Mixed Quantitative Aptitude Questions Set 17

Quantitative Aptitude Questions: Word Problems for Bank PO Exams. Type of Quantitative Aptitude Questions asked in IBPS PO/SBI PO/NIACL/NICL/BoB and other exams

  1. A train starts from station A at 7:00 AM and reaches station B at 9:00 AM. Another train starts from station B at 7:30 AM and reaches station A at 10:30 AM. At what time the two trains will meet?
    A) 8:10 AM
    B) 8:24 AM
    C) 8:48 AM
    D) 9:08 AM
    E) 9:30 AM
    View Answer
     Option B
    Solution:
    Let the distance from A to B is ‘d’ km.
    First train reaches destination in 2 hrs, so speed of first train = d/2 km/hr.
    Speed of second train = d/3 km/hr
    Speed of first train = d/2 km/hr, so after 1/2 hr, i.e. at 7:30 AM, the first train has covered d/4 km.
    The distance left = (d – d/4) km = 3d/4 km. relative speed = d/2 + d/3.
    So time they will meet at = 7:30 AM + (3d/4) / (d/2 + d/3)
    = 7:30 AM + (9/10)*60 = 8:24 AM
    OR Use:
    Time = 7:00AM + [(9:00 – 7:00) * (10:30 – 7:00)]/[(9:00 – 7:00) + (10:30 – 7:30)]
    = 7:00 Am + (2 * 7/2)/(2+5) = 8:24 AM
  2. An article is marked 10% above the cost price with marked price being Rs 120. The profit gets increased by Rs 14.8, when the selling price of this article is increased by 2%. What is the original selling price of the article?
    A) Rs 740
    B) Rs 720
    C) Rs 785
    D) Rs 735
    E) Rs 715
    View Answer
     Option A
    Solution:
    2% of SP= 14.8

    then 100% of SP= 14.8 x 100/2 =Rs 740 
  3. 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 A?
    A) 4 km/hr
    B) 10 km/hr
    C) 9 km/hr
    D) 7 km/hr
    E) None of these
    View Answer
    Option A
    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)
    x2 + 2x – 24 = 0
    solve, x = 4
  4. The amount that A and B earns is in the ratio 4 : 3. After a year, A’s salary got increased by 20% and B’s salary by 25%. If now the total of their salaries become Rs 1,02,600, What was the initial salary of A?
    A) Rs 36,000
    B) Rs 48,000
    C) Rs 54,000
    D) Rs 45,000
    E) Rs 56,000
    View Answer
     Option B
    Solution:
    4x and 3x
    After year, As salary = (120/100)*4x = 480x/100, Bs salary = (125/100)*3x = 375x/100
    So addition of their salaries = (480x/100) +(375x/100) = 855x/100
    So 855x/100 = 1,02,600
    Solve, x = 12,000
    So As initial salary = 4x = 4*12000
  5. How many 4 letter words starting with vowel can be formed by the letters of word UPSTREAM such that no letter repeats?
    A) 460
    B) 680
    C) 630
    D) 850
    E) 720
    View Answer
     Option C
    Solution:
    There are 3 vowels, so we gave 3 choices for the first letter
    There are total 8 letters in UPSTREAM, and one of them has occupied first place, so 7 choices for second, 6 choices for third, and 5 choices for fourth
    So 3*7*6*5 = 630
  6. A boat can travel 15 km upstream in 30 minutes. If the speed of the current is 1/3 of the speed of the boat in still water, how much distance can the boat travel downstream in 30 minutes?
    A) 25 km
    B) 60 km
    C) 48 km
    D) 30 km
    E) 36 km
    View Answer
    Option D
    Solution:
    A boat can travel 15 km upstream in 30 minutes. So Upstream speed = 15 / (30/60) = 30 km/hr
    Let speed of the boat in still water = x
    Speed of the current = x/3
    Upstream speed = x – x/3 = 2x/3
    But from above Upstream speed = 30. So 2x/3 = 30
    Solve, x = 45
    Speed of the current = 45/3 = 15 and speed of boat = 45
    Downstream speed = 45+15 = 60
    Distance travelled in 30min = 60 * (30/60)= 30 km
  7. Find the probability that a two-digit number is a multiple of 3 given that it is also a multiple of 5.
    A) 1/4
    B) 3/7
    C) 5/6
    D) 3/5
    E) 1/3
    View Answer
     Option E
    Solution:
    Given that it is also a multiple of 5, this means that the numbers divisible by 3 are to be taken from these numbers only (which are divisible by 5)
    There are 90 two-digit numbers.
    Out of which 18 are divisible by 5
    LCM of (3 and 5) = 15, Out of 90 numbers, 6 are divisible by 15 (which are also divisible by 3)
    So 6 are divisible by 3 and 18 are total numbers (which are divisible by 5)
    So required probability = 6/18 = 1/3
  8. After 2 years the respective SI and CI obtained on a certain sum of money are Rs 360 and Rs 396 respectively. Find the amount of money invested.
    A) Rs 1680
    B) Rs 1700
    C) Rs 1800
    D) Rs 1850
    E) None of these
    View Answer
     Option C
    Solution:
    Diff = 396-360 = 36
    r = 2*Diff*100/SI
    So r = 2*36*100/360 = 20%
    Now 36 = Pr2/1002
    Solve, P = 900
  9. The ratio of incomes of P and Q is 6 : 5 respectively. After one year their respective income by same percentage. The difference in their new incomes is Rs 14,300. If the difference in the old income of A and new income of B is Rs 6500, what is the % increase in incomes?
    A) 15%
    B) 12%
    C) 10%
    D) 8%
    E) 13%
    View Answer
     Option C
    Solution:
    6x and 5x
    Suppose, y% increased
    So (100+y)/100 * 6x – (100+y)/100 *5x = 14300
    This gives (100+y)*x = 14,30,000 …………..(1)
    Now also given
    6x – [(100+y)/100 * 5x] = 6500
    Solve, x(20-y) = 1,30,000 ………………………(2)
    Divide both equations
    (100+y)/(20-y) = 11
    Solve, y = 10
  10. Two trains which are 200 m long each are moving in opposite directions. They cross each other in 8 seconds. If one train is moving one and a half times as fast the other train, then find the speed of the faster train.
    A) 88km/hr
    B) 122 km/hr
    C) 118 km/hr
    D) 108 km/hr
    E) 96 km/hr
    View Answer
     Option D
    Solution:
    Let speed of slower train = x m/s, then speed of faster train = 1.5x m/s
    Their relative speed becomes = x+1.5x = 2.5x m/s
    So (200+200)/8 = 2.5x
    After solving, speed of slower train, x = 20 m/s
    So speed of faster train = 1.5 × 20 = 30 m/s = 30 × 18/5 km/hr

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