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