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Wednesday, 17 May 2017

IBPS PO 2017 Quantitative aptitude Practice Quiz 21

     
     
    1 .

    A.   16 : 17
    B.   17 : 16
    C.   23 : 21
    D.   21 : 23
    2 . 2.What is the approximate per cent increase in the number of ball bearings manufactured by Company P in the year 2013, from the previous year?

    A.   143%
    B.   157%
    C.   122%
    D.   59%
    3 . 3.What is the average number of ball bearings manufactured by all companies together in the year 2009?

    A.   3360000
    B.   336000000
    C.   3360000000
    D.   None of these
    4 . 4. The number of ball bearings manufactured by Company Q in the year 2009 is approximately what per cent of the total number of ball bearings manufactured by it in all the years together?

    A.   24%
    B.   32%
    C.   11%
    D.   17%
    5 . 5. How many more ball bearings need to be manufactured by Company S in the year 2013 to make the ratio of the number of ball bearings manufactured by Company S to that of those manufactured by Company U in the year 2013 as 54 : 83?

    A.   80000000
    B.   2500000
    C.   34000000
    D.   5000000
    6 . Directions (Q. 6-10): In each of the following questions two equations are given. You have to solve them and give answer as:


    I. ${2p^2}$ - 7p + 3 = 0

    II. ${4q^2}$ + 3q - 1= 0

    A.   if p < q
    B.   if p > q
    C.   if p ≤q
    D.   if p = q or no relation can be established
    7 . I. ${2p^2}$ - 169 = 0

    II.${q^2}$ - 289 + 195 = 0

    A.   if p < q
    B.   if p > q
    C.   if p ≤ q
    D.   if p = q or no relation can be established
    8 . I.${p^2}$-5p+6=0
    II. q${q^2}$ - 3q + 2 = 0

    A.   if p < q
    B.   if p > q
    C.   if p ≤ q
    D.   if p ≥ q
    9 . I. 4${p^2}$ - p - 3 = 0
    II. 2${q^2}$ + q - 1 = 0

    A.   if p < q
    B.   if p > q
    C.   if p ≥ q
    D.   if p = q or no relation can be established
    10 . I. ${p^2}$ + 11p + 30 = 0

    II. ${q^2}$- 11q + 30 = 0

    A.   if p < q
    B.   if p > q
    C.   if p ≥q
    D.   if p = q or no relation can be established
      Answers & Solutions
      1 .    
      Answer : Option B
      Explanation :
      The required ratio=$32\over30.4$=$323\over304$=$17\over16$=17:16
      2 .    
      Answer : Option A
      Explanation :
      The required percent increase=$53.0-21.8\over21.8$x100≈143%
      3 .    
      Answer : Option B
      Explanation :
      The required average (in crores)

      =$36.6+18.1+38.7+43.6+24.1+40.5\over6$=$201.6\over6$

      = 33.6 crores = 33,60,00,000
      4 .    
      Answer : Option D
      Explanation :
      The required per cent
      $36.5\over51.6+36.5+43.5+18.1+23.5+35.35.7$x100

      =$36.5X100\over208.9$≈17.4%
      5 .    
      Answer : Option D
      Explanation :
      To get the ratio (as mentioned in the question part) the total production of the company S in 2013 needs to be
      =$41.5\over83$x54=27 crores
      Hence, (27 cr - 26.5 cr =) 0.5 crore more ball bearings need to be manufactured by the company S in the year 2013.
      6 .    
      Answer : Option B
      Explanation :
      p > q
      I. ${2p^2}$ - 6p - p + 3 = 0
      2p(p - 3) -1(p - 3) = 0; (2p - 1)(p- 3) = 0
      p = $1\over2$,3
      p > q

      II.${4q^2}$+ 4q - q - 1 = 0; 4q(q + 1) -1(q + 1) = 0
      (4q - 1)(q + 1) = 0;
      q = $1\over4$,-1
      7 .    
      Answer : Option C
      Explanation :
      I. p ≤q; ${2p^2}$ = 169
      p = ± 13  ;
      II. q${2q^2}$ - 15q - 13q + 195 = 0
      q(q - 15) - 13(q - 15) = 0
      (q - 13)(q - 15) = 0
      q = 13, 15
      8 .    
      Answer : Option D
      Explanation :
      p ≥q
      I. ${p^2}$ - 3p - 2p + 6 = 0
      p(p - 3) -2(p - 3) = 0
      (p - 2)(p - 3) = 0;
      p = 2, 3
      II. ${q^2}$ - 2q - q + 2 = 0
      q(q - 2) -1(q - 2) = 0
      (q - 1)(q - 2) = 0
      q = 1, 2
      9 .    
      Answer : Option D
      Explanation :
      I. 4${p^2}$ - 4p + 3p - 3 = 0
      4p(p -1)+ 3(p - 1) = 0
      (4p + 3) (p - 1) = 0
      p = 1, $-3\over4$

      II. 2${q^2}$ + 2q - q - 1 = 0
      2q(q + 1) -1(q + 1) = 0
      (q + 1)(2q - 1) = 0;
      q = $1\over2$,-1
      10 .    
      Answer : Option A
      Explanation :
      p < q
      ${p^2}$ + 6p + 5p + 30 = 0
      p(p + 6) +5(p + 6) = 0
      (p + 5) (p + 6) = 0
      p = -5, -6
      II. ${q^2}$ - 6q - 5q + 30 = 0
      q(q - 6) -5(q - 6) = 0
      (q - 6) (q - 5) = 0;
      q = 5, 6

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