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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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