – 880Y + 399 = 0
   A. X > Y
   B. X <Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relationship cannot be established
   
   Answer & Explanation

   A. X > Y
   Solution:
   X = 23/21, 22/21
   Y = 21/22, 19/22
   
   
8. I. 2X² – 42√2X + 440 = 0
   II. 3Y² – 57√3Y + 810 = 0
   A. X > Y
   B. X <Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relationship cannot be established
   
   Answer & Explanation

   B. X <Y
   Solution:
   X = 20/√2, 22/√2
   Y = 27/√3 , 30/√3
   
   
9. I. 49X² + 644X + 2091 = 0
   II. 77Y² + 738Y + 1681 = 0
   A. X > Y
   B. X <Y
   C. X ≥ Y
   D. X ≤ Y
   E. X = Y or relationship cannot be established
   
   Answer & Explanation

   D. X ≤ Y
   Solution:
   X = -41/7, -51/7
   Y = -41/7, -41/11
   
   
10. I. X² + 44X + 483 = 0
    II. Y² + 47Y + 546 = 0
    A. X > Y
    B. X <Y
    C. X ≥ Y
    D. X ≤ Y
    E. X = Y or relationship cannot be established
    
    Answer & Explanation

    E. X = Y or relationship cannot be established
    Solution:
    X = -23,- 21
    Y= -21,- 26