Mixed Quantitative Aptitude Questions Set 9

  1. Population of a town which increases each year by same percentage is 8,75,000. Before one year the population was 7,00,000. Find the population of town 3 years ago.
    A) 456000
    B) 342000
    C) 483000
    D) 324000
    E) 448000
    View Answer
      Option E
    Solution:
    % increase in pop in 1 yr = 875-700)/700 × 100 = 25%
    So each year population increased by 25%.
    So in 2 years population increased by:
    25 + 25 + 25×25/100 = 56.25% (using successive formula)
    Let 3 years ago population was x.
    So 156.25% of x = 7,00,000
    Solve for x, x = 448000
  2. A shopkeeper bought rice at the rate of 6 Rs/kg. While selling he uses false weights of 800 gms instead of 1 kg. If he sells rice making a profit of 66 2/3%, then he sold rice at what rate (per kg)?
    A) Rs 8
    B) Rs 5
    C) Rs 10
    D) Rs 12
    E) Rs 14
    View Answer
      Option A
    Solution:

    When he uses 800 gms instead of 1000 gms. He makes a profit of
    (1000-800)/800 × 100 = 25%
    Let on CP he gains x%.
    So using successive formula:
    25 + x + 25*x/100 = 66 2/3
    Solve x = 100/3 %
    Now CP is Rs 6. Profit % is 100/3 %
    Find SP.
    SP = (100 + 100/3)% of 6 = Rs 8 per kg.
  3. After A and B worked together for 6 days, 7/12 of the work is remaining to be completed by them. If B alone can do 1/9 of work in 4 days, then how much work would be left if A alone works for 6 days?
    A) 1/4
    B) 2/5
    C) 3/4
    D) 4/5      
    E) 3/7
    View Answer
      Option C
    Solution:

    B alone does 1/9 work in 4 days so complete work in 9×4 = 36 days. 
    A and B worked together for 6 days, 7/12 of the work is remaining. Means in 6 days they complete 1 – 7/12 = 5/12 of work.
    Let A does complete work in a days.
    So
    (1/a + 1/36) * 6 = 5/12
    Solve, a = 24 days
    In 24 days A completes 1 work. So in 6 days he completes 6/24 = 1/4 work
    So after 6 days, 1 – 1/4 = 3/4 of work is left.
  4. In a class there are 16 students in a group. If two more students are taken into consideration then the average of the age of group increases by 1. If four more students are taken into consideration then the average of the age of group decreases by 1. If there is a difference of 4 years in the total ages of students added both time, find the average of first 16 students in the group.
    A) 11
    B) 21
    C) 15
    D) 13
    E) 17
    View Answer
      Option E
    Solution:

    Let n is average of group of initial 16 people.
    Cross mutiplication method:
    16………….n
    18…………(n+1)
    So total if ages of 2 people added = 2*(n+1) + 16
    AND
    16………….n
    20…………(n-1)
    So total if ages of 4 people added = 4*(n-1) – 16
    Now given
    2*(n+1) + 16 – 4*(n-1) – 16 = 4
    Solve, n = 17
    Other method: 
    Total of ages of 16 people = 16n
    Total of ages of 18 people becomes = 18*(n+1)
    So total ages if 2 people added = 18*(n+1) – 16n = 2n + 18
    Now
    Total of ages of 20 people becomes = 20*(n-1)
    So total ages if 4 people added = 20*(n-1) – 16n = 4n – 20
    Now given:
    2n + 18 – (4n – 20) = 4
    Solve, n = 17.  
  5. Ratio of ages of B to A is 3 : 2 and also ratio of ages of C to B is 3 : 4. If the average of their ages is 29 years, find ratio of age of A 6 years hence to age of B 4 years hence.
    A) 4 : 9
    B) 7 : 10
    C) 4 : 7
    D) 3 : 4
    E) 2 : 5
    View Answer
      Option D
    Solution:

    A/B = 2/3 and B/C = 4/3
    So A : B : C is
    2*4 : 3*4 : 3*3
    8 : 12 : 9
    Now 8x + 12x + 9x = 3*29 = 84
    Solve, x = 3
    So age of A = 8*3 = 24 and age of B = 12*3 = 36
    So required ratio is (A+6)/(B+4) = 30/40 = 3/4
  6. A and B invested some money in business for different duration. A invested Rs 2700 for 8 months and then increased his investment by Rs 600 for rest months. B invested Rs 2400 for 6 months and then increased his investment by Rs 800 for rest months. After a year they got a profit of Rs 15,960. B was a working partner and got some extra money from the profit. If he got a total of Rs 8710, then what is the extra money that A got?
    A) Rs 1710
    B) Rs 2920
    C) Rs 1740
    D) Rs 2510
    E) Rs 1280
    View Answer
      Option A
    Solution:

    A : B
    2700*8 + 3300*4 : 2400*6 + 3200*6
    9*8 + 11*4 : 8*6 + 32*2 
    9*2 + 11 : 2*6 + 8*2
    29 : 28
    Let A got Rs x extra.
    So
    28/(29+28) * (15960 – x) + x = 8710
    Solve to find x, x = Rs 1710  
  7. The speed of boat A is 3 km/hr less than the speed of boat B. The time taken by boat A to travel a distance of 18 km downstream is 1 hour more than time taken by B to travel the same distance downstream. If the speed of the current is half of the speed of boat A, what is the speed of boat B?
    A) 10 km/kr
    B) 6 km/kr
    C) 9 km/kr
    D) 4 km/kr
    E) 12 km/kr
    View Answer
      Option D
    Solution:

    Speed of A = x, then of B = x+3
    Speed of current = x/2
    So downstream speed from boat A = x + x/2 = 3x/2
    Downstream speed from boat B = x+3 + x/2 = 3x/2 + 3
    So 18/(3x/2) = 18/(3x/2 + 3) + 1
    36/3x = 36/(3x+6) + 1
    12/x = 12/(x+2) + 1
    12(x+2) = 12x + x(x+2)
    x<sup>2</sup> + 2x – 24 = 0
    Solve, x = 4  
  8. P and Q start at the same point on a circular track with speeds 18 km/hr and 27 km/hr respectively in opposite directions. If the circular track is 700 m long, then after how much time both mill meet again?
    A) 43 sec
    B) 25 sec
    C) 31 sec
    D) 44 sec
    E) 13 sec
    View Answer
      Option B
    Solution:

    18 km/hr = 18 * (5/18) = 5 m/s
    27 km/hr = 27 * (5/18) = 15/2 m/s
    In opposite direction, relative speed is 5 + 15/2 = 25/2 m/s
    Time = 700/ (25/2) = 56 sec
  9. An article is marked 15% above the cost price with marked price being Rs 138. The profit gets increased by Rs 25.2, when the selling price of this article is increased by 20%. What is the original selling price of the article?
    A) Rs 168

    B) Rs 156
    C) Rs 126
    D) Rs 134
    E) Rs 194
    View Answer
      Option C
    Solution:

    20% of SP= 25.2
    then 100% of SP= 25.2 x (100/20) =126  
  10. In how many ways all the letters of the word ‘EDITING’ can be arranged?
    A) 2520
    B) 1420
    C) 2340
    D) 3480
    E) 1320
    View Answer
      Option A
    Solution:

    Editing is 7 letters so 7!
    I is 2 times, so, 7!/2! = 2520

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