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IBPS PO 2017 Practice Test – Quant Day 6

By
Aruna
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IBPS PO 2017 : Reasoning Test– 6 PM Every Day
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Q(1-5). Study the following Table containing shapes and dimensions of the field.

table-day-6

  1. What is the difference between the cost of fencing for square and Right angled triangle?
    A. Rs. 3052
    B. Rs. 3092
    C. Rs. 2052
    D. Rs. 2592
    E. None
    Answer – C. Rs. 2052
    Solution:
    Square = 260 * 111 = 28860
    Perimeter of Right angled triangle = 92+69+ sqrt (92 2 + 69 2 ) = 276
    Cost of fencing = 276* 112 = 30912
    Difference = 30912-28860= 2052

  2. Sum of Cost of fencing of all fields together is ?
    A. Rs 144350
    B. Rs 150210
    C. Rs 161280
    D. Rs 174660
    E. None
    Answer – B. Rs 150210
    Solution:
    Square = 260 * 111 = 28860
    Rectangle = 264 * 113 = 29832
    Right Angled Triangle = 276 * 112 =30912
    Parallelogram = 270 * 113 = 30510
    Circle = 264 * 114 = 30096
    Sum = 28860+29832+30912+30510+30096 = 150210

  3. What is the difference between cost of fencing for Rectangle and Parallelogram?
    A. Rs 378
    B. Rs 468
    C. Rs 598
    D. Rs 678
    E. None
    Answer – D. Rs 678
    None
    Solution :
    Rs (30510-29832) = Rs 678

  4. The perimeter of the Right angled triangle is what percent more than Perimeter of Circle?
    A. 7.23
    B. 6.25
    C. 4.54
    D. 3.33
    E. None
    Answer – C. 4.54
    Solution:
    276-264/264 * 100 = 4.54

  5. Cost of fencing for which of the following fields will be highest?
    A. Right Angle Triangle
    B. Parallelogram
    C. Circle
    D. Square
    E. Rectangle
    Answer – A. Right Angle Triangle
    Solution:
    Square = 260 * 111 = 28860
    Rectangle = 264 * 113 = 29832
    Right Angled Triangle = 276 * 112 =30912
    Parallelogram = 270 * 113 = 30510
    Circle = 264 * 114 = 30096

Q(6-10). Each question contains statement followed Quantity I and Quantity II. Find relationship between both the quantities.

  1. Find the Volume
    Quantity I: If the side of cube is 68 metre
    Quantity II: If the diameter of sphere is 84 metre
    A. Quantity I > Quantity II
    B. Quantity I < Quantity II
    C. Quantity I ≥ Quantity II
    D. Quantity I ≤ Quantity II
    E. Quantity I = Quantity II or relationship cannot be established
    Answer – A. Quantity I > Quantity II
    Solution:
    Quantity I:
    V = 68*68*68 = 314432
    Quantity II: 4/3 * 22/7 * 42*42*42 = 310464

  2. No of Days in which A alone can complete work
    Quantity I: B alone can complete a work in 12 days. C can complete it in 24 days. A and B started work and worked for 3 days. They both left now C finished the remaining work in 9 days.
    Quantity II: A and B complete work in 8 days, B and C can complete work in 12 days, C and A can complete work in 8 days.
    A. Quantity I > Quantity II
    B. Quantity I < Quantity II
    C. Quantity I ≥ Quantity II
    D. Quantity I ≤ Quantity II
    E. Quantity I = Quantity II or relationship cannot be established
    Answer – B. Quantity I < Quantity II
    Solution :
    Quantity I:
    3(1/A+1/12) + 9/24 = 1
    A = 8 days
    Quantity II:
    A+B = 8 days
    B+C = 12 days
    A+C = 8 days
    2(1/A+ 1/B+1/C) = 1/8 + 1/12 + 1/8
    A = 12 days

  3. Find profit percentage
    Quantity I: If CP is 39,900 and MP is 46200 and discount offered on MP is 5%
    Quantity II: If difference between MP and SP is 4840 and discount offered is 10% and CP is 39600
    A. Quantity I > Quantity II
    B. Quantity I < Quantity II
    C. Quantity I ≥ Quantity II
    D. Quantity I ≤ Quantity II
    E. Quantity I = Quantity II or relationship cannot be established
    Answer – E. Quantity I = Quantity II or relationship cannot be established
    Solution:
    For Quantity I
    SP = 46200 * 95 /100 = 43890
    43890 = 39900 * (100+x)/100 = 10%
    For Quantity II
    MP –SP =4840
    SP = 43560 since discount is 10%
    Then profit percent is 10%

  4. Find Interest
    Quantity I: If Sum is Rs 1632 lent at simple interest at 5% rate for 2 years
    Quantity II: If sum is Rs 2000 lent at Compound interest at 4% for 2 years
    A. Quantity I > Quantity II
    B. Quantity I < Quantity II
    C. Quantity I ≥ Quantity II
    D. Quantity I ≤ Quantity II
    E. Quantity I = Quantity II or relationship cannot be established
    Answer – E. Quantity I = Quantity II or relationship cannot be established
    Solution:
    Simple Interest rate = 1632* 5*2 /100 = 163.2
    Compound Interest rate = 2000(1+4/100) 2 – 2000 = 163.2

  5. Find X
    Quantity I : X 2 – 84X + 1764 = 0
    Quantity II : X 2 – 86X + 1849 = 0
    A. Quantity I > Quantity II
    B. Quantity I < Quantity II
    C. Quantity I ≥ Quantity II
    D. Quantity I ≤ Quantity II
    E. Quantity I = Quantity II or relationship cannot be established
    Answer – B. Quantity I < Quantity II
    Solution:
    Solving equations X = 42
    X = 43





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