Mixed Quantitative Aptitude Questions Set 20

Bank PO Quantitative Aptitude Questions Answers

  1. Fresh apples contain 70% water by weight, whereas dried apples contain 4% water by weight. How many kg of dried apples can be obtained from 24 kg of fresh apples?
    A) 9.3 kg
    B) 7.1 kg
    C) 8.2 kg
    D) 7.5 kg
    E) 9.8 kg
    View Answer
      Option D
    In 24 kg of apples,  water is 70%, so pulp is 30% = 30% of 24kg = 7.2 kg
    Let total of x kg dried apples is to be obtained from 24 kg of fresh apples.
    So in this x kg dried apples, 96% should be pulp to have 4% water. So
    96% of x = 7.2 kg
    Solve, x = 7.5 kg
  2. There are 420 students residing in a hostel. If the students increase by 100, the expenses of the mess increase by Rs 500 per month, while the average expenditure per head diminishes by Rs 25. What is the original monthly expense of the mess?
    A) Rs 105
    B) Rs 120
    C) Rs 135
    D) Rs 180
    E) Rs 165
    View Answer
      Option C
    Let original monthly expense of the mess = Rs x
    So 520 (x – 25) – 420x = 500  [520 is (420+100)] Solve, x = Rs 135
  3. Points A and B are 45 km apart. Boat P can cover downstream from A to B and back from B to A in a total of 20 hours. Another boat Q can cover the same distance in a total of 8 hours. If the speed of boat P is half the speed of boat Q, then what is the speed of current?
    A) 3 km/hr
    B) 8 km/hr
    C) 6 km/hr
    D) 10 km/hr
    E) 12 km/hr
    View Answer
      Option A
    Speed of boat Q in still water = 2x km/hr, then of boat P = x km/hr.
    Let speed of current = y km/hr
    45/(x + y)  + 45/(x – y)  = 20
    And also
    45/(2x + y)  + 45/(2x – y)  = 8
    Solve both equations, y = 3 km/hr
  4. On selling rice for Rs 120 per kg, a shopkeeper incurs a loss of 10%. What will be the amount of rice sold, if there was a net loss of Rs 40?
    A) 5 kg
    B) 2.5 kg
    C) 4 kg
    D) 3 kg
    E) 4.5 kg
    View Answer
      Option D
    10% of total CP = Rs 40
    So 100% = Rs 400, means total CP = Rs 400
    So SP = 400 – 40 = Rs 360
    So total rice sold at this price = 360/120 = 3kg  
  5. Can A contains 108 l of milk of milk and water in ratio 7 : 1. Another can B contains milk and water in ratio 6 : 1. If the mixtures of both the cans are mixed in another can, the respective ratio becomes 13 : 2. Find the amount of new mixture formed?
    A) 94.5 l
    B) 119.5 l
    C) 143.5 l
    D) 187.5 l
    E) 202.5 l
    View Answer
      Option E
    By rule of allegation:
    (108 l)………..…………..(x l)
    So ratio is 1/7*15 : 1/8*15 = 8 : 7
    So 108/x = 8/7
    Solve, x = 94.5 l
    So total mixture formed is 108 + 94.5 = 202.5 l
  6. A letter is chosen at random from ‘ASPIRANTSZONE’. What is the probability that it is a vowel?
    A) 4/13
    B) 8/13
    C) 6/13
    D) 7/13
    E) 5/13
    View Answer
      Option E
    There are total 13 letters in ASPIRANTSZONE
    And total 5 vowels – 2 A’s, 1 I, 1 E, 1 O
    So probability is 5/13
  7. A shopkeeper bought rice at the rate of 12 Rs/kg. While selling he uses false weights of 750 gms instead of 1 kg. If he sells rice making a profit of 60%, then he sold rice at what rate (per kg)?
    A) Rs 16.9
    B) Rs 15.7
    C) Rs 12.8
    D) Rs 14.4
    E) Rs 15.3
    View Answer
     Option D

    When he uses 750 gms instead of 1000 gms. He makes a profit of
    (1000-750)/750 × 100 = 100/3%
    Let on CP he gains x%.
    So using successive formula:
    100/3 + x + (100/3 * x)/100 = 60
    Solve x = 20 %
    Now CP is Rs 12. Profit % is 20 %
    Find SP.
    SP = 120% of 12 = Rs 14.4 per kg.
  8. A sum of Rs 6800 is deposited in two schemes. One part is deposited in Scheme A which offers 12% rate of interest. Remaining part is invested in Scheme B which offers 10% rate of interest compounded annually. If interest obtained in scheme A after 5 years is Rs 678 more than the interest obtained in scheme B after 2 years, find the part deposited in scheme A.
    A) Rs 4200
    B) Rs 3900
    C) Rs 3400
    D) Rs 2600
    E) Rs 3000
    View Answer
      Option D

    (6800-x)*12*5/100 = x [ (1 + 10/100)<sup>2</sup> – 1] + 678
    68*12*5 – 60x/100 = 21x/100 + 678
    Solve, x = Rs 4200
    So in scheme A = 6800-4200 = Rs 2600
  9. Ratio of age of A 3 years hence to C 4 years hence is 6 : 7. Age of B is 9 years greater than the age of A and the ratio of age of B 2 years ago to that of C 8 years ago is 4 : 3. Find the sum of all of A, B and C’s present ages.
    A) 134
    B) 129
    C) 124
    D) 108
    E) 113
    View Answer
          Option E

    (A+3)/(C+4) = 6/7
    B = 9 + A
    (B-2)/(C-8) = 4/3
    Solve all, A = 33, B = 42,  = 38
  10. A and B started a business by investing Rs 2400 and Rs 2000 respectively. After 6 months A added Rs 900 while B added Rs 700 to their respective investments. If after a year, out of total profit share of A is Rs 2500 more than the share of B, find the total profit.
    A) Rs 34000
    B) Rs 25000
    C) Rs 29000
    D) Rs 31000
    E) Rs 26000
    View Answer
     Option E

    Ratio of shares A : B is
    2400*6 + 3300*6 : 2000*6 + 2700*6
    24 + 33 : 20 + 27
    57 : 47
    So (57-47)/(57+47) * x = 2500
    Solve, x = Rs 26000


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14 Thoughts to “Mixed Quantitative Aptitude Questions Set 20”

  1. Diya

    sir in ques 3

    45/(x + y) + 45/(x – y) = 20

    And also

    45/(2x + y) + 45/(2x – y) = 8
    after this we put the option???
    but in option only the value of y is given then how to put the value of x??

    1. Suraj

      no u dont have to put the value from option, u have to solve these two equation and get the value of y,

  2. Mera bhi din ayega :)))

    Ty team :))

  3. Jasmeen

    Thnk yu AZ :))

  4. Pawan

    please explain ques 2 not getting answer by above equation

    1. Suraj

      Let avg expenditure in case I =x
      thus total expenditure= no of student*avg expenditure=420*x — (i)
      when 100 more students join, new avg= (x-25)
      total no. of students=520
      hence new total expentiture= 520*(x-25) –(ii)
      it is given that the expenses of the mess increase by Rs 500
      which means that,(ii) -(i)=500
      520*(x-25) – 420*x=500
      solve it 100x =500 +520*25

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