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Home IBPS PO 2017 Quantitative aptitude Practice Quiz 2

IBPS PO 2017 Quantitative aptitude Practice Quiz 2

,
    1 . x = ($4\times 5\over55....\infty$)x $4^5$

    A.   4
    B.   5
    C.   $\infty$
    D.   0
    2 . What is the area of the right-angled triangle?
    I.$sec^2 \theta - tan^2 \theta = 1$
    II.$cot(90^o - \theta)$ = $12\over13$

    A.   1
    B.   2
    C.   3
    D.   4
    3 . Solving the simultaneous equations, what will be the value of P?
    I.$7\over2$P+9q=42
    II.$3\over2$P + $27\over7$q = 18

    A.   1
    B.   2
    C.   3
    D.   4
    4 . Direction (Q. 4 to 8): Each of these questions has 2 number series marked as I and II. Study the pattern of series I and apply it to complete series II, whose Ist place is filled up for you.
    $Q$. I. 2160 360 72 18 6
    II. 420 a b c d
    What is the value of b?

    A.   70
    B.   84
    C.   41
    D.   14
    5 . I. 3 19 93 375 1121
    II. 0 a b c d
    Which number must replaced?

    A.   1
    B.   2
    C.   3
    D.   4
    6 . I. 2 0 –6 –18
    II. 8 a b c
    can be represented by:

    A.   -2
    B.   -20
    C.   1
    D.   -12
    7 . I. 4 17 153 2448
    II. 2 p q r
    The number r will be:

    A.   1331
    B.   1234
    C.   1243
    D.   3131
    8 . I. 32 37 47 58 71 79
    II. 23 m n o p q
    What should come in place of o?

    A.   36
    B.   94
    C.   14
    D.   63
    9 . The number 7, expressed as a percentage is:

    A.   0.07
    B.   0.7
    C.   7
    D.   700
    10 . The sum of the series 3 – 3 3 + 9 – 9 3 + ... upto 8 terms is:

    A.   1
    B.   2
    C.   3
    D.   4
      Answers & Solutions
      1 .    
      Answer : Option D
      Explanation :
      $5^{55}....\infty$ $\cong$ $\infty$, Thus, $4\times5\times45\over \infty$ $\cong$0
      2 .    
      Answer : Option D
      Explanation :
      From I: no information
      From II: tan $\theta$ = $12\over13$
      base = 13, highest = 12
      But, tan is only a ratio of base and heights, not necessarily true values.
      3 .    
      Answer : Option D
      Explanation :
      I x $3\over7$ = I.Thus, both equations are same
      To solve for 2 values, at least 2 different equations are needed.
      4 .    
      Answer : Option D
      Explanation :
      2160 ÷ 6 = 360, 360 ÷ 5 = 72, 72 ÷ 4 = 18, 18 ÷ 3 = 6.
      Thus, 420 ÷ 6 ÷ 5 = 14
      5 .    
      Answer : Option C
      Explanation :
      3 x 6 + 1 = 19, 19 x 5 – 2 = 93, 93 x 4 + 3 = 375, 375 x 3 – 4 = 1589, etc
      Thus, 0 x 6 + 1 = 01, 01 x 5 – 2 = 03, 03 x 4 + 3 = 15, 15 x 3 – 4 = 41
      6 .    
      Answer : Option D
      Explanation :
      $2 - 2^ 2 + 2 = 0, 0 - 3^ 2 + 3 = -6, -6 -4^ 2 + 4 = - 18, etc$
      $Thus, 8 - 2^ 2 + 2 = 6, 6 - 3^ 2 + 3 = 0, 0 - 4^ 2 + 4 = -12, etc$
      7 .    
      Answer : Option A
      Explanation :
      The series is: $x 2^2+1, 3^2+2,4^2+3, etc$
      Thus, r =1331
      8 .    
      Answer : Option D
      Explanation :
      The given series is : 32 + (3 + 2) = 37, 37 + (3 + 7) = 47, 47 + (4 + 7) = 58, etc
      Thus, m = 23 + (2 + 3) = 28, n = 28 + (2 + 8) = 38,
      O = 38 + (3 + 8) = 49.
      9 .    
      Answer : Option D
      Explanation :
      10 .    
      Answer : Option D
      Explanation :
      $S_n$(Sum of n terms) = $a(rn-1)\over r-1$ = $3[1-\sqrt{318-1}]\over \sqrt{3-1}$
      =$3(80)\over 0.7$ $\cong$ 343

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