Mixed Quantitative Aptitude Questions Set 19

  1. The area of the rectangle gets reduced by 10 if its length is increased by 2 and breadth decreased by 2. Also it increases by 14, if its breadth is increased by 3 and length decreased by 2. Find the original area of the rectangle.
    A) 126 sq. units
    B) 154 sq. units
    C) 173 sq. units
    D) 187 sq. units
    E) 163 sq. units
    View Answer
      Option B
    Solution
    :
    Let length = x, and breadth = y
    So xy – (x+2)(y-2) = 10
    Gives 2x – 2y + 4 = 10………..(1)
    And (x-2)(y+3) – xy = 14
    Gives 3x – 2y – 6 = 14………………(2)
    Solve equations, x = 14, y = 11
  2. Incomes of Karan, Arjun and Aarti are in ratio 3 : 5 : 7. Their expenditures are in ratio 3 : 6 : 8. If B saves 10% of his income, find the ratio of their savings.
    A) 2 : 4: 5
    B) 5 : 4 : 2
    C) 4 : 3 : 2
    D) 3 : 2 : 4
    E) 3 : 4 : 2
    View Answer
      Option D
    Solution
    :
    Incomes = 3x, 5x, 7x
    Expenditures = 3y, 6y, 8y
    B’s savings = 10/100 * 5x = x/2
    So x/2 = 5x – 6y
    Solve, x = 4y/3
    Now savings ratio is 3x – 3y : 5x – 6y : 7x – 8y
    Put x = 4y/3
    So ratio becomes 1 : 2/3 : 4/3 = 3 : 2 : 4
  3. Karan can complete a work in 60 days and Arjun can complete the same work in 5 days more than the number of days in which they both can complete work together. They both start the work and after 5 days, Karan leaves and Arjun starts working with 4/3 times efficiency as before. In how much time the work will be completed?
    A) 12
    B) 8
    C) 10
    D) 15
    E) 18
    View Answer
      Option D
    Solution
    :
    Let they both complete work together in x days
    So Arjun in (x+5) days
    So 1/60 + 1/(x+5) = 1/x
    Solve, x = 15, So Arjun can complete work in 20 days
    Now they work for 5 days so completed (1/60 + 1/20) * 5 = 1/3 of work
    Remaining work = 2/3.  Arjun works with 4/3 times efficiency. So now he can complete whole work in 20/(4/3) = 15 days
    So 1/15 * y = 2/3
    Solve, y = 10
    So total number of days = 5 + 10 = 15
  4. Kavya borrowed Rs 2460 to be paid back in 2 yearly installments at 5% rate of interest compounded annually. How much will be each installment?
    A) Rs 1323
    B) Rs 1486
    C) Rs 1545
    D) Rs 1139
    E) Rs 1376
    View Answer
      Option A
    Solution
    :
    (1 + 5/100) = 21/20
    So for 2 yearly installments
    20x/21 + 400x/441 = 2460
    Solve, x = Rs 1323  
  5. A cylindrical bucket whose diameter is 14 m is partly filled with water. A cuboid whose dimensions are 33cm × 14cm × 7cm is immersed in the bucket. Find the level of water that rise?
    A) 22 cm
    B) 15 cm
    C) 18 cm
    D) 14 cm
    E) 21 cm
    View Answer
      Option E
    Solution
    :
    Volume of water that rise = Volume of cuboid
    So πr2h = l*b*h
    22/7 * 7 * 7 * h = 33 * 14 * 7
    Solve, h = 21 cm
  6. A loss of 10% is made by selling an article. Had it been sold for Rs 75 more, there would have been a profit of 5%. The initial loss is what percentage of the profit earned, if the article was sold at a profit of 20%?
    A) 44%
    B) 47%
    C) 50%
    D) 54%
    E) Cannot be determined
    View Answer
      Option C
    Solution
    :
    Use shortcut for these type of questions:
    CP of article = 75 × 100/ [5 – (-10)] = Rs 500  (+5 for 5% profit, -10 for 10% loss)
    So loss was = 10/100 * 500 = Rs 50
    and new profit = 20/100 * 500 = Rs 100
    So required % = 50/100 * 100 = 50%
  7. Vanya inquires about a mobile at two shops. One offers 30% discount and the other offers 25%. She has just sufficient amount of Rs 21,000 to purchase mobile at 30% discount, how much amount she has less than the amount required to purchase mobile at 25% discount?
    A) Rs 1600
    B) Rs 1500
    C) Rs 1700
    D) Rs 1300
    E) Rs 1400
    View Answer
     Option B
    Solution:
    Let MP of mobile = Rs x
    So at 30% discount, she gets mobile at Rs 70% of x
    So 7x/10 = 21000
    Solve, x = 30,000
    So SP at 25% discount = 75% of 30,000= Rs 22,500
    Required difference =  22500 – 21000 = Rs 1500
  8. Two containers contain mixture of milk and water. Container A contains 25% water and rest milk. Container B contains 36% water and rest milk. How much amount should be mixed from container A to 50 litres of container B so as to get a new mixture having water to milk in ratio 2 : 5?
    A) 162 l
    B) 104 l
    C) 135 l
    D) 178 l
    E) 150 l
    View Answer
      Option B
    Solution:

    In resultant mixture, water is 2/7 * 100 = 200/7%
    So by method of allegation:
    25%……………………………36%
    …………..200/7%
    52/2%……………………….25/2%
    So ratio is 52/2 : 25/2 = 52 : 25.
    So 52/25 = x/50
    So x = 104 l
  9. A boat travels downstream from point A to B and comes back to point C which is half distance between points A and B in a total of 40 hours. If the speed of boat is still water is 7 km/hr and distance from point A to B is 120 km, then find the upstream speed.
    A) 5 km/hr
    B) 8 km/hr
    C) 2 km/hr
    D) 3 km/hr
    E) 6 km/r
    View Answer
     <strong> Option C
     Solution: </strong>
    A to B is 120, so B to is 120/2 = 60 km
    Let speed of current = x km/hr
    So 120/(7+x) + 60/(7-x) = 40
    Solve, x = 5 km/hr
    So downstream speed = 7 – 5 = 2 km/hr
  10. Ratio of age of A 2 years hence to age of C 5 years ago is 12 : 5. The average of ages of A and B is 20 years. Also age of B 6 years hence will be eight-five times of the present age of C. Find the age of A.
    A) 16 years
    B) 22 years
    C) 29 years
    D) 34 years
    E) 38 years
    View Answer
     Option B
    Solution:
    (A+2)/(C-5) = 12/5
    (A+B)/2 = 20
    (B + 6) = 8/5 * C

 

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