Mixed Quantitative Aptitude Questions Set 188

  1. In a fraction the numerator is increased by 80% and the denominator of the fraction is increased by 120%, then it becomes 5/11. Find the original fraction?
    1/8
    2/5
    5/9
    2/11
    4/7
    Option C
    The original fraction be (x/y)
    [X*(180/100)]/[Y*(220/100)] = 5/11
    9x/11y = 5/11
    x/y = 5/9

     

  2. Akil, Brahma and kannan together have Rs. 33250. If 3/7th of Akil’s amount is equal to 2/3rd of Brahma’s amount and 1/4th of Brahma’s amount is equal to 5/8th of kannan’s amount, then how much amount does Akil have?
    Rs. 17000
    Rs. 47500
    Rs. 17500
    Rs. 27500
    Rs. 15600
    Option C
    A kil+ Brahma + kannan = Rs. 33250
    (3/7)* A = (2/3)* B
    (A/B) = (2/3) * (7/3) = 14/9
    (1/4)* B = (5/8)* k
    (B/k) = (5/8) * (4/1) = 5/2
    The ratio of A, B and k = 70: 45: 18
    133’s = 33250
    1’s = 250
    70’s = Rs. 17500

     

  3. Person 1, 2 and 3 started a business by investing an amount of Rs. 45000, Rs. 65000 and Rs. 70000 respectively. After 3 months, 1 invested Rs. 15000 more and 2 withdraw Rs. 5000. And after 4 months, 3 withdraw Rs. 15000. If the total profit at the end of the year is Rs. 114405, then find the share of 1?
    Rs. 35500
    Rs. 37505
    Rs. 25505
    Rs. 30005
    Rs. 35505
    Option
    The share of 1, 2 and 3,
    = > [45000*3 + 60000*9]: [65000*3 + 60000*9]: [70000*7 + 55000*5]= > 675000: 735000: 765000
    = > 45: 49: 51
    Total profit = Rs. 114405
    145’s = 114405
    1’s = 789
    The share of 1 = 789*45 = Rs. 35505

     

  4. A 90 litres of mixture of acid and water contains acid and water in the ratio of 3: 2. How much water should be added to the mixture to get 45 % acid on it?
    30
    45
    67
    23
    12
    Option A
    Total mixture of acid and water = 90 litres
    5’s = 90
    1’s = 18
    Total acid = 54 litres, water = 36 litres
    According to the question,
    54/(36 + x) = (45/55)
    54/(36 + x) = 9/11
    66 = 36 + x
    X = 30
    30 litres of water should be added to the mixture.

     

  5. A yacht can sail 55 km downstream in 66 min. The ratio of the speed of the yacht in still water to the speed of the stream is 4: 1. How much time will the yacht take to cover 72 km upstream?
    1 hr 24 m
    2hr 24 m
    5 hr 20 m
    6 hr 32 m
    1 hr 15m
    Option B
    Speed of downstream = D/T = 55/(66/60) = 55*(60/66) = 50 km/hr
    The ratio of the speed of the yacht in still water to the speed of the stream
    = > 4: 1 (4x, x)
    5x = 50
    X = 10
    Speed of upstream = 4x – x = 3x = 30 km/hr
    Distance = 752 km
    Time = D/S = 72/30 = 2 2/5 hr = 2 hours 24 mins

     

  6. 79.89% of 1400 + 17.05% of 399.99 – 20% of 520 =? * 2
    430
    220
    541
    123
    450
    Option C
    80% of 1400 + 16.5% of 400 – 20% of 520 =? * 2
    80/100 * 1400 + 16.5/100 * 400 – 20/100 * 520 =? * 2
    1120 + 66 -104 =? * 2
    ? = 541

     

  7. 98.902 * √1681 + 22.87 * √5184 – 48 * 69 =? * 89
    27
    23
    22
    24
    25
    Option A
    99 * √1681 + 23 * √5184 – 48 * 69 =? * 89
    4059 + 1656 – 3312 =? * 89
    ? = 27

     

  8. 34.56 % of 119.9 + 12.5 % of 231.5 + 9 * 29 =? + 132
    200
    150
    250
    300
    350
    Option A
    35 % of 120 + 12.5 % of 232 + 9 * 29 =? + 132
    42 + 29 + 261 =? + 132
    ? = 200

     

  9. (√1023.5 + √3720.5) * √15.89 +? = 482
    120
    130
    110
    100
    90
    Option C
    (√1024 + √3721) * √16 +? = 482
    (32+ 61) *4 +? =482
    372 + ? = 482
    ? = 482 – 372
    ? = 110

     

  10. 19.99 % of 64.67 + 111.5 * √81 -13 * 8 =? + 27 * 15
    340
    230
    123
    540
    512
    Option E
    20 % of 65 + 112 * √81 -13 * 8 =? + 27 * 15
    13 + 1008 – 104 =? + 405
    ? = 512

     

Related posts

Leave a Comment